Physical Chemistry Third Edition

26.1 The Statistical Thermodynamics of a Dilute Gas 1085

To find an expression forPi, we consider a system consisting of a single molecule

in a specified state. From Eq. (4.2-6) we have the thermodynamic relation:

P−

(

∂U

∂V

)

S,n

(26.1-8)

If the single molecule is known to be in molecular stateiwith energy eigenvalueεi,

Ω1 andSst0. Since the statistical entropy is constant and the amount of substance

is fixed at one molecule, we write an expression for the pressure due to a molecule in

state numberi

Pi−

dεi

dV

(26.1-9)

If the molecule is confined in a rectangular box, its translational energy levels depend

on the dimensions of the box. We assume that the shape of the box is unimportant and

proceed as though the box were a cube so that the translational energy levels depend

only on the volume of the system. The rotational, vibrational, and electronic energy

levels are independent of the volume of the system.

We substitute the expression forpifrom Eq. (25.3-24) into Eq. (26.1-7). We apply

a mathematical trick by which we recognize the summand as a derivative:

P

N

z

∑

i

Pie−εi/kBT−

N

z

∑

i

(

dεi

dV

)

e−εi/kBT



NkBT

z

∑

i

(

∂

∂V

e−εi/kBT

)

T

(26.1-10)

If the sum is uniformly convergent, the order of summing and differentiating can be

exchanged to give

P

NkBT

z

(

∂

∂V

∑

i

e−εi/kBT

)

T

NkBT

1

z

(

∂z

∂V

)

T

NkBT

(

∂ln(z)

∂V

)

T

(26.1-11)

Uniform convergence means that the series converges with at least a certain rate for all

values of the variables in the expression.

The translational partition function is the only factor in the partition function that

depends onV.

PNkBT

(

∂ln(z)

∂V

)

T

NkBT

(

∂ln(ztr)

∂V

)

T

NkBT

(

∂ln[(2πmkBT/h^2 )^3 /^2 V]

∂V

)

T

PNkBT

(

dln(V)

dV

)



NkBT

V

(any dilute gas) (26.1-12)

Our dilute gas obeys the ideal gas equation of state.

Exercise 26.3

Verify Eq. (26.1-12) by explicit differentiation.