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Introduction to Chemical Engineering Calculations
Lecture 6.
Real Gas Relationships
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
6
Real Gas Relationships
PVT Relationship for a Real Gas
For a real gas,
PV  nRT
The best method for obtaining the PVT relationship for a
real gas is thru experimentation. In the absence of an
experimental data, the following methods may be used:
1. Compressibility Chart
2. Equations of State
3. Estimated Properties
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Compressibility Factor, Z
The compressibility factor is a factor that compensates for
the nonideality of the gas.
Using this factor, the ideal gas equation becomes a real gas
equation:
PV = nZRT
This is called the generalized equation of state.
The value of Z can be obtained from compressibility charts.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
The Critical and Reduced Properties
Substances have critical properties (critical temperature
critical pressure, and critical volume). Values of critical
properties for common substances are available.
Reduced properties are PVT conditions normalized by their
respective critical conditions:
T
Tr 
TC
P
Pr 
PC
V
Vr 
VC
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-1. Use of the Compressibility Factor
After liquid ammonia (NH3) fertilizer has been injected into
the soil, there is still some ammonia left in the source tank
(V = 120 ft3), but in the form of a gas.
Determine the weight of NH3 remaining in the tank if the
pressure is 292 psig and temperature of the tank is 1250F.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-1. Use of the Compressibility Factor
Solve for n using the generalized equation of state,
PV
n
ZRT
As given, values of the P-V-T are
P = 292 psi + 14.7 psi = 306.7 psi
T = 12560F + 460 = 5850R
V = 120 ft3
And the value of R:
ft3  atm
R  10.73
lbmol 0 R
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-1. Use of the Compressibility Factor
The value of Z is obtained from the generalized
compressibility chart if Tr and Pr are known,
The critical properties for ammonia are:
TC = 405.5K  729.90R
PC = 11.28 Mpa  1636 psia
Solving for the reduced parameters:
T
5850 R
P 306.7 psia
Tr 


0.801
and
P


 0.187
r
0
Tc 729.9 R
Pc 1636 psia
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-1. Use of the Compressibility Factor
From the compressibility chart,
Z  0.855
Solving for n and m:
n
 306.7 psia  120ft 3 
3

ft  atm 
0
585
R
 0.855  10.73

0
lbmol  R 


 6.858lbmol NH 3

 17lbm 
m  6.858lbmol NH 3 
 116.58lbm NH 3

 1lbmol NH 3 
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-2. Use of the Compressibility Factor
Liquid oxygen is used in the steel industry, in the chemical
industry, in hospitals, as rocket fuel, and for wastewater
treatment as well as many other applications.
In a hospital a tank of 0.0284 m3 volume is filled with 3.500
kg of liquid O2 that vaporized at -250C.
Determine whether the pressure in the tank exceed the
safety limit of the tank of 104 kPa.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-2. Use of the Compressibility Factor
Using the generalized equation of state:
nZRT
P
V
Given the following values:
n = 3.500 kg O2 (1/32) = 0.109 kmol
T = - 250C + 273 = 248 K
V = 0.0284 m3
R = 8.314 (kPa)(m3)/(kmol)(K)
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-2. Use of the Compressibility Factor
To determine the value of Z, values of Tr and Pr must be
known.
Critical properties of oxygen:
TC = 154.8 K
PC = 5.08 MPa = 5080 kPa
And the reduced temperature is
Tr 
T
248K

 1.602
Tc 154.8 K
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-2. Use of the Compressibility Factor
The pressure is determined by iteration
Step
P (kPa)
Pr
Z
Pcalc (kPa)
e = |P – Pcalc|
1
10000.00
1.97
0.87
6845.20
3154.80
2
6845.20
1.35
0.91
7201.31
356.11
3
7201.31
1.42
0.88
6983.90
237.41
4
6983.90
1.37
0.89
7043.04
79.14
5
7043.04
1.39
0.90
7122.17
79.14
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Equations of State – Van der Waals Equation
2

n
a
P a  V
ˆ
 b   RT or  P  2   V  nb   nRT

2 
V̂ 
V 


ˆ  specific molar volume (V/n)
where V
27  R 2 TC2

a = 
 64  PC
1  RTC

b  
 8  PC
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Equations of State – Redlich-Kwong Equation
RT
a
P
 1/2
ˆ
 V  b  T Vˆ  Vˆ  b 
R 2 TC2.5
where a = 0.42748
PC
b  0.08664
RTC
PC
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Equations of State – Peng-Robinson Equation
P
RT
a

 Vˆ  b  Vˆ  Vˆ  b   b  Vˆ  b 
R 2 TC2
where a = 0.45724
PC
b  0.07780
  1  

RTC
PC
1  Tr1/2

2
  0.37464  1.5422  0.269922
  accentric factor
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-3. Use of Van der Waals’ Equation
A cylinder 0.150 m3 in volume containing 22.7 kg of
propane (C3H8) stands in the hot sun. A pressure gauge
shows that the pressure is 4790 kPa gauge.
Using the Van der Waals equation of state, determine the
temperature of the propane in the cylinder.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-3. Use of Van der Waals’ Equation
Solving for T using Van der Waals equation

n 2a 
 P  2   V  nb 
V 

T
nR
Given the following values:
n = 22.7 kg (1/44) = 0.516 kmol = 516 gmol
V = 0.150 m3 = 0.150  106 cm3
P = 4790 kPa + 101 kPa = 4891 kPa = 48.3 atm
R = 82.06 (atm)(cm3)/(gmol)(K)
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-3. Use of Van der Waals’ Equation
The values of a and b can be computed using values for the
critical properties or obtained from handbook.
3


cm
6
a  9.24 10 atm 

gmol


2
cm3
and b  90.7
gmol
Solving for T:
2

516  9.24  106 

 48.3 
 0.150  106   516  90.7  


6 2


0.150  10


T
 516  82.06 




T  384 K
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-4. Solution of Van der Waals Equation for V
Given the values in a vessel of
3
P  679.7 psia
n  1.136 lbmol
T  6830 R
 ft 
a  3.49  10 psia 

lbmol


2
4
psia  ft 3
R  10.73
lbmol  0 R
ft 3
b  1.45
lbmol
Solve for the volume of the vessel.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-4. Solution of Van der Waals Equation for V
Since the Van der Waals equation cannot be easily solved
explicitly for volume, it is written in cubic form:
2
3


nRT
n
a
n
ab

3 
2
V   nb 
V
0
V 

P 
P

 P 
(a 3 x 3  a 2 x 2  a1x  a 0  0)
Obtaining the coefficients:
nRT
n 2a
n 3ab
nb 
 13.896 ;
 66.262;
 109.147
P
P
P
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-4. Solution of Van der Waals Equation for V
Rewriting the equation:
f(V) = V3 – 13.896V2 + 66.262V – 109.147 = 0
Solve for the roots using any suitable technique such as the
Newton’s method which is an iterative procedure. Starting
with a guess value for V, a new value is calculated using the
following formula:
f (Vk )
Vk 1  Vk 
f '(Vk )
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-4. Solution of Van der Waals Equation for V
f’(V) is obtained by differentiating f(V):
f’(V) = 3V2 – 27.792V + 66.262
For the initial guess value, the volume is calculate using the
ideal gas law:
nRT 1.136 10.73 683
V0 

 12.26 ft 3
P
679.7
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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Real Gas Relationships
Example 6-4. Solution of Van der Waals Equation for V
Step
VK(ft3)
f(V)
f'(V)
VK+1(ft3)
E
0
12.26
457.32
176.45
9.67
2.59
1
9.67
136.30
77.99
7.92
1.75
2
7.92
40.81
34.34
6.73
1.19
3
6.73
12.26
15.13
5.92
0.81
4
5.92
3.61
6.89
5.40
0.52
5
5.40
0.92
3.66
5.15
0.25
6
5.15
0.13
2.69
5.10
0.05
7
5.10
0.00
2.55
5.10
0.00
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
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