capítulo 4 koresk

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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 for a because the positively charged gas molecules repel each other 
due to coulombic forces. Therefore, the system pressure increases, i.e.,: 
 
 idealA PP >+ 
 
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. 
Therefore, 
 
 
3
1=n 
 
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 and propane, the 
dipole moments are zero. Therefore, the expression for the interactions reduce to 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−=Γ ji
jiji
II
II
r62
3 αα 
 
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 −raa 
 
 [ ]( ) [ ]( ) ( )( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡
×+×
××××−=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 −rab 
 
 [ ]( ) ( )( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡
×+×
×××−=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 −rbb 
 
(b) 
Calculation: 
 
 [ ] [ ]⎟⎟⎠⎞⎜⎜⎝⎛ ⋅×−⎟⎟⎠⎞⎜⎜⎝⎛ ⋅×−=ΓΓ
−−
ergcm 1020.5ergcm 1071.3 66
58
6
6
59
rr
bbaa 
 4
 [ ]ergcm 1039.1 66 58 ⋅×−=ΓΓ −rbbaa 
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 bbaamix 
 
 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. For the mixture, we calculate the 
compressibility factor as follows 
 
 
v
B
z mix+= 1 
where 
 2
2
212211
2
1 2 ByByyByBmix ++= 
 
Since BClH2 is polar and H2 is non-polar and small 
 
 21 BBB mix >>> 
 
Therefore, the plot may look like the following. 
 
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1
yBClH2
z
quadradic, not 
linear
zBClH2 lowest
zH2 close to 1
 
 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 1025 cm3) 
% Difference 
C3H8 63.4 62.9 0.8 
CH3Cl 45.8 45.6 0.3 
CH2Cl2 65.5 64.81.1 
CHCl3 85.3 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 1025 cm3) 
% Difference 
C3H8 63.4 62.9 0.8 
CH3Cl 45.8 45.6 0.3 
CH2Cl2 65.5 64.8 1.1 
CHCl3 85.3 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 
 
 ( ) [ ]m1018.1m 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 is the mass of an argon 
atom. The mass of an argon atom is calculated as follows 
 
 [ ][ ]
[ ]
[ ]
[ ]
[ ] ⎥⎦
⎤⎢⎣
⎡×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×⎟⎟⎠
⎞⎜⎜⎝
⎛=
atom
kg 10633.6
g 1000
kg 1
atoms 10023.6
mol 1
Ar mol 1
Ar g 948.39 26-
23Arm 
 
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. 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: 
 
K 229=
k
ε 
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 
CH4 190.6 46.00 
NH3 405.6 112.77 
H2O 647.3 220.48 
 
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 
CH4 0.230 4.31 0 26 
NH3 0.425 3.74 1.47 22.2 
H2O 0.554 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 
CH4 190.6 46.00 4.306 
C6H6 562.1 48.94 11.94 
CH3OH 512.6 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
πσ = 
 
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 
CH4 3.24 3.8 14.7 
C6H6 4.56 5.27 13.5 
CH3OH 3.74 3.6 3.8 
 
 
 244.17 
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 
CH4 190.6 46.00 0.2303 
C6H6 562.1 48.94 1.883 
CH3OH 512.6 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 
CH4 3.8 1.66 
C6H6 5.27 36.3 
CH3OH 3.6 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
 [ ]6776 mJ 10 ⋅×C Species 
Page 187 Equation 4.13 
Percent Difference 
CH4 1.66 1.021 63 
C6H6 36.3 12.0 202 
CH3OH 5.82 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 = RTv − b −
a
T1/2v(v + b) 
 
Peng-Robinson: P = RTv − 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, where V is the molecular 
velocity. If we relate molecular velocity to temperature 
 
 12 mV
2 = 32 kT 
 
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 
 
 ∫−= dVVnRTW 
 [ ]( ) [ ]( ) [ ][ ]⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅−= 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 ++++=− xxxx 
 
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 critical 
point data (Problem 4.22). We find 
 
 bvc 3= (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(1st value) 
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. 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: 
 
 
⎥⎦
⎤⎢⎣
⎡ ++= ...1 22v
B
v
RTP OH
 
 
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) and add to Equation 2 to get: 
 
 0
22
43
2 =−−
cc v
B
v
OH 
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 TabvRTPsat α 
( )
( ) 19.1100)(
00459.0
000551.0
=⎥⎦
⎤⎢⎣
⎡ −−++−−∫ dvPbvbbvv TabvRT 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 
 ( ) ( ) ( )( ) 22
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
MWρ 
 
( )
( ) ⎥⎦
⎤⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡×
⎥⎦
⎤⎢⎣
⎡
==
− 333
3
3
3 cm
g 0229.0
cm 100
m 1
mol
m 10578.1
mol
g 0694.36
vap
ethanevap
v
MWρ 
 
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 1097.1
3
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) 
 25.0
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
cr
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
Cº0100Cº 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 = 20ºC
T = 100 ºC
Linear (T = 100 ºC)
Linear (T = 20ºC)
 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 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: 
 
 
( )
[ ] ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅=×
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅= 2
3
5
2
2
mol
mJ 40.0
Pa 1073.76
K 2.304
Kmol
J 314.8
45724.0a 
 
( )
[ ] ⎥⎥⎦
⎤
⎢⎢⎣
⎡×=×
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅= −
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. 
 
v
P
vapor
liquid
liquid - vapor
critical�
point
part c
part d 
not enough 
CO2
 
 
 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: 
 
Following the 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 (1): 
[ ]
[ ]bar 76.73
K 2.304
=
=
c
c
P
T
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅⋅==
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 
 
 [ ][ ] ⎥⎥⎦⎤⎢⎢⎣⎡=== molm 002.0mol 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
3
mixB 
 
Therefore, 
 
 mixmix BP
RTP
RT
B
P
RTv +=⎟⎠
⎞⎜⎝
⎛ += 1 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
mol
m 002446.0
mol
cm 6.2445
33
v 
 
(b) 
We can estimate k12 using an EOS where the a parameter is the only unknown. b can be 
calculated as follows 
 
 2211 bybybmix += 
 
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 as follows 
 
 2
2
212211
2
1 2 ayayyayamix ++= 
 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,
o
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