Vander Waal’s Equation and The Viral Equation:

The constants of Van der Waal’s equation can be related to the coefficients of the virialequation.

Z = PV / RT = 1 + BPP + …….. ---------- (1)

PV = RT

P = RT / V

Z = 1 + (BP RT ) 1 / V + …….. ------------- (2)

Z = 1 + (BV ) 1 /V + ………. ------------- (3)

Where

BV = BP RT or BP = BV / RT ---------- (4)

The Van der Waal’s Equation is

------------- (5)

Multiply this equation by V / (RT)

We get,

In most cases b is very much less than V.

The term 1/ ( 1 –b/V)

Can be expanded using the binomial expansion.

RT a P = - (V – b) V2

PV V a = - RT (V – b) RTV

PV 1 a = - RT 1 – b/V RTV

(1- X ) -1 = 1 + X + X2 + …..

The above equation becomes

Compare this with virial equation (3), we get the second virial coefficient BV in terms ofVan der Waal’s constants a and b.

BV = b – a / (RT) ----------- (6)

The Boyle Temperature:

The Boyle temperature is the temperature at which BP = 0.

BV = RTBP.

Thus,

BV = 0

From equation (6)

0= b – a / (RTBoyle)

TBoyle = a /bR

If we substitute the value for a and b in terms of the critical pressure and temperature.

i.e. a = 27 R2Tc2 / ( 64 Pc) and b = RTc / (8Pc)

We get

TBoyle = 27 Tc / 8 or

TBoyle / Tc = 3.375 ----------- (7)

PV b b2 a = 1 + + - +………… RT V V2 RTV

PV a 1 b2

= 1 + b - + +………… RT RT V V2

The Van der Waal’s Equation is semiempirical. The form of the equation is based ontheoretical treatment, while the constants used in the equation are obtained from theexperimental PVT data.

Molecular Interactions:

Energy of the pairwise Molecular Interaction

Let us consider the two atoms are at some “comfortable” intermediate distance.

In order to change the distance between the atoms, we need to supply the energy in orderto over come forces of attraction as well as repulsion.

Case I

As we increase the force to pull the atoms further apart, it has to overcome the forces ofattractions, hence potential energy increases as the distance between the atoms. But atvery large separations the force of attraction between the atoms is negligible, hence thereis no further increase in the potential energy.

Case II

Now as we increase the force to push atoms more closely, it has to overcome the forcesof repulsion, hence potential energy increases. Also, as the atom comes close to eachother, they hit each other, giving a steep increase in the potential energy.

Please Refer to Figure 1-17

J. E. Lennard-Jones gave the generalized equation for the variation of potential energy ofany pair of atoms with intermolecular distance.

This is known as Lennard-Jones, or LJ, potential and is given as follows:

U (r) = 4εL.J [ (σL,J / r )12 – (σL,J / r )

6 ] -------- (8)

Where εL.J and σL,J are adjustable parameter and are characteristics of each interactingmolecules.

The 12th power term increases as the molecular distance decreases. This function showsthe repulsion between the molecules.

The 6th power term has negative sign, thus decreases as the molecular distance decreases.This function represents the attraction between the molecules.

Thus, the potential-energy function, made up of the repulsive and attractive components.

1. At the infinite separation, potential energy is zero.2. The minimum of the net potential-energy curve is at an energy of -εL.J.

3. At an intermolecular separation of σL,J the potential energy is again zero andrising very steeply.

Thus, the values of εL.J shows the attractive forces between the molecules and the valuesof σL,J shows the size of the molecules.

Pairwise Molecular Interactions and the Second Virial Coefficient:

The effect of the pairwise interaction on the second virial coefficient is given by integral

∞

BV = 2 π NA ∫ ( 1 – e – U (r) / (kT) ) r2 dr ---------- (9) 0

First consider the integral from r = 0 to r = σL,J.

Since the potential is very large, the exponential term is, negligibly small compared to 1.

Thus,r = σL,J

BV = 2 π NA ∫ ( 1 – e – U (r) / (kT) ) r2 dr r = 0

r = σL,J

BV = 2 π NA ∫ ( 1) r2 dr since e – U (r) / (kT) is negligible 0

r = σL,J

BV = 2 π NA ∫ r2 dr r = 0

BV = ( 2 π NA σ3L,J ) / 3 ------------------------ (10)

The remaining term depends upon the integration over a region of relatively small U(r)values for beyond σL,J.

We assume this value to be kT and in this region U(r) is negative has magnitude less thankT.

This is justified by the value of kT = 4 × 10-21 J for various experimental values fordifferent gases at room temperature.

Thus we can write

1 – e – U (r) / (kT) = U(r) / kT for r > σL,J

Thus, the equation (9) becomes r = ∞

BV = ( 2 π NA σ3L,J )/ 3 + 2 π NA ∫ r2 U(r) dr / kT

r = σL,J

Since the U(r) is negative, the integration values over this range is also negative.

Comparing this with

BV = b – a / (RT)

Gives

b = ( 2 π NA σ3L,J ) / 3

Thus, the b depend upon the σL,J i.e. molecular diameter.