Physical Chemistry Third Edition

1144 27 Equilibrium Statistical Mechanics. III. Ensembles

Exercise 27.7

a.Show that the vibrational energy given by Eq. (26.2-8) approaches the equipartition value

ofNkBTnRTfor large values ofT. However, remember that this equation is not valid

for large values ofTsince the molecule would dissociate for large values of the vibrational

energy.

b.Calculate the vibrational energy of 1.000 mol of Br 2 at 298.15 K and at 1000.0 K. Compare

your result withRTfor each temperature.

c.Repeat the calculation of part b for H 2.

The Entropy of a Dilute Gas

We apply Eq. (27.5-4) to a dilute gas:

S

U

T

+kBln(Zcl)

U

T

+kBln(zNcl)

U

T

+NkBln(zcl) (27.5-17)

As previously stated, the classical molecular partition function has units of kg m^2 s−^1

raised to some power, so a divisor with units must be included to make the argument

of the logarithm dimensionless. If a divisor of 1 kg m^2 s−^1 is used, values are

obtained for the entropy and the Helmholtz energy that differ from the experimental

values. However, when the classical canonical translational partition function is

divided by h^3 NN! and Stirling’s approximation is used for ln(N!), the same formulas

are obtained as Chapter 26. For a dilute monatomic gas the corrected classical

formula is

S

U

T

+NkBln

(z

cl

Nh^3

)

+NkB

U

T

+NkBln

(z

qm

N

)

+NkB (27.5-18)

which is the same as Eq. (26.1-5).

Equation (27.5-18) also holds for the translational contribution to the entropy of

molecular substances. For a diatomic substance the rotational contribution to the

entropy with a unit divisor would be

Srot,cl

Urot

T

+NkBln(zrot,cl) (27.5-19)

With a divisor ofσh^2 , whereσis the symmetry number,

Srot

Urot

T

+NkBln

(z

rot,cl

σh^2

)



Urot

T

+NkBln(zrot,qm) (27.5-20)

which is also identical to the formula from quantum statistical mechanics. A similar

formula holds for polyatomic dilute gases, with division byσh^3 instead of byσh^2.

EXAMPLE27.4

a.Calculate the rotational contribution to the entropy of 1.000 mol of Cl 2 gas at 298.15 K,

using the formula of Eq. (27.5-19).

b.Repeat the calculation of part a using the corrected formula of Eq. (27.5-20).

c.Show that the difference between the result of part a and part b would be the same for any

temperature.