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Chapter 1 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 1.2 An approximate solution can be found if we combine Equations 1.4 and 1.5: molecular keVm = 2 2 1 molecular kekT =2 3 m kTV 3≈∴ Assume the temperature is 22 ºC. The mass of a single oxygen molecule is kg 1014.5 26−×=m . Substitute and solve: [ ]m/s 6.487=V The molecules are traveling really, fast (around the length of five football fields every second). Comment: We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitative expression for this expression, we have: dvvv kT m kT mdvvf 22 2/3 2 exp 2 4)( − = π π where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by integrating the expression above [ ]m/s 4498 )( )( 0 0 === ∫ ∫ ∞ ∞ m kT dvvf vdvvf V π 3 1.3 Derive the following expressions by combining Equations 1.4 and 1.5: a a m kTV 32 = b b m kTV 32 = Therefore, a b b a m m V V = 2 2 Since mb is larger than ma, the molecules of species A move faster on average. 4 1.4 We have the following two points that relate the Reamur temperature scale to the Celsius scale: ( )Reamurº 0 C,º 0 and ( )Reamurº 80 C,º 100 Create an equation using the two points: ( ) ( ) Celsius º 8.0Reamurº TT = At 22 ºC, Reamurº 6.17 =T 5 1.5 (a) After a short time, the temperature gradient in the copper block is changing (unsteady state), so the system is not in equilibrium. (b) After a long time, the temperature gradient in the copper block will become constant (steady state), but because the temperature is not uniform everywhere, the system is not in equilibrium. (c) After a very long time, the temperature of the reservoirs will equilibrate; The system is then homogenous in temperature. The system is in thermal equilibrium. 6 1.6 We assume the temperature is constant at 0 ºC. The molecular weight of air is kg/mol 0.029g/mol 29 ==MW Find the pressure at the top of Mount Everest: ( ) ( ) [ ]( )( ) ( ) ⋅ − = K 73.152 Kmol J 314.8 m 8848m/s .819kg/mol029.0expatm 1P kPa 4.33atm 330.0 ==P Interpolate steam table data: Cº 4.71=satT for kPa 4.33=satP Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the temperature remains constant. In reality the temperature decreases with height as we go up the mountain. However, a solution in which T and P vary with height is not as straight-forward. 7 1.7 To solve these problems, the steam tables were used. The values given for each part constrain the water to a certain state. In most cases we can look at the saturated table, to determine the state. (a) Subcooled liquid Explanation: the saturation pressure at T = 170 [o (b) Saturated vapor-liquid mixture C] is 0.79 [MPa] (see page 508); Since the pressure of this state, 10 [bar], is greater than the saturation pressure, water is a liquid. Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume of this state, 3 [m3/kg], is in between these values we have a saturated vapor- liquid mixture. (c) Superheated vapor Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since the volume of this state, 0.05 [m3 (d) Superheated vapor /kg], is greater than this value, it is a vapor. Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5 [MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state, 7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go to the superheated water vapor tables for P = 500 [kPa], we see the state is constrained to T = 200 [oC]. 8 1.8 From the steam tables in Appendix B.1: = kg m 003155.0ˆ 3 criticalv ( )MPa 089.22 C,º 15.374 == PT At 10 bar, we find in the steam tables = kg m 001127.0ˆ 3 sat lv = kg m 19444.0ˆ 3 sat vv Because the total mass and volume of the closed, rigid system remain constant as the water condenses, we can develop the following expression: ( ) satvsatl critical vxvxv ˆˆ1ˆ +−= where x is the quality of the water. Substituting values and solving for the quality, we obtain 0105.0=x or 1.05 % A very small percentage of mass in the final state is vapor. 9 1.9 The calculation methods will be shown for part (a), but not parts (b) and (c) (a) Use the following equation to estimate the specific volume: ( ) ( ) ( ) ( )[ ]Cº 250 MPa, 8.1ˆCº 250 MPa, 0.2ˆ5.0Cº 250 MPa, 8.1ˆCº 250 MPa, 9.1ˆ vvvv −+= Substituting data from the steam tables, ( ) = kg m 11821.0Cº 250 MPa, 9.1ˆ 3 v From the NIST website: ( ) = kg m 11791.0Cº 250 MPa, 9.1ˆ 3 NISTv Therefore, assuming the result from NIST is more accurate % 254.0%100 ˆ ˆˆ Difference % = × − = NIST NIST v vv (b) Linear interpolation: ( ) = kg m 13284.0Cº 300 MPa, 9.1ˆ 3 v NIST website: ( ) = kg m 13249.0Cº 300 MPa, 9.1ˆ 3 NISTv Therefore, % 264.0Difference % = (c) Linear interpolation: ( ) = kg m 12406.0Cº 270 MPa, 9.1ˆ 3 v 10 Note: Double interpolation is required to determine this value. First, find the molar volumes at 270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from the previous step to find the molar volume at 270 ºC and 1.9 MPa. NIST website: ( ) = kg m 12389.0Cº 270 MPa, 9.1ˆ 3 NISTv Therefore, % 137.0Difference % = With regards to parts (a), (b), and (c), the values found using interpolation and the NIST website agree very well. The discrepancies will not significantly affect the accuracy of any subsequent calculations. 11 1.10 For saturated temperature data at 25 ºC in the steam tables, = = kg L 003.1 kg m 001003.0ˆ 3 sat lv Determine the mass: [ ] kg 997.0 kg L 003.1 L 1 ˆ = == sat lv Vm Because molar volumesof liquids do not depend strongly on pressure, the mass of water at 25 ºC and atmospheric pressure in a one liter should approximately be equal to the mass calculated above unless the pressure is very, very large. 12 1.11 First, find the overall specific volume of the water in the container: == kg m 2.0 kg 5 m 1ˆ 33 v Examining the data in the saturated steam tables, we find satv sat l vvv ˆˆˆ << at bar 2= satP Therefore, the system contains saturated water and the temperature is Cº 23.120== satTT Let lm represent the mass of the water in the container that is liquid, and vm represent the mass of the water in the container that is gas. These two masses are constrained as follows: kg 5=+ vl mm We also have the extensive volume of the system equal the extensive volume of each phase Vvmvm satvv sat ll =+ ˆˆ ( ) ( ) 333 m1/kgm8857.0kg/m 001061.0 =+ vl mm Solving these two equations simultaneously, we obtain kg 88.3=lm kg12.1=vm Thus, the quality is 224.0 kg 5 kg 12.1 === m mx v The internal energy relative to the reference state in the saturated steam table is satvv sat ll umumU ˆˆ += From the steam tables: 13 [ ]kJ/kg 47.504ˆ =satlu [ ]kJ/kg 5.2529ˆ =satvu Therefore, ( ) [ ]( ) ( ) [ ]( )kJ/kg 529.52kg .121kJ/kg 47.504kg 88.3 +=U kJ 4.4790=U 14 1.12 First, determine the total mass of water in the container. Since we know that 10 % of the mass is vapor, we can write the following expression ( ) ( )[ ]satvsatl vvmV ˆ1.0ˆ9.0 += From the saturated steam tables for MPa 1=satP = kg m 001127.0ˆ 3 sat lv = kg m 1944.0ˆ 3 sat vv Therefore, ( ) ( ) + = + = kg m 1944.01.0 kg m 001127.09.0 m 1 ˆ1.0ˆ9.0 33 3 sat v sat l vv Vm kg 9.48=m and ( ) ( )( )kg/m 1944.0kg 9.481.0ˆ1.0 3== satvv vmV 3m 950.0=vV 15 1.13 In a spreadsheet, make two columns: one for the specific volume of water and another for the pressure. First, copy the specific volumes of liquid water from the steam tables and the corresponding saturation pressures. After you have finished tabulating pressure/volume data for liquid water, list the specific volumes and saturation pressures of water vapor. Every data point is not required, but be sure to include extra points near the critical values. The data when plotted on a logarithmic scale should look like the following plot. The Vapor-Liquid Dome for H2O 0.0010 0.0100 0.1000 1.0000 10.0000 100.0000 0.0001 0.001 0.01 0.1 1 10 100 1000 Specific Volume, v (m3/kg) Pr es su re , P (M Pa ) Vapor-Liquid VaporLiquid Critical Point 16 1.14 The ideal gas law can be rewritten as P RTv = For each part in the problem, the appropriate values are substituted into this equation where ⋅ ⋅ = Kmol barL 08314.0R is used. The values are then converted to the units used in the steam table. (a) = == = − kg m 70.1 L m 10 mol kg 018.0 mol L 7.30 ˆ mol L 7.30 33 3 MW vv v For part (a), the calculation of the percent error will be demonstrated. The percent error will be based on percent error from steam table data, which should be more accurate than using the ideal gas law. The following equation is used: %100 ˆ ˆˆ Error% × − = ST STIG v vv where IGv is the value calculated using the ideal gas law and STv is the value from the steam table. %62.1% kg m 6729.1 kg m 6729.1 kg m 70.1 Error% 3 33 = − = This result suggests that you would introduce an error of around 1.6% if you characterize boiling water at atmospheric pressure as an ideal gas. 17 (b) = = kg m 57.3ˆ mol L 3.64 3 v v =Error% 0.126% For a given pressure, when the temperature is raised the gas behaves more like an ideal gas. (c) = = kg m 0357.0ˆ mol L 643.0 3 v v %Error = 8.87% (d) = = kg m 0588.0ˆ mol L 06.1 3 v v %Error = 0.823% The largest error occurs at high P and low T. 18 1.15 The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall – your answer should vary. The volume of the room is ( )( )( ) 33 m 4.54ft 1920ft 10ft 16ft 12 ===V The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles using the ideal gas law: ( )( ) ( )K 15.295 Kmol J 314.8 m 4.54Pa 1001325.1 35 ⋅ × == RT PVn mol 2246=n Use the molecular weight of air to obtain the mass: ( ) ( )( ) kg 2.65kg/mol 029.0mol 27.2246 === MWnm That’s pretty heavy. 19 1.16 First, find the total volume that one mole of gas occupies. Use the ideal gas law: ( ) ( ) = × ⋅ == mol m 0244.0 Pa 101 K 15.293 Kmol J 314.8 3 5P RTv Now calculate the volume occupied by the molecules: = × = 3 3 4 molcule one of Volume moleper molecules ofNumber rNv A occupied π ( ) × × = − 310 23 m 105.1 3 4 mol 1 molecules 10022.6 πoccupiedv ×= − mol m 1051.8 3 6occupiedv Determine the percentage of the total volume occupied by the molecules: % 0349.0%100Percentage = ×= v voccupied A very small amount of space, indeed. 20 1.17 Water condenses on your walls when the water is in liquid-vapor equilibrium. To find the maximum allowable density at 70 ºF, we need to find the smallest density of water vapor at 40 ºF which satisfies equilibrium conditions. In other words, we must find the saturation density of water vapor at 40 ºF. From the steam tables, = kg m 74.153ˆ 3 sat vv (sat. water vapor at 40 ºF = 4.44 ºC) We must correct the specific volume for the day-time temperature of 70 ºF (21.1 ºC ) with the ideal gas law. 3.294 K 294.3 6.277 K 6.277, ˆˆ T v T vsatv = ( ) = == kg m 163 kg m 74.153 K 6.277 K 3.294ˆˆ33 K 6.277, 6.277 3.294 K 294.3 sat vvT Tv Therefore, ×== − 3 3 m kg 1013.6 ˆ 1ˆ sat vv ρ If the density at 70 ºF is greater than or equal to [ ]33 kg/m 1013.6 −× , the density at night when the temperature is 40 ºF will be greater than the saturation density, so water will condense onto the wall. However, if the density is less than [ ]33 kg/m 1013.6 −× , then saturation conditions will not be obtained, and water will not condense onto the walls. 21 1.18 We consider the air inside the soccer ball as the system. We can answer this question by looking at the ideal gas law: RTPv = If we assume it is a closed system, it will have the same number of moles of air in the winter as the summer. However, it is colder in the winter (T is lower), so the ideal gas law tells us that Pv will be lower and the balls will be under inflated. Alternatively, we can argue that the higher pressure inside the ball causes air to leak out over time. Thus we have an open system and the number of moles decrease with time – leading to the under inflation. 22 1.19 First, write an equation for the volume of the system in its initial state: Vvmvm satvv sat ll =+ ˆˆ Substitute ( )mxml −= 1 and xmmv = : ( )[ ] Vvxvxm satvsatl =+− ˆˆ1 At the critical point Vvm critical =ˆ Since the mass and volume don’t change ( ) satvsatl critical vxvxv ˆˆ1ˆ +−= From the steam tables in Appendix B.1: = kg m 003155.0ˆ 3 criticalv ( )MPa 089.22 C,º 15.374 == PT At 0.1 MPa, we find in the steam tables = kg m 001043.0ˆ 3 sat lv = kg m 6940.1ˆ 3 sat vv Therefore, solving for the quality, we get 00125.0=x or 0.125 % 99.875% of the water is liquid. 23 1.20 As defined in the problem statement, the relative humidity can be calculated as follows saturationat water of mass waterof massHumidity Relative = We can obtain the saturation pressure at each temperature from the steam tables. At 10 [o [kPa] 10.122.19.0 =×=Waterp C], the saturation pressure of water is 1.22 [kPa]. This value is proportional to the mass of water in the vapor at saturation. For 90% relative humidity, the partial pressure of water in the vapor is: This value represents the vapor pressure of water in the air. At 30 [o [kPa] 12.263.55.0 =×=Waterp C], the saturation pressure of water is 4.25 [kPa]. For 590% relative humidity, the partial pressure of water in the vapor is: Since the total pressure in each case is the same (atmospheric), the partial pressure is proportional to the mass of water in the vapor. We conclude there is about twice the amount of water in the air in the latter case. 24 1.21 (a) When you have extensive variables, you do not need to know how many moles of each substance are present. The volumes can simply be summed. ba VVV +=1 (b) The molar volume, v1 ba ba tot nn VV n Vv + + == 11 , is equal to the total volume divided by the total number of moles. We can rewrite the above equation to include molar volumes for species a and b. b ba b a ba a ba bbaa v nn nv nn n nn vnvnv + + + = + + =1 bbaa vxvxv +=1 (c) The relationship developed in Part (a) holds true for all extensive variables. ba KKK +=1 (d) bbaa kxkxk +=1 Chapter 2 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 2.1 There are many possible solutions to this problem. Assumptions must be made to solve the problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the towel when you dry yourself. In other words, let [ ]kg 5.0 2 =OHm Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy balance and neglecting potential and kinetic energy effects reveals hq ˆˆ ∆= Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find the minimum energy required for drying the towel, assume that the temperature of the towel remains constant at K 298.15Cº 25 ==T . In the drying process, the absorbed water is vaporized into steam. Therefore, the expression for heat is l OH v OH hhq 22 ˆˆˆ −= where is v OHh 2 ˆ is the specific enthalpy of water vapor at bar 01.1=P and K 15.298=T and l OHh 2 ˆ is the specific enthalpy of liquid water at bar 01.1=P and K 15.298=T . A hypothetical path must be used to calculate the change in enthalpy. Refer to the diagram below P = 1 [atm] liquid vapor liquid vapor ∆ ?h 1 P ∆ ?h 2 ∆ ?h 3 ∆ ?h Psat 1 atm 3.17 kPa By adding up each step of the hypothetical path, the expression for heat is ( ) ( )[ ] ( ) ( )[ ] ( ) ( )[ ]Cº 25ˆbar 01.1 C,º 25ˆ Cº 25ˆCº 25ˆbar 1.01 C,º 25 ˆCº 25ˆ ˆ , ,,, 321 22 2222 satv OH v OH satl OH satv OH l OH satl OH hh hhhh hhhq −+ −+−= ∆+∆+∆= 3 However, the calculation of heat can be simplified by treating the water vapor as an ideal gas, which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the hypothetical path containing the pressure change of the liquid can be neglected. This leaves ( ) ( )kPa 3.17 C,º 25ˆ kPa .173 C,º 25ˆˆ 22 l OH v OH hhq −= From the steam tables: = kg kJ 2.2547ˆ , 2 satv OHh (sat. H2 = kg kJ 87.104ˆ , 2 satl OHh O vapor at 25 ºC) (sat. H2 = kg kJ 3.2442qˆ O liquid at 25 ºC) which upon substitution gives Therefore, [ ]( ) [ ]kJ 2.1221 kg kJ 442.32kg 5.0 = =Q To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20 minutes (1200 s) to dry the towel. From the definition of electrical work, [ ]( ) [ ]( ) [ ]( ) [ ]kJ 7488s 1200V 208A 30 === IVtW Therefore, the efficiency is [ ][ ] %3.16%100kJ 7488 kJ 221.21%100 = ×= ×= W Qη There are a number of ways to improve the drying process. A few are listed below. • Dry the towel outside in the sun. • Use a smaller volume dryer so that less air needs to be heated. • Dry more than one towel at a time since one towel can’t absorb all of the available heat. With more towels, more of the heat will be utilized. 4 2.3 In answering this question, we must distinguish between potential energy and internal energy. The potential energy of a system is the energy the macroscopic system, as a whole, contains relative to position. The internal energy represents the energy of the individual atoms and molecules in the system, which can have contributions from both molecular kinetic energy and molecular potential energy. Consider the compression of a spring from an initial uncompressed state as shownbelow. Since it requires energy to compress the spring, we know that some kind of energy must be stored within the spring. Since this change in energy can be attributed to a change of the macroscopic position of the system and is not related to changes on the molecular scale, we determine the form of energy to be potential energy. In this case, the spring’s tendency to restore its original shape is the driving force that is analogous to the gravity for gravitational potential energy. This argument can be enhanced by the form of the expression that the increased energy takes. If we consider the spring as the system, the energy it acquires in a reversible, compression from its initial uncompressed state may be obtained from an energy balance. Assuming the process is adiabatic, we obtain: WWQE =+=∆ We have left the energy in terms of the total energy, E. The work can be obtained by integrating the force over the distance of the compression: 2 2 1 kxkxdxdxFW ==⋅−= ∫∫ Hence: 2 2 1 kxE =∆ We see that the increase in energy depends on macroscopic position through the term x. It should be noted that there is a school of thought that assigns this increased energy to internal energy. This approach is all right as long as it is consistently done throughout the energy balances on systems containing springs. 5 2.4 For the first situation, let the rubber band represent the system. In the second situation, the gas is the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the energy balance becomes WU =∆ Since work must be done on the rubber band to stretch it, the value of the work is positive. From the energy balance, the change in internal energy is positive, which means that the temperature of the system rises. When a gas expands in a piston-cylinder assembly, the system must do work to expand against the piston and atmosphere. Therefore, the value of work is negative, so the change in internal energy is negative. Hence, the temperature decreases. In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the rubber band goes to increase its potential energy; however, a part of it goes into stretching the polymer molecules which make up the rubber band, and the qualitative argument given above still is valid. 6 2.5 To explain this phenomenon, you must realize that the water droplet is heated from the bottom. At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The water vapor forms an insulation layer between the skillet and the water droplet. At low temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures. 7 2.6 Apartment System Fr id ge + -W Q HOT Surr. If the entire apartment is treated as the system, then only the energy flowing across the apartment boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the refrigerator is not explicitly accounted for in the energy balance because it is within the system. By neglecting kinetic and potential energy effects, the energy balance becomes WQU +=∆ The Q term represents the heat from outside passing through the apartment’s walls. The W term represents the electrical energy that must be supplied to operate the refrigerator. To determine whether opening the refrigerator door is a good idea, the energy balance with the door open should be compared to the energy balance with the door closed. In both situations, Q is approximately the same. However, the values of W will be different. With the door open, more electrical energy must be supplied to the refrigerator to compensate for heat loss to the apartment interior. Therefore, shutajar WW > where the subscript “ajar” refers the situation where the door is open and the subscript “shut” refers to the situation where the door is closed. Since, shutajar QQ = shutshutshutajarajarajar WQUWQU +=∆>+=∆ shutajar TT ∆>∆∴ The refrigerator door should remain closed. 8 2.7 The two cases are depicted below. Let’s consider the property changes in your house between the following states. State 1, when you leave in the morning, and state, the state of your home after you have returned home and heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2, the state of the system is the same and ∆U must be zero, so 0=+=∆ WQU or WQ =− where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that must be delivered to the heater. The case where more heat escapes will require more work and result in higher energy bills. When the heater is on during the day, the temperature in the system is greater than when it is left off. Since heat transfer is driven by difference in temperature, the heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off when you are gone. 9 2.8 The amount of work done at constant pressure can be calculated by applying Equation 2.57 QH =∆ Hence, hmQH ˆ∆==∆ where the specific internal energy is used in anticipation of obtaining data from the steam tables. The mass can be found from the known volume, as follows: [ ]( ) [ ]kg 0.1 kg m0.0010 L m001.0L1 ˆ 3 3 = == v Vm As in Example 2.2, we use values from the saturated steam tables at the same temperature for subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1: [ ]( ) [ ]( ) = − =−=∆ kg kJ 15.314 kg kJ 87.104 kg kJ 02.419C25at ˆC100at ˆˆ o1, o 2, ll uuu Solve for heat: [ ]( ) [ ]kJ 15.314 kg kJ 05.314kg 0.1ˆ = =∆= umQ and heat rate: [ ] [ ]( ) [ ] [ ]kW 52.0 min s 60min. 01 kJ 15.314 = == t QQ This value is the equivalent of five strong light bulbs. 10 2.9 (a) From Steam Tables: = kg kJ 8.2967ˆ1u (100 kPa, 400 ºC) = kg kJ 8.2659ˆ2u (50 kPa, 200 ºC) −=−=∆ kg kJ 0.308ˆˆˆ 12 uuu (b) From Equations 2.53 and 2.63 ( )∫∫ −==−=∆ 2 1 2 1 12 T T P T T v dTRcdTcuuu From Appendix A.2 )( 322 ETDTCTBTARcP ++++= − [ ]∫ −++++=∆ − 2 1 1322 T T dTETDTCTBTARu Integrating −+−−−+−+−−=∆ )( 4 )11()( 3 )( 2 ))(1( 41 4 2 12 3 1 3 2 2 1 2 212 TT E TT DTTCTTBTTARu The following values were found in Table A.2.1 0 1021.1 0 1045.1 470.3 4 3 = ×= = ×= = − E D C B A Substituting these values and using 11 K 473.15K) 15.273200( K 673.15K) 15.273400( Kmol J 314.8 2 1 =+= =+= ⋅ = T T R provides −= −=∆ −=∆ kg kJ 1.308 J 1000 kJ 1 kg 1 g 1000 O]H g[ 0148.18 O]H [mol 1 mol J 5551ˆ mol J 5551 2 2u u The values in parts (a)and (b) agree very well. The answer from part (a) will serve as the basis for calculating the percent difference since steam table data should be more accurate. ( ) %03.0%100 0.308 1.308308% =× − −−− =Difference 12 2.10 (a) Referring to the energy balance for closed systems where kinetic and potential energy are neglected, Equation 2.30 states WQU +=∆ (b) Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal 0=∆U According to Equation 2.77 = = 1 2 11 1 2 lnln P PRTn P PnRTW From the ideal gas law: 1111 VPRTn = = 1 2 11 ln P PVPW Substitution of the values from the problem statement yields ( )( ) [ ]J 940 bar 8 bar 5lnm 105.2Pa 108 335 −= ××= − W W The energy balance is [ ]J 940 J 0 =∴ += Q WQ (c) Since the process is adiabatic 0=Q The energy balance reduces to WU =∆ 13 The system must do work on the surroundings to expand. Therefore, the work will be negative and 12 0 0 2 1 TT TcnU U v T T <∴ <∆=∆ <∆ ∫ T2 will be less than 30 ºC 14 2.11 (a) (i). 2 P [b ar ] v [m3/mol] 1 2 3 0.01 0.030.02 1 Path A Path B (ii). Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal 0=∆u Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore, 0=∆h Performing an energy balance and neglecting potential and kinetic energy produces 0=+=∆ wqu For an isothermal, adiabatic process, Equation 2.77 states = 1 2ln P PnRTW or == 1 2ln P PRT n Ww Substituting the values from the problem statement gives 15 ( ) + ⋅ = bar 3 bar 1lnK )15.27388( Kmol J 8.314w −= mol J 3299w Using the energy balance above =−= mol J 3299wq (b) (i). See path on diagram in part (a) (ii). Since the overall process is isothermal and u and h are state functions 0=∆u 0=∆h The definition of work is ∫−= dvPw E During the constant volume part of the process, no work is done. The work must be solved for the constant pressure step. Since it is constant pressure, the above equation simplifies to )( 12 vvPdvPw EE −−=−= ∫ The ideal gas law can be used to solve for 2v and 1v ( ) ( ) = × + ⋅ == = × + ⋅ == mol m 010.0 Pa 103 K )15.27388( K molJ 314.8 mol m 030.0 Pa 101 K )15.27388( K molJ 314.8 3 5 1 1 1 3 5 2 2 2 P RTv P RTv Substituting in these values and realizing that 1PPE = since the process is isobaric produces 16 − ×−= mol m 010.0 mol m 030.0Pa) 103( 33 5w −= mol J 6000w Performing an energy balance and neglecting potential and kinetic energy results in 0=+=∆ wqu =−=∴ mol J 6000wq 17 2.12 First, perform an energy balance. No work is done, and the kinetic and potential energies can be neglected. The energy balance reduces to QU =∆ We can use Equation 2.53 to get ∫= 2 1 T T vdTcnQ which can be rewritten as ∫= 2 1 T T PdTcnQ since the aluminum is a solid. Using the atomic mass of aluminum we find mol 3.185 mol kg 0.02698 kg 5 = =n Upon substitution of known values and heat capacity data from Table A.2.3, we get ( ) ( )∫ −×+ ⋅ = K 15.323 K 15.294 3 1049.1486.2 Kmol J 314.8mol 3.185 dTTQ [ ]kJ 61.131=Q 18 2.13 First, start with the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes WQU +=∆ The value of the work will be used to obtain the final temperature. The definition of work (Equation 2.7) is ∫−= 2 1 V V EdVPW Since the piston expands at constant pressure, the above relationship becomes ( )12 VVPW E −−= From the steam tables = kg m 02641.0ˆ 3 1v (10 MPa, 400 ºC) [ ]33111 m 07923.0kg m 02641.0kg) 3(ˆ = == vmV Now 2V and 2v are found as follows [ ]36312 m 4536.0Pa 100.2 J 748740m 07923.0 = × − −=−= EP WVV [ ][ ] === kg m 1512.0 kg 3 m 4536.0ˆ 33 2 2 2 m Vv Since 2vˆ and 2P are known, state 2 is constrained. From the steam tables: [ ]Cº 4002 =T kg m 0.1512 bar, 20 3 Now U∆ will be evaluated, which is necessary for calculating Q . From the steam tables: 19 = kg kJ 2.2945ˆ2u kg m 0.1512 bar, 20 3 = kg kJ 4.2832ˆ1u ( )Cº 400 bar, 001 ( ) [ ]( ) [ ]kJ 4.338 kg kJ 4.2832 kg kJ 2.2945kg 3ˆˆ 121 = − =−=∆ uumU Substituting the values of U∆ and W into the energy equation allows calculation of Q WUQ −∆= [ ]( ) [ ]J 1009.1J 748740 [J] 338400 6×=−−=Q 20 2.14 In a reversible process, the system is never out of equilibrium by more than an infinitesimal amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of 1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible. To solve for the final temperature of the system, the energy balance will be written. The piston- cylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore, potential and kinetic energy effects can be neglected. The energy balance simplifies to WU =∆ Conservation of mass requires 21 nn = Let 21 nnn == The above energy balance can be rewritten as ∫∫ −= 2 1 2 1 V V E T T v dVPdTcn Since vc and EP are constant: ( ) ( )1212 VVPTTnc Ev −−=− 2V and 1T can be rewritten using the ideal gas law 2 2 2 P nRTV = nR VPT 111 = Substituting these expressions into the energy balance, realizing that 2PPE = , and simplifying the equation gives nR VPP T 2 7 2 5 112 2 + = Using the following values 21 [ ] [ ] [ ] [ ] ⋅ ⋅ = = = = = Kmol barL 08314.0 mol 0.1 L 10 bar 1 bar 2 1 2 1 R n V P P results in [ ]K 2062 =T To find the value for work, the energy balance can be used ( )12 TTncUW v −=∆= Before the work can be calculated, 1T must be calculated [ ]( ) [ ]( ) [ ]( ) [ ]K 241 Kmol barL 08314.0mol 1 L 10bar 211 1 = ⋅ ⋅ == nR VPT Using the values shown above [ ]J 727−=W 22 2.15 The maximum work can be obtained through a reversible expansion of the gas in the piston. Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston assembly is well-insulated, so the heat transfer contribution to the energy balance can be neglected, in addition to potential and kinetic energy effects. The energy balance reduces to WU =∆ In this problem, the process is a reversible, adiabatic expansion. For this type of process, Equation 2.90 states [ ]11221 1 VPVP k W − − = From the problem statement (refer to problem 2.13), [ ] [ ] [ ]bar 1 L 10 bar 2 2 1 1 = = = P V P To calculate W, 2V must be found. For adiabatic, reversible processes, the following relationship (Equation 2.89) holds: constPV k = where k is defined in the text. Therefore, kkV P PV 1 1 2 1 2 = Noting that 5 7 == v P c ck and substituting the proper values provides [ ]L 4.162 =V Now all of the needed values are available for calculating the work. [ ] [ ]J 900barL 9 −=⋅−=W From the above energy balance, [ ]J 900−=∆U 23 The change in internal energy can also be written according to Equation 2.53: ∫=∆ 2 1 T T vdTcnU Since vc is constant, the integrated form of the above expression is ( )122 5 TTRnU − =∆ Using the ideal gas law and knowledge of 1P and 1V , [ ]K 6.2401 =T and [ ]K 3.1972 =T The temperature is lower because more work is performed during the reversible expansion. Review the energy balance. As more work is performed, the cooler the gas will become. 24 2.16 Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible. Furthermore, no work is done on the system and potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes 0ˆ =∆u or 12 ˆˆ uu = From the steam tables = kg kJ 2.2619ˆ1u (200 bar, 400 ºC) =∴ kg kJ 2.2619ˆ2u The values of 2uˆ and 2P constrain the system. The temperature can be found from the steam tables using linear interpolation: Cº 5.3272 =T [ ] = kg kJ 2.2619ˆ ,bar 100 2u Also at this state, = kg m 02012.0ˆ 3 2v Therefore, [ ]( ) [ ]332 m 020.0kg m 02012.0kg 0.1 = == mvVvessel 25 2.17 Let the entire tank represent the system. Since no heat or work crosses the system boundaries, and potential and kinetic energies effects are neglected, the energy balance is 0=∆u Since the tank contains an ideal gas K 300 0 12 12 == =− TT TT The final pressure can be found using a combination of the ideal gas law and conservation of mass. 22 2 11 1 VP T VP T = We also know 12 2VV = Therefore, bar 5 2 1 2 == PP 26 2.18 (a) First, as always, simplify the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance is WQU +=∆ Since, this system contains water, we can the use the steam tables. Enough thermodynamic properties are known to constrain the initial state, but only one thermodynamic property is known for the final state: the pressure. Therefore, the pressure-volume relationship will be used to find the specific volume of the final state. Since the specific volume is equal to the molar volume multiplied by the molecular weight and the molecular weight is constant, the given expression can be written constvP =5.1ˆ This equation can be used to solve for 2vˆ . 5.1 1 5.1 1 2 1 2 ˆˆ = v P Pv Using [ ] [ ] [ ] [ ]bar 100 kg m 1.0 kg 10 m 1.0 ˆ bar 20 2 33 1 1 = == = P v P gives = kg m 0.0342 ˆ 3 2v Now that the final state is constrained, the steam tables can be used to find the specific internal energy and temperature. [ ] = = kg kJ 6.3094ˆ K 7.524 2 2 u T To solve for the work, refer to the definition (Equation 2.7). 27 ∫−= 2 1 V V EdVPW or ∫−= 2 1 ˆ ˆ ˆˆ v v E vdPw Since the process is reversible, the external pressure must never differ from the internal pressure by more than an infinitesimal amount. Therefore, an expression for the pressure must be developed. From the relationship in the problem statement, constvPvP == 5.111 5.1 ˆˆ Therefore, the expression for work becomes ∫∫ −=−= 2 1 2 1 ˆ ˆ 5.1 5.1 11 ˆ ˆ 5.1 5.1 11 ˆ ˆ 1ˆˆ ˆ ˆˆ v v v v vd v vPvd v vPw Integration and substitution of proper values provides = ⋅ = kg kJ 284 kg mbar 840.2ˆ 3 w [ ]( ) [ ]kJ 2840 kg kJ 284kg 10 = =∴W A graphical solution is given below: 28 To solve for Q , U∆ must first be found, then the energy balance can be used. ( ) [ ]( ) [ ]kJ 4918 kg kJ 8.2602 kg kJ 6.3094kg 10ˆˆ 12 = − =−=∆ uumU Now Q can be found, [ ] [ ] [ ]kJ 2078kJ 2840kJ 4918 =−=−∆= WUQ (b) Since the final state is the same as in Part (a), U∆ remains the same because it is a state function. The energy balance is also the same, but the calculation of work changes. The pressure from the weight of the large block and the piston must equal the final pressure of the system since mechanical equilibrium is reached. The calculation of work becomes: ∫−= 2 1 ˆ ˆ ˆ v v E vdmPW All of the values are known since they are the same as in Part (a), but the following relationship should be noted 2PPE = Substituting the appropriate values results in [ ]kJ 6580=W Again we can represent this process graphically: 29 Now Q can be solved. [ ] [ ] [ ]kJ 1662kJ 6580kJ 4918 −=−=−∆= WUQ (c) This part asks us to design WU =∆ a process based on what we learned in Parts (a) and (b). Indeed, as is characteristic of design problems there are many possible alternative solutions. We first refer to the energy balance. The value of heat transfer will be zero when For the same initial and final states as in Parts (a) and (b), [ ]kJ 4918=∆= UW There are many processed we can construct that give this value of work. We show two alternatives which we could use: Design 1: If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a reversible compression followed by an irreversible compression. Let the subscript “i" represent the intermediate state where the process switches from a reversible process to an irreversible process. The equation for the work then becomes [ ]J 4918000ˆ ˆ 1ˆ)ˆˆ( ˆ ˆ 5.1 5.1 1122 1 = −−−= ∫ iv v i vd v vPvvPmW Substitutingin known values (be sure to use consistent units) allows calculation of ivˆ : = kg m 0781.0ˆ 3 iv The pressure can be calculated for this state using the expression from part (a) and substituting the necessary values. [ ]bar 0.29 5.1 1 1 = = i i i P v vPP 30 Now that both iP and ivˆ are known, the process can be plotted on a P-v graph, as follows: Design 2: In an alternative design, we can use two irreversible processes. First we drop an intermediate weight on the piston to compress it to an intermediate state. This step is followed by a step similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this case, we again must find the intermediate state. Writing the equation for work: [ ] [ ]J 4918000)ˆˆ()ˆˆ( 221 =−−−−= iii vvPvvPmW However, we again have the relationship: 5.1 1 1 = i i v vPP Substitution gives one equation with one unknown vi [ ]J 4918000)ˆˆ()ˆˆ( 221 5.1 1 1 = −−− −= ii i vvPvv v vPmW : There are two possible values vi = kg m 043.0ˆ 3 iv to the above equation. Solution A: 31 which gives [ ]bar 8.70=iP This solution is graphically shown below: Solution B = kg m 0762.0ˆ 3 iv which gives [ ]bar 0.30=iP This solution is graphically shown below: 32 2.19 (a) Force balance to find k: Piston Fspring=kx Fatm=PatmA Fgas=PgasA Fmass=mg Pgas = Patm + mg A − kx A since ∆V=Ax Pgas = Patm + mg A + k∆V A2 since ∆V is negative. Now solve: ( )( ) ( ) ( )22 3 2 2 55 m 1.0 m 02.0 m 1.0 m/s 81.9kg 2040Pa 101Pa 102 k−+×=× ×=∴ m N1001.5 4k Work can be found graphically (see P-V plot) or analytically as follows: Substituting the expression in the force balance above: WA = Patm + mg A + k∆V A2 ∫ dV ( )∫ ∫ ∆ ∆ ∆ ∆ + += f i f i V V V V atmA Vd A VkdV A mgPW 2 ( )( ) [ ] ( ) ( ) −× +−×= 2 0m02.0 m 1.0 N/m 1001.5m 05.003.0Pa1000.3 223 22 4 25 AW 33 [ ]J 105 3×−=AW V [m 3] 1 2 3 0.01 0.03 0.05 Patm mg A k∆V A2 -W 10squares 0.5kJsquare = 5kJ Work P [b ar ] 2 1 = (b) You need to find how far the spring extends in the intermediate (int) position. Assume PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer slightly). Since you know P and V for each state in Part (a), you can calculate n. PiPf = Vi V f n or 10 5 Pa 2×105 Pa = 0.03 m 3 0.05 m3 n ⇒ n = 1.35 Now using the force balance Pgas,int = Patm + mg A + k∆V A2 with the above equation yields: Pint = Pi Vi V int( ) n = Patm + mg A + k Vint −Vi( ) A2 This last equality represents 1 equation. and 1 unknown (we know k), which gives 3int m 0385.0=V Work can be found graphically (see P-V plot) or analytically using: 34 WB = Pgas Vi V int ∫ dV + Pgas Vint V f ∫ dV expanding as in Part (a) WB = 2 ×10 5 N m2 Vi V int ∫ dV + 5 ×106 N m5 ∆V ∆Vi ∆Vint ∫ d ∆V( )+ 3 ×105 Nm2 Vint Vf ∫ dV + 5 ×106 N m 5 ∆V ∆Vint ∆Vif ∫ d ∆V( ) Therefore, [ ]kJ85.3−=BW just like we got graphically. V [m 3] 1 2 3 0.01 0.03 0.05 Patm mg A P [b ar ] 2 1 Work mg A (c) The least amount of work is required by adding differential amounts of mass to the piston. This is a reversible compression. For our assumption that PVn ∫∫ == f i f i V V V V gasrevC dV V constdVPW 35.1, = const, we have the following expression: Calculate the constant from the initial state 35 ( )( ) 335.13 5 1075.1m05.0Pa101. ×=×=const Therefore, WC,rev = 1.75 ×10 3 .05 .03 ∫ dV V1.35 = − 1.75×10 3 0.35 V −0.35 .05 .03 [ ]J2800−= 36 2.20 Before this problem is solved, a few words must be said about the notation used. The system was initially broken up into two parts: the constant volume container and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” denotes the final state. To begin the solution, the mass of water present in each part of the system will be calculated. The mass will be conserved during the expansion process. Since the water in the rigid tank is saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is constrained to a specific state. From the steam tables, [ ] kPa 8.1553 kg kJ 3.2595ˆ kg kJ 64.850ˆ kg kJ 12736.0ˆ kg kJ 001156.0ˆ ,1 ,1 ,1 ,1 = = = = = sat v i l i v i l i P u u v v (Sat. water at 200 ºC) Knowledge of the quality of the water and the overall volume of the rigid container can be used to calculate the mass present in the container. ( ) ( )vil i vmvmV ,11,111 ˆ95.0ˆ05.0 += Using the values from the steam table and [ ]31 m 5.0=V provides [ ]kg 13.41 =m Using the water quality specification, [ ] [ ]kg 207.005.0 kg 92.395.0 11 11 == == mm mm l v For the piston-cylinder assembly, both P and T are known. From the steam tables 37 = = kg kJ 9.2638ˆ kg m 35202.0ˆ ,2 3 ,2 i i u v (600 kPa, 200 ºC) Enough information is available to calculate the mass of water in the piston assembly. [ ]kg 284.0 ˆ2 2 2 == v Vm Now that the initial state has been characterized, the final state of the system must be determined. It helps to consider what physically happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of the entire system will match the surroundings’. In other words, == kg kJ 9.2638ˆˆ ,2 if uu (600 kPa, 200 ºC) Thus, the change in internal energy is given by l i l i v i v iif l i v i umumumummmU ,1,1,1,1,22,1,12 ˆˆˆˆ)( −−−++=∆ Substituting the appropriate values reveals [ ]kJ 0.541=∆U To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa (weight of the piston was assumed negligible). From Equation 2.7, )( ifE V V E VVPdVPW f i −−=−= ∫ where [ ] [ ] [ ] [ ] [ ]333 3 ,2,1,12 m 6.0m 5.0m 0.1 m 55.1ˆ)( Pa 600000 =+= =++= = i i l i v if E V vmmmV P 38 Note: iv ,2ˆ wasused to calculate fV because the temperature and pressure are the same for the final state of the entire system and the initial state of the piston-cylinder assembly. The value of W can now be evaluated. [ ]kJ 570−=W The energy balance is used to obtain Q. [ ] [ ]( ) [ ]kJ 1111kJ 570kJ 0.541 =−−=−∆= WUQ 39 2.21 A sketch of the process follows: The initial states are constrained. Using the steam tables, we get the following: State 1,A State 1,B p 10 [bar] 20 [bar] T 700 [o 250 [C] oC] v 0.44779 [m3 0.11144 [m/kg] 3/kg] u 3475.35 [kJ/kg] 2679.58 [kJ/kg] V 0.01 m 0.05 m3 3 m = V v 0.11 [kg] 0.090 [kg] All the properties in the final state are equal. We need two properties to constrain the system: We can find the specific volume since we know the total volume and the mass: [ ][ ] == + + = kg m 30.0 kg 0.20 m 06.0 33 ,1,1 ,1,1 2 BA BA mm VV v We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is rigid, W=0. An energy balance gives: ∆U = Q −W = 0 Thus, U2 =U1 = m1,Au1, A + m1,Bu1,B or = + + == kg kJ 3121 ,1,1 ,1,1,1,1 2 2 2 BA BBAA mm umum m Uu We have constrained the system with u2 and v2, and can find the other properties from the steam Tables. Very close to T2 = 500 [oC] and P2 = 1200 [kPa] Thus, 40 [ ]( ) [ ]m 267.0kg 0.09 kg m 30.0 3 ,2,2,2 = == AAA mvV and ( ) [ ]m 167.01.0267.0 =−=∆x 41 2.22 We start by defining the system as a bubble of vapor rising through the can. We assume the initial temperature of the soda is 5 oC. Soda is usually consumed cold; did you use a reasonable estimate for T1? A schematic of the process gives: where the initial state is labeled state 1, and the final state is labeled state 2. To find the final temperature, we perform an energy balance on the system, where the mass of the system (CO2 wu =∆ in the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy effects are negligible, the energy balance is Expressions for work and internal energy can be substituted to provide ( ) ( )1212 vvPdvPTTc EEv −−=−=− ∫ where cv = cP – R. Since CO2 ( ) −−= −−=− 1 12 2 1 1 2 2 12 P TPTR P T P TRPTTc Ev is assumed an ideal gas, the expression can be rewritten as where the equation was simplified since the final pressure, P2, is equal to the external pressure, PE += + 1 2 12 11 Pc RPT c RT vv . Simplifying, we get: or K 2371 1 2 12 = += P v v c c Pc RPTT 42 2.23 The required amount of work is calculated as follows: VPW ∆−= The initial volume is zero, and the final volume is calculated as follows: ( ) 3333 m0436.0ft 54.1ft 5.0 3 4 3 4 ==== ππrV Assuming that the pressure is 1 atm, we calculate that ( )( ) J4417m 0m 0436.0Pa1001325.1 33 5 =−×=W This doesn’t account for all of the work because work is required to stretch the rubber that the balloon is made of. 43 2.24 (a) Since the water is at its critical point, the system is constrained to a specific temperature, pressure, and molar volume. From Appendix B.1 = kg m 003155.0ˆ 3 cv Therefore, [ ] [ ]kg 17.3 kg m 003155.0 m 01.0 ˆ 3 3 = == cv Vm (b) The quality of the water is defined as the percentage of the water that is vapor. The total volume of the vessel can be found using specific volumes as follows ( )[ ] ( ) vlvvll vxmvmxvmvmV ˆˆ1ˆˆ +−=+= where x is the quality of the water. To solve for the quality, realize that starting with saturated water at a pressure of 1 bar constrains the water. From the steam tables, = = kg m 001043.0ˆ kg m 6940.1ˆ 3 3 l v v v (sat. H2 00125.0=x O at P = 1 bar) Now the quality can be found Thus, the quality of the water is 0.125%. (c) To determine the required heat input, perform an energy balance. Potential and kinetic energy effects can be neglected, and no work is done. Therefore, QU =∆ 44 where ( )[ ] ( )[ ]vl uxmumxumU 112 ˆˆ1ˆ +−−=∆ From the steam tables = kg kJ 58.2029ˆ2u (H2 = = kg m 1.2506ˆ kg m 33.417ˆ 3 1 3 1 v l u u O at its critical point) (sat. H2 [ ] [ ]J 1010.5kJ 6.5102 6×==∆U O at P=1 bar) Evaluation of the expression reveals 45 2.25 (a) Consider the air in ChE Hall to be the system. The system is constant volume, and potential and kinetic energy effects can be neglected. Furthermore, disregard the work. The energy balance is q dt du = since the temperature of the system changes over time. Using the given expression for heat transfer and the definition of dU , the expression becomes ( )surrv TThdt dTc −−= We used a negative sign since heat transfer occurs from the system to the surroundings. If vc is assumed constant, integration provides ( ) ChtTTc surrv +−=−ln where C is the integration constant. Therefore, − += t c h surr veCTT 1 where C1 time Te m pe ra tu re T0 Tsurr is a constant. Examining this equation reveals that the temperature is an exponential function of time. Since the temperature is decreasing, we know that the plot of temperature vs. time shows exponential decay. 46 (b) Let time equal zero at 6 PM, when the steam is shut off. At 6 PM, the temperature of the hall is 22 ºC. Therefore, − += t c h surr veCTT 1 )0(1Cº 2 Cº 22 eC+= Cº 201 =∴C After 10 PM, ( hr 4=t ), the temperature is 12 ºC. ( ) − += hr) 4( Cº 20Cº 2 Cº 21 vc h e 1-hr 173.0=∴ vc h At 6 AM, hr 12=t . Substitution of this value into the expression for temperature results in Cº 5.4=T 47 2.26 The gas leaving the tank does flow work as it exits the valve. This work decreases the internal energy of the gas – lowering the temperature. During this process, water from the atmosphere will become supersaturated and condense. When the temperature drops below the freezing point of water, the water forms a solid. Attractive interactions between the compressed gas molecules can also contribute to this phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of these interactions in Chapter 4. 48 2.27 Mass balance inoutin mmmdt dm =−= Separating variables and integrating: ∫∫ = t in m m dtmdm 0 2 1 or ∫=− t indtmmm 0 12 Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is ∑ ∑ ++−= out s in ininoutout sys WQhmhm dt dU The process is adiabatic and no shaft work is done. Furthermore,there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to inin sys hm dt dU = The following math is performed ( ) in t inin t inin U U hmmumumUU dtmhdthmdU ˆˆˆ 12112212 00 2 1 −=−=− == ∫∫∫ where the results of the mass balance were used. Both 2m and 1m can be calculated by dividing the tank volume by the specific volume 49 1 1 2 2 ˆ ˆ v Vm v Vm = = Substitution of these relationships and simplification results in ( ) ( ) 0 ˆ ˆ ˆ ˆ 1 1 2 2 = − − − v hu v hu inin From the steam tables: = = kg m 19444.0ˆ kg kJ 6.2583ˆ 3 1 1 v u (sat. H2 = kg kJ 2.3177iˆnh O vapor at 1 MPa) (6 MPa, 400 ºC) There are still two unknowns for this one equation, but the specific volume and internal energy are coupled to each other. To solve this problem, guess a temperature and then find the corresponding volume and internal energy values in the steam tables at 6 MPa. The correct temperature is the one where the above relationship holds. Cº 600=T : Expression = 4427.6 Cº 500=T : Expression = 1375.9 Cº 450=T : Expression = -558.6 Interpolation between 500 ºC and 450 ºC reveals that the final temperature is Cº 4.4642 =T 50 2.28 We can pick room temperature to be 295 K Tin = T1 = 295 K[ ] Mass balance inoutin nnndt dn =−= Separating variables and integrating: ∫∫ = t in n n dtndn 0 2 1 or ∫=− t indtnnn 0 12 Energy balance Neglecting ke and pe, he unsteady energy balance, written in molar units is written as: WQhnhn dt dU outoutinin sys −+−= The terms associated with flow out, heat and work are zero. inin sys hn dt dU = Integrating both sides with respect to time from the initial state where the pressure is 10 bar to the final state when the tank is at a pressure of 50 bar gives: dtnhdthndU t inin t inin U U ∫∫∫ == 00 2 1 since the enthalpy of the inlet stream remains constant throughout the process. Integrating and using the mass balance above: ( ) inhnnunun 121122 −=− 51 Now we do some math: ( ) inhnnunun 121122 −=− ( ) ( )inin hunhun −=− 1122 By the definition of h 11 RTuRTuvPuh inininininin +=+=+= so ( ) ( ) 1111112122 TRnuunRTnuun −−=−− ( ) 1112122 RTnRTnTTcn v −=−− Since RRcc Pv 2 3 =−= n2 3 2 T2 − T1( )− n2T1 = −n1T1 or 111222 253 TnTnTn −=− dividing by n1 11 1 2 2 1 2 253 TT n nT n n −=− : Using the ideal gas law: 21 12 1 2 TP TP n n = so 11 21 12 2 21 12 253 TT TP TPT TP TP −= − or 52 [K] 434 23 5 1 2 1 12 2 = + = P P P TP T (b) Closed system ∆u = q − w = q ( ) ( ) =−=−=∆ kg kJ 9.28 2 5ˆ 1212 TTMW RTT MW cu v = kg kJ 9.28q (c) P2T2 = P3T3 [ ]bar 34 3 22 3 == T TPP 53 2.29 Mass balance inoutin nnndt dn =−= Separating variables and integrating: ∫∫ = = t in n n dtndn 00 2 1 or ∫= t indtnn 0 2 Energy balance Neglecting ke and pe, the unsteady energy balance, in molar units, is written as: WQhnhn dt dU outoutinin sys ++−= The terms associated with flow out and heat are zero. Whn dt dU inin sys += Integrating both sides with respect to time from the empty initial state to the final state gives: WnhWdtnhdtWdthndU in t inin tt inin U U +=+=+= ∫∫∫∫ 2 000 2 1 since the enthalpy of the inlet stream remains constant throughout the process. The work is given by: 22122 )( vPnvvPnW extext −=−−= 54 [ ]2222 vPhnun extin −= Rearranging, 222 vPvPuvPhu extinininextin −+=−= 22 vPvPuu extininin −=− ( ) 2 2 2 P TPRRTTTc extininv −=− so ( ) [ ]K 333 2 2 = + + = in ext v v T R P Pc RcT 55 2.30 valve maintains pressure in system constant T1 = 200 oC x1 = 0.4 V = 0.01 m3 v l Mass balance outoutin mmmdt dm −=−= Separating variables and integrating: ∫∫ −= t out m m dtmdm 0 2 1 or ∫−=− t outdtmmm 0 12 Energy balance Qhm dt dU outout sys +−= ˆ Integrating [ ]∫ ∫∫∫ +−=+−= t tt outoutoutout um um dtQdtmhdtQhmdU 0 00 ˆ ˆ ˆˆ 22 11 Substituting in the mass balance and solving for Q ( ) outhmmumumQ ˆˆˆ 121122 −−−= 56 We can look up property data for state 1 and state 2 from the steam tables: =×+×=+−= kg m 0.0511274.04.0001.6.0ˆˆ)1(ˆ 3 1 gf vxvxv = kg m 1274.0ˆ 3 2v So the mass in each state is: [ ] [ ]kg 196.0 kg m 0.051 m 01.0 ˆ 3 3 1 1 1 = == v Vm [ ] [ ]kg 0785.0 kg m 0.1274 m 01.0 ˆ 3 3 2 2 2 = == v Vm [ ]kg 1175.012 −=− mm And for energy and enthalpy =×+×=+−= kg kJ 54915.25974.064.8506.0ˆˆ)1(ˆ1 gf uxuxu = kg kJ 3.2595ˆ2u = kg kJ 2.2793oˆuth Solving for heat, we get ( ) [ ]kJ 228ˆˆˆ 121122 =−−−= outhmmumumQ 57 2.31 Consider the tank as the system. Mass balance inoutin mmmdt dm =−= Separating variables and integrating: ∫∫ = t in m m dtmdm 0 2 1 or ∫=− t indtmmm 0 12 Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is ∑ ∑ ++−= out s in ininoutout sys WQhmhm dt dU The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to inin sys hm dt dU = The following math is performed in t inin t inin U U hmumU dtmhdthmdU ˆˆ 2222 000 2 1 == == ∫∫∫ = where the results of the mass balance were used. Thus, inhu ˆˆ2 = From the steam tables 58 = kg kJ 5.3632ˆ2u (9 MPa, 800 ºC) so = kg kJ 5.3632iˆnh We can use the value of inh and the fact that the steam in the pipe is at 9 MPa to find the temperature. Cº 600=inT 59 2.32 (a) First, the energy balance must be developed. Since the problem asks how much energy is stored in the battery after 10 hours of operation, the process is not steady-state. Let the battery be thesystem. Potential and kinetic energy effects can be neglected. Furthermore, heating of the battery as it is charged can be ignored. The energy balance is s sys WQ dt dU += No shaft work is performed, but electrical is supplied to the battery, which must be accounted for in sW . The value of Q is given explicitly in the problem statement. Both of these values remain constant over time, so integration provides ( )tWQU s +=∆ From the problem statement [ ] [ ] [ ]s 36000 kW 1 kW 5 = −= = t Q Ws Substituting these values allows the calculation of the amount of energy stored: [ ] [ ]MJ 144kJ 000,144 ==∆U (b) To calculate the velocity of the falling water, an energy balance must be developed with the water passing through the electricity generator (probably a turbine) as the system, where the water enters with a velocity 1V and leaves with a negligible velocity, which will be approximated as 0. Assume that potential energy changes can be neglected. Furthermore, assume that the temperature of the water does not change in the process, so the change in internal energy is zero. Also, view the process as adiabatic. The energy balance reduces to riverK WE =∆ where riverW is the power of the flowing water. The actual power being provided by the stream can be calculated using the efficiency information. Let η represent the efficiency. 60 [ ] [ ]kW 10 5.0 kW 5 ===∴ = η η s river river s WW W W The value of riverW should be negative since the water is supplying work that is stored electrical energy. Therefore, the energy balance becomes [ ]W 10000−=∆ KE This expression can be rewritten as ( ) [ ]W 10000 2 1 2 1 2 2 −=−VVm From the problem statement and the assumptions made, = = s m 0 s kg 200 2V m Therefore, = s m 101V There are a number of reasons for the low conversion efficiency. A possible potential energy loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is lost to the surroundings during conversion. Some of the energy is also lost due to friction (drag) effects. 61 2.33 Considering the turbine to be the system, rearrangement of the steady-state, open system energy balance provides ∑ ∑ +=++−++ out s in inpkinoutpkout WQeehneehn )()( Performing a mass balance reveals 21 nnnn outin === Assuming the rate of heat transfer and potential energy effects are negligible and realizing that there is one inlet and one outlet allows the simplification of the above equation to ( ) ( )[ ]1,2,122 KKs eehhnW −+−= ( )12 hh − can be rewritten using Equations 2.58 and Appendix A.2 ( ) ( )∫∫ ++++==− − 2 1 322 12 T T p dTETDTCTBTARdTchh Since the quantity ( )1,2, kk ee − is multiplied by n , it is rewritten as follows for dimensional homogeneity ( ) ( )21221,2, )(2 1 VVMWee airkk −=− To solve for n , the ideal gas law is used 2 22 2 21222 RT VPn RTnVP = = To solve for the volumetric flow rate, the fluid velocity must be multiplied by the cross-sectional area = 4 2 2 2 2 VDV π The energy balance is now 62 ( ) ( ) −+ ++++ = ∫ − 21223222 2 2 2 2 )( 2 1 4 2 1 VVMWdTETDTCTBTARVD RT PW air T T s π Substituting values from Table A.2.1 and the problem statement results in [MW] -4.84[W] 1084.4 6 =×−=W 63 2.34 First, a sketch of the process is useful: q 30 bar 100 oC 20 bar 150 oC To find the heat in we will apply the 1st law. Assuming steady state, the open system energy balance with one stream in and one stream out can be written: ( ) Qhhn +−= 210 which upon rearranging is: 12 hhn Q −= Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an expression for the ideal gas heat capacity: 263 10392.410394.14424.1 TT R c p −− ×−×+= with T in (K). Since this expression is limited to ideal gases any change in temperature must be under ideal conditions. From the definition of heat capacity: ( )∫ −− ×−×+=−= 2 1 263 12 10392.410394.14424.1 T T dTTThh n Q By integrating and substituting the temperatures, we obtain: = mol J5590 n Q 64 2.35 A schematic of the process follows: To solve for nWs / we need a first law balance. With negligible eK and eP ( ) sWQhhn ++−= 210 , the 1st law for a steady state process becomes: If heat transfer is negligible, h n Ws ∆= We can calculate the change in enthalpy from ideal gas heat capacity data provided in the Appendix. [ ]∫∫ −− ×−×+==∆= 2 1 2 1 263 10824.810785.28213.1 T T T T p s dTTTRdTch n W Integrate and evaluate: −= mol J 5358 n Ws 65 2.36 (a) First start with the energy balance. Nothing is mentioned about shaft work, so the term can be eliminated from the energy balance. The potential and kinetic energy effects can also be neglected. Since there is one inlet and one outlet, the energy balance reduces to 1122 hnhnQ −= A mass balance shows 12 nn = so the energy balance reduces to ( )121 hhnQ −= Using the expressions from Appendix A.1, the energy balance becomes ( )∫ ++++= − 2 1 322 1 T T dTETDTCTBTARnQ Using 0 10031.0 0 10557.0 376.3 5 3 = ×−= = ×= = − E D C B A ⋅ = Kmol J 314.8R = s mol 201n [ ] [ ]K 15.773 K 15.373 2 1 = = T T gives [ ] [ ]kW 1.245W 245063 ==Q 66 (b) To answer this question, think about the structure of n-hexane and carbon monoxide. N-hexane is composed of 20 atoms, but carbon monoxide has two. One would expect the heat capacity to be greater for n-hexane since there are more modes for molecular kinetic energy (translational, kinetic, and vibrational). Because the heat capacity is greater and the rate of heat transfer is the same, the final temperature will be less. 67 2.37 First start with the energy balance around the nozzle. Assume that heat transfer and potential energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance reduces to 0)()( 1122 =+−+ KK ehnehn A mass balance shows 21 nn = On a mass basis, the energy balance is ( )22212,1,12 2 1ˆˆˆˆ VVeehh KK −=−=− Since the steam outlet velocity is much greater than the velocity of the inlet, the above expression is approximately equal to ( )2212 2 1ˆˆ Vhh −=− The change in enthalpy can be calculated using the steam tables. ×= kg J 109.2827 31h (10 bar, 200 ºC) ×= kg J 105.2675 32h(sat. H2 = s m 5522V O(v) at 100 kPa) Therefore, To solve for the area, the following relationship is used 2 2 vˆ VAm = From the steam tables 68 = kg m 6940.1ˆ 3 2v Now all but one variable is known. [ ]23 m 1007.3 −×=A 69 2.38 First start with the energy balance around the nozzle. Assume that heat transfer and potential energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance reduces to 0)()( 1122 =+−+ kk ehnehn The molar flow rates can be eliminated from the expression since they are equal. Realizing that 1,2, KK ee >> since the velocity of the exit stream is much larger than the velocity of the inlet stream simplifies the energy balance to 2,12 kehh −=− Using Appendix A.2 and the definition of kinetic energy ( ) 2232212 83 2 1 )( 2 1 VMWdTETDTCTBTARhh HC T T −=++++=− ∫ − From Table A.2.1 0 0 10824.8 10785.28 213.1 6 3 = = ×−= ×= = − − E D C B A It is also important that the units for the molecular weight and universal gas constant are consistent. The following values were used ⋅ = Kmol J 314.8R = mol kg 0441.0)( 83HC MW Integration of the above expression and then solving for 2T provides [ ]K 2.4192 =T 70 2.39 First start an energy balance around the diffuser. Assume that heat transfer and potential energy effects are negligible. The shaft work term is also zero. The energy balance reduces to 0)()( 1122 =+−+ kk ehnehn A mass balance reveals 21 nn = The molar flow rates can be eliminated from the expression. Using the definitions of enthalpy and the kinetic energy, the equation can be rewritten as ( )2122)(2 12 1 VVMWdTc air T T P −−=∫ The temperature and velocity of the outlet stream are unknown, so another equation is needed to solve this problem. From the conservation of mass, ( ) ( ) 2 222 1 111 1 11 T VAP T VAP T VP == where A2 1 1 2 2 1 2 2 1 V T T P PV = , the cross-sectional area of the diffuser outlet, is twice the area of the inlet. Therefore, Using Appendix A.2 and the above expression, the energy balance becomes ( ) − −=++++∫ − 21 2 1 1 2 2 1322 2 1)( 2 12 1 VV T T P PMWdTETDTCTBTAR air T T Substituting values from the problem statement provides an equation with one unknown: K 3812 =T Therefore, ( ) = = s m 111m/s300 K15.343 K 381 bar 5.1 bar 1 2 1 2V 71 2.40 To find the minimum power required for the compressor, one must look at a situation where all of the power is used to raise the internal energy of the air. None of the power is lost to the surroundings and the potential and kinetic energy effects must be neglected. Therefore, the energy balance becomes sWhnhn +−= 22110 Performing a mass balance reveals 21 nn = The energy balance reduces to ( )121 hhnWs −= Using Equation 2.58 and Appendix A.2, the equation becomes ( )∫ ++++= − 2 1 322 1 T T s dTETDTCTBTARnW Table A.2.1 and the problem statement provide the following values 0 1600 0 10575.0 355.3 3 = −= = ×= = − E D C B A K 300 s mol 50 1 1 = = T n To find the work, we still need T2. We need to pick a reasonable process to estimate T2 .constPV k = . Since the heat flow is zero for this open system problem, we choose an adiabatic, reversible piston situation. For this situation, Since we are assuming the air behaves ideally, we can rewrite the equation as 72 kk P RTnP P RTnP = 2 22 2 1 11 1 kkkk TPTP 2 1 21 1 1 −− = Substituting values from the problem statement, we obtain ( ) ( ) ( ) ( )( ) K 579 bar 10 bar1K300 7/5 5/71 5/71 5/7 2 = = − − T Substitute this value into the expression for the work and evaluate: [ ]kW 4.417=sW 73 2.41 (a) Perform a mass balance: outnnn =+ 21 Apply the ideal gas law: outnRT VP RT VP =+ 2 22 1 11 Substitute values from the problem statement: ( )( )( )( ) ( )( ) ( )( ) outn = ×× + ×× −− 15.293314.8 105.2102 15.373314.8 105101 3535 [ ]mol/s 366.0=outn (b) No work is done on the system, and we can neglect potential and kinetic energy effects. We will assume the process is also adiabatic. The energy balance reduces to out out outin in in hnhn ∑∑ −= 0 ( ) ( ) ( ) 022112211 =−+−=+− hhnhhnhnhnhn outoutoutout We can calculate the enthalpy difference from the given ideal heat capacity: ( ) ( ) 010324.5267.310324.5267.3 15.293 3 2 15.373 3 1 =×++×+ ∫∫ −− outout TT dTTRndTTRn Again, we must calculate the molar flow rates from the ideal gas law. Upon substitution and evaluation, we obtain [ ]K 329=outT 74 2.42 (a) Since the temperature, pressure, and volumetric flow rate are given, the molar flow rate is constrained by the ideal gas law. Note: [ ] [ ]min/cm 1/sm 1067.1 338 =× − ( ) [ ]( ) [ ]( )( ) [ ]mol/s 1045.7K 15.273KJ/mol 314.8 /sm 1067.1Pa100135.1 7385 −− ×= ⋅ ×× == RT VPn To recap, we have shown [ ] [ ]mol/s 1045.7SCCM 1 7−×= (b) Assumptions: N2 is an ideal gas All power supplied by the power supply is transferred to the N2 Uniform temperature radially throughout sensor tube Kinetic and potential energy effects negligible in energy balance Let x represent the fraction of N2 sn diverted to the sensor tube, and represent the molar flow rate through the sensor tube. Therefore, the total molar flow rate, totaln , is x nn stotal = We can use temperature and heat load information from the sensor tube to find the molar flow rate through the sensor tube. First, perform an energy balance for the sensor tube: ( ) QhhnH inouts =−=∆ The enthalpy can be calculated with heat capacity data. Therefore, ∫ = 2 1 T T P s dTc Qn Now, we can calculate the total molar flow rate. 75 ∫ = 2 1 T T P total dTcx Qn To find the flow rate in standard cubic centimeters per minute, apply the conversion factor found in Part (a) [ ] [ ] × = − ∫ mol/s 1045.7 SCCM 1)SCCM( 72 1 T T P total dTcx Qv (c) To find the correction factor for SiH4, re-derive the expression for flow rate for SiH4 and then divide it by the expression for N2 [ ] [ ] [ ] [ ] ∫ ∫ ∫ ∫ = × × == − − 2 1 4 2 1 2 2 1 2 2 1 4 2 4 , , 7 , 7 , , , mol/s 1045.7 SCCM 1 mol/s 1045.7 SCCM 1 T T SiHP T T NP T T NP T T SiHP Ntotal SiHtotal dTcdTc dTcx Q dTcx Q v v Factor for the same power input, temperatures, and fraction of gas diverted to the sensor tube. If we assume that heat capacities are constant, the conversion factor simplifies: 4 2 , , SiHP NP c c Factor = Using the values in Appendix A.2.2 at 298 K, we get 67.0=Factor 76 2.43 (a) It takes more energy to raise the temperature of a gas in a constant pressure cylinder. In both cases the internal energy of the gas must be increased. In the constant pressure cylinder work, Pv work must also be supplied to expand the volume against the surrounding’s pressure. This is not required with a constant volume. (b) As you perspire, sweat evaporates from your body. This process requires latent heat which cools you. When the water content of the environment is greater, there is less evaporation; therefore, this effect is diminished and you do not feel as comfortable. 77 2.44 From the steam tables at 10 kPa: ( ) ⋅ += ⋅ += = Kmol J 001434.0516.3 Kkg kJ 0006624.06241.1 RTT dT dhc P P Now compare the above values to those in Appendix A.2. T(K) h 323.15 2592.6 373.15 2687.5 423.15 2783 473.15 2879.5 523.15 2977.3 573.15 3076.5 673.15 3279.5 773.15 3489 873.15 3705.4 973.15 3928.7 1073.15 4159.1 1173.15 4396.4 1273.15 4640.6 1373.15 4891.2 1473.15 5147.8 1573.15 5409.7 h vs . T y = 0.0003x 2 + 1.6241x + 2035.7 R2 = 1 0 1000 2000 3000 4000 5000 6000 0 500 1000 1500 2000 T Series1 Poly . (Series1) A B Steam Tables 3.516 0.001434 Appendix A 3.470 0.001450 % difference 1.3 1.4 78 2.45 For throttling devices, potential and kinetic energy effects can be neglected. Furthermore, the process is adiabatic and no shaft work is performed. Therefore, the energy balance for one inlet and one outlet is simplified to 2211 hnhn = which is equivalent to 2211 ˆˆ hmhm = Since mass is conserved 221 ˆˆ hmh = From the steam tables: = kg kJ 3.33981ˆh (8 MPa, 500 ºC) =∴ kg kJ 3.3398ˆ2h Now that we know 2uˆ and 2P , 2T is constrained. Linear interpolation of steam table data gives [ ]Cº 4572 =T 79 2.46 (a) An expression for work in a reversible, isothermal process was developed in Section 2.7. Equation 2.77 is = 1 2ln P PnRTW Therefore, = 1 2ln P PRTw Evaluating the expression with [ ] [ ] [ ]kPa 500 kPa 100 K 300 Kmol J 314.8 1 2 = = = ⋅ = P P T R gives −= kg J 4014w (b) Equation 2.90 states [ ]121 TTk Rw − − = Since the gas is monatomic Rc Rc v P = = 2 3 2 5 and 80 3 5 =k 2T can be calculated by applying the polytropic relation derived for adiabatic expansions. From Equation 2.89 constPv constPV k k =∴ = By application of the ideal gas law P RTv = Since R is a constant, substitution of the expression for P into the polytropic relation results in ( ) ( ) ( ) kkkk kk TPTP constTP 2 1 2 1 1 1 1 −− − =∴ = This relation can be used to solve for 2T . [ ]K 6.1572 =T Now that 2T is known, value of work can be solved. −= kg J 9.1775w 81 2.47 (a) The change in internal energy and enthalpy can be calculated using ( ) ( ) ( ) ( )Cº 0 ,atm 1ˆCº 100 ,atm 1ˆˆ Cº 0 ,atm 1ˆCº 100 ,atm 1ˆˆ ll ll uuu hhh −=∆ −=∆ We would like to calculate these values using the steam tables; however, the appendices don’t contain steam table data for liquid water at 0.0 ºC and 1 atm. However, information is provided for water at 0.01 ºC and 0.6113 kPa. Since the enthalpy and internal energy of liquid water is essentially independent of pressure in this pressure and temperature range, we use the steam table in the following way ( ) = kg kJ 02.419Cº 100 ,atm 1ˆlh ( ) ( ) == kg kJ 0Cº 01.0 ,kPa .61130ˆCº 0 ,atm 1ˆ ll hh ( ) = kg kJ 91.418Cº 100 ,atm 1ˆ lu ( ) ( ) == kg kJ 0Cº 01.0 ,kPa .61130ˆCº 0 ,atm 1ˆ ll uu Therefore, =∆ kg kJ 02.419hˆ =∆ kg kJ 91.418uˆ (b) The change in internal energy and enthalpy can be calculated using ( ) ( ) ( ) ( )Cº 100 ,atm 1ˆCº 100 ,atm 1ˆˆ Cº 100 ,atm 1ˆCº 100 ,atm 1ˆˆ lv lv uuu hhh −=∆ −=∆ From the steam tables 82 = = = kg kJ 5.2506ˆ kg kJ 02.419ˆ kg kJ 0.2676ˆ v l v u h h = kg kJ 91.418ˆ lu Therefore, =∆ =∆ kg kJ 59.2087ˆ kg kJ 99.2256ˆ u h The change in internal energy for the process in Part (b) is 5.11 times greater than the change in internal energy calculated in Part (a). The change in enthalpy in Part (b) is 5.39 times greater than the change in enthalpy calculated in Part (a). 83 2.48 To calculate the heat capacity of Ar, O2, and NH3 322 ETDTCTBTA R cP ++++= − the following expression, with tabulated values in Table A.2.1, will be used, where T is in Kelvin. From the problem statement [ ]K 300=T and from Table 1.1 ⋅ = Kmol J 314.8R To find the A-E values, Table A.2.1 must be referred to. Formula A 310×B 610×C 510−×D 910×E Ar - - - - - O 3.639 2 0.506 0 -0.227 0 NH 3.5778 3 3.02 0 -0.186 0 The values are not listed for Ar since argon can be treated as a monatomic ideal gas with a heat capacity independent of temperature. The expression for the heat capacity is Rc ArP = 2 5 , Now that expressions exist for each heat capacity, evaluate the expressions for [ ]K 300=T . ⋅ = ⋅ = ⋅ = Kmol J 560.35 Kmol J 420.29 Kmol J 785.20 3 2 , , , NHP OP ArP c c c By examining the heat capacity for each molecule, it should be clear that the magnitude of the heat capacity is directly related to the structure of the molecule. 84 Ar • Since argon is monatomic, translation is the only mode through which the atoms can exhibit kinetic energy. O • Translation, rotation, and vibration modes are present. Since oxygen molecules are linear, the rotational mode of kinetic energy contributes RT per mol to the heat capacity. 2 NH • Translation, rotation, and vibration modes are present. Ammonia molecules are non- linear, so the rotation mode contributes 3RT/2 per mole to the heat capacity. 3 The vibration contributions can also be analyzed for oxygen and ammonia, which reveals that the vibration contribution is greatest for ammonia. This is due to ammonia’s non-linearity. 85 2.49 For a constant pressure process where potential and kinetic energyeffects are neglected, the energy balance is given by Equation 2.57: HQ ∆= The change in enthalpy can be written as follows ( )12 hhmH −=∆ From the steam tables: ( ) == kg kJ 6.2584kPa 10at vapor sat.ˆ2 hh ( ) == kg kJ 81.191kPa 10at liquid sat.1ˆ hh Therefore, ( ) − = kg kJ 81.191 kg kJ 6.2584kg 2Q [ ]kJ 6.4785=Q We can find the work from its definition: ∫−= f i V V EdVPW The pressure is constant, and the above equation can be rewritten as follows ( )12 ˆˆ vvmPW E −−= From the steam tables: ( ) == kg m 674.14kPa 10at vapor sat.ˆˆ 3 2 vv ( ) == kg m 00101.0kPa 10at liquid sat.ˆˆ 3 1 vv Therefore, 86 ( )( ) kJ 5.293 kg m 00101.0 kg 674.14kg2Pa10000 3 −= − −=W 87 2.50 First, perform an energy balance on the system. Potential and kinetic energy effects can be neglected. Since nothing is mentioned about work in the problem statement, W can be set to zero. Therefore, the energy balance is UQ ∆= Performing a mass balance reveals 12 mm = where vl mmm 111 += Now the energy balance can be written as ( )vvll umumumuumQ 111121121 ˆˆˆ)ˆˆ( +−=−= Since two phases coexist initially (water is saturated) and 1P is known, state 1 is constrained. From the saturated steam tables = = kg kJ 9.2437ˆ kg kJ 79.191ˆ 1 1 v l u u (sat. H2 vl vvll mm vmvm m V m Vv 11 1111 1 1 2 2 2 ˆˆ + + === O at 10 kPa) As heat is added to the system, the pressure does not remain constant, but saturation still exists. One thermodynamic property is required to constrain the system. Enough information is known about the initial state to find the volume of the container, which remains constant during heating, and this can be used to calculate the specific volume of state 2. From the saturated steam tables 88 = = kg m 674.14ˆ kg m 001010.0ˆ 3 1 3 1 l l v v (sat. H2 = kg m 335.1ˆ 3 2v O at 10 kPa) Therefore, The water vapor is now constrained. Interpolation of steam table data reveals = kg kJ 6.2514ˆ2u Now that all of the required variables are known, evaluation of the expression for Q is possible. [ ]kJ 11652=Q 89 2.51 Let the mixture of ice and water immediately after the ice has been added represent the system. Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are neglected, the energy balance reduces 0=∆H We can split the system into two subsystems: the ice (subscript i) and the water (subscript w). Therefore, 0=∆+∆=∆ wwii hmhmH and wwii hmhm ∆−=∆ We can get the moles of water and ice. [ ] [ ]kg 399.0 kg m 001003.0 m .00040 ˆ 3 3 = == w w w v Vm ( ) [ ]( ) [ ]mol 15.22 mol kg 0180148.0 kg 399.0 2 = == OH w w MW mn ( ) [ ]mol 55.5 2 == OH i i MW mn Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0 ºC, the assumption is correct.) The following expression mathematically represents the change in enthalpy. ( )( ) ( )[ ] ( )Cº 25º 0Cº 100 ,,, −−=−+∆−−− fwPwfwPfusiPi TcnCTchcn Note: Assumed the heat capacities are independent of temperature to obtain this expression. From Appendix A.2.3 Rc iP 196.4, = Rc wP 069.9, = and −=∆ mol kJ0.6fush 90 Substitution of values into the above energy balance allows calculation of Tf Cº 12.3=fT . (Our assumption that all the ice melts is correct.) (b) To obtain the percentage of cooling achieved by latent heat, perform the following calculation ( ) ( )iwfwPw fusi latent TTcn hn Fraction ,, −− ∆− = [ ]( ) [ ]( ) ( ) 911.0 Cº 25Cº 12.3 Kmol J 314.8069.9mol 15.22 mol J 6000mol 55.5 = − ⋅ ⋅− =latentFraction %1.91=latentPercent 91 2.52 A mass balance shows vl nnn 112 += To develop the energy balance, neglect kinetic and potential energy. Also, no shaft work is performed, so the energy balance becomes QU =∆ The energy balance can be expanded to ( ) vvllvl unununnQ 1111211 −−+= If the reference state is set to be liquid propane at 0 ºC and 4.68 bar, the internal energies become ( ) ( ) ( )dTRcuu uu u Pvap vap v l ∫ −+∆= ∆= = 2T K 273 2 1 1 Cº 0 Cº 0 0 Once the change in internal energy for vaporization and temperature of state 2 is determined, Q can be solved. As the liquid evaporates, the pressure increases. At state 2, where saturated propane vapor is present, the ideal gas law states 2 2 2 v RTP = To find 2v , assume that lv vv 11 >> . The volume of the rigid container is === mol m 00485.0 3 1 1 111 P RTnvnV vvv Therefore, ( ) = + = mol m 00243.0 3 11 2 vl nn Vv 92 Also, since the propane is saturated, 2P and 2T are not independent of each other. They are related through the Antoine Equation, ( ) CT BAP sat sat + −=ln where 2PP sat = and 2TT sat = Substitution provides, CT BA v RT + −= 22 2ln Using values from Table A.1.1 and ⋅ ⋅ ×= − Kmol mbar 10314.8 3 5R K 7.3012 =T To find vapu∆ , refer to the definition of enthalpy. ( )lvlvvap PvuPvuhhh +−+=−=∆ )( Since lv vv >> , the above the change in internal energy of vaporization can be written as ( ) RThPvhu vapvvapvap −∆=−∆=∆ Therefore, ( ) =∆ mol kJ 39.14Cº 0vapu Evaluation of the following equation after the proper values have been substituted from Table A.2.1 ( ) ( ) ( ) ( ) ( )Cº 00Cº 0 11 K 301.7 K 273 11 vap vl Pvap vl unndTRcunnQ ∆−− −+∆+= ∫ 93 gives [ ]kJ 1.18=Q 94 2.53 The equation used for calculating the heat of reaction is given in Equation 2.72. It states ( )∑ °° =∆ ifirxn hvh This equation will be used for parts (a)-(e). Since the heat of reaction at 298 K is desired, values from Appendix A.3 can be used. (a) First the stoichiometric coefficient must be determined for each species in the reaction. 1 1 1 1 )( )( )( )( 2 2 2 4 = = −= −= gOH gCO gO gCH v v v v From Tables A.3.1 and A.3.2 ( ) ( ) ( ) ( ) −= −= = −= ° ° ° ° mol kJ 82.241 mol kJ 51.393 mol kJ 0 mol kJ 81.74 )(298, )(298, )(298, )(298, 2 2 2 4 gOHf gCOf gOf gCHf h h h h From Equation 2.72, the equation for the heat of reactionis ( ) ( ) ( ) ( ) )(298,)()(298,)()(298,)()(298,)(298, 22222244 gOHfgOHgCOfgCOgOfgOgCHfgCHrxn hvhvhvhvh °°°°° +++=∆ −+ −+ − −−=∆ ° mol kJ 82.241 mol kJ 51.393 mol kJ 0 mol kJ 81.74298,rxnh −=∆ ° mol kJ 52.560298,rxnh Now that a sample calculation has been performed, only the answers will be given for the remaining parts since the calculation process is the same. 95 (b) −=∆ ° mol kJ 53.604298,rxnh (c) −=∆ ° mol kJ 12.206298,rxnh (d) −=∆ ° mol kJ 15.41298,rxnh (e) −=∆ ° mol kJ 38.905298,rxnh 96 2.54 The acetylene reacts according to the following equation C2H2(g) + (5/2)O2(g) 2CO2(g) + H2 mol 1 22 =HCn O(g) (a) First, choose a basis for the calculations. Calculate the heat of reaction at 298 K using Equation 2.72 and Appendix A.3 ( )∑ °° =∆ ifirxn hvh ( ) ( ) ( ) ( ) OHfCOfOfHCfrxn hhhhh 22222 25.2 °°°°° ++−−=∆ ( ) [ ]J 10255.1 mol J 10255.1 6 298, 6 22 ×−=∆=∆ ×−=∆ ° ° rxnHCrxn rxn hnH h The required amount of oxygen is calculated as follows ( ) mol 5.2)(5.2 11 222 == HCO nn The compositions for both streams are Streams 22 HCn 2On 2Nn 2COn OHn 2 1 (Inlet) 1 2.5 0 0 0 2 (Outlet) 0 0 0 2 1 From Table A.2.2 Species A 310×B 610×C 510−×D 910×E C2H 6.132 2 1.952 0 -1.299 0 O 3.639 2 0.506 0 -0.227 0 CO 5.457 2 1.045 0 -1.157 0 H2 3.470 O 1.45 0 0.121 0 Integration of the following equation provides an algebraic expression where only 2T is unknown. 97 ( ) ( ) 0 2 298 2298, =+∆ ∫ ∑ T i iPirxn dTcnH Substituting the proper values into the expression gives K 61692 =T (b) The calculations follow the procedure used in Part (a), but now nitrogen is present. The basis is mol 1 22 =HCn The heat of reaction is the same as in Part (a), but the gas composition is different. Since stoichiometric amount of air is used, ( ) mol 5.2 12 =On ( ) ( ) mol 40.9 2 2 22 11 = =∴ airO N ON y y nn The composition of the streams are summarized below Streams 22 HCn 2On 2Nn 2COn OHn 2 1 1 2.5 9.40 0 0 2 0 0 9.40 2 1 From Appendix A.2 Species A 310×B 610×C 510−×D 910×E C2H 6.132 2 1.952 0 -1.299 0 O 3.639 2 0.506 0 -0.227 0 CO 5.457 2 1.045 0 -1.157 0 H2 3.470 O 1.45 0 0.121 0 N 3.280 2 0.593 0 0.04 0 Therefore, K 27922 =T 98 (c) Now excess air is present, so not all of the oxygen reacts. The heat of reaction remains the same because only 1 mole of acetylene reacts. Since the amount of air is twice the stoichiometric amount ( ) mol 5 12 =On ( ) ( ) mol 80.18 2 2 22 11 = =∴ airO N ON y y nn The compositions of the streams are summarized below Streams 22 HCn 2On 2Nn 2COn OHn 2 1 1 5 18.8 0 0 2 0 2.5 18.8 2 1 The table of heat capacity data in Part (b) will be used for this calculation. Using the expression shown in Part (a) K 17872 =T 99 2.55 (a) The combustion reaction for propane is C3H8(g) + 5O2(g) 3CO2(g) + 4H2 ( ) ( ) ( ) ( ) OHfCOfOfHCfrxn hhhhh 22283 435 °°°°° ++−−=∆ O(g) For all subsequent calculations, the basis is one mole of propane. The heat of reaction is calculated as follows ( ) [ ]J 10044.2 mol J 10044.2 6 298, 6 83 ×−=∆=∆ ×−=∆ ° ° rxnHCrxn rxn hnH h The required amount of oxygen for complete combustion of propane is ( ) mol 5)(5 11 832 == HCO nn ( ) ( ) mol 81.18 2 2 22 11 = =∴ airO N ON y y nn The stream compositions are listed below Streams 83 HCn 2On 2Nn 2COn OHn 2 1 (Inlet) 1 5 18.8 0 0 2 (Outlet) 0 0 18.8 3 4 From Table (a)2.2 Species A 310×B 610×C 510−×D 910×E N 3.280 2 0.593 0 0.04 0 CO 5.457 2 1.045 0 -1.157 0 H2 3.470 O 1.45 0 0.121 0 Now all of the necessary variables for the following equation are known, except 2T . ( ) ( ) 0 2 298 2298, =+∆ ∫ ∑ T i iPirxn dTcnH Solving the resulting expression provides K 23742 =T 100 (b) The combustion reaction for butane is C4H10(g) + (13/2)O2(g) 4CO2(g) + 5H2 [ ]J 10657.2 6298, ×−=∆ rxnH O(g) For all subsequent calculations, the basis is one mole of butane. The heat of reaction is calculated as shown in Part (a) The moles of nitrogen and oxygen in the feed stream are calculated according to the method in Part (a). The compositions are Streams 104 HCn 2On 2Nn 2COn OHn 2 1 (Inlet) 1 6.5 24.5 0 0 2 (Outlet) 0 0 24.5 4 5 The Pc data listed in Part (a) can also be used for this reaction since there is no remaining butane. K 23762 =T (c) The combustion reaction for pentane is C5H12(g) + 8O2(g) 5CO2(g) + 6H2 [ ]J 10272.3 6298, ×−=∆ rxnH O(g) The basis is one mole of pentane. The heat of reaction is calculated as shown in Part (a). The moles of nitrogen and oxygen in the feed stream are calculated according to the method in Part (a). The compositions are listed below Streams 125 HCn 2On 2Nn 2COn OHn 2 1 1 8 30.1 0 0 2 0 0 30.1 5 6 Substitution of the values into the expression used to find 2T and subsequent evaluation results in K 23822 =T The adiabatic flame temperatures are nearly identical in all three cases. 101 2.56 The equation for the combustion of methane is CH4(g) + 2O2(g) CO2(g) + 2H2 ×−=∆ ° mol J 1002.8 5rxnh O(g) Using Equation 2.72 The basis for this problem is ( ) mol 1 14 =CHn Also, let ξ represent the fractional conversion of methane. Therefore, the composition of the product gas leaving the reactor is ( ) ( ) ( ) ( ) ( ) ( ) ( ) mol 2 mol mol 52.7 mol 12 mol 11 2 2 2 2 2 2 2 2 2 4 ξ ξ ξ ξ = = = −= −= OH CO N O CH n n n n n Furthermore, the heat of reaction is calculated as follows ( ) [ ]J 1002.8 5298, ×−=∆ ξrxnH From Table A.2.2 Species A 310×B 610×C 510−×D 910×E CH 1.702 4 9.081 -2.164 0 0 O 3.639 2 0.506 0 -0.227 0 CO 5.457 2 1.045 0 -1.157 0 H2 3.470 O 1.45 0 0.121 0 N 3.280 2 0.593 0 0.04 0 After substitution of the outlet composition values, heat capacity data, and the heat of reaction into the following equation 102 ( ) ( ) 0 1273 298 2298, =+∆ ∫ ∑ i iPirxn dTcnH integration provides an equation with one unknown: ε . Solving the equation gives ξ =0.42 Since the fractional conversion is 0.42, 58% of the methane passed through the reactor unburned. 103 2.57 For the entire cycle, [ ]kJ 50 0 12 31231211 −=∆∴ =∆+∆+∆=∆ U UUUU From state 1 to state 2 [ ]kJ 40012 121212 −=∴ +=∆ Q WQU From state 2 to state 3 [ ]kJ 023 232323 =∴ +=∆ Q WQU From state 3 to state 1 [ ]kJ 25031 313131 −=∴ +=∆ W WQU Hence, the completed table is Process [ ]kJ U∆ [ ]kJ W [ ]kJ Q State 1 to 2 -50 -400 350 State 2 to 3 800 800 0 State 3 to 1 -750 -250 -500 To determineif this is a power cycle or refrigeration cycle, look at the overall heat and work, 11W and 11Q . [ ] [ ]kJ 150 kJ 150 31231211 31231211 −=++= =++= QQQQ WWWW Since work is done on the system to obtain a negative value of heat, which means that heat is leaving the system, this is a refrigeration cycle. 104 2.58 Refer to the graph of the Carnot cycle in Figure E2.20. From this graph and the description of Carnot cycles in Section 2.9, it should be clear that state 3 has the lowest pressure of all 4 states, and state 1 has the highest pressure. States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since both the temperature and pressure are known for states 1 and 3, the molar volume can be calculated using P RTv = The table below summarizes the known thermodynamic properties. State [ ]K T [ ]bar P [ ]/molm 3v 1 1073 60 0.00149 2 1073 3 298 0.2 0.124 4 298 For each step of the process, potential and kinetic energy effects can be neglected. The step from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in internal energy is 0, and the energy balance becomes 1212 WQ −= From Equation 2.77, = 1 2 112 ln P PnRTW where 12W is the work done from state 1 to state 2. The value of 2P is not known, but recognizing that the process from state 2 to 3 is an adiabatic expansion provides an additional equation. The polytropic relationship can be employed to find 2P . A slight modification of Equation 2.89 provides constPvk = From the ideal gas law v RTP = Combining this result with the polytropic expression and noting that R is constant, allows the expression to be written as 105 constTvk =−1 Therefore, 1 1 1 3 2 3 2 −− = kkv T Tv Substituting the appropriate values (k=1.4) gives = mol m 00504.0 3 2v Applying the ideal gas law [ ]bar 7.17 2 2 2 == v RTP Now, 12W can be calculated. [ ]kJ 89.1012 −=W Calculation of 34W follows a completely analogous routine as calculation for 12W . The following equations were used to find the necessary properties = = −− mol m 0367.0 31 1 1 1 4 1 4 kkv T Tv [ ]bar 675.0 4 4 4 == v RTP Now the following equation can be used = 3 4 334 ln P PnRTW which gives [ ]kJ 01.334 =W 106 For an adiabatic, reversible process, Equation 2.90 states [ ]121 TTk nRW − − = This equation will be used to calculate the work for the remaining processes. [ ] [ ]kJ 11.16 1 2323 −=− − = TT k nRW [ ] [ ]kJ 11.16 1 4141 =− − = TT k nRW To find the work produced for the overall process, the following equation is used 41342312 WWWWWnet +++= Evaluating this expression with the values found above reveals [ ]kJ 88.7−=netW Therefore, 7.88 kilojoules of work is obtained from the cycle. The efficiency of the process can be calculated using Equation 2.98: 72.0 89.10 88.7 === H net Q W η since [ ]kJ 89.1012 =−= WQH . Alternatively, if we use Equation E2.20D. H C T T −=1η where 43 TTTC == and 21 TTTH == . Upon substitution of the appropriate values 72.0=η 107 2.59 Since this is a refrigeration cycle, the direction of the cycle described in Figure 2.17 reverses. Such a process is illustrated below: States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since the both the temperature and pressure are known for states 2 and 4, the molar volume can be calculated using P RTv = The following table can be made State [ ]K T [ ]bar P [ ]/molm 3v 1 1073 2 1073 60 0.00149 3 298 4 298 0.2 0.124 For each step of the process, potential and kinetic energy effects can be neglected. The process from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in internal energy is 0, and the energy balance becomes 1212 WQ −= 108 From Equation 2.77, = 1 2 12 ln P PnRTW H where 12W is the work done from state 1 to state 2. The value of 1P is not known, but recognizing that the process from state 4 to 1 is an adiabatic compression provides an additional relation. The polytropic relationship can be employed to find 1P . A slight modification of Equation 2.89 provides constPvk = From the ideal gas law v RTP = Combining this result with the polytropic expression and noting that R is constant allows the expression to be written as constTvk =−1 Therefore, 1 1 1 4 1 4 1 − − = k kv T Tv Substituting the appropriate values (k=1.4) gives = = − − mol m 00504.0 31 1 1 4 1 4 1 k kv T Tv [ ]bar 7.17 1 1 1 == v RTP Now, 12W can be calculated. [ ]kJ 9.1012 =W Calculation of 34W follows a completely analogous routine as the calculation for 12W . The following equations were used to find the necessary properties: 109 = mol m 0367.0 3 3v Applying the ideal gas law [ ]bar 675.0 3 3 3 == v RTP Now the following equation can be used = 3 4 34 ln P PnRTW C which gives [ ]kJ 0.341 −=W Equation 2.90 can be used to determine the work for adiabatic, reversible processes. This equation will be used to calculate the work for the remaining processes. [ ] [ ]kJ 11.16 123 −=− − = HC TTk nRW [ ] [ ]kJ 11.16 141 =− − = CH TTk nRW To find the work produced for the overall process, the following equation is used 41342312 WWWWWnet +++= Evaluating this expression with the values found above reveals [ ]kJ 88.7=netW Therefore, 7.88 kJ of work is obtained from the cycle. The coefficient of performance is defined in Equation 2.99 as follows net C W QCOP = where CQ is the equal to 34Q . From the energy balance developed for the process from state 3 to state 4 110 [ ]kJ 01.33434 =−= WQ Therefore, [ ][ ] 382.0kJ 88.7 kJ 01.3 ==COP 111 2.60 (a) The Pv path is plotted on log scale so that the wide range of values fits (see Problem 1.13) logP v 1 100 0.075 2 3 4 log v (b) The work required to compress the liquid is the area under the Pv curve from state 3 to state 4. Its sign is positive. The power obtained from the turbine is the area under the curve from state 1 to 2. Its sign is negative. The area under the latter curve is much larger (remember the log scale); thus the net power is negative. (c) First, perform a mass balance for the entire system: mmmmm ==== 4321 Since no work is done by or on the boiler, the energy balance for the boiler is HQhmhm =− 41 ˆˆ Similarly, the energy balance for the condenser is CQhmhm =− 23 ˆˆ To find the necessary enthalpies for the above energy balances, we can use the steam tables: = kg kJ 5.34241ˆh (520 ºC, 100 bar) 112 = + = kg kJ 2.2334 kg kJ 8.257490.0 kg kJ 77.16810.0ˆ2h (sat. liq at 7.5 kPa) (sat. vap. at 7.5 kPa) = kg kJ 77.168ˆ3h (sat. liquid at 0.075 bar) = kg kJ 81.342ˆ4h (subcooled liquid at 80 ºC, 100 bar) Now, we can calculate the heat loads: ( ) [ ]kW 308169 kg kJ 81.342 kg kJ 5.3424kg/s 100 = − =HQ ( ) [ ]kW 216543 kg kJ 2.2334 kg kJ 68.771kg/s 100 −= − =CQ (d) Use Equation 2.96: 0=+ netnet QW From Part (c), we know [ ] [ ] [ ]kW 91626kW 216543kW 308169 =−=netQ Therefore, [ ]kW 91626−=netW (e) Using the results from Parts (c) and (d): [ ][ ] 297.0kW 081693 kW 91626 ==η Chapter 3 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 3.1 Since entropy is a state function 2,1, stepsysstepsyssys sss ∆+∆=∆ Step 1 is a constant volume process. Therefore, no work is done. After neglecting potential and kinetic energy effects, the energy balance for a reversible process becomes qu =∆ qdu δ=∴ qdTcv δ= Using the definition of entropy, ∫ ∫==∆ final initial T T vrev stepsys T dTc T qs 2 1 1, δ Step 2 is an isothermal process. For and ideal gas ∆u is zero and the energy balance is (PE and KE neglected) wq −= wdq δ−=∴ For an ideal gas, the following can be shown dv v RTPdvwrev −=−=δ Therefore, ===∆ ∫ ∫ 1 2 2, ln 2 1 v vRdv v R T qs final initial v v rev stepsys δ Combination of both steps yields +=∆ ∫ 1 2ln 2 1 v vR T dTc s T T v sys 3 3.2 Equation 3.62 states ∫ −=∆ 2 1 1 2ln T T P sys P PRdT T cs Substituting the equation for Pc yields ∫ − ++ =∆ 2 1 1 2 2 ln T T sys P PRdT T CTBTAs ( ) ( ) −−+−+ =∆ 1 22 1 2 212 1 2 ln 2 ln P PRTTCTTB T TAssys 4 3.3 Equation 3.3 states surrsysuniv SSS ∆+∆=∆ where )ˆˆ( 12 ssmSsys −=∆ surr surr surr T QS =∆ (the temperature of the surroundings is constant) First, we will calculate sysS∆ . From the steam tables: ⋅ = Kkg kJ 1228.71ˆs (300 ºC, 10 bar) Since the container is rigid and mass is conserved in the process, 12 ˆˆ vv = = kg m 25794.0ˆ 3 1v (300 ºC, 10 bar) =∴ kg m 25794.0ˆ 3 2v To find the number of phases present in state 2, compare the specific volume of state 2 to the specific volume for saturated water and saturated water vapor at 1 bar. From the steam tables, = kg m 001043.0ˆ 3 , 2 satlv (sat. H2 = kg m 6940.1ˆ 3 , 2 satvv O(l) at 1 bar) (sat. H2 satvsatl vvv ,22 , 2 ˆˆˆ << O(v) at 1 bar) Since , two phases are present. The quality of the water can be calculated as follows ( ) satvsatl vxvxv ,2 , 22 ˆˆ1ˆ +−= Therefore, 5 152.0=x From the steam tables: ⋅ = Kkg kJ 3025.1ˆ ,2 satls (sat. H2 ⋅ = Kkg kJ 3593.7ˆ ,2 satvs O(l) at 1 bar) (sat. H2 ( ) satvsatl sxsxs ,2 , 22 ˆˆ1ˆ +−= O(v) at 1 bar) The entropy of state 2 can be calculated using the following equation: ⋅ = Kkg kJ 2231.2ˆ2s Now the entropy change of the system can be calculated. [ ]( ) −= ⋅ − ⋅ =∆ K kJ 0.49 Kkg kJ 1228.7 Kkg kJ 2231.2kg 10sysS To find the change in entropy of the surroundings, an energy balance will be useful. Since no work is done, the energy balance is QU =∆ We also know QQsurr −= To following expression is used to solve for the internal energy: ( )( )[ ]1,2,2 ˆˆ1 uxuuxmU satvsatl −+−=∆ From the steam tables = kg kJ 2.2793ˆ1u (300 ºC, 10 bar) = kg kJ 33.417ˆ ,2 satlu (sat. H2O(l) at 1 bar) 6 = kg kJ 1.2506ˆ ,2 satvu (sat. H2 [ ]kJ 77.20583−=∆U O(v) at 1 bar) The expression for internal energy yields Therefore, [ ]kJ 77.20583−=Q and [ ][ ] ===∆ K kJ 25.70 K 293 kJ 77.20583 surr surr surr T QS Now the change in entropy of the universe can be calculated = + −=∆+∆=∆ K kJ 25.21 K kJ 25.70 K kJ 0.49surrsysuniv SSS Since ∆Suniv > 0, this process is possible. 7 3.4 Entropy Balance: surrsysuniv SSS ∆+∆=∆ Considering the copper block to be the system, no work is done on the system; thus, the energy balance is qdu δ= (neglecting PE and KE) qdTcv δ= This can be used in the following expression for entropy ∫=∆ 2 1 T T rev sys T q s δ ∫=∆ 2 1 T T v sys dTT cs From Table A.2.3 RcP 723.2= Rcc Pv 723.2== (for liquids and solids) Therefore, [ ][ ] ⋅ = ==∆ ∫ K 15.373 K 280ln Kmol J 314.8723.2ln723.2723.2 1 2 2 1 T TRdT T Rs T T sys ⋅ −=∆ Kmol J 50.6syss [ ] −= ⋅ − =∆ K J 1024 Kmol J 50.6 mol kg 063465.0 kg 10 sysS Since the temperature of the lake remains constant, the change in entropy of the surroundings can be calculated as follows ( ) ( ) surr Cu surr PCu surrsurr surr surr T TTRn T TTcn T Q T QS 1212 723.2 −−=−−=−==∆ 8 [ ] [ ] [ ]( ) [ ] − ⋅ −=∆ K 280 K 15.373K 280 Kmol J 314.8723.2 mol kg 0.063546 kg 10 surrS =∆ K J 1184surrS From the definition of entropy: = + −=∆+∆=∆ K J 160 K J 1184 K J 1024surrsysuniv SSS 9 3.5 (a) Since the process is adiabatic and reversible, 0=∆ sysS (b) The change in entropy is calculated by ( ) −= − = ∆− ==∆ ∫ K J 9.73 K111 mol kJ2.8mol 1liquid vapor b vap rev sys T hn T qnS δ The sign is negative because there is less randomness in the liquid phase. (c) For this situation dTcq P=δ Therefore, ( ) ⋅ ===∆ ∫ ∫ K 15.373 K 15.273ln Kg J2.4g 0148.18 K 15.273 K 373.15 f i T T Prev sys dTT cT qmS δ −=∆ K J6.23sysS The sign is negative because as the water cools, less translational energy states are occupied by the molecules. Therefore, the randomness decreases. (d) First, calculate the temperature at which the blocks (block A and block B) equilibrate. The energy balance for the process is ( ) ( ) 0K 15.473K 15.373 22 =−+− TcnTcn PBPA Since the heat capacities number of moles of A and B are equal, we find that K 15.4232 =T Now, calculate the change in entropy: 10 BsysAsyssys SSS ,, ∆+∆=∆ + =∆ B PB A PAsys T Tcn T TcnS ,1 2 ,1 2 lnln Substituting values, we obtain =∆ K J 337.0sysS The sign is positive because two objects at different temperatures will spontaneously equilibrate to the same temperature when placed together. 11 3.6 The solution below compares problems 2.14 and 2.15, the calculation of 2.13 was erroneously included in the problem statement of the first printing and is shown at the end of this problem. Problem 2.14 Since the system is well-insulated no heat is transferred with the surroundings. Therefore, the entropy change of the surroundings is zero and sysuniv SS ∆=∆ The gas in the piston-cylinder system is ideal and Pc is constant, so − =∆ 1 2 1 2 lnln P PR T TcnS Psys From the problem statement, we know [ ] [ ]bar 1 bar 2 2 1 = = P P The ideal gas law can be used to solve 1T . [ ]( )( ) [ ]( ) [ ]K 6.240 mol 1 Kmol barL 0.08314 L 10bar 211 1 = ⋅ ⋅ == nR VPT Solving for 2T is slightly more involved. The energy balance for this system where potential and kinetic energy effects are neglected is WU =∆ Conservation of mass requires 21 nn = Let 21 nnn == The energy balance can be rewritten as ∫∫ −= 2 1 2 1 V V E T T v dVPdTcn 12 Since vc and EP are constant ( ) ( )1212 VVPTTnc Ev −−=− 2V and 1T can be rewritten using the ideal gas law 2 2 2 P nRTV = nR VPT 111 = Substituting these expressions into the energy balance, realizing that 2PPE = , and simplifying the equation gives nR VPP T 2 7 2 5 112 2 + = Using the following values [ ] [ ] [ ] [ ] ⋅ ⋅ = = = = = Kmol barL 08314.0 mol 0.1 L 10 bar 1 bar 2 1 2 1 R n V P P results in [ ]K 2062 =T Since both states are constrained, the entropy can be calculated from Equation 3.63: [ ]( ) [ ][ ] [ ] [ ] ⋅ − ⋅ ×=∆=∆ bar 2 bar 1ln Kmol J 314.8 K 6.240 K 206ln Kmol J 314.85.3mol 0.1sysuniv SS =∆=∆ K J 24.1sysuniv SS 13 Problem 2.15 Since the initial conditions in the piston-cylinder assembly are equal to the initial conditions of Problem 2.14, 1T and 1P are known. Moreover, 2P is known, so we only need to find 2T in order to calculate the entropy change. For adiabatic, reversible processes, the following relationship (Equation 2.89) holds: constPV k = This can be used to find 2V . kkV P PV 1 1 2 1 2 = Noting that 5 7 == v P c ck and substituting the proper values results in [ ]L 4.162 =V Th polytropic expression can also be manipulated to yield .1 constTV k =− Therefore, 1 2 1 11 2 − − = k k V VT T Substitution of the appropriate variables provides [ ]K 4.1972 =T Now the entropy can be calculated, [ ]( ) [ ][ ] [ ] [ ] ⋅ − ⋅ ×=∆=∆ bar 2 bar 1ln Kmol J 314.8 K 6.240 K 97.41ln Kmol J 314.85.3mol 0.1sysuniv SS =∆=∆ K J 004.0sysuniv SS The value of the entropy shown represents round-off error. Since the process is reversible and adiabatic, we know from Table 3.1 and the related discussion in Section 3.3 that the entropy changes of the system, surroundings, and universe will be zero. 14 Problem 2.13 Equation 3.3 states surr surr surrsysuniv T QssmSSS +−=∆+∆=∆ )ˆˆ( 12 where the temperature of the surroundings is constant. First, we will determine 2s . The first law can be applied to constrain state 2. With potential and kinetic energy effects neglected, the energy balance becomes WQU +=∆ The value of the work will be used to obtain the final temperature. The definition of work (Equation 2.7) is ∫−= 2 1 V V EdVPW Since the piston expands at constant pressure, the above relationship becomes ( )12 VVPW E −−= From the steam tables ⋅ = Kkg kJ 2119.61ˆs (10 MPa, 400 ºC) = kg m 02641.0ˆ 3 1v (10 MPa, 400 ºC) [ ]33111 m 07923.0kg m 02641.0kg) 3(ˆ = == vmV Now 2V and 2v are found as follows [ ]36312 m 4536.0Pa 100.2 J 748740m 07923.0 = × − −=−= EP WVV [ ][ ] === kg m 1512.0 kg 3 m 4536.0ˆ 33 2 2 2 m Vv 15 Since 2vˆ and 2P are known, state 2 is constrained. From the steam tables: ⋅ = Kkg kJ 1270.7ˆ2s kg m 0.1512 bar, 20 3 Now U∆ will be evaluated, which is necessary for calculating surrQ . From the steam tables: = kg kJ 2.2945ˆ2u kg m 0.1512 bar, 20 3 = kg kJ 4.2832ˆ1u ( )Cº 400 bar, 001 ( ) [ ]( ) [ ]kJ 4.338 kg kJ 4.2832 kg kJ 2.2945kg 3ˆˆ 121 = − =−=∆ uumU Substituting the values of U∆ and W into the energy equation allows calculation of Q WUQ −∆= [ ]( ) [ ] surrQQ −=×=−−= J 1009.1J 748740 [J] 338400 6 so [ ]( ) [ ][ ] = × − − = +−=∆ K kJ 126.1 K 15.673 kJ 1009.1 K kg kJ 2119.6 K kg kJ .12707kg 3 )ˆˆ( 3 12 surr surr univ T QssmS Therefore, this process is irreversible. 16 3.7 (a) From Equation 3.63: − =∆ 1 2 1 2 lnln P PR T Tcs Psys [ ][ ] [ ] [ ] ⋅ = − ⋅ =∆ Kmol J 63.20 bar 1 bar 5.0ln K 300 K 500ln 2 7 Kmol J 314.8syss (b) From Equation 3.65: + =∆ 1 2 1 2 lnln v vR T Tcs vsys [ ][ ] ⋅ = + ⋅ =∆ Kmol J 85.4 mol m 05.0mol m .0250 ln K 300 K 500ln 2 5 Kmol J 314.8 3 3 syss (c) First, we can find the molar volume of state 1 using the ideal gas law. == mol m 025.0 3 1 1 1 P RTv Now, we can use Equation 3.65 + =∆ 1 2 1 2 lnln v vR T Tcs vsys [ ][ ] ⋅ = + ⋅ =∆ Kmol J 62.10 mol m 025.0 mol m .0250 ln K 300 K 500ln 2 5 Kmol J 314.8 3 3 syss 17 3.8 (i). We wish to use the steam tables to calculate the entropy change of liquid water as it goes from its freezing point to its boiling point. The steam tables in Appendices B.1 – B.5 do not have data for subcooled water at 1 atm. However, there is data for saturated water at 0.01 ºC and a pressure of 0.6113 kPa. If we believe that the entropy of water is weakly affected by pressure, then we can say that the entropy of water at 0.01 ºC and 0.6113 kPa is approximately equal to the entropy at 0 ºC and 1 atm. The molar volumes of most liquids do not change much with pressure at constant temperature. Thus, the molecular configurations over space available to the water molecules do not change, and the entropy essentially remains constant. We do not need to consider the molecular configurations over energy since the temperature difference is so slight. So, from the steam tables: ( ) ( ) ⋅ =≅ Kkg kJ 0kPa .61130 ,º 01.0ˆatm 1 ,º 0ˆ CsCs ( ) ⋅ = Kkg kJ 3068.1atm 1 ,º 100ˆ Cs Therefore, ( ) ( ) ⋅ − ⋅ =−=∆ Kkg kJ 0 Kkg kJ 3068.1atm 1 ,º 0ˆatm 1 ,º 100ˆˆ CsCss ⋅ =∆ Kkg kJ 3068.1sˆ (ii). From the steam tables: ( ) ⋅ = Kkg kJ 3068.1atm 1 ,º 100ˆ Csl ( ) ⋅ = Kkg kJ 3548.7atm 1 ,º 100ˆ Csv Therefore, ( ) ( ) ⋅ − ⋅ =−=∆ Kkg kJ 3068.1 Kkg kJ 3548.7atm 1 ,º 100ˆatm 1 ,º 100ˆˆ CsCss lv ⋅ =∆ Kkg kJ 048.6sˆ The change in entropy for process (ii) is 4.63 times the change in entropy for process (i). There are many ways to reconcile this difference, but think about it from a molecular point of view. In process (i), the available molecular configurations over energy are increased as the temperature increases. As the temperature increases, the molar volume also increases slightly, so the 18 available molecular configurations over space also increase. Now consider process (ii), where the molecules are being vaporized and entering the vapor phase. The molecular configuration over space contribution to entropy is drastically increased in this process. In the liquid state, the molecules are linked to each other through intermolecular interactions and their motion is limited. In the vapor state, the molecules can move freely. Refer to Section 3.10 for a discussion of entropy from a molecular view. 19 3.9 Before calculating the change in entropy, we need to determine the final state of the system. Let the mixture of ice and water immediately after the ice has been added represent the system. Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are neglected, the energy balance reduces 0=∆H We can split the system into two subsystems: the ice (subscript i) and the water (subscript w). Therefore, 0=∆+∆=∆ wwii hmhmH and wwii hmhm ∆−=∆ We can get the moles of water and ice. [ ] [ ]kg 399.0 kg m 001003.0 m .00040 ˆ 3 3 = == w w w v Vm ( ) [ ]( ) [ ]mol 15.22 mol kg 0180148.0 kg 399.0 2 = == OH w w MW mn ( ) [ ]mol 55.5 2 == OH i i MW mn Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0 ºC, the assumption is correct.) The following expression mathematically represents the change in internal energies (cP=cv ( )( ) ( )[ ] ( )Cº 25º 0Cº 100 ,,, −−=−+∆−−− fwPwfwPfusiPi TcnCTchcn ). Note: Assumed the heat capacities are independent of temperature to obtain this expression. From Appendix A.2.3 Rc iP 196.4, = Rc wP 069.9, = and −=∆ mol kJ0.6fush 20 Substitution of values into the above energy balance allows calculation of Tf Cº 12.3=fT . (Our assumption that all the ice melts is correct.) Now, we can calculate the change in entropy. From Equation 3.3 surrsysuniv sss ∆+∆=∆ Since the glass is considered adiabatic, 0=∆ surrs sysuniv ss ∆=∆ We will again break the system into two subsystems: the ice and the water. The change in entropy of the universe can be calculated as follows wwiiuniv snsnS ∆+∆=∆ The definition of change in entropy is ∫=∆ final initial rev T qs δ Assuming the heat capacities of ice and water are independent of temperature, the expressions for the change in entropy of the subsystems are + ∆ + =∆ K 15.273 K 27.276ln K 15.273K 15.263 K 15.273ln ,, wP fus iPi c h cs =∆ K 98.152 K 27.276ln,wPw cs Therefore + + ∆ + =∆ K 98.152 K 27.276ln K 15.273 K 27.276ln K 15.273K 15.263 K 15.273ln ,,, wPwwP fus iPiuniv cnc h cnS Substituting the values used before, we obtain 21 =∆ mol J 59.6univS 22 3.10 (a) The maximum amount of work is obtained in a reversible process. We also know the entropy change for the universe is zero for reversible processes. From the steam tables ⋅ = Kkg kJ 2119.61ˆs (400 ºC, 100 bar) ⋅ = Kkg kJ 5434.8ˆ2s (400 ºC, 1 bar) Using the entropy criterion, ( ) surruniv SssmS ∆+−==∆ 12 ˆˆ0 ( ) [ ]( ) ⋅ − ⋅ −=−−=∆ Kkg kJ 2119.6 Kkg kJ 5434.8kg 5.0ˆˆ 12 ssmSsurr −=∆ K kJ 1658.1surrS Since the process is isothermal, we know that the temperature of the surroundings is constant at 400 ºC. Therefore, −=∆= K kJ 1658.1surr surr surr S T Q [ ]kJ 76.784−=surrQ To find the work obtained, perform an energy balance. An energy balance where potential and kinetic energy effects are neglected is WQU +=∆ We can calculate the change in internal energy from the steam tables, and we also know that surrQQ −= . From the steam tables: = kg kJ 4.2832ˆ1u (400 ºC, 100 bar) = kg kJ 8.2967ˆ2u (400 ºC, 1 bar) Therefore, 23 ( ) QuumW −−= 12 ˆˆ [ ]( ) [ ]kJ 76.784 kg kJ 4.2832 kg kJ 8.2967kg 5.0 − − =W [ ]kJ 06.717−=W (b) If steam is modeled as an ideal gas, the change in internal energy is zero. Therefore, the energy balance is WQ −= The work can be found using Equation 2.77 which is developed for reversible, isothermal processes. = 1 2ln P PnRTW [ ]( ) ( ) ⋅ = bar 100 bar 1lnK 15.673 Kmol J 3145.8 mol kg 0180148.0 kg 5.0W [ ] [ ]kJ 3.715J 715332 −=−=W Therefore, [ ]kJ 3.715=Q and [ ]kJ 3.715−=surrQ Since the temperature of the surroundings is constant, −=∆ K kJ 06.1surrS 24 3.11 (a) The initial pressure is calculated as follows ( )( ) Pa 106.24 m 05.0 m/s 81.9kg50002 5 2 2 1 ×=+= surrPP Similarly, the final pressure is ( )( ) Pa 104.34 m 05.0 m/s 81.9kg 50003 5 2 2 2 ×=+= surrPP (b) The temperature should rise, which can be understood by considering an energy balance. Because the system is insulated, the work done by the mass being added to the piston is transformed into molecular kinetic energy. (c) The final temperature can be calculated with an energy balance: WU =∆ ( ) WTTncv =− 12 Since the pressure of the surroundings is constant after the block is added to the piston, the work is calculated as follows: ( )122 VVPW −−= Assume ideal gas behavior: 2 2 2 P nRTV = 3 1 1 1 m 00169.0== P nRTV Now, we can create one equation with one unknown: ( )( ) −−=− 1 2 2 2122/5 VP nRTPTTRn Substitute values and solve for T2 K 5572 =T : 25 (d) Since the system is well-insulated 0=∆ surrs Use Equation 3.65 to calculate the change in entropy: ( ) + = + =∆=∆ 21 12 1 2 1 2 1 2 lnln2/5lnln PT PTR T TR V VR T Tcss vsysuniv Substituting values, we obtain, ⋅ =∆ Kmol J 354.0syss (e) Because the change in entropy of the universe is equal to change in entropy of the system, which is positive in this situation, the second law is not violated. 26 3.12 (a) To calculate ∆ssys State 1 24.6 bar 500 K State 2 34.4 bar 557 K State I PI TI = 557 K rev. adiabatic rev. isothermal , we need to take the gas from state 1 to state 2 using a reversible process. The process in part a can be drawn as follows: We need to find the intermediate pressure, PI where we end up at the temperature in state 2, T2 ( ) ( ) ( ) kk I kk II kk PTPTPT /12 /1/1 11 −−− == . To find it, we can use the results from pages 78-79 of the text: Therefore, ( ) [bar] 9.351 1 2 1 = = − P T TP k k I (I) If we draw the process on a PT diagram, we get: 27 The change in entropy can be represented as follows: isothermaladiabaticsys sss ∆+∆=∆ For the reversible adiabatic process, the change in entropy is zero. Therefore, isothermalsys ss ∆=∆ For an isothermal process, dP P RTwq revrev −=−= δδ Therefore, ⋅ = −=−==∆ ∫∫ Kmol J 354.0ln 2 2 I P P final initial sys P PRdP P R T qs I δ If we substitute relation (I) in the expression above, we get the same expression in the book: ( ) − = − + −=∆ 1 2 1 2 1 2 1 2 lnlnln 1 ln P PR T Tc T TR k k P PRs Psys (b) For this construction, we use the path diagrammed on the following PT diagram: We write 28 isothermalisobarsys sss ∆+∆=∆ First, find an expression for the isobaric heating dTcdhq Prev ==δ Therefore, ==∆ ∫ 1 2ln 2 1 T TcdT T cs P T T P isobar Now, we need an expression for the isothermal, reversible expansion Pdvwq revrev =−= δδ Therefore, −= ===∆ ∫∫ = = = = 1 2 / / 2 1 / / lnln 222 121 222 12 P PR P PRdv v Rdv T Ps PRTv PRTv PRTv PRTv isothermal I Combine the two steps: ⋅ = − =∆ Kmol J 354.0lnln 1 2 1 2 P PR T Tcs Psys (c) For this construction, we use the path diagrammed on the following PT diagram: 29 For isochoric heating followed by an isothermal expansion, the entropy can be expressed as follows: isothermalisochoricsys sss ∆+∆=∆ For the reversible, isothermal expansion, we obtain (refer to Parts (a) and (b) to see how this is derived) −=∆ I isothermal P PRs 2ln However, we can relate the intermediate pressure to the initial pressure through the ideal gas law: 1 1 2 T P T PI = or [bar] 4.27=IP − = −= −=∆∴ 1 2 1 2 1 1 2 22 lnlnlnln P PR T TR P T T PR P PRs I isothermal For the isochoric process: dTcq vrev =δ Therefore, 30 ==∆ ∫ 1 2ln 2 1 T TcdT T cs v T T v isochoric Combining these results: ( ) ⋅ = − = − +=∆ Kmol J 354.0lnlnlnln 1 2 1 2 1 2 1 2 P PR T Tc P PR T TRcs Pvsys Using any of these three reversible paths, we get the same answer! 31 3.13 (a) Since the process is reversible and adiabatic, the entropy change for the process is zero. 0=∆ syss Furthermore, since the process is adiabatic, the changes in entropy of the surroundings and the universe are also zero. (b) Since this process is isentropic (∆s=0), we can apply an expression for the entropy change of an ideal gas. ∆s = 0 = cP TT1 T2 ∫ dT − Rln P2 P1 Insert the expression for cP − ×−×+ ==∆ ∫ −− 1 2 263 ln10227.010506.0639.30 2 1 P PRdT T TTRs T T from Appendix A which upon substituting [ ] [ ] [ ]bar 06.12 bar 1 K 250 2 1 1 = = = P P T yields [ ]K 4822 =T (c) An energy balance gives ( )∫ ∫ −==∆= 2 1 2 1 1 T T T T Pv dTcdTcuw 32 ( ) =×−×+= ∫ −− mol J 539010227.010506.0639.2 2 1 263 T T dTTTRw (d) Reversible processes represent the situation where the minimum amount of energy is required for compression. If the process were irreversible, more work is required, and since the process is adiabatic, the change in internal energy is greater. Since the change in internal energy is greater, so too is the change in temperature. Therefore, the final temperature would be higher than the temperature calculated in Part (b). 33 3.14 The problem statement states that the vessel is insulated, so we can assume that heat transfer to the surroundings is negligible. Therefore, the expression for the entropy change of the universe is sysuniv ss ∆=∆ An energy balance will help us solve for the entropy change. Neglecting potential and kinetic energy effects, the energy balance is WQU +=∆ Since the vessel isinsulated, the heat term is zero. Furthermore, no work is done, so the energy balance is ( ) 0ˆˆ 12 =−=∆ uumU 21 uu =∴ The values for the initial pressure and temperature constrain the value of specific energy at state 1. From the steam tables, = kg kJ 2.2619ˆ1u (400 ºC, 200 bar) =∴ kg kJ 2.2619ˆ2u The problem statement provides the pressure of state 2 (100 bar). Since we know the pressure and internal energy at state 2, the entropy is constrained. Information given in the problem statement also constrains state 1. From the steam tables, ⋅ = Kkg kJ 5539.51ˆs (400 ºC, 200 bar) ⋅ = Kkg kJ 7754.5ˆ2s = kg kJ 2.2619ˆ bar, 100 2u Therefore, ⋅ =−=∆=∆ Kkg kJ 2215.0ˆˆˆˆ 12 ssss sysuniv =∆=∆ K kJ 2215.0sysuniv SS 34 3.15 The subscript “2” refers to the final state of the system. “1” refers to the gas initially on one side of the partition. If you take the system to be the entire tank (both sides of the partition), then no net work is performed as the gas leaks through the hole. Furthermore, the tank is well insulated, and kinetic and potential energy effects can be neglected. Thus, the energy balance is 0=∆u and 12 TT = (ideal gas) Initially, the partition is divided into two equal parts. The gas fills the entire volume for the final state ( 12 2VV = ). The ideal gas law can be used to calculate the moles of gas present. [ ]( ) [ ]( ) [ ]( ) mol 5.200 K 003 Kmol J 314.8 m .50Pa 1010 35 1 11 = ⋅ × == RT VPn Now, the change in entropy can be solved using a slight modification of Equation 3.65: + = + =∆ 1 2 1 2 1 2 1 2 lnln 2 3lnln nv nvR T TRn v vR T TcnS vsys ( ) ( ) ( ) + ⋅ =∆ 2ln1ln 2 3 Kmol J 314.8mol 5.200sysS =∆ K J 1155sysS 35 3.16 (a) A schematic of the process is shown below: 1 bar, 298 K Mixing Process N2 N2 N2 N2 O2 O2 N2 N2 N2 N2 N2 N2 N2 N2 O2 N2 N2 N2 N2 O2 1 bar, 298 K P2, T2 The tank is insulated. The change in entropy of the universe can be rewritten as 22 ,, OsysNsysuniv SSS ∆+∆=∆ since the tank is well-insulated. After the partition ruptures, the pressure and temperature will remain constant at 1 bar and 298 K, respectively. This can be shown by employing mass and energy balances. Therefore, the partial pressures in the system are bar79.0 2,2 =Np bar21.0 2,2 =Op Use Equation 3.62 to determine the entropies: ⋅ −= −=∆ ∫ bar 1 bar79.0ln Kmol J314.8ln 2 2 1 2 K 298 K 298 , N P Nsys P PRdT T cs ⋅ =∆ Kmol J96.1 2,Nsyss ⋅ −= −=∆ ∫ bar 1 bar21.0ln Kmol J314.8ln 2 2 1 2 K 298 K 298 , O P Osys P PRdT T cs ⋅ =∆ Kmol J98.12 2,Osys s Therefore, ( ) ( ) ⋅ + ⋅ =∆ Kmol J98.12mol21.0 Kmol J96.1mol79.0univS 36 =∆ K J27.4univS (b) A schematic of the process is shown below: Before calculating the change in entropy, we need to find how the temperature and pressure change during the process. The energy balance simplifies to 0=∆U which can be rewritten as ( ) ( ) ( ) 0K298K298 2222 2 =−−+ vOvNvNO cncnTcnn Assuming the heat capacities are equal, we can show that K 2982 =T We can find the pressure after the rupture by recognizing that the tank is rigid. Therefore, totNO VVV =+ 22 By employing the ideal gas law, we get the following equation (it has been simplified): ( ) 2,1,1 22 2 2 2 2 P nn P n P n NO O O N N + =+ Substitute values and solve to obtain bar65.12 =P Now, use Equation 3.62 to calculate the entropies as was done in Part (a): 37 ⋅ −= −=∆ ∫ bar 2 bar30.1ln Kmol J314.8ln 2 2 1 2 K 298 K 298 , N P Nsys P PRdT T cs ⋅ =∆ Kmol J 58.3 2,Nsyss ⋅ −= −=∆ ∫ bar 1 bar347.0ln Kmol J314.8ln 2 2 1 2 K 298 K 298 , O P Osys P PRdT T cs ⋅ =∆ Kmol J8.8 2,Osys s Therefore, ( ) ( ) ⋅ + ⋅ =∆ Kmol J8.8mol21.0 Kmol J58.3mol79.0univS =∆ K J68.4univS 38 3.17 First start with the energy balance for the throttle. Potential and kinetic energy effects can be neglected. During the throttling process, no shaft work is performed and the rate of heat transfer is negligible. Therefore, 1122 ˆˆ hmhm = A mass balance allows the energy balance to be simplified to 12 ˆˆ hh = From the steam tables: = kg kJ 3.33981ˆh (500 ºC, 8 MPa) =∴ kg kJ 3.3398ˆ2h Now calculate entropy: surrsysuniv sss ˆˆˆ ∆+∆=∆ Since the process is adiabatic, ⋅ =∆ Kkg kJ 0ˆsurrs The steam tables can be used to calculate the change in entropy of the system. ⋅ = Kkg kJ 7098.8ˆ2s = kg kJ 3.3398ˆ kPa, 100 2h ⋅ = Kkg kJ 7239.61ˆs (8 MPa, 500 ºC) Thus, ⋅ =−=∆=∆ Kkg kJ 9859.1ˆˆˆˆ 12 ssss sysuniv 39 3.18 For this process to work, conservation of mass and the first and second laws of thermodynamics must hold. The subscript “1” refers to the inlet stream, “2” refers to the cold outlet, and “3” refers to the hot outlet. To test the conservation of mass, perform a mass balance 321 mmm += + = s kg 5.1 s kg 5.0 s kg 2 Clearly, the conservation of mass holds. Now test the first law of thermodynamics by writing an energy balance. Since there is no heat transfer or work, the energy balance becomes 0ˆˆˆ 113322 =−+ hmhmhm (PE and KE effects neglected) Using the conservation of mass we can rewrite the mass flow rate of stream 1 in terms of streams 2 and 3: ( ) ( ) 0ˆˆˆˆ 133122 =−+− hhmhhm If we assume that the heat capacity of the ideal gas is constant, the equation can be written as follows: ( ) ( ) 0ˆˆ 133122 =−+− TTcmTTcm PP Therefore, 0K) 20(ˆ s kg 5.1K) 60(ˆ s kg 5.0 = +− PP cc This proves that the first law holds for this system. For the second law to be valid, the rate of change in entropy of the universe must greater than or equal to 0, i.e., dS dt univ ≥ 0 Assuming the process is adiabatic, we can write the rate of entropychange of the universe for this steady-state process using Equations 3.48-3.50: ( ) ( )133122113322 ˆˆˆˆˆˆˆ ssmssmsmsmsmdt dS univ −+−=−+= 40 where a mass balance was used. For constant heat capacity, we can calculate the entropy differences using Equation 3.63: − = − =− 1 2 1 2 1 2 1 2 12 lnln2 5lnln P P T TR P PR T Tcss P − = − =− 1 3 1 3 1 3 1 3 13 lnln2 5lnln P P T TR P PR T Tcss P Applying these relations, we get: ( )P univ cR MWdt dS 057.077.21 −= Therefore, the second law holds if RcP 6.48≤ ; this value is far in excess of heat capacity for gases, so this process is possible. 41 3.19 (a) T1 = 640 oC V1= 20 m/s Nozzle V2=? P1 = 4 MPa P2 = 100 kPa (b) Since the process is reversible and adiabatic, =∆ K J 0sysS (c) The steam tables can be used to determine the final temperature. From the interpolation of steam table data ( ) ⋅ == Kkg kJ 4692.7MPa 4 C,º 640ˆˆ 12 ss Now two thermodynamic properties are known for state 2. From the steam tables: Cº 4.1212 =T ⋅ == Kkg kJ 4692.7ˆ MPa, 1.0 22 sP (d) An energy balance is required to calculate the exit velocity. The energy balance for the nozzle is ( ) ( )1122 ˆˆˆˆ0 KK ehmehm +−+= Realizing that 21 mm = allows the energy balance to be written as ( ) 212, ˆˆˆˆ hehe KK −+= which is equivalent to ( )[ ]2122 ˆ5.0ˆ2 hVhV −+= Substituting the following values 42 = s m 201V = kg J 2719100ˆ2h (0.1 MPa, 121.4 ºC) = kg J 37670001ˆh (4 MPa, 640 ºC) yields = s m 14482V Note: the velocity obtained is supersonic; however, the solution does not account for this type of flow. 43 3.20 A schematic of the process follows: Since this process is isentropic (∆s=0), we can apply an expression for the entropy change of an ideal gas. We must be careful, however, to select an expression which does not assume a constant heat capacity. ∆s = 0 = cP TT1 T2 ∫ dT − Rln P2 P1 Insert the expression for cP − ×−×+ ==∆ ∫ −− 1 2 263 ln10824.810785.28213.10 2 1 P PRdT T TTRs T T from Appendix A We can also relate P1 to T1 and v1 (which are known) through the ideal gas law: P1 = RT1 v1 This leaves us with 1 equation and 1 unknown (T2). Integrating: ( ) −−×−−×+ ==∆ −− 1 122 1 2 2 6 12 3 1 2 ln10412.4)(10785.28ln213.10 RT vPTTTT T Ts Using T1 = 623 K, v1 = 600 cm3/mol, P2 = 1 atm 44 R=82.06 cm3 atm/mol K, we get: T2 = 454 K To solve for n WS we need a first law balance, with negligible ke and pe, the 1st law for a steady state process becomes: ( ) SWQhhn ++−= 210 If heat transfer is negligible (isentropic), ( )∫∫ −− ×−×+==−= 2 1 2 1 263 12 10824.810785.28213.1 T T T T p S dTTTRdTchh n W integrating: ( ) ( ) ( )[ ]313262122312 10941.210393.14213.1 TTTTTTRn WS −×−−×+−= −− Substituting K 623T K 454 Kmol J 314.8 1 2 = = ⋅ = T R results in −= mol J 19860 n WS 45 3.21 A reversible process will require the minimum amount of work. For a reversible process, 0=∆ univs surrsys ss ∆−=∆∴ For this process, the change in entropy of the system can be calculated as follows 22 NOsys sss ∆+∆=∆ where ( ) 2 2 1 2 2 1 2ln O T T OP O P PRdT T c s ∫ −=∆ ( ) 2 2 1 2 2 1 2ln N T T NP N P PRdT T c s ∫ −=∆ Since the temperature of the streams do not change, the expressions reduce to 2 2 1 2ln O O P PRs −=∆ 2 2 1 2ln N N P PRs −=∆ Therefore, + =∆ 22 1 2 1 2 lnln NO surr P P P PRs Since we are assuming the temperature of the surroundings remain constant at 20 ºC, surr surr surr T qs =∆ where qqsurr =− 46 Therefore, + =∆= 22 1 2 1 2 lnln NO surrsurrsurrsurr P P P PRTsTq ( ) + ⋅ = 22 bar 79.0 bar 1ln bar 21.0 bar 1lnK 15.293 Kmol J 314.8 NO surrq = mol J 4378.2surrq and −= mol J 4378.2q Energy Balance: 0=++−∑∑ out Soutout in inin WQhnhn Because the temperature of the oxygen and nitrogen doesn’t change in the process, the energy balance per mole of feed becomes qwS −= = mol J 4378.2Sw 47 3.22 (a) Take the entire container to be the system and assume no heat or work crosses the system boundary. Energy Balance: 0=∆U After the oscillation cease, the temperature and pressure on both sides of the piston will be the same assuming the metallic piston is very thin and the thermal conductivity coefficient is large. Now, let’s rewrite the energy balance. 0B gas,1A gas,1 =∆+∆=∆ ununU BA or 0B gasA gas ,1 ,1 =∆+∆ uu n n B A 41.2 2 1,1, 1,1, 1,1,1, 1,1,1, ,1 ,1 === AB BA ABB BAA B A TP TP TVP TVP n n since 1,1,2 BA VV = so ( ) ( ) 0101.55/2K 15.773 2 341.2 2T K 15.373 3- 2 =×++ − ∫ dTTRTR K 5852 =∴T Mass Balance: 2,2,1,1, BABA nnnn +=+ Using the ideal gas law, we get 48 2 2 2, 2,2, 2, 2,2, 1, 1,1, 1, 1,1, RT VP RT VP RT VP RT VP RT VP tot B BB A AA B BB A AA =+=+ since the pressure and temperature of state 2 are equal on both sides. But 1,1,1, 3 ABAtot VVVV =+= Using the volume relationships and simplifying, we get [ ] [ ][ ] [ ] [ ] += += K15.373 bar 12 K15.773 bar 10 3 K 8552 3 1, 1, 1, 1,2 2 B B A A T P T PTP [ ]bar 56.32 =P (b) The container is well-insulated, so sysuniv ss ∆=∆ The entropy change of the system can be split into two subsystems: B gas ,1,1 ,1 A gas ,1,1 ,1 s nn n s nn n s BA B BA A sys ∆+ +∆ + =∆ Using Equation 3.63 and realizing that Rcc vP += , ⋅ = − =∆ K mol J 96.1lnln5.2 41.3 41.2 1, 2 1, 2 A gas AA P PR T TRs ⋅ = − ×+ =∆ ∫ − K mol J 51.1ln105.15.3 41.3 1 1, 2 3 B gas 2 1, B T T P PRdT T TRs B Therefore, ⋅ = ⋅ + ⋅ =∆=∆ K mol J 47.3 K mol J 51.1 K mol J 96.1sysuniv ss The process is possible. 49 3.23 Equation 3.3 can be modified to show surrsysuniv SSS ∆+∆=∆ Let’s first calculate sysS∆ . Before this problem is solved, a few words must be said about the notation used. The system was initially broken up into two parts: the constant volume container and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant volume container; “2” refers the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” denotes the final state. First, the mass of water present in each part of the system will be calculated. The mass will be conserved during the expansion process. Since the water in the rigid tank is saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water’s entropy is constrained. From the steam tables, [ ] kPa 8.1553 kg kJ 4322.6ˆ kg kJ 3308.2ˆ kg kJ 12736.0ˆ kg kJ 001156.0ˆ ,1 ,1 ,1 ,1 = = = = = sat v i l i v i l i P s s v v (Sat. water at 200 ºC) Knowledge of the quality of the water and the overall volume of the rigid container can be used to calculate the mass present in the container. ( ) ( )vil i vmvmV ,11,111 ˆ95.0ˆ05.0 += Using the values from the steam table and [ ]31 m 5.0=V provides [ ]kg 13.41 =m Using the water quality specification, 50 [ ] [ ]kg 207.005.0 kg 92.395.0 11 11 == == mm mm l v For the piston-cylinder assembly, both P and T are known. From the steam tables ⋅ = = Kkg kJ 9665.6ˆ kg m 35202.0ˆ ,2 3 ,2 i i s v (600 kPa, 200 ºC) Now enough information is available to calculate the mass of water in the piston assembly. [ ]kg 284.0 ˆ2 2 2 == v Vm Now the final state of the system must be determined. It helps to consider what physically happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of the entire system will match the surrounding’s. In other words, == kg kJ 9665.6ˆˆ ,2 if ss (600 kPa, 200 ºC) Thus, the change in entropy is given by l i l i v i v iif l i v isys smsmsmsmmmS ,1,1,1,1,22,1,12 ˆˆˆˆ)( −−−++=∆ Substituting the appropriate values reveals =∆ K kJ 05.3sysS Now we calculate the change in entropy of the surroundings. Since the temperature of the surroundings is constant, surrsurr surr surr T Q T QS −==∆ After neglecting potential and kinetic energy effects, the energy balance becomes 51 WQU +=∆ The change in internal energy and work will be calculated in order to solve for Q. The following equation shows how the change in internal energy can be calculated. l i l i v i v iif l i v i umumumummmU ,1,1,1,1,22,1,12 ˆˆˆˆ)( −−−++=∆ From the steam tables = = kg kJ 3.2595ˆ kg kJ 64.850ˆ ,1 ,1 v i l i u u (sat. H2 == kg kJ 9.2638ˆˆ ,2 if uu O at 200 ºC) (600 kPa, 200 ºC) Using these values and the values of mass calculated above, [ ]kJ 0.541=∆U Calculating the work is relatively easy since the gas is expanding against a constant pressure of 600 kPa (weight of the piston was assumed negligible). From Equation 2.7, )( ifE V V E VVPdVPW f i −−=−= ∫ where [ ] [ ] [ ] [ ] [ ]333 3 ,2,1,12 m 6.0m 5.0m 0.1 m 55.1ˆ)( Pa 600000 =+= =++= = i i l i v if E V vmmmV P Note: iv ,2ˆ was used to calculate fV because the temperature and pressure are the same for the final state of the entire system and the initial state of the piston-cylinder assembly. The value of W can now be evaluated. [ ]kJ 570−=W 52 The energy balance can be used to obtain Q. [ ] [ ]( ) [ ]kJ 1111kJ 570kJ 0.541 =−−=−∆= WUQ Therefore, [ ][ ] −= − =∆ K kJ 35.2 K 15.473 kJ 1111 surrS and = − =∆+∆=∆ K kJ 70.0 K kJ 35.2 K kJ 05.3surrsysuniv SSS 53 3.24 Let the subscript “1” represent the state of the system before the partition is removed. It has two components: component a and component b. Subscript “2” represents the system after the partition is removed. Mass balance: 2,1,1 nnn ba =+ Energy balance for the adiabatic process: WU sys =∆ ( ) ( ) ( )[ ]babbvbaava VVVPTTcnTTcn ,1,12,12,,1,12,,1 +−−=−+−∴ The external pressure, P, for the above energy balance is equal to P1,a [ ]( ) ( ) [ ]( ) [ ]( ) [ ] [ ]3 2 2-,1 m 05.0 m 098.0 sm 8.9kg 1000 K300 Kmol J 314.8mol 2 = ⋅ ⋅ =aV . To find the final temperature, first find the volumes using the ideal gas law. [ ]( ) [ ]( ) [ ]( ) [ ] [ ]324 2 2- 2 2 m 103256.3 m 098.0 sm 8.9kg 1000 Kmol J 314.8mol 4 T T V −×= ⋅ ⋅ = Substitute the volumes into the energy balance and solve for 2T : ( ) ( ) ( ) ( ) ( )[ ]324522 m 15.0103256.3Pa 101K 3002 3mol 2K 300 2 3mol 2 −××−=− +− − TTRTR K 4.3602 =T To calculate the change in entropy, we can use the following relationship b Pb a Pasys P PR T Tcn P PR T TcnS − + − =∆ 1 2 1 2 1 2 1 2 lnlnlnln The only unknown in the above equation is P1,b, so we can calculate it with the ideal gas law: 54 ( ) ( ) Pa 105.0 m 1.0 K 300 Kmol J 314.8mol 2 5 3,1 ×= ⋅ =bP Substitute values into the expression for entropy and solve: ( ) ( ) ba sys RRS − + − =∆ bar .50 bar 1ln K 300 K4.360ln 2 5mol 2 bar 1 bar 1ln K 300 K 4.360ln 2 5mol 2 =∆ K J 72.3sysS 55 3.25 First perform an energy balance on the process: WUsys =∆ The change in internal energy can be written: unununU iAfBfBfAfA ,,,,, −+=∆ A mass balance gives: fBfAiA nnn ,,, += These two expressions can be substituted to give: ( ) ( ) WTTcnTTcnU ifBvfBifAvfA =−+−=∆ ,,,, Using the ideal gas law and Rcv 2 5 = , we obtain WVPVPVP iAiAfAfAfBfB =−+ ,,,,,, 2 5 2 5 2 5 Now, find an expression for the work∫−= f i V V EdVPW The pressure is not constant; its value is given by ( ) A mg A VVk P A mg A kxPP iBBatmatmE + − +=++= 2 , [ ]Pa 10284.8 1006.1 55 BE VP ×+×= Integrating 10142.4 1006.1 2, 5 , 5 fBfB VVW ×−×−= Substituting this expression into the energy balance gives: 2, 5 , 5 ,,,,,, 10142.4 1006.12 5 2 5 2 5 fBfBiAiAfAfAfBfB VVVPVPVP ×−×−=−+ 56 This expression can be simplified by recognizing that fAfB PP ,, = and iAfA VV ,, = : ( ) 2,5,5,,,,, 10142.4 1006.12 5 2 5 fBfBiAiAiAfBfB VVVPVVP ×−×−=−+ We still have two unknowns: fBP , and fBV , . We can eliminate fBP , by writing a force balance for the final state: [ ]Pa 10284.8 1006.1 ,55, fBfB VP ×+×= Substituting this into the energy balance, we obtain ( )( ) 2,5,5,,,,,55 10142.4 1006.12 510284.8 1006.1 2 5 fBfBiAiAiAfBfB VVVPVVV ×−×−=−+×+× Solving for the final volume: 3, m 268.0=fBV and for the final pressure. ( ) Pa 1028.3268.010284.81006.1 555,, ×=×+×=== ffAfB PPP Since the gas in A has undergone an adiabatic, reversible expansion 0, =∆ AsysS Therefore, − = 1 2 1 2 lnln0 P PR T TcP × × − ⋅ = Pa 107.0 Pa 1028.3ln K 15.313 ln 2 7 Kmol J 314.80 5 5 2T Solve for T2 K 2522 =T : 57 3.26 A schematic of the process is drawn: (a) and (b) The maximum work occurs for a reversible process. Applying the second law, we get: 0= + = surrsysuniv dt dS dt dS dt dS or ( ) 0ˆˆ 12 =+− surr surr T Qssm Calculate the change in entropy of the surroundings: We can look up property values of state 1 from the steam tables = kgK kJ 8802.61ˆs and = kg kJ 13.422,31ˆh . Converting the units of mass flow rate gives: [ ] [ ] = = s kg 25.1 s ,6003 hr 1 hr kg 500,4m so =−= kgK kJ 6939.6ˆˆ 12 surr surr Tm Qss We now know two properties of steam ( 2sˆ and P2 T2 = 200 oC ) From the steam tables: Using the steam tables for superheated steam at 1 MPa, we find that when water has this value of entropy 58 = kg kJ 9.2827ˆ2h Energy balance: ( ) sWQhhm +=− 12 ˆˆ Calculate Ws ( ) ( ) ( ) [ ]kW 86.69 kg kJ1.3422 kg kJ9.2827kg/s25.1ˆˆ 12 −− − =−−= QhhmWs : [ ]kW 673−=sW (c) The isentropic efficiency is defined as follows: ( ) ( )reversibleW W S actualS =η For this situation, ( ) [ ]( ) [ ]kW5.447kW 673665.0 −=−=actualSW (d) The real temperature should be higher since not as much energy is converted into work. (e) Use the energy balance: ( ) sWQhhm +=− 12 ˆˆ = + +− =+ + = kg kJ 2.3008 kg kJ1.3422 kg/s 25.1 kW -447.5kW 86.69ˆˆ 12 hm WQh s At 1 MPa, water has this value of enthalpy when ( ) Cº2.2802 =actualT 59 3.27 Since the compressor is adiabatic, the energy balance after neglecting potential and kinetic energy becomes ( ) SWhhn =− 12 Using the ideal gas law and Appendix A.2, the above equation becomes ( )∫ ++++= − 2 1 322 1 11 T T S dTETDTCTBTAT VPW Substituting the following values [ ] [ ]Pa 101bar 1 51 ×==P = s m 1 3 1V [ ]K 15.2931 =T [ ]K 15.4732 =T ;355.3=A ;10575.0 3−×=B ;0=C ;10016.0 5×−=D 0=E (Table A.2.2) gives [ ]kW 8.218=SW This is the work of our real turbine (with 80% isentropic efficiency). We can use the isentropic efficiency to calculate the work of an equivalent 100% efficient process. ( ) ( )compressorS reversibleS compressor W W =η ( ) ( ) ( ) [ ]( )kW 8.2188.0== compressorcompressorSreversibleS WW η ( ) [ ]kW 75.41=reversibleSW Since the process is adiabatic, we can use the following equation again to calculate what the final temperature would be in a reversible process. ( ) 2 1 322 1 11 ∫ ++++= − T T S dTETDTCTBTAT VPW 60 [ ] ( )∫ ++++=∴ − 2 1 W 175400 322 1 11 T T dTETDTCTBTA T VP Substituting the values from above and solving for 2T , we obtain K 8.164,2 =revT To obtain the pressure the final pressure for the real compressor, we can calculate the final pressure for the reversible process because the final pressure is the same in both cases. For the isentropic expansion −==∆ ∫ 1 2ln T 0 2 1 P PRdT c s T T p − ++++ ==∆ ∫ − 1 2 322 ln 0 2 1 P PRdT T ETDTCTBTARs T T ( ) [ ] − −×+ ==∆ ∫ = = −− bar 1 ln 160010575.0355.30 2 K 438 K 15.293 232 1 PdT T TTRs T T Solve for P2 [ ]bar 16.42 =P : 61 3.28 Isentropic efficiency for a turbine is defined as ( ) ( )revs actuals w w =η If the rate of heat transfer is assumed negligible, the energy balance for this process is ( ) sWhhm +−= 21 ˆˆ0 For a reversible, adiabatic process, 0=∆ syss ( )MPa 10 C,º 500ˆˆ 1,2 ss rev = From the steam tables ⋅ == Kkg kJ 5965.6ˆˆ 1,2 ss rev Since satvsatl sss ,22 , 2 ˆˆˆ ≤≤ , a mixture of liquid and vapor exists. The quality of the water can be calculated as follows ( ) satvsatlrev sxsxs ,2,2,2 ˆˆ1ˆ +−= ( ) ⋅ + ⋅ −= ⋅ Kkg kJ .35937 Kkg kJ .302511 Kkg kJ 5965.6 xx 874.0=x Therefore, ( ) satvsatlrev hxhxh ,2,2,2 ˆˆ1ˆ +−= ( ) + −= kg kJ 675.52874.0 kg kJ 17.444874.01ˆ ,2 revh = kg kJ 0.2391ˆ ,2 revh Also, from the steam tables, = kg kJ 6.33731ˆh (500 ºC, 10 MPa) 62 The reversible work is calculated as: ( ) −= − =−= kg kJ 6.982 kg kJ 6.3373 kg kJ 0.2391ˆˆ 12 1 hh m W revS and the actual work as ( ) ( ) −== kg kJ 835 11 m W m W revSactualS η The exit temperature is calculated by determining the enthalpy of the actual exit state: ( ) =+= kg kJ 4.2538ˆˆ 1 1,2 m W hh actualSactual This state is still saturated (although the quality is higher), so the temperature is T2 = 99.6 oC 63 3.29 The subscripts “2” and “3” represent the two outline streams, and “1” represents the inlet stream. First, perform a mass balance: 11321 3 1 3 2 nnnnn +=+= Now writethe energy balance: ( ) SWQhnhnhn +++−= 3322110 Since the system is insulated and there is no shaft work, the energy balance can be rewritten as: 0 3 1 3 2 113121 =−+ hnhnhn ( ) ( ) 0 3 1 3 2 1312 =−+− hhhh Substituting expressions for heat capacity A.2.2, we obtain the following expression: ( ) ( ) 0400010593.0280.3 3 1400010593.0280.3 3 2 3 K 300 2 3 K 400 K 300 2 3 = +×++ +×+ ∫∫ −− T dT T TRdT T TR Solving, we find K 6.1003 =T Now set up an entropy balance for the process. The minimum pressure is required for a reversible process. 0 3 1 3 2 113121 =−+ snsnsn ( ) ( ) 0 3 1 3 2 1312 =−+− ssss Assuming ideal gas behavior, we can express the changes in entropies using Equation 3.62: ( ) − +×+=− ∫ − 1 K 400 K 300 3 3 12 bar1ln400010593.0280.3 P dT TT Rss ( ) − +×+=− ∫ − 1 K 6.001 K 300 3 3 13 bar1ln400010593.0280.3 P dT TT Rss 64 Substitute these expressions into the entropy balance and solve for P1 bar85.11 =P : 65 3.30 A schematic of the process is illustrated below: (a) Since the process is adiabatic and reversible, 0 0 0 =∆ =∆ =∆ surr sys univ s s s (b) We can obtain the final temperature using the steam tables. ⋅ = Kkg kJ 999.61ˆs (540 ºC, 60 bar) ⋅ =∴ Kkg kJ 999.6ˆ2s (20 bar) The pressure and entropy of state 2 can be used to back out T2 Cº 5.3622 =T from the steam tables. (c) To obtain the value of work, perform an energy balance. The process is adiabatic, and potential and kinetic energy effects can be neglected. Therefore, the energy balance is ( ) WuumU =−=∆ 12 ˆˆ From the steam tables, = kg kJ 15.2881ˆ2u (362.5 ºC, 20 bar) 66 = kg kJ 12.3156ˆ1u (540 ºC, 60 bar) Hence, [ ]( ) [ ]kJ 1373 kg kJ 12.3156 kg kJ 15.2881kg 5 −= − =W (d) The specific volume can be found in the steam tables = kg m 14173.0ˆ 3 2v Therefore, [ ]( ) [ ]332 m 709.0kg m 14173.0kg 5 = =V 67 3.31 A schematic of the process is shown below: well-insulated very tiny leak hole P1 = 60 bar m = 5 kg Pure H2O T1 = 540 oC P2 = 20 bar Pure H2O T'2 = ? oC Tsurr = 25 oC Psurr = 1 bar In order to leave the system, the gas must do flow work on the surroundings. The initial state is the same as for Problem 3.30 and the final pressures are the same. Since the water only expands against 1 bar, the work is lower than that for the differential process described in Problem 3.30. Thus, this adiabatic process looses less energy, leading to a higher final temperature. Another way to view this argument is to look at this process as a closed system. This depiction is the expansion analog of the compression process depicted for Example 2.5 in Figure E2.5B (page 57). We can represent this process in terms of two latches, one that keeps the process in its initial state at 60 bar and one that stops the expansion after the pressure has reduced to 20 bar. The process is initiated by removal of the first latch and ends when the piston comes to rest against the second latch. Such a process is depicted as “Problem 3.31’ below. The corresponding reversible process of Problem 3.30 is shown next to it for comparison. Clearly the process on the left does less work, resulting in a greater final temperature. Problem 3.31 Problem 3.30 68 3.32 (a) Consider the tank as the system. Mass balance inoutin mmmdt dm =−= Separating variables and integrating: ∫∫ = t in m m dtmdm 0 2 1 or ∫=− t indtmmm 0 12 Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is ∑ ∑ ++−= out s in ininoutout sys WQhmhm dt dU The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to inin sys hm dt dU = The following math is performed in t inin t inin U U hmumU dtmhdthmdU ˆˆ 2222 000 2 1 ==− == ∫∫∫ = where the results of the mass balance were used. Thus, inhu ˆˆ2 = 69 From the steam tables, = kg kJ 5.3456iˆnh (3 MPa, 773 K) Now the water in the tank is constrained. From the steam tables: = ⋅ = kg m14749.0ˆ Kkg kJ 743.7ˆ 3 2 2 v s = kg kJ 5.3456ˆ MPa, 3 2u Compute the change in entropy. An entropy balance gives: inin sysuniv sm dt dS dt dS − = Integrating with sin ( )in t ininuniv ssmdtmssmS ˆˆˆˆ 22 0 22 −=−=∆ ∫ constant From the steam tables: ⋅ = Kkg kJ 2337.7iˆns (3 MPa, 773 K) Therefore, = ⋅ − ⋅ =∆ K kJ 173.0 Kkg kJ 2337.7 Kkg kJ 743.7 kg m 14749.0 m 05.0 3 3 univS (b) If it tank sits in storage for a long time and equilibrates to a final temperature of 20 ºC, some or all of the vapor will condense and exchange heat with the surroundings. Let the subscript “3” designate the final state of the water when it has reached a temperature of 20 ºC. The change in entropy between state 2 and state 3 is given by the following equation 70 ( ) T QssmS surruniv +−=∆ 232 ˆˆ Let the subscript “3” designate the final state of the water when it has reached a temperature of 20 ºC. First, find quality of the water. From the steam tables: = = kg m79.57ˆ kg m001002.0ˆ 3 ,3 3 ,3 sat v sat l v v (sat. water at 20 ºC) Calculate the quality as follows: ( ) satv sat l vxvxvv ,3,323 ˆˆ1ˆˆ +−== ( ) + −= kg m79.57 kg m 001002.01 kg m14749.0 333 xx 0025.0=x Now calculate the entropy of state 3 ( ) satvsatl sxsxs ,3,33 ˆˆ1ˆ +−= Substitute values from the steam tables: ( ) ⋅ + ⋅ −= Kkg kJ6671.80025.0 Kkg kJ 2966.00025.01ˆ3s ⋅ = Kkg kJ .31750ˆ3s Now calculate the amount of heat transferred to the surroundings: ( )232 ˆˆ uumQQsurr −−=−= Calculate the internal energy of state 3: ( ) satv sat l uxuxu ,3,33 ˆˆ1ˆ +−= Substituting values from the steam tables: 71 = kg kJ 74.89ˆ3u Therefore, [ ]kJ 1141 kg kJ 5.3456 kg kJ 89.74 kg m 14749.0 m 05.0 3 3 = − −=surrQ Now calculate the change in entropy of the universe [ ] K293 kJ 1411 Kkg kJ 743.7 Kkg kJ .31750 kg m 14749.0 m 05.0 3 3 + ⋅ − ⋅ =∆ univS =∆ K kJ 1.38univS The entropy change for both ( ) =∆ K kJ1.55(b) and (a)univS processes can be fount by adding together the entropy change from Part (a) and Part (b): 72 3.33 A schematic is given below valve maintains pressure in system constant T1 = 200 oC x1 = 0.4 V = 0.01 m3 v l Mass balance outoutin mmmdt dm −=−= Separating variables and integrating: ∫∫ −= t out m m dtmdm 0 2 1 or ∫−=− t outdtmmm 0 12 Entropy Balance: surr outout syssurr surr outout sys univ T Qsm dt dS T Qsm dt dSS −+ =++ =∆ Integrating with sout ( ) surr outuniv T QsmmsmsmS −−−−=∆ 121122 constant (1) From steam tables: 73 =×+×=+−= K kg kJ 9715.34323.64.03309.26.0)1(1 gf xuuxs = K kg kJ 4323.62s is the same as 2K kg kJ 4323.6 ssout = = Thus Equation 1 simplifies to surr univ T QssmS −−=∆ )( 121 (2) Energy Balance to find Q : Qhm dt dU outout sys +−= ˆ Integrating [ ]∫ ∫∫∫ +−=+−= t tt outoutoutout um um dtQdtmhdtQhmdU 0 00 ˆ ˆ ˆˆ 22 11 Substituting in the mass balance and solving for Q ( ) outhmmumumQ ˆˆˆ 121122 −−−= To find the mass in each state: =×+×=+−= kg m 0.0511274.04.0001.6.0)1( 3 1 gf xvvxv = kg m 1274.0 3 2v [ ] [ ]kg 196.0 kg m 0.051 m 01.0 3 3 1 1 1 = == v Vm and [ ] [ ]kg 0785.0 kg m 0.1274 m 01.0 3 3 2 2 2 = == v Vm From the seam tables: 74 =×+×=+−= kg kJ 54915.25974.064.8506.0)1(1 gf xuuxu = kg kJ 3.25952u and = kg kJ 2.2793outh Plugging in: ( ) [ ]kJ 228ˆˆˆ 121122 =−−−= outhmmumumQ Back into Equation 2 ( ) 0 s K 473 s kJ 228 K kg kJ 9715.34323.6 s kg 196.0)( 121 = − − =−−=∆ surr univ T QssmS Note: Vaporization is reversible if Tsurr = Tsys. 75 3.34 The process can be represented as: State 1 State 2 V1 = 1 m3 V2 = 10 m3 P1 = 10 bar T1 = 1000 K Work out Process Q = 0 P2 = ? T2 = ? Solving for the number of moles: n = P1V1 RT1 =120.3 mol The maximum work is given by a reversible process. Since it is also adiabatic, the entropy change of the system is zero: ∆s = 0 = cP ln T2 T1 − Rln P2 P1 Since cP = 5 2 R 12 2 1 2 2/5 1 2 PV nRT P P T T == Solving for T2 ( ) K 215 3/2 12 2/5 1 2 = = PV TnRT gives An energy balance on this closed system gives: ∆U = Q +W =W = n cvdT = n cP − R( )dT T1 T2 ∫ T1 T2 ∫ Solving for work, we get W = n 3 2 R T2 − T1( )= −1.18 ×106 J 76 3.35 The maximum efficiency is obtained from a Carnot cycle. From Equation 3.32 H C T Tn −=1 where temperature is in Kelvin. Hence, 614.0 K 15.773 K 15.2981 =−=n 77 3.36 The labeling described in Figure 3.8 will be used for this solution. First, a summary of known variables is provided. State P (bar) T (ºC) 1 30 500 2 0.1 3 4 From our knowledge of ideal Rankine cycles, the table can be expanded as follows State Thermodynamic State P (bar) T (ºC) 1 Superheated Steam 30 500 2 Sat. Liq. and Vapor 0.1 3 Sat. Liquid 0.1 4 Subcooled Liquid 30 The saturation condition constrains state 3. First, start with the turbine. Since the rate of heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, the following is known 21 ˆˆ ss = and ( )12 ˆˆ hhmWs −= From the steam tables: ⋅ = Kkg kJ 2337.71ˆs (500 ºC, 30 bar) ⋅ = ⋅ = Kkg kJ 1501.8ˆ Kkg kJ 6492.0ˆ 2 2 v l s s (sat. H2 = kg kJ 5.34561ˆh O at 0.1 bar) (500 ºC, 30 bar) = = kg kJ 6.2584ˆ kg kJ 81.191ˆ 2 2 v l h h (sat. H2O at 0.1 bar) 78 The quality of the water can be calculated as follows vl sxsxs 221 ˆˆ)1(ˆ +−= By substituting the steam table data, we find 878.0=x Now the quality can be used for the following expression vl hxhxh 222 ˆˆ)1(ˆ +−= Substituting the steam table data and the quality calculated above, we get = kg kJ 7.2292ˆ2h Therefore, [ ]MW 38.116 kg kJ 5.3456 kg kJ 7.2292 s kg 100 −= − =sW At state 3, = = kg m 001010.0ˆ kg kJ 81.191ˆ 3 3 3 v h (sat. H2 ( ) [ ]MW 09.210ˆˆ 23 −=−= hhmQC O(l) at 0.1 bar) Therefore, Since the molar volume of water does not change noticeably with pressure and the compressor is adiabatic, the work required for compressing the fluid can be calculated as follows ( )343ˆ PPvmWc −= [ ]( ) [ ] [ ]MW 302.0W 1002.3Pa 101.01030 kg m 00101.0 s kg 100 555 3 =×=×−× =cW 79 Now we find the rate of heat transfer for the evaporator. For the entire Rankine cycle, Equation 3.84 gives 0=+++=+ CsHCnetnet WWQQWQ We have CCsH QWWQ −−−= Using the values calculated above: [ ]MW 2.326=HQ Equation 3.84: CSnet WWW −= [ ]MW 08.116=netW The efficiency can be calculated with Equation 3.82 [ ][ ] 356.0MW 02.326 MW 08.116 === H net Q W η 80 3.37 To make the Rankine cycle more efficient, we need to increase the area that represents the net work in Figure 3.8. This can be done in a variety of ways: 1. Increase the degree of superheating of steam in the boiler. This process is sketched in the upper left hand Ts diagram below. This change reduces the moisture content of steam leaving the turbine. This effect is desirable since it will prolong the life of the turbine; however, if the steam is heated too high, materials limitations of the turbine may need to be considered. 2.Lower the condenser pressure. Lowering the pressure in the condenser will lower the corresponding saturation temperature. This change will enlarge the area on the Ts diagram, as shown on the upper right below. Thus, we may want to consider lowering the pressure below atmospheric pressure. We can achieve this since the fluid operates in a closed loop. However, we are limited by the low temperature off the heat sink that is available. 81 3. Increase the boiler pressure. This will increase the boiler temperature which will increase the area as shown on the bottom left Ts diagram. 4. We can be more creative about how we use the energy available in the boiler. One way is to divide the turbine into two stages, Turbine 1 and Turbine 2, and reheat the water in the boiler between the two stages. This process is illustrated below. The pressure in Turbine 1 will be higher than the pressure in turbine 2. This process is schematically shown on the bottom right Ts diagram. This process leads to less moisture content at the turbine exit (desirable) and limits the temperature of the superheat (desirable). 82 3.38 The labeling described in Figure 3.8 will be used for this solution. First, a summary of known variables is provided. State P (bar) T (ºC) 1 100 500 2 1 3 4 From our knowledge of ideal Rankine cycles, the table can be expanded as follows State Thermodynamic State P (bar) T (ºC) 1 Superheated Steam 100 500 2 Sat. Liq. and Vapor 1 3 Sat. Liquid 1 4 Subcooled Liquid 100 The saturation condition constrains state 3. First start with the turbine (state 2). Since the rate of heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, the following is known 21 ˆˆ ss = From the steam tables, ⋅ = Kkg kJ 5965.61ˆs (500 ºC, 100 bar) ⋅ = ⋅ = Kkg kJ 3593.7ˆ Kkg kJ 3025.1ˆ 2 2 v l s s (sat. H2 vl sxsxs 221 ˆˆ)1(ˆ +−= O at 1 bar) The quality of the water can be calculated as follows By substituting the steam table data, we find 874.0=x 83 States 1, 2, and 3 are constrained; therefore, the enthalpies can be determined using the steam tables. The enthalpy of state 4 can be calculated from Equation 3.80: ( )3434 ˆˆˆ PPvhh l −+= where == kg m 001043.0ˆˆ 3 3vvl since specific volume of liquids are insensitive to pressure changes. From the steam tables: State 1 2 3 4 hˆ kg kJ 240.83 kg kJ 391.02 kg kJ 17.444 kg kJ 8.724 Note: vl hxhxh 222 ˆˆ)1(ˆ +−= Equation 3.84: CHnet QQW −= Therefore, ( ) ( ) ( ) ( )[ ]23412341 ˆˆˆˆˆˆˆˆ hhhhmhhmhhmWnet −+−=−+−= and ( ) ( )[ ] [ ] − + − × = −+− = kg kJ 0.2391 kg kJ 27.84 kg kJ 5.417 kg kJ 240.83 kW 10100 ˆˆˆˆ 3 2341 hhhh Wm net = s kg 3.116m Now consider the non-ideal turbine and compressor. Use the definition of isentropic efficiencies: ( ) ( )reversibles actuals turbine W W =η ( ) ( ) ( )( )reversibleturbinereversibleSturbineactualS hhmWW 12 ˆˆ −==∴ ηη 84 ( ) ( )actualC reversibleC compressor W W =η ( ) ( ) ( ) compressor reversible compressor reversibleC actualC hhmW W ηη 34 ˆˆ −==∴ Equation 3.84 is used to find the mass flow rate Csnet WWW −= ( ) − +−= compressor reversible reversibleturbinenet hh hhmW η η 3412 ˆˆ ˆˆ Substituting the enthalpy values from the table shown above and the efficiency values gives = s kg 147m A higher flow rate is needed as compared to the reversible process. 85 3.39 The labeling shown in Figure 3.8 will be used for this solution. First, a summary of known variables is provided. State Thermodynamic State P (MPa) 1 Sat. Vapor 1.7 2 0.7 3 4 From our knowledge of ideal Rankine cycles, the table can be expanded as follows State Thermodynamic State P (MPa) 1 Sat. Vapor 1.7 2 Sat. Liq. And Vapor 0.7 3 Sat. Liquid 0.7 4 Subcooled Liquid 1.7 The enthalpy of states 1 and 3 can be found using the NIST website. For states 2 and 4, use the following expressions to find the enthalpies vl hxhxh 222 ˆˆ)1(ˆ +−= ( ) ( )34333434 ˆˆˆˆˆ PPvhPPvhh l −+=−+= To find x, use the fact that the turbine is isentropic. 21 ˆˆ ss = vl sxsxs 221 ˆˆ)1(ˆ +−= From the NIST website: ⋅ = Kmol J 38.1461s (sat. vapor at 1.7 MPa) ⋅ = ⋅ = Kmol J 07.150 Kmol J 45.90 2 2 v l s s (sat. mixture at 0.7 MPa) By substituting the above values into the entropy relationship, we find 86 938.0=x Also, from the NIST website: = mol J 360461h (sat. vapor at 1.7 MPa) = = mol J 35353 mol J 18416 2 2 v l h h (sat. mixture at 0.7 MPa) ×= = − mol m 10954.6 mol J 18416 3 5 3 3 v h (sat. liquid at 0.7 MPa) Substituting these values into the expression for 4h and 2h yields = mol J 9.343022h = mol J 5.184854h Equation 3.82 states ( ) 41 3412 hh hhhh rankine − −−− =η − − − − =∴ mol J 5.18485 mol J 36046 mol J 18416 mol J 5.18485 mol J 36046 mol J 9.34302 rankineη 095.0=rankineη This efficiency is significantly lower than a conventional power plant. 87 3.40 (a) Using the information in the problem statement and our knowledge of ideal vapor compression cycles, the following table can be created State Thermodynamic State P (MPa) 1 Sat. Mixture 0.12 2 Sat. Vapor 0.12 3 Superheated Vapor 0.7 4 Sat. Liquid 0.7 Note: Refer to Figure 3.9 to review labeling convention. Equation 3.85 states ( )12 hhnQC −= which upon combination with Equation 3.89 gives ( )42 hhnQC −= From the NIST website: = mol kJ 295.392h (sat. vapor at 0.12 MPa) = mol kJ 181.244h (sat. liquid at 0.7 MPa) Therefore, [ ]kW 557.7 mol kJ181.24 mol kJ 295.39 s mol 5.0 = − =CQ (b) The power input to the compressor can be calculated with Equation 3.86: ( )23 hhnWC −= Equation 3.87 states 32 ss = 88 From the NIST website: ( ) ⋅ == Kmol J 89.177MPa 0.12at vapor sat.23 ss Now, state 3 is constrained. = mol kJ 031.433h ⋅ = Kmol J 89.177 MPa, 7.0 3s Therefore, [ ]kW 87.1 mol kJ 295.39 mol kJ 031.43 s mol 5.0 = − =CW (c) Equation 3.90 states:05.4 mol kJ 295.39 mol kJ 031.43 mol kJ 181.24 mol kJ 295.39 23 42 23 12 = − − = − − = − − = hh hh hh hhCOP 89 3.41 Using the information in the problem statement and our knowledge of ideal vapor compression cycles, the following table can be created State Thermodynamic State P (MPa) 1 Sat. Mixture 0.12 2 Sat. Vapor 0.12 3 Superheated Vapor 0.7 4 Sat. Liquid 0.7 Note: Refer to Figure 3.9 to review labeling convention. From Equation 3.90 23 12 hh hh W QCOP C C − − == The enthalpy of state 2 can be found directly from the NIST website, but the enthalpies of states 1 and 3 require the use of additional information. = mol kJ 295.392h (sat. vapor at 0.12 MPa) For the process between state 2 and state 3, 32 ss = From the NIST website: ⋅ == Kmol J 89.17723 ss (sat. vapor at 0.12 MPa) Now, state 3 is constrained. = mol kJ 031.433h ⋅ = Kmol J 89.177 MPa, 7.0 3s The process between state 4 and state 1 is also isentropic. ⋅ == Kmol J 09.11541 ss (sat. liquid at 0.7 MPa) The quality of the R134a can be calculated as follows 90 ( ) vl xssxs 111 1 +−= where ⋅ = Kmol J 649.901 ls (sat. R134a(l) at 0.12 MPa) ⋅ = Kmol J 89.1771 vs (sat. R134a(v) at 0.12 MPa) Therefore, ( ) ⋅ + ⋅ −= ⋅ Kmol J 89.177 Kmol J 649.901 Kmol J 09.115 xx 280.0=∴ x Using the quality of R134a, the enthalpy of state 1 can be calculated as follows ( ) vl xhhxh 111 1 +−= From the NIST website: = mol kJ 412.171 lh (sat. R134a(l) at 0.12 MPa) = mol kJ 295.391 vh (sat. R134a(v) at 0.12 MPa) Therefore, ( ) + −= mol kJ 295.3928.0 mol kJ 412.1728.011h = mol kJ 74.201h Now, everything needed to calculate COP is available. Using Equation 3.90, 97.4 mol kJ 295.39 mol kJ 031.43 mol kJ 74.20 mol kJ 295.39 23 12 = − − = − − = hh hhCOP 91 Is this modification practical? No. An isentropic turbine adds significant level of complexity to the cycle. Turbines are expensive and wear over time. Furthermore, the real turbine added to the cycle will not be 100% efficient, so the COP will not increase as much. The cost of the turbine is not justified by the increase in COP. 92 3.42 The subscript “h” refers to the hotter cycle, while “c” refers to the cooler cycle. The (a) The flow rate of the cooler cycle can be found by performing an energy balance on the condenser/evaporator shared between the two cycles. An energy balance shows: QQQ hCcH == ,, Written in terms of enthalpy, the equation is ( ) ( )1278 hhnhhn hc −−=− We can find the enthalpies for positions 2 and 8 directly from the thermodynamic tables because the fluid is saturated in at these positions. To find the other enthalpies we must use the following relationships: 76 ss = 41 hh = Using the NIST website: = mol J 241811h (sat. R134 liquid at 0.7 MPa) = mol J 409672h (sat. R134 vapor at 0.35 MPa) = mol J 415107h (R134 vapor at 0.35 MPa with ⋅ = Kmol J89.177s ) = mol J 210998h (sat. R134 liquid at 0.35 MPa) Now the flow rate can be calculated: [ ]( ) − − − = mol J 41510 mol J 21099 mol J 24181 mol J 40967mol/s 5.0 cn = s mol 411.0cn 93 (b) Performing an energy balance we find ( )56 hhnQ cC −= We can also use the following relationship == mol J 2109985 hh From the NIST website: = mol J392956h (sat. R134 vapor at 0.12 MPa) Therefore, [ ]kW 48.7 mol J 21099 mol J39295 s mol 411.0 = − =CQ (c) The power input is calculated as follows: hCcCtotalC WWW ,,, += where ( )67, hhnW ccC −= ( )23, hhnW hhC −= We have all of the required enthalpies except h3 ⋅ == Kmol J95.17523 ss . State 3 is constrained because (sat. R134a vapor at 0.35 MPa) MPa 7.03 =P From the NIST website: = mol J424283h (R134a vapor at 0.7 MPa with ⋅ = Kmol J95.175s ) Now compute the power for each unit: 94 [ ] [ ] − + − = mol J 40967 mol J 42428mol/s 5.0 mol J 39295 mol J 41510mol/s 411.0,totalCW [ ]kW 64.1, =totalCW (d) The coefficient of performance is calculated using the following equation: 56.4 kW64.1 kW 48.7 , === totalC C W QCOP (e) The COP for the cascade is 4.56, while the COP is 4.05 in Problem 3.40. The cascade system’s COP is 12.6% greater. 95 3.43 One design follows; your design may differ: In order to cool a system to -5 ºC, the temperature of the fluid must be colder than -5 ºC so that heat transfer will occur. We arbitrarily specify that the working fluid evaporates at -10 ºC. Similarly, in order to eject heat to the 20 ºC reservoir, the fluid must condense at a temperature greater than 20 ºC. Arbitrarily, we choose 25 ºC. One possible refrigeration cycle is presented below. A number of fluids will work sufficiently for this system, but the design process will be illustrated using R134a. For states 3 and 4, the pressure is constant, and for states 1 and 2, the pressure is constant at a different value. From the NIST website, we find MPa 201.021 == PP (T sat MPa 665.043 == PP = -10 ºC) (Tsat ( )12 hhnQC −= = 25 ºC) (Note: The temperatures of each state are not constant. The listed saturation temperatures are the temperatures at which the fluid evaporates and condenses.) Now that the pressures are known, we can compute the required flow rate required in order to provide 20 kW of cooling. We can get h2 from the thermodynamic tables for saturated R134a. In order to find h1 41 hh = , we can use the following relationship: From the NIST website: = mol J 400642h (sat. R134 vapor at 0.201 MPa) 96 = mol J 239314h (sat. R134 liquid at 0.665 MPa) Now, calculate the required flow rate of R134a. ( ) [ ] − = − = mol J 23931 mol J 40064 J/s 000,20 12 hh Qn C = s mol24.1n 97 3.44 The schematic of the process and the corresponding Ts is presented below A number of refrigerants will work for this system, but the design process will be illustrated for R134a only. To define each state, we need to thermodynamically constrain each state. You should note that the problem doesn’t statewhat temperature the fluid condenses at. Therefore, we can assume the temperature is 25 ºC. By using information in the Ts diagram and from the NIST website, we find MPa 34966.021 == PP (T1=T2=T sat MPa 16394.043 == PP = 5 ºC) (T3=T4=Tsat MPa 66538.065 == PP = -15 ºC) (T6 State = 25 ºC) Before solving for additional pressures and temperatures, we will list the known temperatures and pressures. Temperature (ºC) Pressure (MPa) 1 5 0.24334 2 5 0.24334 3 -15 0.16394 4 -15 0.16394 5 25 0.66538 6 25 0.66538 States 4, 5, and 6 are completely constrained as confirmed by Gibbs phase rule. Now, we need to find the liquid and vapor compositions of states 1, 2, and 3 to completely constrain the states. Since the heat duties are equal for the refrigerator and the freezer, we have the following relationship: RCFC QQ ,, = 98 which upon performing an energy balance becomes 1234 hhhh −=− Energy balances around the valves provide: 61 hh = 32 hh = Substituting these relationships into the expression that equates the heat loads results in 2 64 32 hhhh +== From the NIST website: = mol J 397554h = mol J 239316h Now constrain states 1, 2, and 3 by determining the enthalpies: == mol J 2393161 hh = + == mol J 31843 2 mol J 23931 mol J 39755 32 hh Technically, the states are now all constrained, but we would like also like to know the vapor and liquid compositions of each state. The compositions of states 4, 5, and 6 are already known; we can calculate the compositions of states 1, 2, and 3 as follows: ( ) satv sat l hxhxh ,11,111 1 +−= ( ) satv sat l hxhxh ,22,222 1 +−= ( ) satv sat l hxhxh ,33,333 1 +−= From the NIST website, 99 = mol J 21095,1 sat lh = mol J 40965,1 sat vh = mol J 21095,2 sat lh = mol J 40965,2 sat vh = mol J 18380,3 sat lh = mol J 39755,3 sat vh Now, we can solve the composition of vapor for each state: 143.01 =x 541.02 =x 630.03 =x The following table presents a summary of our results: State Temperature (ºC) Pressure (MPa) Phases Liquid Composition Vapor Composition 1 5 0.24334 Saturated Liquid and Vapor 0.857 0.143 2 5 0.24334 Saturated Liquid and Vapor 0.459 0.541 3 -15 0.16394 Saturated Liquid and Vapor 0.37 0.630 4 -15 0.16394 Saturated Vapor 0 1 5 25 0.66538 Superheated Vapor 0 1 6 25 0.66538 Saturated Liquid 1 0 100 3.45 In this case, the working material is a solid. The four states of the magnetic material are shown of the sT diagram below. Note the axis are shifted from the usual manner. 1 2 3 4 QC QH (a) The heat expelled by the cold reservoir can be approximated by: ( )12 ssTq CC −≈ where T is the average temperature between states 1 and 2, which is approximately 1 K. Taking values of S/R from the sT diagram, we get: ( ) ⋅ =−≈ Kmol J0.53.19.1RqC (b) Similarly, the heat absorbed by the hot reservoir can be approximated by: ( ) ( ) ⋅ =−=−≈ Kmol J519.12.18.844 RssTq HH (b) The coefficient of performance is given by: 11.0 , = − == CH C totalC C qq q w qCOP (d) The value of COP is much lower than a typical refrigeration process (COP= 4-6); in general, refrigeration processes at these low temperatures are much less efficient. F. Work is supplied to magnetize the material and to spin the wheel. 101 3.46 When the polymer is unstretched it is in a more entangled state. When stretched the polymer chains tend to align. The alignment decreases the spatial configurations the polymer can have, and therefore, reduces that component of entropy. If the process is adiabatic, the entropy of the system cannot decrease. Consequently, its thermal entropy must increase. The only way this can be accomplished is by increased temperature. 102 3.47 Assume the temperature is 298 K. The following data was taken from Table A.3.2: Species ºfh∆ (kJ/mol) º fg∆ (kJ/mol) Cu2 -170.71 O (s) -147.88 O2 0 (g) 0 CuO (s) -156.06 -128.29 The data listed above were used to create the next table using T gh s fff ºº º ∆−∆=∆ Species ºfs∆ (kJ/mol K) Cu2 -0.0766 O (s) O2 0 (g) CuO (s) -0.0932 Now the change in entropy of the reaction can be calculated in a method analogous to Equation 2.72 ( ) ⋅ + ⋅ −=∆=∆ ∑ Kmol kJ 0.0932-4 Kmol kJ 0.0766-2ºº ifirxn svs −=∆ mol kJ 22.0ºrxns Does this violate the second law of thermodynamics? This problem shows that the entropy change of the system is negative, but nothing has been said about the entropy change of the universe. We must look at the change in entropy of the surroundings to determine if the second law is violated. By looking at the enthalpies, we see that the reaction is exothermic, which means that heat is transferred from the system to the surroundings. Therefore, the entropy of the surroundings will increase during this reaction. 103 3.48 The temperature and pressure terms in the equation for entropy do not contribute anything to the entropy change because they are constant. Therefore, the only remaining entropy contribution, the randomness of the atomic arrangements, must be considered. The randomness does not increase when CdTe forms. In pure crystals of Cd and Te, the location of each atom is known because the crystal lattice constrains the atomic locations. In CdTe, the crystal lattice still defines the location of each atom, so the randomness has not increased. Therefore, the change in entropy is zero. 3.49 This argument is not scientifically sound. Morris is arguing that since evolution results in more order, the second law of thermodynamics is violated, so evolution must be impossible. However, the flaw in this argument is caused by ignoring the entropy change of the entire universe. The second law states that the entropy of the universe will remain constant or increase for any process. Morris’ argument was based on the entropy of the system undergoing evolution – not the entropy of the entire universe. A system can decrease in entropy if the entropy of the surroundings increases by at least that much. 104 3.50 Entropy is related to the number of configurations that a state can have. The greater the number of configurations, the more probable the state is and the greater the entropy. We can qualitatively relate this concept to the possible hands in a game of poker, but it is more interesting to quantify the results using some basic concepts of probability. We consider a hand of poker containing 5 cards randomly draw from a deck of 52 cards.. There are a finite number of permutations in which we can arrange a 52 card deck in 5 cards. For the first card in the hand, we select from 52 cards, the second card can be any other card so we select from 51 cards, the third card has 50 cards, and so on. Thus the number of permutations of 5 cards is: P = 52 × 51×50 × 49 × 48 = 311,875,200 However, we do not care the order in which the cards are dealt, so we must divide this number by the number of ways we can come up with the same hand. We do this math in a similar way. For a given handthere are five cards we can pick first, four we can pick second, …. So the number of ways we can make the same hand is: N = 5 × 4 × 3 ×2 ×1=120 The number of unique configurations can be found by dividing P by N . Thus, the cards can display C = P N = 2,598,960 or 2,598,960 unique configurations. To find the “entropy” of a given hand, we need to find out how many of these unique configurations belong to the hand Consider a four of a kind. There are 13 different possible ranks of four of a kind, one for each number A, 2, 3, 4, 5, 6, 7,8, 9, 10, J, Q, K. The fifth card in the hand could be any of the other 48 cards. Therefore, the number of combinations of four of a kind is: 13 x 48 = 624 The probability of a four of a kind is: 624 / 2598960 = 1 / 4165 which is very unlikely. Thus the “entropy” of this hand is low. In contrast, there are 1,098,240 to have a hand that has one pair; therefore the probability of getting this hand is much greater, 1/2.4, and its “entropy” is high. The probability of having nothing is 1/2 which represents the most likely hand in poker, or the hand with the highest “entropy”. In fact, the rules of poker are defined so the hand of lower “entropy” always beats the hand of higher “entropy.” 1 Chapter 4 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 4.1. (a) Yes, the form of the equation is reasonable. This can be rewritten: n AA A v a v RTP ++ + −= It is equivalent to RTPv = , except the pressure term has been corrected to account for the ions’ intermolecular forces. The coulombic forces between the gas molecules affect the system pressure. This modification is similar to the van der Waals equation. Since we are limited to 1 parameter, we need to choose the most important interaction. Since net electric point charges exert very strong forces, this effect will be more important than size. (b) The sign should be negative idealA PP >+ for a because the positively charged gas molecules repel each other due to coulombic forces. Therefore, the system pressure increases, i.e.,: Coulombic forces are much stronger than van der Waals interactions, so a will be large – much larger than a values for the van der Waals EOS. (c) Coulombic repulsion is the primary intermolecular force present in the gas. Coulombic potential energy is proportional to r-1. v is proportional to r3, so the coulombic potential goes as v-1/3 3 1 =n . Therefore, a must have must have the following units to maintain dimensional homogeneity with the pressure term: = = 323/1 m J m N v a so [ ] ⋅ = 1/32 molm Ja 3 4.2 The attractive interactions are described by van der Waals forces. For both O2 + −=Γ ji jiji II II r62 3 αα and propane, the dipole moments are zero. Therefore, the expression for the interactions reduce to From Table 4.1: [ ] [ ]erg 10933.1eV 07.12 11−×==aI [ ] [ ]erg 10753.1eV 94.10 11−×==bI Now obtain the requested expressions: [ ]( ) ( )( ) ×+× ××× −=Γ −− −−− erg 10933.1erg 10933.1 erg 10933.1erg 10933.1cm 1016 2 3 1111 1111 6 2325 r aa [ ]ergcm 1071.3 66 59 ⋅ ×− =Γ − r aa [ ]( ) [ ]( ) ( )( ) ×+× ×××× −=Γ −− −−−− erg 10753.1erg 10933.1 erg 10753.1erg 10933.1 cm 109.62cm 1016 2 3 1111 1111 6 325325 r ab [ ]ergcm 1039.1 66 58 ⋅ ×− =Γ − r ab [ ]( ) ( )( ) ×+× ××× −=Γ −− −−− erg 10753.1erg 10753.1 erg 10753.1erg 10753.1cm 109.62 2 3 1111 1111 6 2325 r bb [ ]ergcm 1020.5 66 58 ⋅ ×− =Γ − r bb (b) Calculation: [ ] [ ] ⋅ ×− ⋅ ×− =ΓΓ −− ergcm 1020.5ergcm 1071.3 66 58 6 6 59 rr bbaa 4 [ ]ergcm 1039.1 66 58 ⋅ ×− =ΓΓ − r bbaa Note: Disregarded the positive value. The value of bbaaΓΓ is equal to abΓ . The values are equal because the ionization energies are similar. (c) An expression for the average intermolecular attraction in the mixture can be found using the mixing rules bbbabbaaaamix yyyy Γ+Γ+Γ=Γ 22 2 ( ) [ ]ergcm 1020.51039.11071.3 66 25858259 ⋅ ×+×+×− =Γ −−− r yyyy bbaa mix 5 4.3 (a) 300 K, 10 atm. The intermolecular distance of molecules is greater at lower pressures. Therefore, fewer intermolecular interactions exist, which cause less deviation from ideality. (b) 1000 K, 20 atm. At higher temperatures, the kinetic energy of the molecules (speed) is greater. The molecules interact less; thus, the compressibility factor is closer to unity. (c) Let subscript “1” denote BClH2 and “2” denote H2 v B z mix+= 1 . For the mixture, we calculate the compressibility factor as follows where 2 2 212211 2 1 2 ByByyByBmix ++= Since BClH2 is polar and H2 21 BBB mix >>> is non-polar and small Therefore, the plot may look like the following. 6 4.4 (a) The intermolecular attractions and volume occupied by the styrene monomers will contribute to the deviations from ideality. Since styrene monomers are essentially non-polar, the order of importance is as follows inductiondipole-dipoledispersion ≅> The van der Waals EOS is appropriate since it accounts for the occupied molar volume and intermolecular forces, but it should be noted that more modern EOSs will give more accurate results. (b) Use critical data to calculate the a and b parameters: ( ) ( ) [ ] ⋅ = × ⋅ == 2 3 5 2 2 mol mJ13.3 Pa 1039 K 15.647 Kmol J 314.8 64 27 64 27 c c P RTa ( ) [ ]( ) ×= × ⋅ == − mol m 1072.1 Pa 10398 K 15.647 Kmol J 314.8 8 3 4 5c c P RTb Now we can use the van der Waals EOS to solve for temperature. 2v a bv RTP − − = [ ] 23 4 2 3 3 4 3 4 5 mol m 100.3 mol mJ13.3 mol m 1072.1 mol m 100.3 Kmol J 314.8 Pa 1010 × ⋅ − ×− × ⋅ =× −−− T Cº 65.277K 8.550 ==∴T The styrene will not decompose at this temperature. (c) The a parameter is related to attractive intermolecular forces. Dispersion is the controlling intermolecular force in this system, and its magnitude is directly related to the size of the molecules (polarizability component of dispersion). For the 5-monomer long polymer chain, a is 5 times the a value in Part (b). The b parameter is also related to the size of the molecule since it accounts for the volume occupied by the molecules. Again, the b parameter for the reduced polymer chain is 5 times the b value from Part (b). 7 ⋅ = 2 3 mol mJ 65.15a ×= − mol m 106.8 3 4b(d) We must realize that if we initially believed there were 100 moles of styrene in the reactor, then there can only be 20 moles of the 5-monomer long polymer chain. Therefore, ×= − mol m 105.1 3 3v and [ ] 23 3 2 3 3 4 3 3 5 mol m 105.1 mol mJ65.15 mol m 106.8 mol m 105.1 Kmol J 314.8 Pa 1010 × ⋅ − ×− × ⋅ =× −−− T Cº26.393K41.612 ==T Decomposition will occur. 8 4.5 There are many ways to solve this problem, and the level of complexity varies for each method. To illustrate the principles in Chapter 4, two of the simplest solution methods will be illustrated. Method 1. Polarizability of each atom The polarizability of a molecule scales with the number of atoms; the polarizabilities of individual atoms are additive. Using the first two molecules, solution of the following system of equations 4 141 CHHC ααα =+ 62 162 HCHC ααα =+ gives [ ]325 cm 104.11 −×=Cα [ ]325 cm 1065.3 −×=Hα Now, the polarizability of the chlorine atom can be found. For chloroform, 4 141 CClClC ααα =+ Therefore, [ ]325 cm 104.23 −×=Clα The values of 83HC , ClCH3 , 22ClCH , and 3CHCl are calculated with the polarizabilities found above as follows, and compared to the values given. Species α calculated (x 1025 cm3 α reported (x 10) 25 cm3 % Difference ) C3H 63.4 8 62.9 0.8 CH3 45.8 Cl 45.6 0.3 CH2Cl 65.5 2 64.8 1.1 CHCl 85.3 3 82.3 3.6 All agree reasonably well. Now, the polarizabilities of 104HC and ClHC 52 will be calculated. [ ]( ) [ ]( ) [ ]325325325 cm 101.82cm 1065.310cm 104.114 104 −−− ×=×+×=HCα 9 [ ]( ) [ ]( ) [ ]( ) [ ]325325325325 cm 1045.64cm 104.231cm 1065.35cm 104.112 52 −−−− ×=×+×+×=ClHCα Method 2. Bond Polarizability For this method, we calculate the molecules’ polarizabilities by adding the polarizability of each bond, instead of the atoms. For the methane molecule 4 14 CHHC αα =− [ ] [ ]325325 cm 105.6 4 cm 1026 −− − ×= × =∴ HCα To calculate the polarizability of a C-C bond, use ethane as follows: 62 16 HCCCHC ααα =+ −− [ ] [ ]( ) [ ]325325325 cm 107.5cm 105.66cm 107.44 −−−− ×=×−×=∴ CCα The C-C and C-H polarizability calculated above predict the given polarizability of propane well. The polarizability of C-Cl bonds is calculable with the polarizability of chloroform. 4 14 CClClC αα =− [ ] [ ]325325 cm 1025.26 4 cm 10105 −− − ×= × =∴ ClCα The values of 83HC , ClCH3 , 22ClCH , and 3CHCl are calculated with the polarizabilities found above as follows, and compared to the values given. Species α calculated (x 1025 cm3 α reported (x 10) 25 cm3 % Difference ) C3H 63.4 8 62.9 0.8 CH3 45.8 Cl 45.6 0.3 CH2Cl 65.5 2 64.8 1.1 CHCl 85.3 3 82.3 3.6 All agree reasonably well. This value predicts the polarizabilities of the other species in the table reasonably well. Now, the polarizabilities of 104HC and ClHC 52 will be calculated. HCCCHC −− += ααα 1031 104 [ ]( ) [ ]( ) [ ]325325325 cm 101.82cm 105.610cm 107.53 104 −−− ×=×+×=HCα 10 ClCHCCCClHC −−− ++= αααα 1511 52 [ ]( ) [ ]( ) [ ]( ) [ ]325325325325 cm 1045.64cm 1025.261cm 105.65cm 107.51 52 −−−− ×=×+×+×=ClHCα Note both the atom method and the bond method yield identical results. More accurate values for the polarizabilities can be calculated using more of the data given in the problem. 11 4.6 (a) σ increases with molecular size. Therefore, 222 OSI σσσ >> ε scales with magnitude of van der Waals interactions. Because these are non-polar, diatomic molecules, only dispersion forces are present. Dispersion forces depend on the first ionization potential and polarizability. Ionization energy is approximately equal for each molecule. The polarizability scales with molecular size, so 222 OSI εεε >> (b) σ increases with molecular size. Diethylether and n-butanol have the same atomic formula and similar spatial conformations. Therefore, they should be about equal in size. Methyl ethyl ketone has fewer atoms, but has two exposed electron pairs on the double-bonded oxygen. The size of the molecular electron orbital of methyl ethyl ketone is approximately equal to the sizes of diethyl ether and n-butanol, so ketone ethyl methylether diethylbutanol-n σσσ ≅≅ Methyl ethyl ketone and n-butanol are much more polar than diethyl ether due to their greater asymmetry, so their ε values are greater than diethyl ether’s. Now we must determine if there is greater attraction in n-butanol or methyl ethyl ketone. ε scales with the magnitude of van der Waals interactions. Since induction and dispersion forces are similar in these molecules, we must consider the strength of dipole-dipole forces. There is greater charge separation in the double bond of ketone, so ether diethylbutanol-nketone ethyl methyl εεε >> 12 4.7 (a) At 30 bar, the water molecules are in closer proximity than they are at 20 bar. Intermolecular attractions are greater, so the magnitude of molecular potential energy is greater. The potential energy has a negative value for attractive interactions. The molecular kinetic energy is identical since the temperature is the same. Hence, the internal energy, the sum of kinetic and potential energies, is less at 30 bar. (b) The key to this phenomenon is hydrogen bonding. At 300 K and 30 bar, isopropanol and n- pentane are both liquids. The hydrogen bonding and dipole-dipole interactions are present in isopropanol, and dispersion is present in n-pentane. The intermolecular forces are greater in the isopropanol, so the compressibility factor is smaller for isopropanol. At 500 K and 30 bar, both species are gases. In the gas phase, hydrogen bonding does not play a significant role. The dispersion forces in n-pentane are stronger than the dipole-dipole forces of isopropanol. Therefore, the compressibility factor is smaller for n-pentane. 13 4.8 (a) Ideal: For ideal NH3, the compressibility factor is equal to one. For real NH3, the strong intermolecular forces (dipole-dipole and dispersion) cause the molar volume to decrease. They outweigh the volume displaced by the physical size of NH3; thus, z will be less than one. (b) Internal energy will be greater for the ideal gas. In the real gas, intermolecular attractions are present. Internal energy value is the sum of potential and kinetic energies of the molecules. The absolute values of the kinetic energies are identical at identical temperature: however, the potential energy decreases for real NH3 due to attractive interactions - so the internal energy is less for the real NH3. (c) The entropy will be greater for the ideal gas. Entropy is a measure of possible molecular configurations or “randomness”. Ammonia has an electric dipole in which positive and negative charge are separated. The intermolecular forces in the real gas cause the molecules to align so that the positive charge in one molecule is adjacent to a negative charge in a neighboring molecule to reduce potential energy. Therefore, fewer possible configurations exist, which creates less randomness and lower entropy. 14 4.9 (a) We can determine which case has the higher compressibility factor by comparing the molar volumes at constant T and P. With Ne, very weak intermolecular attractions are present,so volume displacement becomes important. The compressibility factor will be slightly greater than one. In NH3, the strong intermolecular attractions decrease the molar volume, so z is less than one. The compressibility factor is greater for Ne. (b) Entropy is a measure of randomness. Both species are gases at these conditions. The intermolecular attractions present in NH3 reduce the number of possible configurations. The weak forces present in Ne have a much smaller effect. However, NH3 is asymmetrical, while Ne is symmetrical. The asymmetry of NH3 results in more possible configurations that NH3 can have. Therefore, it is difficult to qualitatively show for which case the entropy is greater. Since both species are gases, intermolecular interactions are relatively weak, and we can guess that entropy is greater in NH3. 15 4.10 (a) To find the average distance between the two atoms of Ar, we can find the volume that each atom occupies. The molar volume can be found from the compressibility factor. At 300 K and 25 bar, 513.0 99.1 = = r r P T From the generalized compressibility charts: 9859.0=z Therefore, ( ) [ ] ×= × ⋅ = − mol m 1084.9 Pa 1025 K 300 Kmol J 314.8 9859.0 3 4 5v ×= − atom m 1063.1 3 27v The second number was found by dividing the molar volume by Avogadro's number. To estimate the distance between each atom, we note that the distance between molecules can be related to the volume by: vr ∝3 Consider the geometry shown below: r v3 Ar Ar A rough estimate of r is ( ) [ ]m 1018.1m 1063.1 93/1327 −− ×=×=r 16 [ ]A8.11=r (b) The potential energy due to gravity can be calculated as follows r mG Ar G 2⋅− =Γ where G is the gravitation constant ⋅ ×= − 2 2 11 skg m 1067.6G and mAr [ ] [ ] [ ] [ ] [ ] [ ] ×= × = atom kg 10633.6 g 1000 kg 1 atoms 10023.6 mol 1 Ar mol 1 Ar g 948.39 26- 23Arm is the mass of an argon atom. The mass of an argon atom is calculated as follows Using the distance calculated in Part (a), we get [ ]J1049.2 -52×−=ΓG (c) Equation 4.13 quantifies the potential energy due to London interactions + −=Γ − ArAr ArArArAr ArAr II II r62 3 αα From Table 4.1 [ ] [ ]erg 1052.2eV 76.15 -4×==ArI [ ]325 cm 106.16 −×=Arα Using the distance calculated in Part (a) [ ]( )cm1018.1 7−×=r , we get [ ] [ ]J 1093.1erg 1093.1 1710 −−− ×−=×−=Γ ArAr (d) The potential energy due to London interactions is around 3510 times greater than the potential energy due to gravity. Clearly, London interactions are much more important, and gravitational effects can be neglected. 17 4.11 We want to compare the Lennard-Jones potential to one with an exponential repulsion term. As provided in the text, Equation 4.19, the Lennard-Jones potential is − =Γ 612 4 rr σσε To simplify further analysis, we can rewrite the equation in dimensionless quantities: ( ) ( ) −= Γ =Γ 612 * 1 * 1 4 * rrε where σ rr =* We want to compare this to an exponential repulsive function ( ) − =Γ 12** 2 1 * exp 1exp rr c c We have two adjustable parameters, c1 and c2, to match the first (LJ) potential to the second (exp) potential. We need to choose reasonable criteria to specify. For this solution we choose equal well depths and equal values at Γ=1. Other choices may be just as valid; you should realize that you have two parameters to fit and so must specify two features. Using the above criteria, we iterate on a spreadsheet, to get the solution: c1 = 143,000 and c2 = 11.8 This solution is shown at two magnifications in the plots on the following page: 18 Potential functions -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 Lennard-Jones exponential Potential functions -5 0 5 10 15 20 0.7 1 1.3 1.6 1.9 Lennard-Jones exponential We draw the following conclusions: 1. The most stable configuration (the bottom of the well) occurs at a greater separation for the exp model. 2. The Lennard-Jones potential increases more steeply at small radii, i.e., it behaves more like the hard-sphere potential. 3. The two models are in reasonable qualitative agreement 19 4.12 (a) The bond strength of a sodium ion can be viewed as the amount of energy it would take to remove the sodium ion from the crystal lattice. The interaction between the chlorine and sodium ions is Coulombic attraction. ( )( )( ) [ ]erg 1001.5cm 1076.2 esu10803.4esu10803.466 11-8 1010 − −− ×−= × ×−× ==Γ r QQ ClNa [ ]eV 3.31−=Γ (b) r QQ r QQ NaNaClNa 126 +=Γ [ ] ( )( ) cm 1090.3 esu10803.4esu10803.412erg 1001.5 -8 1010 11 × ×× +×−=Γ −− − [ ] [ ]eV 0.13erg 10089.2 11 =×=Γ − (c) r QQ r QQ r QQ ClNaNaNaClNa 8126 ++=Γ [ ] ( )( ) cm 1078.4 esu10803.4esu10803.48erg 10089.2 -8 1010 11 × ×−× +×=Γ −− − [ ] [ ]eV 1.11erg 1077.1 11 −=×−=Γ − (d) r QQ r QQ r QQ r QQ NaNaClNaNaNaClNa 68126 +++=Γ [ ] ( )( ) cm 1052.5 esu10803.4esu10803.46erg 1077.1 -8 1010 11 × ×× +×−=Γ −− − [ ] [ ]eV 60.4erg 1037.7 12 =×=Γ − 20 4.13 The boiling points of the halides depend upon the strength of intermolecular attractions. The stronger the intermolecular attraction, the higher the boiling point. Dispersion and dipole-dipole interactions are present in all five species listed. The magnitude of the dipole-dipole interactions is similar so the pertinent intermolecular force in these molecules is dispersion. The molecular size increases from left to right. Polarizabilities are greater in larger molecules, which manifests in larger dispersion forces. Therefore, the boiling point increases from left to right. 21 4.14 van der Waals forces hold the Xe atoms together in a molecule of Xe2 K 229= k ε . The potential energy can be quantified with the Lennard-Jones potential function. The bond length is the r value where the potential is a minimum. From Table 4.2: and A 1.4=σ We start with Equation 4.19: to get − =Γ 612 4 rr σσε Differentiation with respect to r yields 06124 612 = + −= Γ rrrrdr d σσε where we set this derivative equal to zero to find the minimum. Solving gives 2 16 = r σ or A 60.412.126 === σσr 22 4.15 The data in the following table were taken from Table A.1.1 Species [ ]K cT [ ]Pa 10-5×cP He 5.19 2.27 CH 190.6 4 46.00 NH 405.6 3 112.77 H2 647.3 O 220.48 The van der Waals a parameter can be calculated using Equation 4.39. ( ) c c P RT a 2 64 27 = The van der Waals b parameter can be calculated using Equation 4.40.c c P RT b 8 = Using the data table and equations listed above, the following table was created; the values of dipole moment and polarizability reported in Table 4.1 are also included. Species ⋅ 2 3 mol mJ a × mol m 10 3 5b µ [D] α [cm3 x 1025] He 0.00346 2.38 0 2.1 CH 0.230 4 4.31 0 26 NH 0.425 3 3.74 1.47 22.2 H2 0.554 O 3.05 1.85 14.8 The values of a for helium is two orders of magnitude less than the other species since it only has weak dispersion forces (small atom, small α). The values of a for methane, ammonia, and water are of the same magnitude because the sums of the intermolecular attractions for each molecule are similar. All three molecules have comparable dispersion forces; although slightly weaker in the ammonia and water. However, unlike methane, these two molecules also have dipole-dipole and induction forces. In fact, the strong dipole in water gives it the largest value The values of b are all of the same magnitude, as expected since b scales with size according to the number of atoms in the molecule. Size of Molecules: HeOHNHCH >>> 234 ∴Value of b: HeOHNHCH bbbb >>> 234 23 4.16 The van der Waals b parameter can be calculated using Equation 4.40. c c P RTb 8 = Critical point data can be found in Table A.1.1. The following table was made: Species [ ]K cT [ ]Pa 10-5×cP × mol m 10 3 5b CH 190.6 4 46.00 4.306 C6H 562.1 6 48.94 11.94 CH3 512.6 OH 80.96 6.58 Note: Used ⋅ = Kmol J 314.8R for calculation of a. The equation from page 186 can be rewritten to calculate σ . 3 2 3 AN b π σ = The table listed below was created using this equation, and data from Table 4.2 are included along with the percent difference. Species [ ]m 1010×σ Table 4.2 Value [ ]m 1010×σ Percent Difference CH 3.24 4 3.8 14.7 C6H 4.56 6 5.27 13.5 CH3 3.74 OH 3.6 3.8 24 4.17 The van der Waals a parameter can be calculated using Equation 4.39. The values for the above equation were taken from Table A.1.1, and the following table was made: Species [ ]K cT [ ]Pa 10-5×cP ⋅ 2 3 mol mJ a CH 190.6 4 46.00 0.2303 C6H 562.1 6 48.94 1.883 CH3 512.6 OH 80.96 0.9464 Note: Used ⋅ = Kmol J 314.8R for calculation of a. The equation from page 187 can be used to find C6 3 6 2 3 2 σ π CNa a= . 2 3 6 2 3 aN aC π σ =∴ The σ values can be found in Table 4.2. The following table can now be created Species [ ]m 1010×σ [ ]6776 mJ 10 ⋅×C CH 3.8 4 1.66 C6H 5.27 6 36.3 CH3 3.6 OH 5.82 To compare the values obtained from Equation 4.13, first calculate the potential energy with Equation 4.13: + − ≈Γ ji jiji II II r 62 3 αα Therefore + − = ji ji ji II II C αα 2 3 6 We obtain the following values using this equation, and the corresponding percent differences were calculated. 25 Species [ ]6776 mJ 10 ⋅×C Percent Difference Page 187 Equation 4.13 CH 1.66 4 1.021 63 C6H 36.3 6 12.0 202 CH3 5.82 OH 1.36 328 The values for the van der Waals a constant have the correct qualitative trends and order of magnitude; however, those values predicted from basic potential theory vary significantly from corresponding states. The basic potential result presented in the text assumes that the species are evenly distributed throughout the volume. It does not take into account the structure given to the fluid through intermolecular forces. In fact, a more careful development includes a radial distribution function, which describes how the molecular density of the fluid varies with r. The radial distribution function depends on pressure and temperature of the fluid. 26 4.18 van der Waals: 2v a bv RTP − − = Redlich-Kwong P = RT v − b − a T1/2v(v + b) Peng-Robinson: P = RT v − b − a(T ) v(v + b) + b(v − b) As you may have discovered, these equations are largely empirical with no theoretical justification. They simply represent experimental data better. We can use our knowledge of intermolecular forces, however, to explain why these may work better. If we compare these equations to the van der Waals equation, we note that the first term on the right hand side of all three equation is identical; we accounted for this term as a correction for finite molecular size (or alternatively repulsive interaction due to the Pauli Exclusion Principle). This form represents a hard sphere model. The second term, that which deals with intermolecular attraction, is different in all three models. Both of the later equations include a temperature dependence in this term. We have seen that if attractive forces depend on orientation (dipole-dipole), they fall off with T as a result of the averaging process (recall discussion of Equation 4.11). Another explanation goes as follows: as T increases, the molecules move faster, reducing the effect of intermolecular forces. If we say that the potential energy between two molecules depends on the amount of time that they spend close to each other, then it would be inversely related to velocity (The faster molecules are moving, the less time they spend in the vicinity of other species). In this case, the correction term would go as V-1 1 2 mV 2 = 32 kT , where V is the molecular velocity. If we relate molecular velocity to temperature Therefore the correction term goes as T-1/2, as shown in the Redlich-Kwong equation. The inclusion of a "b" term in the second term may help relax van der Waal's "hard sphere" model with a more realistic potential function, i.e., something closer to a Lennard Jones potential (Figure 4.8) than the Sutherland potential (Figure 4.7) upon which the van der Waals equation is based. It makes sense that this should be included in the force correction since this is taking into account repulsive forces. One example of a more detailed explanation follows: If we look at the Redlich-Kwong equation, it says that if we have 2 species with the same attractive strength (same a -> same magnitude of van der Waals forces), the larger species will have less of an effect on P. The following sketch illustrates how 2 species could have the same van der Waals attractive forces: 27 +- Species 1 smaller and polar (smaller b, same a) Species 2 larger and non-polar (larger b, same a) London London + Dipole = In the case above, when the two species are the same distance apart, they have the same attractive force; however, the smaller species (1) can get closer before its electron cloud overlaps. Thus it has more opportunity for attractive interactions than the larger species. The Peng-Robinson equation exhibits the most complicated form in an attempt to better fit experimental data. 28 4.19 (a) Work is defined as follows ∫−= PdVW Substitution of the ideal gas law for P yields ∫−= dVV nRTW [ ]( ) [ ]( ) [ ][ ] ⋅ −= L 10 L 1lnK 0001 Kmol J .3148mol 0.2W kJ 29.38=W (b) Instead of substituting the ideal gas law into the definition of work, the Redlich-Kwong equation is used: dv bvvT a bv RTnW v v ∫ + − − −= 2 1 )(2/1 Substituting ⋅ =Kmol J .3148R [ ]K 1000=T [ ]mol 2=n ⋅⋅ = 2 31/2 mol mKJ 24.14a ×= − mol m 1011.2 3 5b = mol m 0005.0 3 1v = mol m 005.0 3 2v 29 and evaluating the resulting formula gives kJ 35.37=W (c) Energy balance: wqu +=∆ Since the process is reversible sTq ∆= and sTuw ∆−∆= To use the steam tables conveniently, we need the initial and final pressures. We can calculate these with the Redlich-Kwong EOS: MPa 65.11 =P MPa 6.152 =P From the steam tables: = kg kJ 6.3522ˆ1u ⋅ = Kkg kJ 101.81ˆs = kg kJ 2.3462ˆ2u ⋅ = Kkg kJ 00.7ˆ2s Therefore, ( ) ⋅ − ⋅ − − = Kkg kJ 101.8 Kkg kJ 00.7K 1000 kg kJ 6.3522 kg kJ 2.3462wˆ = kg kJ 1040wˆ [ ]kJ 5.37=W The answers from the three parts agree very well. Part (a) is not as accurate as Part (b) and Part (c) because water is not an ideal vapor. The value from Part (b) is 1.1 % smaller than the value from Part (c). Clearly, the Redlich-Kwong EOS or steam tables are appropriate for this calculation. 30 4.20 First, we can obtain an expression for the compressibility factor with the van der Waals equation. RTv a v bRTv a bv v RT Pv − − =− − = 1 1 To put this equation in virial form, we can utilize a series expansion: ...1 1 1 32 ++++= − xxx x Therefore, ...1 1 1 2 + + += − v b v b v b and ...1 2 + + − += v b v RT ab RT Pv This expression can also be expanded in pressure. From Equation 4.58, we know that RT BB =' ( )2 2 ' RT BCC −= where ....''1 2 +++= PCPB RT Pv Substituting B and C found above, we get ( ) ( )2 ' RT aRTbB −= and ( ) ( )4 22' RT aRTabC −= Therefore, 31 ( ) ( ) ( ) ( ) ....21 24 2 2 + − + − += P RT aRTabP RT aRTb RT Pv 32 4.21 Using the virial expansion, the pressure can be written as follows +++= ...1 32 v C v B v RTP The virial expansion in pressure is [ ]...''1 2 +++= PCPB v RTP If we substitute the first expression into the second, we obtain +++ +++= +++ 2 323232 1'1'1...1 v C v B v RTC v C v B v RTB v RT v C v B v RT If we combine like terms on the right side and set them equal to terms on the left, we find RTBB '= ( )2'' RTCRTBBC += Substitute the expression for B’ into the expression for C and solve for C’: ( )2 ' RT BCC −= and RT BB =' 33 4.22 At the critical point we have: 2 ccc c c vT a bv RTP − − = (1) ( ) 32 20 ccc c T vT a bv RT v P c + − −== ∂ ∂ (2) ( ) 432 2 620 ccc c T vT a bv RT v P c − − == ∂ ∂ (3) If we multiply Equation 2 by 2 and Equation 3 by (v-b) and add them together: ( ) 43 640 c c c v bva v a − −= this can be solved to give: bvc 3= If we plug this back into Equation 2 and solve for a, we get: 2 8 9 ccRTva = Finally, if we plug this back into Equation 1 we can solve for the Berthelot constants in terms of the critical temperature and the critical pressure: c c P TRa 22 64 27 = and c c P RTb 8 = 34 4.23 Before we can find the reduced form, we need to find expressions for a and b in terms of bvc 3= critical point data (Problem 4.22). We find (1) c c P TR a 32 64 27 = (2) c c P RT b 8 = (3) The Berthelot equation is 2Tv a bv RTP − − = First, substitute Equation 1 into the first term on the right-hand side and rearrange to get 213 Tv a v b RT P r − − = Now, substitute Equation 3: 213 8 Tv a v TP P r rc − − = Substitute Equation 2: 2 32 64 27 13 8 Tv P TR v TP P c c r rc − − = From Equation 3: 2222 64 bPTR cc = Substituting this result, we get 2 227 13 8 Tv bTP v TP P cc r rc − − = 35 Finally, substitute Equation 1: 2 23 13 8 Tv vTP v TP P ccc r rc − − = This can be rewritten as 2 3 13 8 rrr r r vTv TP − − = 36 4.24 (a) The virial equation is: z = Pv RT =1 + B v + C v2 + D v3 +. .. (1) We can rewrite this equation: z −1( )v = Pv RT −1 v = B + C v +. .. (2) PvT data from the steam tables are given in the steam tables and corresponding values of (z-1)v and 1/v can be calculated. An illustrative plot of (z-1)v vs. 1/v for T = 300 oC is shown below. -135 -130 -125 -120 -115 (z -1 )v (c m 3 / m ol ) 0 0. 5 1 1. 5 2 1/v (mol/l ) Plot to determine the second virial coeff at 300 C (z -1)v We see that at pressures above 1.5 MPa (15 bar) we see that the plot reaches a constant value of around -118 cm3/mol. We may choose to report this value as B. A more careful examination of Equation 2 suggests another possibility. If we plot (z-1)v vs. 1/v, we should get a straight line at low to moderate pressures with an intercept equal to the second virial coefficient, B, and a slope equal to the third virial coefficient, C. At very low pressures, the curve is indeed linear and gives an intercept of -140 cm3/mol, as shown on the next page. The slope of this region would yield the third virial coefficient, C. Can you calculate C? 37 The first value uses much more data while the second method uses limited data in a better range. What value would you be more apt to use? Water exhibits this “odd” general behavior at all temperatures. -135 -130 -125 -120 -115 (z -1 )v (c m 3 / m ol ) 0 0. 5 1 1. 5 2 1/v (mol/l ) y = 177.058x - 140.373 Plot to determine the second virial coeff at 300 C lin (z -1)v If we do this at other values of T, we can compile a set of second virial coefficients vs. temperature and get an idea of the differences in the two approaches. Temperatures of 200, 250, 300, 350, 400, 500, 600, 700, 800, 900, 1000, and 1200 oC analyzed in this manner. The results are reported in the table and figure below. Also reported were the average value and the value at the lowest pressure used. Note that the average value is indicative of the 1st method above while the value at the lowest pressure used is indicative of the second value used. T B (Level) B (Linear) B (AVG) B(1stvalue) 200 -214 -226 -215 -220 250 -156 -173 -157 -168 300 -118 -140 -120 -134 350 -89 -102 -92 -99 400 -72 -103 -76 -95 500 -49 -83 -54 -75 600 -34 -73 -39 -63 700 -24 -68 -30 -57 800 -17 -66 -24 -54 900 -13 -51 -19 -52 1000 -9 -50 -16 -52 1200 -3 -51 -11 -52 38 -250 -200 -150 -100 -50 0 B (w at er ) c m 3 /m ol 25 0 50 0 75 0 10 00 12 50 15 00 T (K) Second Virial Coefficient of Water by 4 Methods CRC B (Linear) B (Level) B (lowest P) B (average) For comparison, values of -142.2 cm3/mol and -7160 cm6/mol2 are reported for B and C, respectively, at 260 oC1 ++= ...1 2 2 v B v RTP OH . Values of B reported in the CRC are also shown on the summary plot. They agree most closely with the first (level) method. (b) There are several alternatives how to solve this problem, each of which comes up with a slightly different result: Alternative 1: Rewrite the virial equation: Take the derivative: −−== ∂ ∂ 32 2 210 ccc v B v RT v P OH c T so −=−= mol cm 28 2 3 2 c OH vB 1 J.H. Dymond and E.B. Smith, The Virial Coefficients of Pure Gases and Mixtures, A Critical Compilation, Oxford University Press, Oxford, 1980. 39 Alternative 2: From the virial equation: −= −= mol cm 2.431 3 2 c c cc OH vRT vPB Alternative 3: +++= ...1 32 22 v C v B v RTP OHOH so −−−== ∂ ∂ 432 22 3210 cccc v C v B v RT v P OHOH c T (1) and ++== ∂ ∂ 5432 2 22 12620 ccc c v C v B v RT v P OHOH c T (2) Multiply Equation 1 by (4/vc 0 22 43 2 =−− cc v B v OH ) and add to Equation 2 to get: so −=−= mol cm 56 3 2 cOH vB 40 4.25 A trial-and-error is the easiest method for solving this problem. The general method is as follows 1. Guess satP 2. Calculate values of molar volumes that result at the chosen satP : highmidlow vvv ,, 3. Find areas under the curves with the following expressions ( )[ ]dvvfPmid low v v sat∫ − ( )[ ]dvPvfhigh mid v v sat∫ − [ ( )vf represents the particular EOS implicit in molar volume being used.] 4. If the values of the expressions are not equal, repeat the process until they are. (a) 1. Guess satP : [ ]bar 6=satP 2. Calculate molar volume solutions for: )(2/1 bvvT a bv RTPsat + − − = [ ] [ ]( ) ( ) + ⋅⋅ − − ⋅ =× mol m 0001.0K 15.363 mol mKJ 851.41 mol m 0001.0 K 15.363 Kmol J314.8 Pa 106 3 2/1 2 31/2 3 5 vvv = mol m 000152.0 3 lowv = mol m 000551.0 3 midv = mol m 00459.0 3 highv 3. 19.1157 )( 000551.0 000152.0 2/1 = + − − −∫ dvbvvT a bv RTPsat 78.1325 )( 00459.0 000551.0 2/1 = − + − −∫ dvPbvvT a bv RT sat 41 4. Repeat this process until the areas are equal. This occurs approximately at [ ]bar 38.6=satP (b) 1. Guess satP : [ ]bar 6=satP 2. Calculate molar volume solutions for ( ) ( )bvbbvv Ta bv RTPsat −++ − − = )( α using ⋅ = 2 3 mol mJ 066.2a , ×= − mol m109 3 6b , ( ) 188.1=Tα = mol m 000152.0 3 lowv = mol m 000551.0 3 midv = mol m 00459.0 3 highv 3. ( ) ( ) 48.1568)( 000551.0 000152.0 = −++ − − −∫ dvbvbbvv Ta bv RTPsat α ( ) ( ) 19.1100)( 00459.0 000551.0 = − −++ − −∫ dvPbvbbvv Ta bv RT satα 4. Repeat this process until the areas are equal. This occurs approximately at [ ]bar 945.4=satP The value calculated with the Redlich-Kwong EOS is 11.9% greater than the measured value of 5.7 bar. The Peng-Robinson EOS results in a value that is 13.25% less than the measured value. 42 4.26 First, calculate a , b , and ( )Tα . From Table A.1.1: [ ]K 4.305=cT [ ]Pa 84.48=cP 099.0=w Now we can calculate the required parameters. [ ]( ) [ ] ⋅ = × ⋅ = 2 3 5 2 mol mJ 6036.0 Pa 1084.48 K 305.4 Kmol J 314.845724.0 a [ ]( ) [ ] ×= × ⋅ = − mol m 10045.4 Pa 1084.48 K 305.4 Kmol J 314.807780.0 3 5 5b ( ) ( ) ( )( ) 2 2 K 05.43 K 15.2431099.026992.0099.054226.137464.01 −−++=Tα ( ) 12.1=Tα Now, we can find the three solutions to the Peng-Robinson EOS: ( )( )bvbbvv Ta bv RTPsat −++ − − = )( α Using the values calculated above and [ ]Pa106.10 5×=satP ⋅ = Kmol J 314.8R we get ×= − mol m 10199.7 3 5v , ×= − mol m 10167.2 3 4v , ×= − mol m 10578.1 3 3v The molar volume of saturated ethane liquid is the smallest value from the list above, while the molar volume of saturated ethane vapor is the largest value. Therefore, 43 ( ) ( ) = × == − 333 3 3 5 cm g 501.0 cm 100 m 1 mol m 10199.7 mol g 0694.36 liq ethaneliq v MW ρ ( ) ( ) = × == − 333 3 3 3 cm g 0229.0 cm 100 m 1 mol m 10578.1 mol g 0694.36 vap ethanevap v MW ρ Both of the densities calculated with the Peng-Robinson EOS are larger than the reported values. The liquid density is 7.05% larger, and the vapor density is 18.7% greater. 44 4.27 (a) We will use the Redlich-Kwong EOS in order to obtain an accurate estimate. First, calculate the a and b parameters: ⋅⋅ == 2 2/135.22 mol KmJ 56.1 42748.0 c c P TR a ×== − mol m 1069.2 08664.0 35 c c P RT b (Note: Critical data for nitrogen obtained in Table A.1.1.) Now use the EOS to find the molar volume: ( )bvvT a bv RTP + − − = 2/1 Assume the temperature of the gas is 22 ºC, substitute values, and find the molar volume with a solver function: ×= − mol m 1097.13 4v Now calculate the total volume of gas in the 30,000 units: ( ) [ ]( ) 36 m 1290liters 1029.1liter/unit 43units30000 =×==totalV Therefore, the number of moles is: mol 1055.6 mol m 1097.1 m1290 6 3 4 3 ×= × = − n and the mass ( )( ) kg1083.1kg/mol02801.0mol 1055.6 56 ×=×=m Now, calculate the value of the gas: ( )( ) 300,116,1$kg1083.1kg/1.6$ 5 =×=Value 45 If we use the ideal gas law, we find: ( ) ×= ⋅ == − mol m 1098.1 Pa12400000 K295 Kmol J314.8 3 4 P RTv Following the steps above, 200,113,1$=Value The value calculated using the ideal gas law is $3100 less than the value calculated using the Redlich-Kwong EOS. (b) First, calculate the a and b parameters: ⋅⋅ == 2 2/135.22 mol KmJ 74.1 42748.0 c c P TR a ×== − mol m 1021.2 08664.0 35 c c P RT b (Note: Critical data for oxygen obtained in Table A.1.1.) Assume the temperature of the gas is 22 ºC, substitute values, and find the molar volume with a solver function: ×= − mol m 1053.1 3 4v Following the steps presented in Part (a), we find that $2,428,000=Value With the ideal gas law, we find ×= − mol m 1064.1 3 4v which provides $2,265,000=Value 46 The value found with the ideal gas law is $163,000 less than the value found using the Redlich- Kwong EOS. 47 4.28 First, rewrite the Redlich-Kwong equation in cubic form. 05.0 2 2/1 23 =− −−+− PT abvbb P RT PT av P RTv or at the critical point 05.0 2 5.0 23 =− −−+− ccc c ccc c TP abvbb P RT TP av P RT v Now expand ( ) 03 =− cvv 033 3223 =−+− ccc vvvvvv Setting the coefficients equal we get the following expressions. c c c P RT v =3 (1) 2 5.0 23 bb P RT TP av c c cc c −−= (2) 5.0 3 cc c TP abv = (3) Using Equation 2, find an expression for a: 5.0223 cc C C c TPbP RT va ++= Substitute the above expression into Equation 3 to get 033 3223 =−++ ccc vbvbvb The one real root to this equation is ( ) cvb 12 3/1 −= (4) Substitute Equation 4 into Equation 1: 48 c c P RT b 08664.0 = (Equation 4.48) Now, substitute Equation 1 and Equation 4.48 into the expression for a to get c c P RT a 5.242748.0 = (Equation 4.47) To verify Equation 4.50, substitute Equation 1 into c cc c RT vP z = 3 13 = = c c c c c RT P RT P z (Equation 4.50) The Redlich-Kwong equation is ( )bvvT a bv RTP + − − = 5.0 We can use Equation 4 to get ( )2599.02599.0 2599.0 5.0 + − − = rcr vvvT a v b RT P Now, substitute Equation 4.48 ( )2599.02599.0 3 5.0 + − − = rcr cr vvvT a v PT P Replacing a with Equation 4.47 yields ( )2599.0 42748.0 2599.0 3 5.0 5.22 + − − = rcC c r cr vvvTP TR v PT P From Equation 4.48 ( )2 22 22 08664.0 bP TR cc = 49 which upon substitution yields ( ) ( )2599.008664.0 42748.0 2599.0 3 5.0 2 2 + − − = rcr c r cr vvvT bP v PT P Now, use Equation 4: ( ) ( ) ( )2599.008664.0 2599.042748.0 2599.0 3 5.02 2 + ⋅ − − = rrr c r cr vvT P v PT P Therefore, ( )2599.02599.0 1 2599.0 3 5.0 + − − = rrrr r r vvTv TP (Equation 4.49) 50 4.29 Since we are concerned with liquid water, we can base all our calculations on saturated liquid water since the molar volume of liquids are weakly dependent on pressure. Therefore, our results are also applicable to sub-cooled water (the pressure is greater than the saturation pressure). To find the thermal expansion coefficient, we will use the following approximation ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) −−+ = −−+ −−+ = K 10 Cº ˆCº 5ˆ ˆ 1 Cº 5Cº 5 Cº ˆCº 5ˆ ˆ 1 TvTv TvTT TvTv Tv satsat sat satsat satβ This approximation is valid even though the saturation pressure changes because molar volume is weakly dependent on pressure. From the steam tables ( ) = kg m 001001.0Cº 15ˆ 3 satv ( ) = kg m 001040.0Cº 59ˆ 3 satv ( ) = kg m 001002.0Cº 20ˆ 3 satv ( ) = kg m 001044.0Cº 100ˆ 3 satv ( ) = kg m 001003.0Cº 25ˆ 3 satv ( ) = kg m 001047.0Cº 105ˆ 3 satv Using these values, we can calculate the thermal expansion coefficients ( ) − = − K 10 kg m 001001.0 kg m 001003.0 kg m 001002.0Cº 20 33 13 β ( ) − = − K 10 kg m 001040.0 kg m 001047.0 kg m 001044.0Cº 100 33 13 β ( ) [ ] ( ) [ ]1-4 1-4 K 1071.6Cº 100 K 100.2Cº 20 − − ×= ×= β β The accuracy of these values may be limited since they are based on the small differences between liquid volumes. To calculate the isothermal compressibility, a similar approximation will be used. It is 51 ( ) T sat sat dP vd Tv − = ˆ ˆ 1κ The derivatives are determined from the following graph. It is clear that ⋅ ×= = − == MPakg m 105 ˆˆ 37 Cº 0 100Cº 0 20 T sat T sat dP vd dP vd Therefore, ( ) ⋅ × = − − MPakg m 105 kg m 001002.0Cº 20 3 7 13 κ ( ) ×= ×= −− Pa 1 1099.4 MPa 11099.4Cº 20 104κ ( ) ⋅ × = − − MPakg m 105 kg m 001044.0Cº 100 3 7 13 κ ( ) ×= ×= −− Pa 1 1079.4 MPa 11079.4Cº 100 104κ Specific Volume of Liquid Water vs. Pressure v = -5E-07*P + 0.001044 R2 = 0.9949 v = -5E-07*P + 0.001002 R2 = 0.9979 0.000990 0.001000 0.001010 0.001020 0.001030 0.001040 0.001050 0 5 10 15 20 25 Pressure [MPa] Sp ec ifi c Vo lu m e [m 3 /k g] T = 20篊 T = 100 篊 Linear (T = 100 篊) Linear (T = 20篊) 52 4.30 (a) Substitute critical data into the Rackett equation: ( )( ) ( )[ ] ( )[ ] 7/2190/111115 008.008775.029056.01046 6.190314.8 −+− × =calcv = mol cm 38 3 calcv Now, calculate the error. %100 exp exp × − = v vv Error calc %8.0=Error (b) = mol cm 2.54 3 calcv %1.1=Error (c) = mol cm 2.161 3 calcv %8.10=Error (d) = mol cm 5.20 3 calcv %5.13=Error (e) = mol cm 5.68 3 calcv %7.19=Error 1-hexanol had the largest percent error, while ethane and methane had the smallest percent errors. The alkenes have low percentile errors due to their nonpolar nature, which the alcohols and acids had large percentile errors due to being polar substances and exhibiting hydrogen bonds. 53 4.31 (a) Since we are given the temperature and pressure, we can make use of the generalized compressibility charts. Using data from Table A.1.1, the required quantities can be found. [ ][ ] 12.1K 4.305 K 15.343 === c r T TT [ ][ ] 616.0bar 8.744 bar 03 === c r P PP 099.0=w By double interpolation of the charts ( ) 8376.00 =z ( ) 0168.01 =z and ( ) ( ) ( )( ) 8393.00168.0099.08376.010 =+=+= wzzz Therefore, ( ) [ ][ ] [ ]( ) [ ][ ]Pa 1030 K 15.343 Kmol J 314.8 kg/mol 03007.0 kg 308393.0 5× ⋅ == P znRTV [ ]3m 796.0=V (b) The Redlich-Kwong EOS should give reasonably accurate results. Room temperature was assumed to be 25 ºC. The molar volume is required for the calculations, so [ ]( ) [ ] [ ] ×= === − mol m 10518.7 kg/mol 03007.0 kg 40 m 1.0 353 n VVv Using data from Table A.1.1 and Equations 4.47 and 4.48, ⋅⋅ = 2 2/13 mol KmJ 88.9a 54 ×= − mol m 1051.4 3 5b Substituting these values into the Redlich-Kwong EOS and evaluating gives [ ]bar 3.191=P 55 4.32 (a) For propane ( ) = mol kg 0441.0propaneMW [ ] [ ]mol 1130 mol kg 0.0441 kg 50 = =∴n Now calculate the volume: [ ]( ) ( ) [ ]Pa 1035 K 15.27350 Kmol J .3148mol 1130 5× + ⋅ == P nRTV [ ]3m 870.0=V (b) The Redlich-Kwong EOS is ( )bvvT a bv RTP + − − = 2/1 where ⋅⋅ == 2 2/135.22 mol KmJ 33.18 42748.0 c c P TR a ×== − mol m 1028.6 08664.0 35 c c P RT b (Note: Critical data for propane obtained in Table A.1.1.) Now that these values are known, there is only one unknown in the Redlich-Kwong EOS: v . Using a numerical technique, e.g., the solver function on a graphing calculator = mol m 000109.0 3 v [ ]3m 124.0=∴V 56 (c) The Peng-Robinson EOS is ( )( ) ( )bvbbvv Ta bv RTP −++ − − = α where ( )[ ]211 rT−+= κα ( ) 873.0 K 370 K 15.27350 = + == c r T TT ( ) ( ) 605.01152.026992.0152.054226.137464.0 2 =−+=κ 081.1=α ( ) ⋅ == mol mJ 02.1 45724.0 32 c c P RT a ×== − mol m 1064.5 07780.0 35 c c P RT b Now every variable in the Peng-Robinson EOS is known, except v. = mol m 0000942.0 3 v [ ]3m 107.0=∴V (d) We must calculate the reduce temperature and pressure to use the compressibility charts: 873.0 K 370 K 15.323 === c r T TT 825.0 bar 42.44 bar 35 === c r P PP By double-interpolation on the compressibility charts (Appendix C), ( ) 1349.00 =z ( ) 052.01 −=z Therefore, 57 ( ) ( ) ( ) 127.0052.0152.01349.000 =−+=+== wzzz RT Pv == mol m 00009749.0127.0 3 P RTv 3m 111.0=V (e) From ThermoSolver Using the Peng-Robinson equation, = mol m 00009429.0 3 v [ ]3m 107.0=∴V Using the generalized compressibility charts: = mol m 0000971.0 3 v Therefore, [ ]3m 110.0=V 58 4.33 Note: Multiple possibilities exist for which substance to use in the vial. The solution using one possibility is illustrated below. (a) The substance must have a critical temperature above room temperature but below the temperature of one’s hand. From the Appendix A.1.2, we that for carbon dioxide Cº 31.1 K 2.304 ==cT Clearly, CO2 ( ) [ ] ⋅ = × ⋅ = 2 3 5 2 2 mol mJ 40.0 Pa 1073.76 K 2.304 Kmol J 314.8 45724.0a is a suitable substance, and it is safe to use. (b) The vial must be able to withstand the pressure of the substance at its critical point. Therefore, the vial must withstand 73.76 bar (the critical pressure of carbon dioxide). (c) Since the substance passes through its critical point, the molar volume at that state is constrained by the critical temperature and pressure. To estimate the molar volume, we can use the Peng- Robinson EOS. Calculate the necessary parameters for the EOS: ( ) [ ] ×= × ⋅ = − mol m 104.3 Pa 1073.76 K 2.304 Kmol J 314.8 07780.0 3 4 5b 1=α Substitute the above parameters and solve for the molar volume: = ×= − mol cm 118 mol m 1018.1 33 4 cv Now, we can calculate the required amount of CO2 [ ] mol 847.0 mol cm 118 cm 100 3 3 = == cv Vn . 59 (d) If the vial contains less substance than needed, the molar volume will be greater. Therefore, the substance will not pass through the critical point. As the substance is heated, it will go through a transition from a saturated liquid and vapor mixture to superheated vapor. Then, it will become a supercritical fluid once the critical temperature is exceeded. 60 4.34 Methane: The following quantities are required to calculate the molar volume with the Peng-Robinson equation. ( ) ( )[ ]211 rTT −+= κα ( ) 9626.0=Tα ( ) ⋅ == mol mJ 25.0 45724.0 32 c c P RT a ×== − mol m 1068.2 07780.0 35 c c P RT b [ ]K 66.209== crTTT [ ]Pa 102.55 5×== cr PPP The molar volume is the only unknown in the following equation. ( )( ) ( )bvbbvv Ta bv RTP −++ − − = α ×= − mol m 1083.1 3 4v Therefore, [ ]( ) [ ]( ) 58.0 K 66.209 Kmol J 8.314 mol m 1083.1Pa 102.55 3 45 = ⋅ ×× = − z From the compressibility charts, ( ) 5984.00 =z ( ) 0897.01 =z Therefore, ( ) ( ) ( ) 0897.0008.05984.010 +=+= wzzz 599.0=z 61 Methanol: Followingthe procedure outlined above, the Peng-Robinson equation produces ×= − mol m 1014.3 3 4v 651.0=∴ z Using the compressibility charts: ( ) 5984.00 =z ( ) 0897.01 =z From Table A.1.1, 559.0=w Therefore, ( ) ( ) ( ) 0897.0559.05984.010 +=+= wzzz 649.0=z Summary: Methane 58.0=z (Peng-Robinson) 599.0=z (Compressibility charts) The value from the Peng-Robinson EOS is 3.2% smaller than the value from the charts. Methanol 651.0=z (Peng-Robinson) 649.0=z (Compressibility charts) The value from the Peng-Robinson EOS is 0.31% smaller than the value from the charts. 62 4.35 From Table A.1.1: K 6.405=cT bar 77.112=cP 25.0=w Calculate reduced temperature and pressure: ( ) 9.0 K 6.405 K 15.27392 = + == c r T TT 718.2 bar 77.112 bar 306.5 === c r P PP From interpolation of the compressibility charts: ( ) 4133.00 =z ( ) 1351.01 −=z Therefore, ( ) ( ) 38.010 =+== wzzz RT Pv and ×== − mol m 1076.338.0 3 5 P RTv Since the temperature of the ammonia in this system is below the critical temperature, we know the ammonia is not a supercritical fluid. The pressure is greater than the critical pressure, so the ammonia is a liquid. 63 4.36 Let the subscript “ace” represent acetylene and “but” represent n-butane. First, convert the given quantities of acetylene and n-butane into moles. [ ] [ ]mol 2.1152 mol kg10038.62 kg 30 3- = × =acen [ ] [ ]mol 2.860 mol kg10123.58 kg 50 3- = × =butn Therefore, 573.0=acey 427.0=buty We can also calculate a and b parameters for pure species using the following equations c c P TRa 5.2242748.0 = c c P RTb 08664.0= Substituting critical data from Table A.1.1 gives ⋅⋅ = 2 31/2 mol mKJ 03.8acea ×= − mol m 1062.3 3 5 aceb ⋅⋅ = 2 31/2 mol mKJ 07.29buta ×= − mol m 10081.8 3 5 butb Now, we can use mixing rules to calculate the parameters for the mixture. 2 2 212211 2 1 2 ayayyayamix ++= ( ) ⋅⋅ =−⋅= 2 31/2 12 mol mKJ 87.13092.0107.2903.8a ⋅⋅ =∴ 2 31/2 mol mKJ 72.14mixa 64 ×=+= − mol m 1053.5 3 5 2211 bybybmix Substitution of the mixture parameters into the Redlich-Kwong EOS results in an equation with one unknown. = mol m 00111.0 3 v ( ) [ ] [ ]( ) +=+=∴ mol m 00111.0mol 2.860mol 2.1152 3 vnnV butace [ ]3m 23.2=V 65 4.37 First, we need to calculate the a and b parameters for species (1) and (2). From Table A.1.1 and A.1.2: CO2 [ ] [ ]bar 76.73 K 2.304 = = c c P T (1): ⋅⋅ == mol KmJ 466.642748.0 1/235.22 1 c c P TRa ×== − mol m 1097.2 08664.0 35 1 c c P RT b Toluene (2): [ ] [ ]bar 14.41 K 7.591 = = c c P T ⋅⋅ == mol KmJ 17.61 42748.0 1/235.22 2 c c P TR a ×== − mol m 10036.1 08664.0 34 1 c c P RT b We can use the mixing rules to calculate bmix 2211 bybybmix += . × + × = −− mol m 10036.1 mol 5 mol 3 mol m 1097.2 mol 5 mol 2 3435 mixb ×= − mol m 1040.7 3 5 mixb Now we can substitute the known values into the Redlich-Kwong EOS. Note that the molar volume can be calculated as follows [ ][ ] === mol m 002.0 mol 5 m 01.0 33 totn Vv We can then solve for amix ⋅⋅ = mol KmJ 0.31 1/23 mixa . 66 From the mixing rules, we know 2 2 212211 2 1 1/23 2 mol KmJ 0.31 ayayyayamix ++= ⋅⋅ = Therefore, ⋅⋅ = mol KmJ 55.16 1/23 12a Equation 4.79 states ( )122112 1 kaaa −= Substitution of the values provides: 17.012 =k 67 4.38 (a) We can match the species by looking at the b parameter alone. The b parameter is directly related to the size of the molecule. Because 2262 HOHHC sizesizesize >> , 2262 HOHHC bbb >> . Therefore, ⋅ mol mJ 3 a mol m 3 b Species 0.564 51038.6 −× C2H6 (3) 0.025 51066.2 −× H2 (1) 0.561 51005.3 −× H2O (2) These values are also consistent with what we would expect for the magnitude of van der Waals interactions given by a. (b) The van der Waals parameters for the mixture can be calculated according to the mixing rules. From Equations 4.81 and 4.82 3 2 33223311323322 2 22112133112211 2 1 ayayyayyayyayayyayyayyayamix ++++++++= 3 2 323322 2 2133112211 2 1 222 ayayyayayyayyayamix +++++= 332211 bybybybmix ++= Calculate mole fractions: 5.0 mol 10 mol 5 321 1 1 ==++ = nnn ny 4.02 =y 1.03 =y Since binary interaction parameters are not available, we must use Equation 4.78 to calculate 231312 , , aaa . ⋅ == mol mJ 118.0 3 2112 aaa ⋅ = mol mJ 119.0 3 13a 68 ⋅ = mol mJ 562.0 3 23a Now we can find the numerical values for the van der Waals parameters. ⋅ = mol mJ 206.0 3 mixa ×= − mol m 1019.3 3 5 mixb (c) For the mixture, the van der Waals equation is 2v a bv RTP mix mix − − = The molar volume is calculated as follows [ ] == ++ = mol m 00125.0 mol 10 m 0125.0 33 321 nnn Vv Therefore, ( ) 23 3 3 5 3 mol m 00125.0 mol mJ 206.0 mol m 1019.3 mol m 00125.0 K 300 Kmol J 314.8 ⋅ − ×− ⋅ = − P MPa 92.1=P 69 4.39 (a) To calculate the molar volume of the mixture, we will use the virial expansion in pressure since it is more accurate at moderate pressures. But first, we must calculate the second virial coefficient for the mixture. 2 2 212211 2 1 2 ByByyByBmix ++= ( ) ( )( ) ( ) −+ −+ −= mol cm 11075.0 mol cm 15375.025.02 mol cm 62525.0 3 2 33 2 mixB −= mol cm 3.158 3mixB Therefore, mixmix BP RTP RT B P RTv += += 1 = = mol m 002446.0 mol cm 6.2445 33 v (b) We can estimate k12 2211 bybybmix += using an EOS where the a parameter is the only unknown. b can be calculated as follows Using the Redlich-Kwong EOS, ×= ×+ ×= −−− mol m1069.3 mol m 1097.275.0 mol m 1083.525.0 3 5 3 5 3 5 mixb Substitute the known properties into the Redlich-Kwong EOS and solve for a. ⋅⋅ = 2 31/2 mol mKJ 693.8mixa Now we can calculate k12 2 2 212211 2 1 2 ayayyayamix ++= as follows 70 ⋅⋅ = ⋅⋅ = 2 31/2 2 2 31/2 1 mol mKJ 466.6 mol mKJ 03.28 a a (Calculated using Equation 4.47) ⋅⋅ =∴ 2 31/2 12 mol mKJ 81.8a From Equation 4.79 ( )122112 1 kaaa −= 35.0 03.28 466.6 .818112 = ⋅ −=k 71 4.40 During this process the gas in Tank A undergoes an expansion from 7 bar to 3.28 bar. During this isentropic process, the gas cools. If attractive forces are present, they will manifest themselves more in state 1 at the higher pressure than in state 2. (While T has some effect to counter this trend, we expect it will be secondary to the effect of pressure). Thus, in the non- ideal case, additional energy must be supplied to “pull” the molecules apart. Since the tank is insulated, the only place that this can come from is the kinetic energy of the molecules. Thus, they slow down even more than in the ideal gas case, and the final temperature is lower. If we had an equation of state that appropriately described the non-ideal behavior, we could apply the concepts of the thermodynamic web that we have learned in this chapter to solve for the final temperature. 72 4.41 Select the “Equation of State Solver” from the main menu. Enter the pressure and temperature and solve for the molar volume using the Lee-Kesler EOS. The program provides ×= − mol m 1007.1 3 4v To calculate the volume occupied by 10 kilograms of butane, we need to know the number of moles present in 10 kg. [ ][ ] [ ]mol 1.172kg/mol 0.05812 kg 10 ==n Therefore, [ ]( ) [ ]334 m 018.0 mol m 1007.1mol 1.172 = ×= −V The answers provided in Example 4.9 agree well with this answer. The answer found using the Redlich-Kwong EOS is 16.7% larger than the answer from ThermoSolver, but the answer from the compressibility charts is only 5.56% larger. 73 4.42 Using ThermoSolver should be straightforward; thus, only the answers are provided. (a) [ ] [ ] −=∆ = = mol kJ68.84 bar 84.48 K 4.305 298, f c c h P T (b) [ ]K 42.299=satT (c) The percent difference is calculated as follows %100or % × −− = LK PRLK LK PRLK z zz v vv Difference (i). Quantity Lee Kessler EOS Peng Robinson EOS % Difference v × − mol m 1034.8 3 5 × − mol m 1070.8 3 5 4.3 z 0.1383 0.1444 4.4 (i). Quantity Lee Kessler EOS Peng Robinson EOS % Difference v × − mol m 1068.3 3 4 × − mol m 1058.3 3 4 2.7 z 0.5859 0.5711 2.5 Chapter 5 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 5.1 (a) Following the example given by Equation 5.5a in the text dP P udT T udu TP ∂ ∂ + ∂ ∂ = (b) ds s udT T udu Ts ∂ ∂ + ∂ ∂ = (c) ds s udh h udu hs ∂ ∂ + ∂ ∂ = 3 5.2. The internal energy can be written as follows dv v udT T udu Tv ∂ ∂ + ∂ ∂ = Substituting Equations 5.38 and 5.40 v v c T u = ∂ ∂ and − ∂ ∂ = ∂ ∂ P T PT v u vT into the above expression yields dvP T PTdTcdu v v − ∂ ∂ += From the ideal gas law, we have v R T P v = ∂ ∂ Therefore, dvP v RTdTcdu v −+= which upon noting that v RTP = for an ideal gas, becomes dTcdu v= 4 5.3 The heat capacity at constant pressure can be defined mathematically as follows ( ) PvPP P T vP T u T Pvu T hc ∂ ∂ + ∂ ∂ = ∂ ∂+∂ = ∂ ∂ = For an ideal gas: P R T v P = ∂ ∂ Therefore, R T uc v P + ∂ ∂ = One mathematical definition of du is dP P udT T udu TP ∂ ∂ + ∂ ∂ = We can now rewrite vT u ∂ ∂ : v vTvPv c T P P u T T T u T u = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ For an ideal gas: 0= ∂ ∂ TP u so P v T uc ∂ ∂ = Substituting this result into our expression for Pc gives Rcc vP += 5 5.4 In terms of P, v, and T, the cyclic equation is TPv P v v T T P ∂ ∂ ∂ ∂ ∂ ∂ =−1 For the ideal gas law: RTPv = so the derivatives become: v R T P v = ∂ ∂ R P v T P = ∂ ∂ P v P RT P v T − = − = ∂ ∂ 2 Therefore, 1−= − = ∂ ∂ ∂ ∂ ∂ ∂ P v R P v R P v v T T P TPv The ideal gas law follows the cyclic rule. 6 5.5 For a pure species two independent, intensive properties constrains the state of the system. If we specify these variables, all other properties are fixed. Thus, if we hold T and P constant h cannot change, i.e., 0 , = ∂ ∂ PTv h 7 5.6 Expansion of the enthalpy term in the numerator results in ss T PvsT T h ∂ ∂+∂ = ∂ ∂ ss T Pv T h ∂ ∂ = ∂ ∂ ∴ Using a Maxwell relation Ps v sv T h ∂ ∂= ∂ ∂ PPs v T T sv T h ∂ ∂ ∂ ∂ = ∂ ∂ ∴ We can show that T c T s P P = ∂ ∂ (use thermodynamic web) +− − = ∂ ∂ 32 221 v ab v a bv RT Rv T P (differentiate van der Waals EOS) Therefore, −− − = +− − = ∂ ∂ v b vRT a bv vc v ab v a bv RT RT vc T h P P s 1222 32 8 5.7 : TP h ∂ ∂ v P sT P PvsT P h TTT + ∂ ∂ = ∂ ∂+∂ = ∂ ∂ ( )2''1 PCPB P R T v P s PT ++−= ∂ ∂ −= ∂ ∂ ( ) vvvPCPB P RT P h T +−=+++−= ∂ ∂ 2''1 0= ∂ ∂ TP h : sP h ∂ ∂ v P PvsT P h ss = ∂ ∂+∂ = ∂ ∂ ( )2''1 PCPB P RT P h s ++= ∂ ∂ : PT h ∂ ∂ P P c T h = ∂ ∂ (Definition of cP : sT h ∂ ∂ ) sss T Pv T PvsT T h ∂ ∂ = ∂ ∂+∂ = ∂ ∂ ( )2''1 PCPBR P T c v T T c s P T s T P P P P TPs ++ = ∂ ∂ = ∂ ∂ ∂ ∂ −= ∂ ∂ 9 ( ) ( ) ( )2 2 2 ''1 ''1 ''1 1 PCPB PCPBc PCPBRT Pvc T h PP s ++ ++ = ++ = ∂ ∂ P s c T h = ∂ ∂ 10 5.8 (a) A sketch of the process is provided below ∂mwell insulated ∆s = 0T1 P1 T2 P2 The diagram shows an infinitesimal amount of mass being placed on top of the piston of a piston-cylinder assembly. The increase in mass causes the gas in the piston to be compressed. Because the mass increases infinitesimally and the piston is well insulated, the compression is reversible and adiabatic. For a reversible, adiabatic process the change in entropy is zero. Therefore, the compression changes the internal energy of the gas at constant entropy as the pressure increases. (b) To determine the sign of the relation, consider an energy balance on the piston. Neglecting potential and kinetic energy changes, we obtain WQU +=∆ Since the process is adiabatic, the energy balance reduces to WU =∆ As the pressure increases on the piston, the piston compresses. Positive work is done on the system; hence, the change in internal energy is positive. We have justified the statement 0> ∂ ∂ sP u 11 5.9 (a) By definition: PT v v = ∂ ∂β 1 and TP v v ∂−= ∂ κ 1 Dividing, we get: TP T P v P T v P v T v −= −= ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ κ β where derivative inversion was used. Applying the cyclic rule: vTP P T v P T v =− ∂ ∂ ∂ ∂ ∂ ∂1 Hence, vT P = ∂ ∂ κ β (b) If we write T = T(v,P), we get: dP P Tdv v TdT vP + = ∂ ∂ ∂ ∂ (1) From Equations 5.33 and 5.36 dP T vdT T cdv T PdT T cds P P v v −= += ∂ ∂ ∂ ∂ We can solve for dT to get: 12 dP T v cc Tdv T P cc TdT PvPvvP − + − = ∂ ∂ ∂ ∂ (2) For Equations 1 and 2 to be equal, each term on the left hand side must be equal. Hence, vvPP T P cc T v T − = ∂ ∂ ∂ ∂ or PPv vP T vT T v T PTcc = =− ∂ ∂ κ β ∂ ∂ ∂ ∂ where the result from part a was used. Applying the definition of the thermal expansion coefficient: κ β ∂ ∂ ∂ ∂ 2Tv T v T PTcc Pv vP = =− 13 5.10 We need data for acetone, benzene, and copper. A table of values for the molar volume, thermal expansion coefficient and isothermal compressibility are taken from Table 4.4: Species × mol m 10 3 6v [ ]1-3 K 10×β [ ]1-10 Pa 10×κ Acetone 73.33 1.49 12.7 Benzene 86.89 1.24 9.4 Copper 7.11 0.0486 0.091 We can calculate the difference in heat capacity use the result from Problem 5.9b: κ β 2vTcc vP =− or ( ) [ ]( ) [ ] ⋅ = × × × =− − −− Kmol J 6.37 Pa 107.12 K 1049.1K 293 mol m 1033.73 1-10 21-3 3 6 vp cc Species ⋅ − Kmol J vp cc ⋅ Kmol J pc % difference Acetone 37.6 125.6 30% Benzene 41.6 135.6 31% Copper 0.5 22.6 2% We can compare values to that of the heat capacity given in Appendix A2.2. While we often assume that cP and cv are equal for condensed phases, this may not be the case. 14 5.11 We know from Equations 4.71 and 4.72 PT v v ∂ ∂ = 1β and TP v v ∂ ∂ −= 1κ Maxwell relation: vT T P v s ∂ ∂ = ∂ ∂ Employing the cyclic rule gives PTv T v v P T P ∂ ∂ ∂ ∂ −= ∂ ∂ which can be rewritten as T P vT v P v T v v T P v s ∂ ∂ − ∂ ∂ = ∂ ∂ = ∂ ∂ 1 1 Therefore, κ β = ∂ ∂ Tv s Maxwell Relation: PT T v P s ∂ ∂ −= ∂ ∂ From Equation 4.71: v T v P β= ∂ ∂ Therefore, v P s T β−= ∂ ∂ 15 5.12 (a) An isochor on a Mollier diagram can be represented mathematically as vs h ∂ ∂ This can be rewritten: vvv s PT s PvsT s h ∂ ∂ += ∂ ∂+∂ = ∂ ∂ Employing the appropriate Maxwell relation and cyclic rule results in Tvv v s s TvT s h ∂ ∂ ∂ ∂ += ∂ ∂ We know vv c T s T = ∂ ∂ and vT T P v s ∂ ∂ = ∂ ∂ For an ideal gas: v R T P v s vT = ∂ ∂ = ∂ ∂ Therefore, +=+= ∂ ∂ vvv c RT v R c TvT s h 1 (b) In Part (a), we found vvv T P c TvT s h ∂ ∂ += ∂ ∂ For a van der Waals gas: 16 bv R T P v − = ∂ ∂ Therefore, − += ∂ ∂ bv v c RTT s h vv 17 5.13 (a) The cyclic rule can be employed to give TPs P s s T P T ∂ ∂ ∂ ∂ −= ∂ ∂ Substitution of Equations 5.19 and 5.31 yields PPs T v c T P T ∂ ∂ = ∂ ∂ For an ideal gas: P R T v P = ∂ ∂ Therefore, PPs c v cP RT P T == ∂ ∂ 1 (b) Separation of variables provides P P c R T T P ∂ = ∂ Integration provides Pc R P P T T = 1 2 1 2 lnln which can be rewritten as Pc R P P T T = 1 2 1 2 The ideal gas law is now employed 18 Pc R P P vP vP = 1 2 11 22 1 1 12 1 2 vPvP PP c R c R − − = where kc c c Rc c R P v P P P 11 == − =− If we raise both sides of the equation by a power of k, we find kk vPvP 1122 = .constPvk =∴ (c) In Part (a), we found PPs T v c T P T ∂ ∂ = ∂ ∂ Using the derivative inversion rule, we find for the van der Waals equation ( ) ( )23 3 2 bvaRTv bvRv T v P −− − = ∂ ∂ Therefore, ( ) ( )23 3 2 1 bvaRTv bvRTv cP T Ps −− − = ∂ ∂ 19 5.14 The development of Equation 5.48 is analogous to the development of Equation E5.3D. We want to know how the heat capacity changes with pressure, so consider T P P c ∂ ∂ which can be rewritten as PTTPT P P h TT h PP c ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ Consider the TP h ∂ ∂ term: v T vTv P sT P PvsT P h PTTT + ∂ ∂ −=+ ∂ ∂ = ∂ ∂+∂ = ∂ ∂ Substitution of this expression back into the equation for T P P c ∂ ∂ results in PPT P v T vT TP c + ∂ ∂ − ∂ ∂ = ∂ ∂ PPPT P T v T vT T v T T P c ∂ ∂ + ∂ ∂ − ∂ ∂ ∂ ∂ −= ∂ ∂ 2 2 PT P T vT P c ∂ ∂ −= ∂ ∂ 2 2 Therefore, ∫∫ ∂ ∂ −= real ideal real P ideal P P P P c c P dP T vTdc 2 2 and ∫ ∂ ∂ −= real ideal P P P ideal P real P dP T vTcc 2 2 20 5.15 In order to solve this problem we need to relate the change in entropy from 10 to 12 bar to the change in molar volume (for which we have complete data). First, we can rewrite the change in entropy as ∫ ∂ ∂ =−=∆ bar 12 bar 10 12 dPP ssss T Applying a Maxwell relation, we can relate the above equation to the change in molar volume: ∫∫ ∂ ∂ −+= ∂ ∂ += bar 12 bar 10 1 bar 12 bar 10 12 dPT vsdP P sss PT As 10 bar: ⋅ ×= ∆ ∆ ≅ ∂ ∂ − Kkg m 1060.5 3 4 PP T v T v At 12 bar: ⋅ ×= ∆ ∆ ≅ ∂ ∂ − Kkg m 1080.4 3 4 PP T v T v To integrate the above entropy equation, we need an expression that relates PT v ∂ ∂ to pressure. Thus, we will fit a line to the data. We obtain ⋅ ×+ ⋅⋅ ×−= ∂ ∂ −− Kkg m106.9 PaKkg m 100.4 3 4 3 10 P T v P Now integrate the equation to find the entropy: ( )[ ] ⋅ = ⋅ −=×−×+= ∫ × × −− Kkg kJ 392.5 Kkg kJ 104.04960.5106.9100.4 Pa 101.2 Pa 100.1 410 12 6 6 dPPss 21 5.16 A schematic of the process follows: We also know the ideal gas heat capacity from Table A.2.1: 263 10824.810785.28213.1 TT R cP −− ×−×+= Since this process is isentropic (∆s=0), we can construct a path such that the sum of ∆s is zero. (a) T, v as independent variables Choosing T and v as the independent variables, (and changing T under ideal gas conditions), we get: Temperature vo lu m e Ideal Gas st ep 1 step 2 v1,T1 ∆s=0 v2,T2 ∆s1 ∆s2 or in mathematical terms: 0= ∂ ∂ + ∂ ∂ = dv v sdT T sds Tv However, From Equation 5.33: 22 ds = cv T dT + ∂P ∂T v dv To get ∆s P = RT v − b − a v2 1 so ∂P ∂T v = R v − b and ∆s1 = d∫ s = ∂P ∂T v dv v1 v2 ∫ = R v − b dv = R ln v1 v2 ∫ v2 − b v1 − b or, using the ideal gas law, we can put ∆s1 in terms of T2 ∆s1 = R ln RT2 P2 − b v1 − b : For step 2 ∫ ∫ −− ×−×+ ==∆ 2 1 2 K 15.623 263 2 10824.810785.28213.0 T T T v dT T TTRdT T c s Now add both steps ( ) ( )[ ]222 6 2 32 1 2 2 21 K 15.623 2 10824.8K 15.62310785.28 15.623 ln213.0ln 0 − × −−×+ + − − = =∆+∆=∆ − − TTT bv b P RT sss Substitute K 15.6231 =T [ ]/molcm 600 31 =v atm12 =P ⋅ ⋅ = Kmol atmcm 06.82 3 R 23 and solve for T2: [ ]K 3.4482 =T (b) T, P as independent variables Choosing T and P as the independent variables, (and changing T under ideal gas conditions), we get: Temperature Pr es su re Ideal Gas st ep 1 step 2 P1,T1 ∆s=0 P2,T2 ∆s1 ∆s2 Mathematically, the entropy is defined as follows 0= ∂ ∂ + ∂ ∂ = dP P sdT T sds TP Using the appropriate relationships, the expression can be rewritten as 0= ∂ ∂ −= dP T vdT T cds P P For the van der Waals equation ( ) ( ) + − − − = ∂ ∂ 32 2 v a bv RT bv R T v P Therefore, 24 ( ) ( ) 0 2 2 1 2 1 32 = + − − − −=∆ ∫∫ dP v a bv RT bv R dT T cs P P T T P We can’t integrate the second term of the expression as it is, so we need to rewrite dP in terms of the other variables. For the van der Waals equation at constant temperature: ( ) dv bv RT v adP − −= 23 2 Substituting this into the entropy expression, we get ( ) 0 10824.810785.28213.1 2 1 2 1 263 = − − ×−×+ =∆ ∫∫ −− dv bv RdT T TTs v v T T Upon substituting ⋅ ⋅ = = = = = Kmol atmcm 06.82 mol cm 91b atm) 1at ideally acts (gas cm 600 K 15.623 3 3 2 2 2 3 1 1 R P RT v v T we obtain one equation for one unknown. Solving, we get K 3.4482 =T 25 5.17 (a) Attractive forces dominate. If we examine the expression for z, we see that at any absolute temperature and pressure, .1<z The intermolecular attractions cause the molar volume to deviate negatively from ideality and are stronger than the repulsive interactions. (b) Energy balance: qhh =− 12 Alternative 1: path through ideal gas state Because the gas is not ideal under these conditions, we have to create a hypothetical path that connects the initial and final states through three steps. One hypothetical path is shown below: Choosing T and P as the independent properties: dP P hdT T hdh TP ∂ ∂ + ∂ ∂ = or using Equation 5.46 dPv T vTdTcdh P P + ∂ ∂ −+= The given EOS can be rewritten as += 2/1 1 aT P Rv 26 Taking the derivative gives: 5.05.0 −+= ∂ ∂ aRT P R T v P so ( )dPaRTdTcdh P 5.05.0+= For step 1 ( ) =−==∆ ∫ mol J 2525.05.0 0 bar 50 5.0 1 5.0 11 PaRTdPaRTh For step 2 ( ) =−×+=∆ ∫ −− mol J 7961875.01002.358.3 K 500 K 300 5.03 2 dTTTRh For step 3: ( ) −===∆ ∫ mol J 3235.05.0 bar 50 0 5.0 2 5.0 23 PaRTdPaRTh Finally summing up the three terms, we get, =∆+∆+∆= mol J 7888321 hhhq Alternative 2: real heat capacity For a real gas realPch =∆ From Equation 5.48: ∫ ∂ ∂ −= real ideak P P P ideal P real P dPT vTcc 2 2 For the given EOS 27 += 2/1 1 aT P Rv Therefore, 5.12 2 25.0 −−= ∂ ∂ aRT T v P and [ ]( ) 5.01/2bar 50 bar 0 5.0 2 2 K 875.025.0 − = = − =−= ∂ ∂ ∫∫ RTdPaRTdPT vT real ideak real ideak P P P P P We can combine this result with the expression for realPc and find the enthalpy change. ( )dTTTRh ∫ −− −×+=∆ K 500 K 300 5.03 875.01002.358.3 =∆= mol J 7888hq The answers is equivalent to that calculated in alternative 1 28 5.18 (a) Calculate the temperature of the gas using the van der Waals equation. The van der Waals equation is given by: 2v a bv RTP − − = First, we need to find the molar volume and pressure of state 1. [ ]( ) [ ]( )[ ] ==== mol m 00016.0 mol 250 m 4.0m 1.0 321 1 n Al n Vv [ ]( ) [ ] [ ] [ ]Pa 1008.1Pa 1001325.1m 1.0 s m 81.9kg 10000 65 2 2 1 ×=×+ =+= atmPA mgP Substituting these equations into the van der Waals equation above gives [ ] 23 3 3 5 3 1 6 mol m 00016.0 mol mJ .50 mol m 104 mol m 00016.0 Kmol J 314.8 Pa 1008.1 ⋅ − ×− ⋅ =× − T K 5.2971 =T Since the process is isothermal, the following path can be used to calculate internal energy: Thus, we can write the change in internal energy as: 29 dv v udv v udT T udu TTv ∂ ∂ = ∂ ∂ + ∂ ∂ = Using Equation 5.40 ∫ − ∂ ∂ =∆ 2 1 v v v dvP T PTu For the van der Waals EOS: 2v a bv RTP − − = so bv R T P v − = ∂ ∂ Therefore, ∫=∆ 2 1 2 v v dv v au We can assume the gas in state 2 is an ideal gas since the final pressure is atmospheric. Therefore, we calculate 2v , == mol m 0244.0 3 2 2 2 P RTv and = ⋅ =∆ ∫ mol J 5.3104 mol mJ 5.00244.0 00016.0 2 3 dv v u or [ ]kJ 1.776=∆U 30 (b) From the definition of entropy: surrsysuniv sss ∆+∆=∆ First, let’s solve for syss∆ using the thermodynamic web. dv v sdT T sds Tv sys ∂ ∂ + ∂ ∂ = Since the process is isothermal, dv v sds T sys ∂ ∂ = ∫ ∂ ∂ =∆∴ 2 1 v v v sys dvT Ps Again, for the van der Waals equation, bv R T P v − = ∂ ∂ Substitution of this expression into the equation for entropy yields ∫ −=∆ 2 1 v v sys dvbv Rs ⋅ = ×− ⋅=∆ ∫ − Kmol J 17.44 mol m 104 Kmol J 314.80244.0 00016.0 3 5 dv v ssys K J 5.11042 =∆ sysS The change in entropy of the surroundings will be calculated as follows surr surr surr T Q s =∆ where 31 QQsurr −= (Q is the heat transfer for the system) Application of the first law provides WUQ −∆= We know the change in internal energy from part a, so let’s calculate W using ∫−= 2 1 v v PdvnW Since the external pressure is constant, [ ]( ) [ ]( ) − ×−= mol m 00016.0 mol m 0244.0Pa 1001325.1mol 250 33 5W [ ]J 614030−=W Now calculate heat transfer. [ ] ( ) [ ] [ ]J 1039.1J 614030J 776100 6×=−−=Q Therefore, [ ][ ] −= ×− =∆ K J 4672 K 5.297 J 1039.1 6 surrS and the entropy change of the universe is: = − =∆ K J 5.6370 K J 4672 K J 5.11042univS 32 5.19 First, calculate the initial and final pressure of the system. [ ] [ ]( ) [ ]( )[ ] [ ]Pa 1092.4m05.0 m/s81.9kg 20000Pa1010 62 2 5 ×=+×=iP [ ] [ ]( ) [ ]( )[ ] [ ]Pa 1089.6m05.0 m/s81.9kg 30000Pa1010 62 2 5 ×=+×=fP To find the final temperature, we can perform an energy balance. Since the system is well- insulated, all of the workdone by adding the third block is converted into internal energy. The energy balance is wu =∆ To find the work, we need the initial and final molar volumes, which we can obtain from the given EOS: [ ]/molm 1037.8 34−×=iv ( ) [ ]/molm 103.2 2511089.6 314.8 35- 6 ×+ +× = f f f T T v Now, calculate the work ( ) ( ) ( ) ×−×+ +× ×−=−−= −45- 6 6 1037.8103.2 2511089.6 314.8 Pa 1089.6 f f iff T T vvPw We also need to find an expression for the change in internal energy with only one variable: Tf. To find the change in internal energy, we can create a hypothetical path shown below: 33 For step 1, we calculate the change in internal energy as follows dvP T PTdv v uu low i low i PRTv v v PRTv v T ∫∫ == − ∂ ∂ = ∂ ∂ =∆ // 1 ( ) ( ) ( ) − − + = − + =∆ ∫ = bv bPRT aT aRT bv dv aT aRTu i lowi i i PRTv v i i low i /ln2 2/ 2 2 1 Similarly, for step 3: dvP T PTdv v uu f low f low v PRTv v v PRTv T ∫∫ == − ∂ ∂ = ∂ ∂ =∆ // 3 ( ) ( ) ( ) − − + = − + =∆ ∫ = bPRT bv aT aRT bv dv aT aRT u lowf f f f v PRTv f f f low / ln2 2 / 2 2 3 Insert the expression for the final molar volume into the equation for 3u∆ : ( ) ( )( )( ) −+×+ =∆ bPRTT T aT aRT u lowff f f f //2511089.6 314.8 ln 62 2 3 Since the pressure is low (molar volume is big) during the second step, we can use the ideal heat capacity to calculate the change in internal energy. ( )∫∫ == +−==∆ f i f i T T T T v dTTRdTcu K 500K 500 2 05.020 ( ) ( )222 500025.0500686.11 −+−=∆ ff TTu If we set the sum of the three steps in the internal energy calculation equal to the work and choose an arbitrary value for Plow, 100 Pa for example, we obtain one equation with one unknown: 34 ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ×−×+ +× ×− = −+×+ +−+−+ − − + −45- 6 6 62 2 22 2 2 1037.8103.2 2511089.6 314.8 Pa 1089.6 //2511089.6 314.8 ln 500025.0500686.11/ln f f lowff f f f ff i lowi i i T T bPRTT T aT aRT TT bv bPRT aT aRT Solving for Tf K 2.536=fT we get The piston-cylinder assembly is well-insulated, so sysuniv ss ∆=∆ Since the gas in the cylinder is not ideal, we must construct a hypothetical path, such as one shown below, to calculate the change in entropy during this process. For steps 1 and 3 ∫∫ ∂ ∂ −= ∂ ∂ =∆ low i low i P P P P P T dP T vdP P ss1 35 ∫∫ ∂ ∂ −= ∂ ∂ =∆ f low f low P P P P P T dP T vdP P ss3 We can differentiate the given EOS as required: ( ) ( ) ( ) ( ) + +− = + +− =∆ ∫ i low i ii P P i ii P P Ta TaRTdP PTa TaRTs low i ln22 221 ( ) ( ) ( ) ( ) + +− = + +− =∆ ∫ low f f ff P P f ff P P Ta TaRT dP PTa TaRT s f low ln 22 223 For step 2 ∫∫∫ +== ∂ ∂ =∆ f i f i f i T T T T P T T P dT T dT T cdT T ss 05.0202 ( )if i f TT T T s −+ =∆ 05.0ln202 Sum all of the steps to obtain the change in entropy for the entire process 321 sssss sysuniv ∆+∆+∆=∆=∆ ( ) ( ) ( ) ( )( ) + + −−+ + + +− =∆ low f f ff if i f i low i ii univ P P Ta TaRT TT T T P P Ta TaRTs ln 2 05.0ln20ln2 22 Arbitrarily choose Plow ( ) = ⋅ =∆ ⋅ =∆ K J 766.0 Kmol J 388.0mol 2 Kmol J 388.0 univ univ S s (try 100 Pa), substitute numerical values, and evaluate: 36 5.20 A schematic of the process is given by: V = 1 L T = 500 K well insulated Vacuum V = 1 L nCO=1 mole State i V = 2 L T = ? nCO=1 mole State f (a) The following equation was developed in Chapter 5: dv T PTcc v v v ideal v real v ideal ∫ ∂ ∂ += 2 2 For the van der Waals EOS 02 2 = ∂ ∂ T P Therefore, idealv real v cc = From Appendix A.2: ( ) ⋅ =− −×+= − Kmol J 0.22 500 31005001057.5376.3 2 4 RRcrealv (b) As the diaphragm ruptures, the total internal energy of the system remains constant. Because the volume available to the molecules increases, the average distance between molecules also increases. Due to the increase in intermolecular distances, the potential energies increase. Since the total internal energy does not change, the kinetic energy must compensate by decreasing. Therefore, the temperature, which is a manifestation of molecular kinetic energy, decreases. 37 (c) Because the heat capacity is ideal under these circumstances we can create a two-step hypothetical path to connect the initial and final states. One hypothetical path is shown below: For the first section of the path, we have ∫∫ ==∆ f i f i T T ideal v T T real v dTcdTcu1 ( ) ( ) ( ) 4.105074.2577375.191032.2 31001057.5376.2 23 1 K 500 2 4 1 −++×=∆ −×+=∆ − −∫ f ff T T TTu dT T TRu f i For the second step, we can use the following equation ∫ ∂ ∂ =∆ f i v v T dv v uu2 If we apply Equation 5.40, we can rewrite the above equation as ∫ − ∂ ∂ =∆ f i v v v dvP T PTu2 For the van der Waals EOS, 2v a bv RTP − − = , 38 bv R T P v − = ∂ ∂ Therefore, == − − =∆ ∫∫ = = = = mol J7.73 002.0 001.0 2 002.0 001.0 1 f i f i v v v v dv v advP bv RTu Now set the sum of the two internal energies equal to zero and solve for Tf ( ) 07.734.105074.2577375.191032.2 2321 =+−++×=∆+∆ − f ff T TTuu : K 497=fT (d) Since the system is well-insulated sysuniv ss ∆=∆ To solve for the change in entropy use the following development: dv v sdT T sds Tv sys ∂ ∂ + ∂ ∂ = Using the thermodynamic web, the following relationships can be proven T c T s v v = ∂ ∂ vT T P v s ∂ ∂ = ∂ ∂ For the van der Waals EOS ( )bv R T P v − = ∂ ∂ Now we can combine everything and calculate the change in entropy 39 ( )∫∫ −+=∆ 002.0 001.0 K 497 K 500 dv bv RdT T c s vsys ( ) ( ) ⋅ =∆=∆ ×− + −×+=∆ ∫∫ − − Kmol J 80.5 1095.3 31001057.5376.2 002.0 001.0 5 K 497 K 500 3 4 sysuniv sys ss v dvdT TT Rs 40 5.21 A schematic of the process is given by: V = 0.1 m3 T = 300 K well insulated Vacuum nA=400 moles State i T = ? State f V = 0.1 m3 nA=400 moles V = 0.2 m3 Energy balance: 0=∆u Because the gas is not ideal under these conditions, we have to create a hypothetical path that connects the initial and final states through three steps. One hypothetical path is shown below: For the first section of the path, we have ∫ ∞= ∂ ∂ =∆ v v Ti dv v uu1 If we apply Equation 5.40, we can rewrite the above equation as ∫ ∞= − ∂ ∂ =∆ v v vi dvP T PTu1 41 For the van der Waals EOS 22vT a bv R T P v + − = ∂ ∂ Therefore, == −+ − =∆ ∫∫ ∞= ×= ∞= ×= −− mol J11202 44 105.2 2 105.2 21 v v i v v ii dv vT advP Tv a bv RTu Similarly for step 3: ⋅ − == −+ − =∆ ∫∫ −− ×= ∞= ×= ∞= Kmol J1680002 44 105 2 105 23 f v v f v v T dv vT advP Tv a bv RTu ff For step 2, the molar volume is infinite, so we can use the ideal heat capacity given in the problem statement to calculate the change in internal energy: ( )K 300 2 3 2 −=∆ fTRu If we set sum of the changes in internal energy for each step, we obtain one equation for one unknown: ( ) 0168000K 300 Kmol J314.8 2 3 mol J 1120321 = − +− ⋅ + =∆+∆+∆ f f T Tuuu Solve for Tf K 6.261=fT : 42 5.22 A schematic of the process is shown below: Ethane 3 MPa; 500K Initially: vacuum Tsurr = 293 K (a) Consider the tank as the system. Since kinetic and potential energy effects are negligible, the open system, unsteady-state energy balance (Equation 2.47) is ∑∑ ++−= out soutout in inin sys WQhnhn dt dU The process is adiabatic and no shaft work is done. Furthermore, there is one inlet stream and no outlet stream. The energy balance reduces to inin sys hn dt dU = Integration must now be performed ∫∫ = t inin U U dthndU 0 2 1 ininin t inin hnnhndtnhunun )( 12 0 1122 −===− ∫ Since the tank is initially a vacuum, n1 inhu =2 =0, and the relation reduces to: 43 As is typical for problems involving the thermodynamic web, this problem can be solved in several possible ways. To illustrate we present two alternatives below: Alternative 1: path through ideal gas state Substituting the definition of enthalpy: ininin vPuu +=2 or ( ) ( ) ininin vPuTu =− K 500 MPa, 3at MPa, 3at 2 (1) From the equation of state: ( ) ( ) ( )[ ] =××− ⋅ =+= − mol J 800,3Pa 103108.21K 552 Kmol J 314.8'1 68PBRTvP inin (2) The change in internal energy can be found from the following path: For steps 1 and 3, we need to determine how the internal energy changes with pressure at constant temperature: From the fundamental property relation and the appropriate Maxwell relation: TPTTT P vP T vT P vP P sT P u ∂ ∂ − ∂ ∂ −= ∂ ∂ − ∂ ∂ = ∂ ∂ From the equation of state ( ) RTB P RTPB'P P RT P u ' T −= −−+−= ∂ ∂ 21 44 So for step 1: [J/mol] 349 0 ' 0 ' 1 −==−= ∂ ∂ =∆ ∫ ∫ in Pin PinT PRTBRTdPBdP P uu (3) and for step 3: TPRTBRTdPBdP P uu P P T 7.02 2 0 ' 2 0 ' 3 =−=−= ∂ ∂ =∆ ∫ ∫ (4) For step 2 [ ]dTRcdT T Pv T hdT T uu T P T PP T P ∫∫∫ −= ∂ ∂ − ∂ ∂ = ∂ ∂ =∆ 500500500 2 or [ ]∫ = −− ×−×+=∆ 2 1 K 500 263 2 10561.510225.19131.0 T T dTTTRu (5) Substituting Equations 2, 3, 4, and 5 into 1 and solving for T gives: K 5522 =T Alternative 2: real heat capacity Starting with: inhu =2 The above equation is equivalent to inhvPh =− 222 222 vPhh in =−∴ To calculate the enthalpy difference, we can use the real heat capacity dP T vTcc P P P ideal P real P ideal ∫ ∂ ∂ −= 2 2 For the truncated viral equation, 45 02 2 = ∂ ∂ PT v Therefore, idealP real P cc = Now, we can calculate the change in enthalpy and equate it to the flow work term. ∫ = = 2 1 K 500 22 T T ideal P vPdTc [ ] ( )∫ = −− +==×−×+ 2 1 K 500 2222 263 '110561.510225.19131.1 T T PBRTvPdTTTR Integrate and solve for T2 K 5522 =T : (b) In order to solve the problem, we will need to find the final pressure. To do so, first we need to calculate the molar volume. Using the information from Part (a) and the truncated virial equation to do this ( ) ( ) ( )[ ]Pa 103108.21 Pa 103 K 552 Kmol J 314.8 '1 686 ××−× ⋅ =+= −PB P RTv = mol m 0014.0 3 v This quantity will not change as the tank cools, so now we can calculate the final pressure. ( ) ( ) 28 3 2 108.21 K 293 Kmol J 314.8 mol m 0014.0 P P −×−= ⋅ Solve for 2P : 46 Pa 1066.1 62 ×=P The entropy change of the universe can be expressed as follows: surrsysuniv SSS ∆+∆=∆ To solve for the change in entropy of the system start with the following relationship: dP P sdT T sds TP sys ∂ ∂ + ∂ ∂ = Alternative 1: path through ideal gas state Using the proper relationships, the above equation can be rewritten as dP T vdT T cds P P sys ∂ ∂ += We can then use the following solution path: Plow T step 1 st ep 3 step 2 ideal gas 500 K T2 ∆s1 3 MPa ∆s3 ∆s2 P 1.66 MPa Choosing a value of 1 Pa for Plow ( )∫∫ +−= ∂ ∂ =∆ Pa 1 MPa 66.1 ' Pa 1 MPa 66.11 1 dPPBP RdP T vs P , for step 1: For step 3, ( )∫∫ +−= ∂ ∂ =∆ MPa 3 Pa 1 ' MPa 3 Pa 1 1 1 dPPBP RdP T vs P 47 For step 2: ∫∫ ×−×+==∆ −− K 293 K 552 63 K 293 K 552 2 10561.510225.19 131.1 dTT T RdT T cs P Adding together steps 1, 2 and 3: ⋅ −=∆ Kmol J9.46syss ______________________________________________________________________________ Alternative 2: real heat capacity Using the proper relationships, the above equation can be rewritten as dP T vdT T cds P real P sys ∂ ∂ += For the truncated virial equation += ∂ ∂ '1 B P R T v P Now, substitute the proper values into the expression for entropy and integrate: ∫∫ × × −− ++ ×−×+=∆ Pa 1066.1 Pa 103 K 293 K 552 63 6 ^ '110561.510225.19131.1 dPB P RdTT T Rssys ⋅ −=∆ Kmol J9.46syss ______________________________________________________________________________ In order to calculate the change in entropy of the surroundings, first perform an energy balance. qu =∆ Rewrite the above equation as follows ( ) qPvh =∆−∆ 48 Since the real heat capacity is equal to ideal heat capacity and the molar volume does not change, we obtain the following equation ( )∫ =−− f i T T if ideal P qPPvdTc [ ] ( )∫ = = −− =×× −×−×+ K293 K 552 66 3 263 Pa 103-Pa 1066.1 mol m 0014.010561.510225.19131.1 f i T T qdTTTR −= mol J 15845q Therefore, = mol J 15845surrq and ⋅ =∆ Kmol J 08.54surrs Before combining the two entropies to obtain the entropy change of the universe, find the number of moles in the tank. [ ] mol 7.75 mol m 0014.0 m 05.0 3 3 = =n Now, calculate the entropy change of the universe. ( ) ⋅ −+ ⋅ =∆ Kmol J 9.46 Kmol J08.54mol 7.75univS =∆ K J 544univS 49 5.23 First, focus on the numerator of the second term of the expression given in the problem statement. We can rewrite the numerator as follows: ( ) ( )idealvTidealvTidealvTvTidealvTvT rrrrrrrrrrrr uuuuuu ∞=∞= −−−=− ,,,,,, For an ideal gas, we know 0,, =− ∞= ideal vT ideal vT rrrr uu Therefore, idealvTvT ideal vTvT rrrrrrrr uuuu ∞=−=− ,,,, Substitute this relationship into the expression given in the problem statement: c ideal vTvT c ideal vTvT c dep vT RT uu RT uu RT u rrrrrrrrrr ∞= − = − = ∆ ,,,,, Now, we need to find an expression for idealvTvT rrrr uu ∞=− ,, . Note that the temperature is constant. Equation 5.41 reduces to the following at constant temperature: dvP T PTdu v T − ∂ ∂ = The pressure can be written as v zRTP = and substituted into the expression for the differential internal energy dv T z v RTdv v zRT v RT T z v RTTdu vvv T ∂ ∂ = − + ∂ ∂ = 2 Applying the Principle of Corresponding States 50 r vrr r c T dv T z v T RT du r r ∂ ∂ = 2 If we integrate the above expression, we obtain ∫∫ ∞=∞= ∞= ∂ ∂ = − = r r vrrrr v v r vrr r v v c ideal vTvT c T dv T z v T RT uu RT du 2, , Therefore, ∫ ∞= ∞= ∂ ∂ = − = − = ∆ r r rrrrrrrrrr v v r vrr r c ideal vTvT c ideal vTvT c dep vT dv T z v T RT uu RT uu RT u 2,,,,, 51 5.24 We write enthalpy in terms of the independent variables T and v: dv v hdT T hdh Tv ∂ ∂ + ∂ ∂ = using the fundamental property relation: vdPTdsdh += At constant temperature, we get: dv v Pv T PTdh Tv T ∂ ∂ + ∂ ∂ = For the Redlich-Kwong EOS ( )bvvT a bv R T P v + + − = ∂ ∂ 2/32 1 ( ) ( ) ( )22/122/12 bvvT a bvvT a bv RT v P T + + + + − − = ∂ ∂ Therefore, ( ) ( ) ( ) dv bvT a bvvT a bv RTv bv RTdhT + + + + − − − = 22/12/12 2 3 To find the enthalpy departure function, we can integrate as follows ( ) ( ) ( )∫∫ ∞=∞= + + + + − − − ==∆ v v v v T dep dv bvT a bvvT a bv RTv bv RTdhh 22/12/12 2 3 Since temperature is constant, we obtain ( )bvT a bv v bT a bv RTbhdep + + + + − =∆ 2/12/1 ln2 3 To calculate the entropy departure we need to be careful. From Equation 5.64, we have: ( ) ( )gas ideal 0,gas ideal,gas ideal 0,,gas ideal,, == −−−=− PTPTPTPTPTPT ssssss 52 However, since we have a P explicit equation of state, we want to put this equation in terms of v. Let’s look at converting each state. The first two states are straight -forward vTPT ss ,, = and gas ideal, gas ideal 0, ∞== = vTPT ss For the third state, however, we must realize that the ideal gas volume v’ 'v RTP = at the T and P of the system is different from the volume of the system, v. In order to see this we can compare the equation of state for an ideal gas at T and P to a real gas at T and P ( )bvvT a bv RTP + − − = The volume calculated by the ideal gas equation, v’ −+== gas ideal, gas ideal , gas ideal , gas ideal , gas ideal , '' vTvTvTvTPT sssss , is clearly different from the volume, v, calculated by the Redlich-Kwong equation. Hence: Thus, ( ) ( ) −−−−−=− ∞=∞= gas ideal , gas ideal , gas ideal , gas ideal , gas ideal ,, gas ideal ,, ' vTvTvTvTvTvTPTPT ssssssss Using a Maxwell relation: vT T P dv ds ∂ ∂ = Therefore, dv T Pds v T ∂ ∂ = 53 For the Redlich-Kwong EOS ( )bvvT a bv R T P v + + − = ∂ ∂ 2/32 1 so ( ) ( )∫∞= ∞= + + − =− v v vTvT dvbvvT a bv Rss 2/3 gas ideal ,, 2 1 For an ideal gas v R T P v = ∂ ∂ so ( ) ∫ ∞= ∞= =− v v vTvT dvv Rss gas ideal, gas ideal , Finally: Pv RTR v vR v dvRss v v vTvT lnln ' gas ideal , gas ideal , ' ' ===− ∫ Integrating and adding together the three terms gives:( ) Pv RTR bv v bT a v bvRsdep lnln 2 ln 2/3 − + + − =∆ 54 5.25 Calculate the reduced temperature and pressure: [ ] [ ] 344.0 bar 48.220 K 3.647 = = = w P T c c (Table A.1.2) [ ] [ ] 04.1 K 647.3 K 15.673 36.1 bar 20.482 bar300 == == r r T P By double interpolation of data from Tables C.3 and C.4 921.2 )0( , −= ∆ c dep PT RT h rr 459.1 )1( , −= ∆ c dep PT RT h rr From Tables C.5 and C.6: 292.2 )0( , −= ∆ R sdepPT rr 405.1 )1( , −= ∆ R sdepPT rr Now we can calculate the departure functions ∆ + ∆ =∆ )1( , )0( , c dep PT c dep PT c dep RT h w RT h RTh rrrr ( ) ( )( ) −=−+− ⋅ =∆ mol J 18421459.1344.0921.23.647 Kmol J 314.8deph ∆ + ∆ =∆ )1( , )0( , R s w R s Rs dep PT dep PTdep rrrr ( )( ) ⋅ −=−+− ⋅ =∆ Kmol J 07.23405.1344.0292.2 Kmol J 314.8deps 55 To use the steam tables for calculating the departure functions, we can use the following relationships. idealPTPT dep hhh ,, −=∆ idealPTPT dep sss ,, −=∆ From the steam tables = kg kJ 0.2151,PTh and ⋅ = Kkg kJ 4728.4,PTs We need to calculate the ideal enthalpies and entropies using the steam tables’ reference state. ( ) ∫+∆= K 673.15 K 273.16 , C.01? 0 dTchh ideal p vapideal PT We can get =∆ mol kJ 1.45vaph from the steam tables and heat capacity data from Table A.2.2. Using this information, we obtain ∫ × +×+ ⋅ + = − K 673.15 K 273.16 2 5 3 , 10121.01045.147.3 Kmol kJ 008314.0 mol kJ 1.45 dT T ThidealPT = mol kJ 14.59, ideal PTh Now, calculate the ideal entropy. ( ) −+∆= ∫ 1 2 K 673.15 K 273.16 , lnC.01? 0 P PRdT T c ss ideal pvapideal PT From the steam tables: ( ) ⋅ =∆ Kmol kJ 165.0Cº 01.0vaps Substitute values into the entropy expression: 56 − × +×+ ⋅ += ∫ − 000613.0 30ln10121.01045.147.3 Kmol kJ 008314.0 165.0 K 673.15 K 273.16 3 5 3 , dT TT sidealPT ⋅ = Kmol J 107, ideal PTs Now, calculate the departure functions: [ ]( ) −= − =∆ mol kJ 4.20 mol kJ 14.59kg/mol 0180148.0 kg kJ 0.2151deph [ ]( ) ⋅ −= ⋅ − ⋅ =∆ Kmol kJ 0264.0 Kmol kJ 107.0kg/mol 0180148.0 Kkg kJ 4728.4deps Table of Results Generalized Tables Steam Tables Percent Difference (Based on steam tables) ∆ mol kJ deph -18.62 -20.4 9.9 ⋅ ∆ Kmol kJ deps -0.0231 -0.0264 12.5 57 5.26 State 1 is at 300 K and 30 bar. State 2 is at 400 K and 50 bar. The reduced temperature and pressures are [ ] [ ] 982.0 K 305.4 K 300 616.0 bar 74.48 bar30 ,1 ,1 == == r r T P [ ] [ ] 31.1 K 305.4 K 400 026.1 bar 74.48 bar50 ,2 ,2 == == r r T P and 099.0=ω By double interpolation of data in Tables C.3 and C.4 825.0 )0( , ,1,1 −= ∆ c dep PT RT h rr 799.0 )1( , ,1,1 −= ∆ c dep PT RT h rr 711.0 )0( , ,2,2 −= ∆ c dep PT RT h rr 196.0 )1( , ,2,2 −= ∆ c dep PT RT h rr Therefore, ( ) 904.0799.0099.0825.0,1,1 , −=−+−= ∆ c dep PT RT h rr ( ) 730.0196.0099.0711.0,2,2 , −=−+−= ∆ c dep PT RT h rr The ideal enthalpy change from 300 K to 400 K can be calculated using ideal cP RdTTTRhidealTT 39.71710561.510225.19131.1 2 K 400 K 300 63 21 =×−×+=∆ ∫ −−→ data from Table A.2.1. The total entropy change is 58 depPT ideal TT dep PT rrrr hhhh ,2,221,1,1 ,, ∆+∆+∆−=∆ → ( )[ ]CC TTRh 730.039.717904.0 −+−−=∆ ( ) ( )[ ]K 4.305730.0K 39.717K 4.305904.0 Kmol J314.8 −+ ⋅ =∆h =∆ mol J 2.6406h Using the data in Table C.5 and C.6 ( ) 676.0756.0099.0601.0,1,1 , −=−+−= ∆ R sdepPT rr ( ) 416.0224.0099.0394.0,2,2 , −=−+−= ∆ R sdepPT rr Substituting heat capacity data into Equation 3.62, we get − ×−×+ =∆ ∫ −− bar 30 bar 50ln10561.510225.19131.1 K 400 K 300 263 dT T TTRsideal Rsideal 542.1=∆ Therefore, ( )416.0542.1676.0 ,2,2,1,1 ,, −+=∆+∆+∆−=∆ Rssss depPT idealdep PT rrrr ⋅ =∆ Kmol J 98.14s 59 5.27 The turbine is isentropic. Therefore, we know the following 0 ,2,2,1,1 ,, =∆+∆+∆−=∆ depPT idealdep PT rrrr ssss Using the van der Waals EOS, we can find P1,r, which leaves one unknown in the above equation: T2 ( ) 23 2 3 5 33 3 1 mol cm 600 mol cmatm 1091 mol cm 19 mol cm 600 K 15.623 Kmol atmcm 06.82 ⋅ × − − ⋅ ⋅ =P . [ ] [ ]bar 19.76atm 19.751 ==P Calculate reduced temperature and pressures using data from Table A.1.1 [ ] [ ] 8.1bar 44.42 bar19.76 ,1 ==rP [ ] [ ] 024.0bar 2.444 bar013.1 ,2 ==rP 68.1 K 370.0 K 23.156 ,1 ==rT Also, 152.0=ω From Tables C.5 and C.6: ( ) 343.0102.0152.0327.0,1,1 , −=−+−= ∆ R sdepPT rr Substituting heat capacity data into Equation 3.62, we get − ×−×+ =∆ ∫ −− atm 5.197 atm 1ln10824.810785.28213.1 2 K 23.156 263 dT T TTRs T ideal Therefore, 60 s R s dT T TTR dep PT T rr ∆= ∆ ++ ×−×+ + ∫ −− ,2,2 2 , K 23.156 263 32.410824.810785.28213.1343.0 We can solve this using a guess-and-check method K 6002 =T : 62.1,2 =rT ⋅ =∆ Kmol J 84.33s K 4502 =T : 22.1,2 =rT ⋅ =∆ Kmol J 77.0s K 6.4462 =T : 21.1,2 =rT ⋅ ≅∆ Kmol J 0s Therefore, K 6.4462 =T 61 5.28 A reversible process requires the minimum amount of work. Since the process is reversible and adiabatic 0=∆s which can be rewritten as 0 ,2,2,1,1 ,, =∆+∆+∆−=∆ depPT idealdep PT rrrr ssss Calculate reduced temperature and pressures using data from Table A.1.1 [ ] [ ] 0217.0bar 0.46 bar1 ,1 ==rP [ ] [ ] 217.0bar 6.04 bar10 ,2 ==rP 57.1 K 190.6 K 003 ,1 ==rT From Tables C.5 and C.6: ( ) 0046.00028.0008.000457.0,1,1 , −=−+−= ∆ R s depPT rr Substituting heat capacity data into Equation 3.62, we get − ×−×+ =∆ ∫ −− bar 1 bar 01ln10164.210081.9702.1 2 K 003 263 dT T TTRs T ideal Therefore, ∆ +− ×−×+ +=∆ ∫ −− R s dT T TTRs dep PT T rr ,2,2 2 , K 003 263 303.210824.810785.28213.10046.0 We can solve using a guess-and-check method K 4002 =T : 10.2,2 =rT ⋅ =∆ Kmol J 98.4s K 3852 =T : 02.2,2 =rT 62 ⋅ =∆ Kmol J 42.1s K 3792 =T : 99.1,2 =rT ⋅ −=∆ Kmol J 018.0s Therefore, K 3792 ≅T An energy balance reveals that swhhh =∆=− 12 We can calculate the enthalpy using departure functions. From Tables C.3 and C.4: ( ) 0966.0011.0008.00965.0,1,1 , −=−+−= ∆ c dep PT RT h rr ( ) 0613.0015.00089.00614.0,2,2 , −=+−= ∆ c dep PT RT h rr Ideal heat capacity data can be used to determine the ideal change in enthalpy ×−×+=∆ ∫ −− dTTTRhideal K 379 K 003 263 10164.210081.9702.1 Therefore, ( )( ) ×−×++− ⋅ =∆ ∫ −− dTTTh K 379 K 003 263 10164.210081.9702.10613.00966.0K 6.190 Kmol J 314.8 =∆= mol J 2.3034hws and W1.101 mol J 2.3034 s mol 30/1 = =SW 63 5.29 Equation 4.71 states PT v v ∂ ∂ = 1β PT vv ∂ ∂ =∴ β This can be substituted into Equation 5.75 to give ( ) P JT c Tv 1− = βµ 64 5.30 For an ideal gas P R T v P = ∂ ∂ Therefore, ( ) 0=−= − = PP JT c vv c v P RT µ This result could also be reasoned from a physical argument. 65 5.31 The van der Waals equation is given by: 2v a bv RTP − − = (1) The thermal expansion coefficient PP v T vT v v = = ∂ ∂ ∂ ∂β 11 is given by: (2) Solving Equation 1 for T: − += R bv v aPT 2 Differentiating by applying the chain rule, 3 3 32 221 Rv abavPv v a R bv Rv aP v T P +− = −− += ∂ ∂ (3) Substitution into Equation 2 gives abavPv Rv 23 2 +− =β Substituting Equation 1 for P gives b in terms of R, T, v, a , and b: ( ) ( )23 2 2 bvaRTv bvRv −− − =β The isothermal compressibility TT v P vP v v −= −= ∂ ∂ ∂ ∂κ 11 is given by: From the van der Waals equation: ( ) ( ) ( )23 23 32 22 bvv bvaRTv v a bv RT v P T − −+− =+ − −= ∂ ∂ so 66 ( ) ( )23 22 2 bvaRTv bvv −− − =κ For the Joule-Thomson coefficient ∫ − − = real ideal P P P ideal P P JT dP T vTc v T vT 2 2 ∂ ∂ ∂ ∂ µ , we can use Equation 5.75: Substituting the van der Waals equation into Equation 3 gives ( )( ) ( ) ( ) 33 23 22 Rv bva bv T Rvbv bvaRTv v T P − − − = − −− = ∂ ∂ (4) Thus, the second derivative becomes: ( ) ( ) 4322 2 621 Rv bva Rv a v T bvbv T v T PP − +− − + − −= ∂ ∂ ∂ ∂ or simplifying using Equation 4, ( )42 2 32 Rv bva v T P − = ∂ ∂ (5) Substituting Equations 5 and 4 into Equation 5.75 gives: ( ) ( ) ( )∫ − − −− −+− = real ideal P P ideal P JT dP bva RTvc bvaRTv bvavbRTv 32 2 2 4 23 23 µ At a given temperature the integral in pressure can be rewritten in terms of volume using the van der Waals equation to give: 67 ( ) ( ) ( ) ( ) ( )∫ − −− − + −− −+− = real ideal v v ideal P JT dv bv bvaRTv bva RTvc bvaRTv bvavbRTv 2 23 23 23 2 32 2 2 µ 68 5.32 We can solve this problem by using the form of the Joule-Thomson coefficient given in Equation 5.75. The following approximation can be made PP T v T v ∆ ∆ ≅ ∂ ∂ ˆˆ At 300 ºC, ( ) ( ) Cº 250350 C,1MPaº250ˆC,1MPaº350ˆˆ − − = ∆ ∆ vv T v P Cº 250350 kg m 23268.0 kg m 28247.0 ˆ 33 − − = ∆ ∆ PT v ⋅ = ⋅ = ∂ ∂ ∴ K kg m 0005.0 Cº kg m 0005.0 ˆ 33 PT v A similar process was followed to find cP PP P T h dT hc ∆ ∆ ≅ ∂ = ˆˆ ˆ . At 300 ºC, ( ) ( ) Cº 250350 C,1MPaº250ˆC,1MPaº350ˆˆ − − = ∆ ∆ hh T h P Cº 250350 kg kJ 6.2942 kg kJ 7.3157ˆ − − = ∆ ∆ PT h ⋅ = ⋅ = ∆ ∆ ≅ ∂ = K kg kJ 15.2 Cº kg kJ 15.2 ˆˆ ˆ PP P T h dT hc Now, JTµ can be found. 69 ( ) ⋅ − ⋅ = − ∂ ∂ = K kg kJ 15.2 kg m 25794.0 K kg m 0005.0K 15.573 ˆ ˆˆ 33 P P JT c v T vT µ ⋅ = kJ K m 0133.0 3 JTµ 70 5.33 At the inversion line, the Joule-Thomson coefficient is zero. From Equation 5.75: 0 2 2 = − − = ∫ real idealP P P ideal P P JT dP T vTc v T vT ∂ ∂ ∂ ∂ µ This is true when the numerator is zero, i.e., 0= − v T vT P∂ ∂ For the van der Waals equation, we have 2v a bv RTP − − = Solving for T: − += R bv v aPT 2 so 3 3 32 221 Rv abavPv v a R bv Rv aP v T P +− = −− += ∂ ∂ Substituting for P: ( )( ) 3 23 2 Rvbv bvaRTv v T P − −− = ∂ ∂ Hence, ( ) ( )23 23 2 20 bvaRTv bvavbRTvv T vT P −− −+− ==− ∂ ∂ Solving for T: ( )3 22 bRv bvavT −= (1) 71 Substituting this value of T back into the van der Waals equation gives ( ) ( )223 322 bv bva v a bv bvavP −=−−= (2) We can solve Equations 1 and 2 by picking a value of v and solving for T and P. For N2, the critical temperature and pressure are given by Tc = 126.2 [K] and Pc = 2 32 mol Jm 0.137= 64 27 c c P RT a = 33.84 [bar], respectively. Thus, we can find the van der Waals constants a and b: ×= mol m 103.88=8 3 5- c c P RTb Using these values in Equations (1) and (2), we get the following plot: Joule-Thomson inversion line 0 200 400 600 800 1000 0 100 200 300 400 T [K] 72 5.34 We can solve this problem using departure functions, so first find the reduced temperatures and pressures. [ ][ ] 99.0bar 0.365 bar50 ,1 ==rP [ ] [ ] 2.0bar 0.365 bar10 ,2 ==rP 967.0 K 282.4 K 15.273 ,1 ==rT Since the ethylene is in two-phase equilibrium when it leaves the throttling device, the temperature is constrained. From the vapor-liquid dome in Figure 5.5: 6.214 76.0 2 ,2 =∴ ≅ T T r The process is isenthalpic, so the following expression holds 0 ,2,221,1,1 ,, =∆+∆+∆−=∆ → dep PT ideal TT dep PT rrrr hhhh Therefore, idealTT dep PT dep PT hhh rrrr 21,1,1,2,2 ,, → ∆−∆=∆ From Table A.2.1: [ ]∫ −−→ ×−×+=∆ K 6.214 K 15.273 263 10392.410394.14424.1 21 dTTTRhidealTT From Tables C.3 and C.4 ( )085.0=ω : ( ) 976.351.3085.0678.3,1,1 , −=−+−= ∆ c dep PT RT h rr Now we can solve for the enthalpy departure at state 2. 73 ×−×+−−= ∆ ∫ −− K 6.214 K 15.273 263, 10392.410394.14424.1 K 4.282 1976.3,2,2 dTTT RT h c dep PT rr 01.3,2,2 , −= ∆ c dep PT RT h rr We can calculate the quality of the water using the following relation ( ) c vapdep PT c liqdep PT c dep PT RT h x RT h x RT h rrrrrr , , , ,, ,2,2,2,2,2,2 1 ∆ + ∆ −= ∆ where x represents the quality. From Figures 5.5 and 5.6: ( ) 068.55.5085.06.4 , , ,2,2 −=−+−= ∆ c liqdep PT RT h rr ( ) 464.075.00859.04.0 , , ,2,2 −=−+−= ∆ c vapdep PT RT h rr Thus, 447.0=x 55.3% of the inlet stream is liquefied. 74 5.35 Density is calculated from molar volume as follows: v MW =ρ Substitute the above into the expression for 2soundV : ( ) s s sound v P MW v MW PV ∂ ∂ = ∂ ∂ = /1 12 The following can be shown using differentials: 2 1 v v v ∂ −= ∂ Therefore, ss sound v P MW vPV ∂ ∂ −= ∂ ∂ = 2 2 ρ 75 5.36 From Problem 5.35: ss sound v P MW vPV ∂ ∂ −= ∂ ∂ = 2 2 ρ The thermodynamic web gives: PTTvssPvs T s s P v s s T T P v T v s s P v P ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ −= ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ −= ∂ ∂ Pvv PP TTvs v T T P c c T c s P v s c T v P ∂ ∂ ∂ ∂ −= ∂ ∂ ∂ ∂ −= ∂ ∂ If we treat air as an ideal gas consisting of diatomic molecules only 5 7 = v P c c v R T P v = ∂ ∂ R P v T P = ∂ ∂ Therefore, −= ∂ ∂ v P v P s 5 7 and = = MW RT v P MW vVsound 5 7 5 72 [ ]m/s 343=soundV The lightening bolt is 1360 m away. 76 5.37 From Problem 5.35: ss sound v P MW vPV ∂ ∂ −= ∂ ∂ = 2 2 ρ The thermodynamic web gives: PTTvssPvs T s s P v s s T T P v T v s s P v P ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ −= ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ −= ∂ ∂ so ∂ ∂ ∂ ∂ −= ∂ ∂ T c v T T P c T v P P Pvvs For liquids vP cc ≈ water at 20 ºC, so Pvs v T T P v P ∂ ∂ ∂ ∂ −= ∂ ∂ However, the cyclic rule gives: TPv P v v T T P ∂ ∂ ∂ ∂ ∂ ∂ =−1 So Ts v P v P ∂ ∂ = ∂ ∂ From the steam tables, for saturated water at 20 o = kg m 001002.ˆ 3 v C: P = 2.34 kPa and For subcooled water at 20 o = kg m 0009995.ˆ 3 v C: P = 5 MPa and 77 So − − = ∆ ∆ ≈ ∂ ∂ = ∂ ∂ 3m kPa kg 001002.0009995. 34.25000 ˆˆˆ v P v P v P Ts and [ ]m/s 1414 ˆ ˆ2 = ∂ ∂ = s sound v PvV 78 5.38 (a) The fundamental property relation for internal energy is revrev WQdU δδ += Substituting the proper relationships for work and heat, we obtain FdzTdSdU += The fundamental property relation for the Helmholtz energy is ( ) SdTTdSdUTSddUdA −−=−= Substitute the expression for the internal energy differential: SdTFdzdA −= (b) First, relate the entropy differential to temperature and length. dz Z SdT T SdS Tz ∂ ∂ + ∂ ∂ = Now we need to find expressions for the partialderivatives. zzz z T ST T zFST T unnc ∂ ∂ = ∂ ∂+∂ = ∂ ∂ = Therefore, +== ∂ ∂ b T an T nc T S z z The following statement is true mathematically (order of differentiation does not matter): zTTz Z A TT A Z ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ Furthermore, 79 TTz Z S T A Z ∂ ∂ −= ∂ ∂ ∂ ∂ zzT T F Z A T ∂ ∂ = ∂ ∂ ∂ ∂ ( )0zzkT F Z S zT −−= ∂ ∂ −= ∂ ∂ Substituting the expressions for the partial derivatives into the expression for the entropy differential, we obtain ( )dzzzkdTb T andS 0−− += (c) First, start with an expression for the internal energy differential: dz Z UdT T UdU Tz ∂ ∂ + ∂ ∂ = From information given in the problem statement: ( )bTan T U z += ∂ ∂ Using the expression for internal energy developed in Part (a) and information from Part (b) ( )( ) ( ) 000 =−+−−=+ ∂ ∂ = ∂ ∂ zzkTzzkTF z ST Z U TT Therefore, ( )[ ] ( )[ ]dTbTandzdTbTandU +=++= 0 (d) We showed in Part (c) that 0= ∂ ∂ = T U z UF 80 Using the expression for Tz S ∂ ∂ developed in Part B, we obtain ( )0zzkTz STF T S −= ∂ ∂ −= (e) First, perform an energy balance for the adiabatic process. WdU δ= Substitute expressions for internal energy and work. ( )[ ] ( )dzzzkTFdzdTbTan 0−==+ Rearrangement gives ( ) ( )[ ]bTan zzkT dz dT + − = 0 The right-hand side of the above equation is always positive, so the temperature increases as the rubber is stretched. 81 5.39 The second law states that for a process to be possible, 0≥∆ univs To see if this condition is satisfied, we must add the entropy change of the system to the entropy change of the surroundings. For this isothermal process, the entropy change can be written dv T Pdv T PdT T cds vv v = += ∂ ∂ ∂ ∂ Applying the van der Waals equation: dv bv Rds − = Integrating ==∆ K mol J 5.11ln 1 2 v vRssys For the entropy change of the surroundings, we use the value of heat given in Example 5.2: =−= mol J 600surrqq Hence the entropy change of the surroundings is: −= − ==∆ K mol J 6.1 373 600 surr surr surr T qs and =∆+∆=∆ K mol J 9.9surrsysuniv sss Since the entropy change of the universe is positive we say this process is possible and that it is irreversible. Under these conditions propane exhibits attractive intermolecular forces (dispersion). The closer they are together, on average, the lower the energy. That we need to put work into this system says that the work needed to separate the propane molecules is greater than the work we get out during the irreversible expansion. 82 5.40 A schematic of the process is given by: The energy balance for this process is provided below: Swh =∆ Because the gas is not ideal under these conditions, we have to create a hypothetical path that connects the initial and final states through three steps. One hypothetical path is shown below: For the first section of the path, we have ∫ = ∂ ∂ =∆ 0 1 1 P P T dP P hh If we apply Equation 5.45 we can rewrite the above equation as ∫ = + ∂ ∂ −=∆ 0 1 P P P i i dPv T vTh For the given EOS: 83 2 iP T aP P R T v −= ∂ ∂ Therefore, −= += ++−=∆ ∫∫ = ×= = ×= mol J 24672 0 10100 0 10100 1 55 v P i P P i ii dPb T aPdPv T aPPh Similarly for step 3 = +=∆ ∫ ×= = mol J 2502 51020 0 3 fP P f dPb T aPh For step 2, the pressure is zero, so we can use the ideal heat capacity given in the problem statement to calculate the enthalpy change. ( ) −=+==∆ ∫∫ mol J 627002.030 K 445 K 600 2 dTTdTch f i T T P Now sum each part to find the total change in enthalpy: −=∆+∆+∆=∆ mol J 8487321 hhhh −= mol J 8487sw In other words, for every mole of gas that flows through the turbine, 8487 joules of work are produced. ch01 (b) Saturated vapor-liquid mixture (d) Superheated vapor The ideal gas law can be rewritten as ch02 Mass balance Mass balance Mass balance Mass balance Energy balance Mass balance A ch03 Since entropy is a state function Step 1 is a constant volume process. Therefore, no work is done. After neglecting potential and kinetic energy effects, the energy balance for a reversible process becomes Equation 3.62 states Equation 3.3 states From the definition of entropy: Problem 2.14 Since the system is well-insulated no heat is transferred with the surroundings. Therefore, the entropy change of the surroundings is zero and Equation 3.3 states From Equation 3.63: From Equation 3.65: Mass balance Mass balance Equation 3.84: The efficiency can be calculated with Equation 3.82 Now consider the non-ideal turbine and compressor. Use the definition of isentropic efficiencies: Equation 3.84 is used to find the mass flow rate Equation 3.90 states: From Equation 3.90 The process between state 4 and state 1 is also isentropic. Assume the temperature is 298 K. The following data was taken from Table A.3.2: ch04 Methyl ethyl ketone and n-butanol are much more polar than diethyl ether due to their greater asymmetry, so their values are greater than diethyl ether’s. Now we must determine if there is greater attraction in n-butanol or methyl ethyl ketone. s... The potential energy due to gravity can be calculatedas follows Equation 4.13 quantifies the potential energy due to London interactions From Table 4.1 where we set this derivative equal to zero to find the minimum. Solving gives 4.18 ch05 The heat capacity at constant pressure can be defined mathematically as follows For an ideal gas: One mathematical definition of is Expansion of the enthalpy term in the numerator results in An isochor on a Mollier diagram can be represented mathematically as For an ideal gas: In Part (a), we found The cyclic rule can be employed to give Separation of variables provides In Part (a), we found Mathematically, the entropy is defined as follows From Equation 5.48: Substitution of this expression into the equation for entropy yields Application of the first law provides We write enthalpy in terms of the independent variables T and v: Therefore, so By double interpolation of data from Tables C.3 and C.4 By double interpolation of data in Tables C.3 and C.4 Using the data in Table C.5 and C.6 The turbine is isentropic. Therefore, we know the following Calculate reduced temperature and pressures using data from Table A.1.1 Calculate reduced temperature and pressures using data from Table A.1.1 The process is isenthalpic, so the following expression holds ch06 ch07 ch08 ch09
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