Solution Koretsky ch03

Prévia do material em texto

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 
 ( ) ( ) ⎟⎟⎠⎞⎜⎜⎝⎛−−+−+⎟⎟⎠⎞⎜⎜⎝⎛=∆ 1221221212 ln2ln PPRTTCTTBTTAssys 
 
 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. H2O(l) at 1 bar) 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 6940.1ˆ
3
,
2
satvv (sat. H2O(v) at 1 bar) 
 
Since satvsatl vvv ,22
,
2 ˆˆˆ << , 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. H2O(l) at 1 bar) 
 ⎥⎦
⎤⎢⎣
⎡
⋅= Kkg
kJ 3593.7ˆ ,2
satvs (sat. H2O(v) at 1 bar) 
 
The entropy of state 2 can be calculated using the following equation: 
 
 ( ) satvsatl sxsxs ,2,22 ˆˆ1ˆ +−= 
 ⎥⎦
⎤⎢⎣
⎡
⋅= 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. H2O(v) at 1 bar) 
 
The expression for internal energy yields 
 
 [ ]kJ 77.20583−=∆U 
 
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 280lnKmolJ 314.8723.2ln723.2723.2 12
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 
 
 
( )
⎥⎦
⎤⎢⎣
⎡−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−
=∆−==∆ ∫ KJ 9.73K111 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 15.273lnKg J2.4g 0148.18
K 15.273
K 373.15
f
i
T
T
Prev
sys dTT
c
T
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 07923.0kgm 02641.0kg) 3(ˆ =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡== vmV 
 
Now 2V and 2v are found as follows 
 
 [ ]36312 m 4536.0Pa 100.2 J 748740m 07923.0 =×−−=−= EPWVV 
 [ ][ ] ⎥⎥⎦⎤⎢⎢⎣⎡=== kgm 1512.0kg 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 05.0
mol
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 ,º100ˆatm 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, 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: 
 
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 find the intermediate pressure, PI where we end up at the temperature in state 2, T2. 
To find it, we can use the results from pages 78-79 of the text: 
 
 ( ) ( ) ( ) kkIkkIIkk PTPTPT /12/1/111 −−− == 
 
Therefore, 
 
 
( )
[bar] 9.351
1
2
1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛= − PT
TP
k
k
I (I) 
 
If we draw the process on a PT diagram, we get: 
 
20
40
490 560
1
2
T
P
I
Adiabatic compression Isothermal 
expansion
 
 
 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, 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅=⎟⎟⎠
⎞
⎜⎜⎝
⎛−=−==∆ ∫∫ KmolJ 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: 
 
20
40
490 560
1
2
T
P
Isobaric heating
Isothermal 
compression
 
 
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
20
40
490 560
1
2
T
P
Isochoric heating
Isothermal 
compression
I
 
 
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 from Appendix A 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛−×−×+==∆ ∫ −−
1
2
263
ln10227.010506.0639.30
2
1
P
PRdT
T
TTRs
T
T
 
 
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
 ( ) ⎥⎦⎤⎢⎣⎡=×−×+= ∫ −− molJ 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 is insulated, 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 
 ( ) ( ) ( )⎥⎦
⎤⎢⎣
⎡ +⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=∆ 2ln1ln2
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 systemare 
 
 bar79.0
2,2
=Np 
 bar21.0
2,2
=Op 
 
Use Equation 3.62 to determine the entropies: 
 
 ⎟⎠
⎞⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−=∆ ∫ bar 1 bar79.0lnKmolJ314.8ln
2
2
1
2
K 298
K 298
,
N
P
Nsys P
PRdT
T
cs 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J96.1
2,Nsyss 
 
 ⎟⎠
⎞⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−=∆ ∫ bar 1 bar21.0lnKmolJ314.8ln
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: 
 
2 bar, 298 K
Mixing 
Process
N2
N2
N2
N2
N2
N2 N2
N2
O2
N2
N2
N2
N2
N2
N2 N2
N2
O2
N2
N2
N2
N2
N2
N2
N2
N2
O2
N2
N2
N2
N2
N2
N2
N2
N2
O2
1 bar, 298 K P2, T2
 
 
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 bar30.1lnKmolJ314.8ln
2
2
1
2
K 298
K 298
,
N
P
Nsys P
PRdT
T
cs 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 58.3
2,Nsyss 
 
 ⎟⎠
⎞⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−=∆ ∫ bar 1 bar347.0lnKmolJ314.8ln
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 entropy change 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 −+= rr 
 
Substituting the following values 
 
 42
 ⎥⎦
⎤⎢⎣
⎡=
s
m 201V
r
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
J 2719100ˆ2h (0.1 MPa, 121.4 ºC) 
 ⎥⎦
⎤⎢⎣
⎡=
kg
J 37670001ˆh (4 MPa, 640 ºC) 
 
yields 
 
 ⎥⎦
⎤⎢⎣
⎡=
s
m 14482V
r
 
 
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: 
 
Propane in
v1 = 600 cm3/mol 
T1= 350 oC
Turbine
P2 = 1 atm
ws
Propane
out 
 
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 from Appendix A 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛−×−×+==∆ ∫ −−
1
2
263
ln10824.810785.28213.10
2
1
P
PRdT
T
TTRs
T
T
 
 
We can also relate P1 to T1 and v1 (which are known) through the ideal gas law: 
 
 P1 = RT1v1 
 
This leaves us with 1 equation and 1 unknown (T2). Integrating: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛−−×−−×+⎟⎟⎠
⎞
⎜⎜⎝
⎛==∆ −−
1
122
1
2
2
6
12
3
1
2 ln10412.4)(10785.28ln213.10RT
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 TTTTTTRnWS −×−−×+−= −−&& 
 
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 molJ 51.1ln105.15.341.31 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 followingequation 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. H2O at 200 ºC) 
 ⎥⎦
⎤⎢⎣
⎡==
kg
kJ 9.2638ˆˆ ,2 if uu (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. To find the final 
temperature, first find the volumes using the ideal gas law. 
 
 
[ ]( ) ( )
[ ]( ) [ ]( )[ ]
[ ]3
2
2-,1
m 05.0
m 098.0
sm 8.9kg 1000
K300
Kmol
J 314.8mol 2
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=aV 
 
 
[ ]( )
[ ]( ) [ ]( )[ ]
[ ]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 30023mol 2K 30023mol 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 1006.12525 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 1006.12510284.8 1006.125 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: 
 
Steam 
in
P1 = 60 bar 
T1 = 500 oC
Turbine
P2 = 10 bar
Steam 
out
Qsurr=69.86 kW
Tsurr=300 K
W
 
 
(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) From the steam tables: 
 
 T2 = 200 oC 
 
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: 
 
 
( )( )reversibleWWS 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×−=D0=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 reversible
S
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 write the 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: 
 
P1 = 60 bar
m = 5 kg
Pure H2O
T1 = 540 oC
P2 = 20 bar
m = 5 kg
Pure H2O
T2 = ? oC
Tsurr = 25 oC 
Psurr = 1 bar
well-insulated
reversible
 
 
(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 from the steam tables. 
 
 Cº 5.3622 =T 
 
(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 709.0kgm 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 
P1 = 60 bar
m = 5 kg
Pure H2O
T1 = 540 oC
P2 = 20 bar
m = 5 kg
Pure H2O
Tsurr = 25 oC 
Psurr = 1 bar
irreversible
latches
T'2 = ? oC
 
P1 = 60 bar
m = 5 kg
Pure H2O
T1 = 540 oC
P2 = 20 bar
m = 5 kg
Pure H2O
T2 = ? oC
Tsurr = 25 oC 
Psurr = 1 bar
well-insulated
reversible
 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 sin ininoutoutsys WQhmhmdt
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 constant 
 
 ( )in
t
ininuniv ssmdtmssmS ˆˆˆˆ 22
0
22 −=−=∆ ∫ & 
 
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: 
 
 ( ) satvsatl 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: 
 
 ( ) satvsatl 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 processes can be fount by adding together the entropy change from 
Part (a) and Part (b): 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡=∆
K
kJ1.55(b) and (a)univS 
 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 constant 
 
( )
surr
outuniv T
QsmmsmsmS −−−−=∆ 121122 (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 outoutoutoutum
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 T2T1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ − Rln
P2
P1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
Since 
 
cP = 52 R 
 
12
2
1
2
2/5
1
2
PV
nRT
P
P
T
T ==⎟⎟⎠
⎞
⎜⎜⎝
⎛
 
 
Solving for T2 gives 
 
( )
K 215
3/2
12
2/5
1
2 =⎥⎥⎦
⎤
⎢⎢⎣
⎡=
PV
TnRT 
 
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. H2O at 0.1 bar) 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 5.34561ˆh (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. H2O(l) at 0.1 bar) 
 
Therefore, 
 
 ( ) [ ]MW 09.210ˆˆ 23 −=−= hhmQC && 
 
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. 
 
s
1
2
3
4
wnet
T
s
1
2new3new
4new
wnet
T
lower the condenser pressure
s
1new
2new3
4
wnet
T
increase the superheating of steam in the boiler
s
1
23
4
wnet
T
increase the boiler pressure
Turbine 1
Turbine 2
5
6
2 stage turbine with reheat
 
 
 
 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. 
 
turbine 
1
cooling 
water
Fuel 
Air
Boiler
compressor
condenser
1
4
5
6
Tur- 
bine 
2
2
3
Rankine cycle with reheat
Ws Ws
Wc
QH
QC
 
 
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. H2O at 1 bar) 
 
The quality of the water can be calculated as follows 
 
 vl sxsxs 221 ˆˆ)1(ˆ +−= 
 
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 actual
s
turbine W
W
&
&
=η 
 ( ) ( ) ( )( )reversibleturbinereversibleSturbineactualS hhmWW 12 ˆˆ −==∴ &&& ηη 
 
 84
 
( )( )actualCreversible
C
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 ηη
34
12
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. State 3 is constrained because 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅== Kmol
J95.17523 ss (sat. R134a vapor at 0.35 MPa) 
 MPa 7.03 =P 
 
From the NIST website: 
 
 ⎥⎦
⎤⎢⎣
⎡=
mol
J424283h (R134a vapor at 0.7 MPa with ⎥⎦
⎤⎢⎣
⎡
⋅= KmolJ95.175s ) 
 
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. 
 
evaporator
compressor
condenser
1
4
3
2
Ý 
T
s
3
21
4
valve
refrigerated 
unit at -5 C
20 C 
reservoir
Wc
QH
QH
QC
QC
 
 
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 (Tsat = -10 ºC) 
MPa 665.043 == PP (Tsat = 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. 
 
 ( )12 hhnQC −= && 
 
We can get h2 from the thermodynamic tables for saturated R134a. In order to find h1, we can 
use the following relationship: 
 
 41 hh = 
 
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 
 
 
evaporator
compressor
condenser
1
63
2
T
s
2
1
4
valve
high T 
reservoir
valve
Fridge
Freezer evaporator
5
4
T = -15 oC
T = 5 oC
3
5
6
5 oC
-15 oC
QC,R
QC,F
QH
QH
QC,F
QC,R
Wc
Wc
 
 
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 state what 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=Tsat = 5 ºC) 
 MPa 16394.043 == PP (T3=T4=Tsat = -15 ºC) 
 MPa 66538.065 == PP (T6 = 25 ºC) 
 
Before solving for additional pressures and temperatures, we will list the known temperatures 
and pressures. 
 
State 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: 
 
 ( ) satvsatl hxhxh ,11,111 1 +−= 
 ( ) satvsatl hxhxh ,22,222 1 +−= 
 ( ) satvsatl 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) ºfg∆ (kJ/mol) 
Cu2O (s) -170.71 -147.88 
O2 (g) 0 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) 
Cu2O (s) -0.0766 
O2 (g) 0 
CuO (s) -0.0932 
 
Now the change in entropy of the reaction can be calculated in a method analogous to Equation 
2.72 
 
 ( ) ⎟⎟⎠⎞⎜⎜⎝⎛ ⎥⎦⎤⎢⎣⎡ ⋅+⎟⎟⎠⎞⎜⎜⎝⎛ ⎥⎦⎤⎢⎣⎡ ⋅−=∆=∆ ∑ KmolkJ 0.0932-4KmolkJ 0.0766-2ºº ifirxnsvs 
 ⎥⎦
⎤⎢⎣
⎡−=∆
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 hand there 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.”