Applications of Intermolecular Potentials

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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 isused. This mixture is prepared by starting with an evacuated steel cylinder andadding 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

jjiiij

jiij

ijijijij rr

ru

;2

1

4)(612

What is the exact composition of thegas in the cylinder?

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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 gasmixture containing exactly 0.7 mole fraction of methane and 0.3 mole fractionof 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-wellpotential and Lorentz-Berthelot combining rules

The following procedures will be considered for making the mixture of the desiredcomposition 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?

ijijijijij r if 0 ; if - ;r if )( SWijSWijij RRrru

molecule s(Å) e/k (K) RSW

CH4 3.400 88.8 1.85

CF4 4.103 191.1 1.48

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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

2,2 TB

RT

Pmix

TBxxTBxTBxTB CFCHCFCH

CFCF

CHCHmix

4,4244

42

24

42

24,2 2

The second virial coefficient for a square well potential is,

1113

2 3/2 i

kTeii ieTB

TB CFCH 4,42 combining rules for the square-well parameters:

244 CFCH

mix

244 CFCH

mix

44 CFCHmix

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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 r = N, the initial pressure with only CH4 added can be solved using eqn. :

22 TB

RT

P i

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 propertiesand the LJ parameters are given by:

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

142.0P

;35.0N

;35.13

c3

av

c

c

V

kT

Critical Properties L-J parametersSpecies Tc Pc σ ε/k

(K) (bar) (A) (K)CH4 190.4 45.96 3.918 141.04CF4 227.6 37.4 4.454 168.59Ar 150.7 48.98 3.548 111.63

CO2 304.1 73.825 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 Ar CO2

200 -169.055 -324.061 -76.78 -390.538300 -66.918 -144.13 -21.396 -189.512400 -21.563 -65.414 3.477 -104.884500 3.646 -21.839 17.302 -58.717600 19.47 5.524 25.937 -29.878700 30.191 24.117 31.742 -10.289800 37.848 37.454 35.847 3.797

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Example 4. The triangular well potential is:

TWTWTW

TW Rr if 0 ;R ifR

R- ;r if )( r

rru

a. Obtain an expression for the 2nd virial coefficient for this potentialb. Does the 2nd virial coefficient for the triangular well potential have a maximumas 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.