Applications of Intermolecular Potentials. Example 1. A gas chromatograph is to be used to analyze CH 4 -CO 2 mixtures. To calibrate the response of the.

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Presentation transcript:

Applications of Intermolecular Potentials

Example 1. A gas chromatograph is to be used to analyze CH 4 -CO 2 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 CO 2 until the pressure is exactly 2.5 atm at 25 o C. Then CH 4 is added until the pressure reaches exactly 5 atm at 25 o C. Assuming that both CO 2 and CH 4 are represented by the LJ 12-6 potential with the parameters given below molecule  (Å)  /k (K) CH CO 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 CO 2, N1= mol (we assume the volume is 1 m 3 ). 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(CO 2 )=x1=0.5 3

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 CF 4 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 2 nd virial coefficient, and the molecular interactions are described by the square-well potential and Lorentz-Berthelot combining rules The following procedures will be considered for making the mixture of the desired composition at the specified conditions: 1.CH 4 will be added isothermally to the initially evacuated cylinder until a pressure P 1 is obtained. Then CF 4 will be added isothermally until 25 bar are obtained at 300K. What should P 1 be to obtain exactly the desired composition? 2. CF 4 will be added isothermally to the initially evacuated cylinder until a pressure P 2 is obtained. Then CH 4 will be added isothermally until 25 bar are obtained at 300K. What should P 2 be to obtain exactly the desired composition? molecule  (Å)  /k (K) R SW CH CF

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 2 nd virial coefficient. we know that The second virial coefficient for a square well potential is, 5 combining rules for the square-well parameters:

6 Using SI units and assuming a vessel volume of 1 m 3 the final number of moles is obtained by using Eqn. (1)  N f = 1065 moles for the final mixture ( a) The amount of CH 4 initially added would be 0.7N f Since the vessel retains the same volume  = N, the initial pressure with only CH 4 added can be solved using eqn. : P i = 18.0 bar ( b) Similarly for CF 4 P i = 7.7 bar

7 Example 3. The best estimates of the relations between the critical properties and the LJ parameters are given by: use these expressions to obtain the LJ parameters of CH 4, CF 3, Ar, and CO 2 and then compute the 2 nd virial coefficients for these gases over the temperature range from 200 to 800 K The second virial coefficient as a function of temperature can then be estimated by the L-J potential where y = (r/ σ ) 3.

8 Carrying out this integration at various temperatures along the range T= 200 to 800K

9 Example 4. The triangular well potential is: a.Obtain an expression for the 2 nd virial coefficient for this potential b.Does the 2 nd 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):

10 Region 3 (r > R σ ): For the 2 nd region: Integrating by parts:

11 Then, the complete expression for B 2 (T) is: b)To determine if there is a temperature at which B 2 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.

12 If values for R, α and β were available, a more complete analysis could be done.