1070 25 Equilibrium Statistical Mechanics. I. The Probability Distribution for Molecular States
The excited vibrational states are not significantly populated at room temperature,
as shown by the closeness of the value ofzvibin the previous example to unity. For
a typical diatomic molecule at room temperature, very many translational states are
effectively accessible, a few hundred rotational states are effectively accessible, but
only one vibrational state and one electronic state are accessible.EXAMPLE25.9
Calculate the vibrational partition functions forH 2 and forI 2 at 298.15 K.
Solution
For H 2 ,v ̃e 4401 .2cm−^1 , and for I 2 ,v ̃e214.5 cm−^1. For H 2 ,hcv ̃e
kBT
(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 cm s−^1 )(4401.2cm−^1 )
(1. 3807 × 10 −^23 JK−^1 )(298.15 K) 21. 239zvib1
1 −e−^21.^239 1. 0000000006For I 2 ,hcv ̃e
kBT
(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 cm s−^1 )(214.5cm−^1 )
(1. 3807 × 10 −^23 JK−^1 )(298.15 K)
1. 03510zvib
1
1 −e−^1.^03510 1. 5508The vibrational partition function of H 2 is very close to unity because its vibrational energy
levels are very widely spaced, and only the ground vibrational state is significantly popu-
lated at room temperature. The vibrational energy levels of I 2 are much closer, and there is
some population of the first few excited vibrational states. However, the effective number of
occupied vibrational states is still smaller than 2.Exercise 25.19
Show that if the vibrational energy is written asε′′vibhν(
v+1
2)
(25.4-19)the vibrational partition function isz′′vibe−hν/ 2 kBT
1 −e−hν/kBTe−hνe/^2 kBT
1 −e−hνe/kBT(25.4-20)Calculate the value of this vibrational partition function for^35 Cl 2 at 298.15 K. Compare your
result with that of Example 25.8, and explain the difference.Corrections to the Product Partition Function
Writing the partition function as a product of four factors as in Eq. (25.4-6) depends
on the assumption that only the ground electronic level is needed in the electronic