Applications of
Intermolecular Potentials
Example 1.
A gas chromatograph is to be used to analyze CH4-CO2 mixtures. To calibrate
the response of the GC, a carefully prepared mixture of known composition is
used. This mixture is prepared by starting with an evacuated steel cylinder and
adding CO2 until the pressure is exactly 2.5 atm at 25oC. Then CH4 is added until
the pressure reaches exactly 5 atm at 25 oC. Assuming that both CO2 and CH4
are represented by the LJ 12-6 potential with the parameters given below
molecule
s (Å)
e/k (K)
CH4
4.010
142.87
CO2
4.416
192.25
 s ij 12  s ij
uij (r )  4e ij    
 r 
 r
1
s ij  s i  s j ;
2
e ij  e iie jj



6



What is the exact composition of the
gas in the cylinder?
2
According to the Lennard-Jones (12-6) potential for non-polar molecules,
the second virial Coefficient is
First, based on these two equations, we can calculate the second virial coefficient of the pure CO2 species.
And then we can calculate the number of molecules of CO2, N1=102.19 mol (we assume the volume is 1 m3).
Using the combining rule as shown to obtain the B12
Then we can use Matlab to solve the
equation of x1 so as to find the exact
composition of the gas in the cylinder.
Then we get x(CO2)=x1=0.5
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Example 2. For calibration of a gas chromatograph we need to prepare a gas
mixture containing exactly 0.7 mole fraction of methane and 0.3 mole fraction
of CF4 at 300 K and 25 bar in a steel cylinder that is initially completely evacuated.
Assume this mixture can be described by the virial EOS up to the 2nd virial
coefficient, and the molecular interactions are described by the square-well
potential and Lorentz-Berthelot combining rules
u (rij )   if r  s ij ; - e ij if s ij  r  RSWijs ij ; 0 if r  RSWijs ij
molecule
s(Å)
e/k (K)
RSW
CH4
3.400
88.8
1.85
CF4
4.103
191.1
1.48
The following procedures will be considered for making the mixture of the desired
composition at the specified conditions:
1. CH4 will be added isothermally to the initially evacuated cylinder until a
pressure P1 is obtained. Then CF4 will be added isothermally until 25 bar are
obtained at 300K. What should P1 be to obtain exactly the desired composition?
2. CF4 will be added isothermally to the initially evacuated cylinder until a
pressure P2 is obtained. Then CH4 will be added isothermally until 25 bar are
obtained at 300K. What should P2 be to obtain exactly the desired composition?
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The first step is to calculate the number of moles with a unit volume of
the mixture at 25 bar.
For the virial equation truncated to the 2nd virial coefficient. we know
that
P
   B 2, mix T  2
RT
B2, mix T    x CH 4  B2CH 4 T    x CF 4  B2CF 4 T   2 x CH 4 x CF 4 B2CH 4,CF 4 T 
2
2
The second virial coefficient for a square well potential is,
 


2s i
3
B T  
1  1  e ei / kT i  1
3
i
2
B2CH 4,CF 4 T 
s mix 
combining rules for the square-well parameters:
s CH 4  s CF 4
2
 mix
CH 4  CF 4

2
e mix  e CH 4 e CF 4
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Using SI units and assuming a vessel volume of 1 m3 the final number of
moles is obtained by using Eqn. (1)  Nf = 1065 moles for the final
mixture
(a) The amount of CH4 initially added would be 0.7Nf
Since the vessel retains the same volume  = N, the initial pressure with
only CH4 added can be solved using eqn. :
P
   B2i T  2
RT
Pi= 18.0 bar
(b) Similarly for CF4
Pi= 7.7 bar
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Example 3. The best estimates of the relations between the critical properties
and the LJ parameters are given by:
N avs 3
Pcs 3
 1.35;
 0.35;
 0.142
e
Vc
e
kTc
use these expressions to obtain the LJ parameters of CH4, CF3, Ar, and CO2
and then compute the 2nd virial coefficients for these gases over the
temperature range from 200 to 800 K
Critical Properties
Species
Tc
Pc
(K)
(bar)
CH4
190.4
45.96
CF4
227.6
37.4
Ar
150.7
48.98
CO2
304.1
73.825
L-J parameters
σ
ε/k
(A)
(K)
3.918
141.04
4.454
168.59
3.548
111.63
3.91
225.26
The second virial coefficient as a function of temperature can then be
estimated by the L-J potential where y = (r/σ)3.
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Carrying out this integration at various temperatures along the range T= 200
to 800K
Second Virial Coefficients (A^3)
T(K)
CH4
CF4
200 -169.055 -324.061
300
-66.918
-144.13
400
-21.563
-65.414
500
3.646
-21.839
600
19.47
5.524
700
30.191
24.117
800
37.848
37.454
Ar
-76.78
-21.396
3.477
17.302
25.937
31.742
35.847
CO2
-390.538
-189.512
-104.884
-58.717
-29.878
-10.289
3.797
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Example 4. The triangular well potential is:


R s r
u (r )   if r  s ; - e TW
if s  r  R TWs ; 0 if r  R TWs 
R TWs  s


a. Obtain an expression for the 2nd virial coefficient for this potential
b. Does the 2nd virial coefficient for the triangular well potential have a maximum
as a function of temperature?
Region 1 (0 < r < σ):
Region 2 (σ < r < Rσ):
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Region 3 (r > Rσ):
For the 2nd region:
Integrating by parts:
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Then, the complete expression for B2(T) is:
b)
To determine if there is a temperature at which B2 is at a
maximum, take the derivative of this expression with respect to
temperature and set the result equal to zero. Note that regions 1
and 3 have no temperature dependence, so there is no need to
search them for a maximum.
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If values for R, α and β were available, a more complete analysis could be
done.
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