Physical Chemistry Third Edition

27.5 Thermodynamic Functions in the Classical Canonical Ensemble 1145

Solution

a.From Eq. (27.4-14)

zrot,cl 8 π^2 IekBT

The moment of inertia of the Cl 2 molecule was calculated in Example 25.7 and is equal

to 1.147× 10 −^45 kg m^2.

zrot,cl 8 π^2 (1. 147 × 10 −^45 kg m^2 )(1. 3807 × 10 −^23 JK−^1 )(298.15 K)

 3. 728 × 10 −^64 kg^2 m^4 s−^2

Srot,cl

Urot

T

+NkBln(zrot,cl)n[R+Rln(zrot,cl)]

(1.000 mol)[8. 31455 +(8.3145 J K−^1 mol−^1 )ln(3. 728 × 10 −^64 )]

−1206 J K−^1

where we have divided by argument of the logarithm by 1 kg^2 m^4 s−^2 to make it dimen-

sionless. This result is obviously not useful, showing the deficiency of classical statistical

thermodynamics, which ignores the relationship between quantum states and volumes in

phase space as well as the indistinguishability of the nuclei.

b.The quantum mechanical partition function of Cl 2 has already been calculated in

Example 25.7 and is equal to 424.7.

Srotn[R+Rln(zrot)(1.000 mol)(8.3145 J K−^1 mol−^1 )[1+ln(424.7)]

 58 .63 J K−^1

c.The difference between these two values is

Difference−nRln(σh^2 )

−(1.000 mol)8.3145 J K−^1 mol−^1 )ln([2][6. 6261 × 10 −^34 Js]^2 )

−1264 J K−^1

which is the difference between the results of parts a and b, and is temperature-independent.

After division of the classical vibrational partition function by Planck’s constant,

the expression for the vibrational contribution to the entropy of a diatomic substance is

Svib

Uvib,cl

T

+NkBln

(z

vib,cl

h

)

NkB−NkBln

(

kBT

hv

)

(27.5-21)

where we have used the classical equipartition of energy result forUvib,cl.

EXAMPLE27.5

Calculate the vibrational contribution to the entropy of 1.000 mol of Cl 2 at 298.15 K, using

Eq. (27.5-21). The vibrational frequency is 1.6780× 1013 s−^1. Compare your result with

that of Example 26.8, which is 2.194 J−^1 K−^1.