Physical Chemistry Third Edition

1146 27 Equilibrium Statistical Mechanics. III. Ensembles

Solution

kBT

hv



(1. 3807 × 10 −^23 JK−^1 )(298.15 k)

(6. 6261 × 10 −^34 J s)(1. 6780 × 1013 s−^1 )

 0. 37023

The vibrational energy is given by equipartition of energy, so that

Svib,cl(1.000 mol)NAvkB+(1.000 mol)NAvkBln(0.37020)

(1.000 mol)(8.3145 J K−^1 mol−^1 )(1+ln[0.37020])

 0 .05298 J K−^1

This result compares poorly with the quantum statistical result, 2.194 J K−^1.

Exercise 27.8

Repeat the calculation of Example 27.5 for I 2 at 1000.0 K.

The classical translational and rotational contributions to thermodynamic functions

must be corrected by choosing the correct divisors. This yields the same results as in

quantum statistical mechanics. The classical formulas for vibration are numerically

inadequate, even with the correct divisors, and we do not attempt to use classical

statistical mechanics for electronic motion. There is nothing to be gained by using

classical statistical mechanics for a dilute gas.

PROBLEMS

Section 27.5: Thermodynamic Functions in the Classical
Canonical Ensemble

27.17Calculate the vibrational contribution to the molar heat
capacity of CO at 500.0 K, using classical statistical
mechanics. Compare your result with that obtained by
using Eq. (26.2-10e). Find the minimum temperature at
which the two results agree within 5%, and the minimum
temperature at which they agree within 1%. Do you think
that either one of these equations gives realistic results at
these temperatures? Why or why not?

27.18Calculate the vibrational contribution to the molar heat
capacity of H 2 at 500.0 K, using classical statistical
mechanics. Compare your result with that obtained with
Eq. (26.2-10e). Find the minimum temperature at which
the two results agree within 5% and the minimum
temperature at which they agree within 1%. Do you think

that either one of these equations gives realistic results at

these temperatures? Why or why not?

27.19 a.Assuming no electronic contribution, calculate the

entropy of 1.000 mol of argon gas atT 298 .15 K

andV 0 .02479 m^3 using Eq. (27.5-17) with a unit

divisor. This volume corresponds to the

thermodynamic standard state ofP100,000 Pa.

b.Recalculate the entropy of part b using

Eq. (27.5-18).

c.Compare your results of parts b and c with the

experimental standard-state value, 154.849

JK−^1 mol−^1.

d.Show that the same difference between the results of

parts b and c would result for a different temperature

and a different volume.