Statistical Mechanics - Physics at Oregon State University

78 CHAPTER 4. STATISTICS OF INDEPENDENT PARTICLES.
In our discussion of the ideal gas we have assumed that the orbitals are
characterized by three quantum numbers nx, ny, and nz. These three quantum
numbers describe the motion of the center of mass of the particle, but did not
include any internal degrees of freedom. Therefore, the previous formulas only
are valid for a gas of mono-atomic molecules. A gas of poly-atomic molecules is
easily treated if we assume that the energy associated with the internal degrees
of freedom does not depend on (1) the presence of other particles and (2) the
motion of the center of mass. Hence the rotation of a diatomic molecule is not
hindered by neighboring molecules or changed by its own motion. In that case
we can write, using int for the collective internal degrees of freedom:
ɛ(nx, ny, nz, int) = 2
π
2 (n
2M L
2 x + n 2 y + n 2 z) + ɛint (4.43)
Then internal degrees of freedom represent the rotational quantum numbers
(of which there are at most three) and the vibrational quantum numbers. If
we have a molecule with N atoms, there are 3N internal degrees of freedom.
Three are used for the center of mass, r for the rotations (r is 2 or 3) and hence
3(N − 1) − r for the vibrational state.
Changes in the partition function.
In the classical regime µ ≪ −kBT and hence λ = e µ
k B T ≪ 1. In the partition
function for a given orbital terms with λ 2 and higher powers can be neglected
for bosons and in all cases we find
Zo(T, µ, V ) = 1 + λ
where the internal partition function is defined by
int
Zint(T ) =
e − ɛo+ɛint ɛo
kB T − k = 1 + λZinte B T (4.44)
int
e − ɛ int
k B T (4.45)
The average number of particles in orbital o, independent of its internal
state, is therefore given by
< no >=
int
λe − ɛ int
k B T e − ɛo
k B T
1 + λ
int e− ɛ int
k B T e − ɛo
k B T
ɛo − k ≈ λZinte B T (4.46)
where we used the fact that the denominator is approximately equal to one
because λ is very small. Hence in the Boltzmann distribution function we have
to replace λ by λZint(T ) and we find
N = λZint(T )nQ(T )V (4.47)
n
µ = kBT log( ) − log(Zint)
nQ(T )
(4.48)