Physical Chemistry Third Edition

1094 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics

EXAMPLE26.5

CalculateCVandCPfor 1.000 mol of^35 Cl 2 at 298.15 K.

Solution

From Table A.22 in Appendix A,ν ̃e 559 .7cm−^1. The rotational constant is not needed.

CV,m,tr

3 NAvkB

2



3 R

2



3(8.3145 J K−^1 mol−^1 )

2

 12 .472 J K−^1 mol−^1

CV,m,rotNAvkBR 8 .3145 J K−^1 mol−^1

Letx

hν

kBT



hcν ̃e

kBT



(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 cm s−^1 )(559.7cm−^1 )

(1. 3807 × 10 −^23 JK−^1 )(298.15 K)

 2. 701

CV,m,vibRx^2

ex

(ex−1)^2

R(2.701)^2

e^2.^7009

(e^2.^701 −1)^2

 4 .680 J K−^1 mol−^1

CV,m,el≈ 0

CV,mCV,m,tr+CV,m,rot+CV,m,vib+CV,m,el

 12 .472 J K−^1 mol−^1 + 8 .3145 J mol−^1 + 4 .68 J K−^1 mol−^1

 25 .466 J K−^1 mol−^1

CP,mCV,m+R 25 .466 J K−^1 mol−^1 + 8 .3145 J K−^1 mol−^1

 33 .781 J K−^1 mol−^1

This compares with the experimental, value 33.949 J K−^1 mol−^1. The discrepancy is pre-

sumably due to the use of the harmonic oscillator–rigid rotor approximation.

EXAMPLE26.6

Derive a formula for the electronic contribution to the heat capacity for a system like NO for

which two electronic states must be included.

Solution

Take the energy of the ground electronic state as zero:

zelg 0 +g 1 e−ε^1 /kBT

εel

1

zel

g 1 ε 1 e−ε^1 /kBT