Physical Chemistry I
Recommended Problems
Problem 7.3
P  V m − b T   = RT
=
 
1 ∂V
V ∂T
P
=
⇒
VM =

1 R db

V m P dT
RT
bT 
P

R
1
=
.
PV m T
R
1 P V m − bT   1
b T 
=
=
1−
PV m T P V m
T
Vm
For an ideal gas,  =
Here,




=
bT 
R
1 db
1
db /dT

=
1−

P V m V m dT
T
Vm
Vm
=
bT 
1
1 db
1−

T
Vm
V m dT
=
b T 
1
1 db
−

T
TV m
V m dT

 
 
1 ∂V m
 =−
V m ∂P
T
=
1 RT
V m P2
For an ideal gas,  =
Here,

1
.
P


PV − b  1
RT
b
1
b
= 2 m
=
1−
= −
2
P
Vm
P
PV m
P Vm
P Vm
b T 
1
= −
P
PVm
Chapter 7
Physical Chemistry I
Problem 7.7
Recommended Problems
Z = 1 − 9.00×10−3
P
 4.00×10−5
Chapter 7, continued
2
 
P
. Calculate the fugacity coefficient, γ, at
Po
Po
P= 150, 250, 350, 450 and 550 bar. Note that when P is in units of bar, Po=1.
P
Z−1
Recall that ln  = ∫
dP . Applying that equation,
P
0
P
2
ln  = ∫  −9.00×10 4.00×10 P  dP = −9.00×10 P2.00×10 P
−3
−5
−3
−5
0
P (bar) 150
ln(γ)
250
350
450 550
-0.90 -1.00 -0.70 0.00 1.10
γ
0.41 0.37 0.50 1.00 3.00
The fugacity coefficient is greater than one for pressures greater than 450 bar.
Problem 7.11 Derive the virial coefficient for a van der Waals gas.
B T  C T 
1
As Engel and Reid state in equation 7.4, P = RT V  2  3  ...
V
V
m
[
m
m
]
.
Multiplying through by Vm/(RT) gives the virial equation for Z, where Z=PVm/(RT):
B T  C T 
Z = 1
 2  ... .
Vm
V
m


∂Z
From that equation, one sees that BT  = lim
.The limit is taken to remove higher-order
V ∞ ∂1 /V m  T
terms, such as the term involving C(T).
Vm
RT
a
a
−
For the van der Waals equation, P = V −b − 2 , so Z =
. To make
V m −b
RT V m
Vm
m
differentiation simpler, one may rewrite Z as an explicit function of (1/Vm).
m
Z=
1
1 −b
 
1
Vm
−1
  [  ]
a 1
1
−
= 1 −b
RT V m
Vm
−
 
a 1
RT V m
One may directly take the derivative with respect to (1/Vm), holding T constant.
−2
∂Z
1
= − 1−b
−b  − a
∂1 /V m  T
Vm
RT

 [  ]
In the limit as V m ∞ , the first term on the right-hand-side goes to b. Hence,
∂Z
a
BT  = lim
=b −
, as required.
∂1/V

RT
V ∞
m T
m


Physical Chemistry I
Recommended Problems
Chapter 7, continued
Problem 7.19 Calculate the critical volume for ammonia using the data for Tc and Pc in Table 7.2
assuming (a) ideal gas and (b) the van der Waals equation, iterating from the ideal-gas volume.
Table 7.2 shows Tc=405.40 K, Pc=113.53 bar, and Vc=0.07247 L/mol.
(a) If ammonia were ideal at the critical temperature and pressure, then its Vc=RTc/PC = 0.297 L/mol.
RT
Vm =
b
a
P
V
−b
a


(b) The van der Waals equation,
can be rearranged to
.
m
2
P 2
Vm
Vm
That equation can be solved iteratively. Start by substituting the ideal-gas Vm into the right-hand-side,
and thus calculating Vm. Take that result, substitute it into the right-hand-side, and so on.

iteration
0
1
2
3
4
5
6
7
8
9
10
20
100
500
1000
2000
5000
Vm
0.2969
0.2459
0.2209
0.2055
0.1949
0.1871
0.1810
0.1761
0.1721
0.1686
0.1657
0.1493
0.1255
0.1027
0.0983
0.0983
0.0983
b

RT
Pa /V 2m
0.2459
0.2209
0.2055
0.1949
0.1871
0.1810
0.1761
0.1721
0.1686
0.1657
0.1632
0.1483
0.1254
0.1026
0.0983
0.0983
0.0983
The van der Waals critical volume is 0.0983 L/mol.
The error is (0.0983-0.07247)/(0.07247) = 0.36 = 36%.
Physical Chemistry I
Recommended Problems
Chapter 7, continued
Problem 7.24 35.0 grams of Argon. V=0.0950L. T=450. Kelvin.
moles Argon = 35.0 / 39.95 = 0.876 mol. Vm = 0.0950L / 0.876 mol = 0.1084 L/mol. Calculate P.
(a) ideal P = RT/Vm = 0.083145 L·bar/(mol·K) × 450K / (0.1084 L/mol) = 345 bar.
(b) van der Waals
0.083145 L⋅bar /mol⋅K ×450K 1.355 L 2 bar⋅mol−2
RT
a
P=
− 2 =
−
2
V m−b V m 0.1084 L/mol−0.0320 L/mol
 0.1084 L /mol 
P = 489.73 − 115.31 = 374 bar
(c) Redlich-Kwong
RT
a
P=
−
V m−b  T V m V mb
0.083145 L⋅bar /mol⋅K ×450K
16.86 K 1 /2 L2 bar⋅mol −2
P=
−
0.1084 L/mol−0.02219 L/mol
 450 K 0.1084 L/mol  0.10840.02219  L /mol
P = 434.0 − 56.1 = 378 bar
The non-ideal pressures are larger than the ideal pressure. That indicates that repulsions dominate.