Free energy of a rotational system.

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SUMMARY

The discussion focuses on deriving the Helmholtz free energy for a diatomic molecule at low temperatures, specifically when the condition \( h^2 / IkT \) is large. The energy levels are defined by \( \epsilon_l = \frac{h^2 l(l+1)}{2I} \) with degeneracy \( g_l = (2l+1) \). The partition function is expressed as \( Z = \sum_{l=0}^{\infty} (2l+1)e^{-h^2 l(l+1)/2IkT} \). The Helmholtz free energy is derived as \( F = -3kT e^{-h^2 / IkT} + ... \) by taking the logarithm of the partition function and approximating the terms.

PREREQUISITES
  • Understanding of quantum mechanics and molecular energy levels
  • Familiarity with statistical mechanics concepts, particularly partition functions
  • Knowledge of thermodynamic potentials, specifically Helmholtz free energy
  • Basic calculus for handling infinite series and logarithmic approximations
NEXT STEPS
  • Study the derivation of partition functions in statistical mechanics
  • Explore the implications of low-temperature behavior in thermodynamic systems
  • Learn about the relationship between Helmholtz free energy and entropy
  • Investigate the effects of molecular rotational states on thermodynamic properties
USEFUL FOR

This discussion is beneficial for physicists, chemists, and students studying thermodynamics and statistical mechanics, particularly those interested in molecular behavior at low temperatures.

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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