Free energy of a rotational system.

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Narcol2000
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If one has a diatomic molecule with energy levels

[tex] \epsilon_l = \frac{h^2 l(l+1)}{2I}[/tex]

l = 0,1,2,3,4,5...

if the degneracy is given by [tex]g_l = (2l+1)[/tex]

How does one show that the Helmholtz free energy at low temperature ([tex]h^2/Ikt[/tex] large)
is given by

[tex] F = -3kT e^{-h^2 / IkT} + ...[/tex]

I got as far as getting the partition function to be

[tex] Z = \sum_{l=0}^{\inf} (2l+1)e^{-h^2 l(l+1)/2IkT}[/tex]
 
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Narcol2000 said:
If one has a diatomic molecule with energy levels

[tex] \epsilon_l = \frac{h^2 l(l+1)}{2I}[/tex]

l = 0,1,2,3,4,5...

if the degneracy is given by [tex]g_l = (2l+1)[/tex]

How does one show that the Helmholtz free energy at low temperature ([tex]h^2/Ikt[/tex] large)
is given by

[tex] F = -3kT e^{-h^2 / IkT} + ...[/tex]

I got as far as getting the partition function to be

[tex] Z = \sum_{l=0}^{\inf} (2l+1)e^{-h^2 l(l+1)/2IkT}[/tex]


Take the first two terms of the summation and then use that
F = - k T Log(Z). The term Log(Z) is of the form

Log(1 + small term) = small term - small term^2/2 = approximately small term