Vibration of a diatomic molecule Helmholtz free energy

31.10.08
Boltzmann
Partition function for one molecule
z =
∞∑
e −βεn = Z(T, v, i)
n=0
β = 1
k B T
for N molecules, Z = z n if distinguishable, Z(T, v, N) = 1 N! ZN if indistinguishable.
Helmholtz free energy:
F = −k B T ln Z
Vibration of a diatomic molecule
Using the harmonic oscillator approximation
(
ε n = hν n + 1 )
2
z =
∞∑
e −βhν(n+ 1 2 ) = e −βh ν 2
n=0
∞∑ (
e
−βhν ) n
let’s change the variable, e −βhν = x. We know that ∑ ∞
n=0 xn = 1
1−x , so
Helmholtz free energy
z = e −βh ν 2 ·
n=0
1
1 − e −βhν
1
1
=
e βh ν 2 − e −βh = ν
2
2 sinh( βhν
2 )
F = −k B T ln(Z n )
= − 1 β N (
−βh ν 2 − ln ( 1 − e −βhν))
= N 2 hν + N β ln ( 1 − e −βhν)
(where N 2
hν is the zero point energy of N molecules)
Combined 1. and 2. law
dF = −SdT − P dV + µdN
1

So S = − ( ∂F
∂T
So
so
)v,N = − (
∂F
∂β
) ( )
∂β
βT
[
]
βhν
S = Nk
e βhν − 1 − ln(1 − e−βhν
F = U − T S
U = F + T S
[ ]
1
U = Nhν
2 + 1
e βhν − 1
clearly,
[ ]
1
〈ε〉 = hν
2 + 1
e βhν − 1
Testing the limits
As T approaches 0, β → ∞
, the ZPE.
As T approaches infinity, β → 0
U → N hν
2
U = N hν
2 + Nhν 1
βhν + (βhν)2
2
+ . . .
= N hν
2 + N β
= N hν
2 + N β
= N β
= Nk B T
= nRT
1
1 + βhν
2
+ . . .
(
1 + βhν )
2 + . . .
which is the classical result. Equipartition principle.
Heat Capacity
C v =
( ) ∂U
∂T
V,N
e βhν
C v = Nk(βhν) 2
(e βhν − 1) 2
2