Free energy of a rotational system.

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The discussion focuses on deriving the Helmholtz free energy for a diatomic molecule at low temperatures, where energy levels are defined by ε_l = (h^2 l(l+1))/(2I) and degeneracy by g_l = (2l+1). The partition function is expressed as Z = ∑(2l+1)e^(-h^2 l(l+1)/2IkT). To find the Helmholtz free energy, F, it is suggested to take the first two terms of the summation and apply F = -kT Log(Z). The logarithm of the partition function can be approximated using the series expansion for small terms. This leads to the conclusion that F = -3kT e^(-h^2/IkT) plus additional terms.
Narcol2000
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If one has a diatomic molecule with energy levels

<br /> \epsilon_l = \frac{h^2 l(l+1)}{2I}<br />

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

if the degneracy is given by g_l = (2l+1)

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

<br /> F = -3kT e^{-h^2 / IkT} + ...<br />

I got as far as getting the partition function to be

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

<br /> \epsilon_l = \frac{h^2 l(l+1)}{2I}<br />

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

if the degneracy is given by g_l = (2l+1)

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

<br /> F = -3kT e^{-h^2 / IkT} + ...<br />

I got as far as getting the partition function to be

<br /> Z = \sum_{l=0}^{\inf} (2l+1)e^{-h^2 l(l+1)/2IkT}<br />


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
 
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