Physical Chemistry Third Edition

27.6 The Classical Statistical Mechanics of Dense Gases and Liquids 1149

EXAMPLE27.6

Obtain a formula for the second virial coefficient of a hard-sphere gas with molecular

diameterd.

Solution

The pair potential function is

u(r)

{

∞ ifr<d

0ifr>d

so that

[e−u(r)/kBT−1]

{

−1ifr<d

0ifr>d

The virial coefficient of the hard-sphere gas is equal to a constant

B 2 −

NAV

2

∫d

0

[−1]4πr^2 dr

2 πNAvd^3

3

(27.6-11)

Exercise 27.10

The effective hard-sphere diameter of argon at 273 K is equal to 3.65× 10 −^10 m. Using this

value fordin the formula of Eq. (27.6-11), calculateB 2 for argon at this temperature. Compare

your result with the experimental value at 273 K,− 2. 15 × 10 −^5 m^3 mol−^1. How do you explain

the difference?

Exercise 27.11

a.Obtain a formula for the second virial coefficient of a “square-well” gas, for which

u(r)

⎧

⎪⎨

⎪⎩

∞ ifr<d

−u 0 ifd<r<c

0ifr>c

whereu 0 is a positive constant. Explain why this virial coefficient depends on temperature

while that of a hard-sphere gas does not.

b.Take the following parameters for a square-well representation of the argon intermolecular

potential:d 3. 162 × 10 −^10 m,c 5. 850 × 10 −^10 m, andu 0  9. 58 × 10 −^22 J. Find the

value of the second virial coefficient of argon at 0◦C. Compare it with the experimental value,

− 2. 15 × 10 −^5 m^3 mol−^1.

The second virial coefficient of a hard-sphere gas is positive, illustrating the fact

that repulsive forces correspond to a raising of the pressure of the gas over that of an

ideal gas at the same molar volume and temperature. The second virial coefficient of

the square-well gas has a constant positive part that is identical with that of the hard-

sphere gas, and a temperature-dependent negative part due to the attractive part of the

potential, illustrating the fact that attractive forces contribute to lowering the pressure

of the gas at fixed volume and temperature.