Physical Chemistry Third Edition

27.4 Classical Statistical Mechanics 1139

The quantum rotational partition function of a nonlinear polyatomic gas is related

to the classical version as follows:

zrot,qm

√

π

σ

(

8 π^2 kBT

h^2

) 3 / 2

(IAIBIC)^1 /^2 

zrot,cl

σh^3

(

dilute nonlinear

polyatomic gas

)

(27.4-26)

The division byh^3 corresponds to the relation of a quantum state to the phase space

of the three angular coordinates of a nonlinear object. The division by the symmetry

numberσagain relates to the indistinguishability of the nuclei.

The fact that the classical translational and rotational partition functions are related

to the quantum mechanical partition functions in this simple way arises from the fact

that the spacing between energy levels is relatively small compared with the energy

eigenvalues. Because of this we were able to replace sums by integrals in deriving

the quantum partition functions in Chapter 25, thus effectively removing the effects of

energy quantization.

The relationship between the classical and quantum partition functions is not so

simple for the vibrational factor. If we divide the classical vibrational factor for a

diatomic molecule by Planck’s constant, we obtain

zvib,cl

h



kBT

hν

(diatomic molecule) (27.4-27)

which does not match the quantum vibrational partition function of Eq. (25.4-18).

However, if we consider the quantum vibrational partition function for large values

ofT, we obtain

zvib,qm

e−hν/^2 kBT

1 −e−hν/kBT

≈

1

1 −(1−hν/kBT)

≈

kBT

hν

(for large values ofT) (27.4-28)

Division of the classical vibrational partition function by Planck’s constant gives this

result, and it would appear that we have a result that is compatible with the correspon-

dence principle. However, the limit of large temperature is physically unreasonable,

since it corresponds to large values of the vibrational quantum number that would cause

almost any molecule to dissociate.

EXAMPLE27.3

a.Evaluate the quantum mechanical vibrational partition function for Cl 2 at 298.15 K and

at 1000.0 K.

b.Compare your result in part a withkBT /hνfor the two temperatures.

Solution

a.From Example 25.8,zvibat 298.15 K is equal to 1.0720. At 1000.0 K, letxhν/kBT:

x

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

(1. 3807 × 10 −^23 JK−^1 )(1000.0K)

 0. 8053

zvib

1

1 −e−x



1

1 −e−^0.^8053

 1. 8082