Statistical Relationship Between Helmoltz Free Energy and Entropy

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SUMMARY

The discussion centers on the derivation of the relationship between Helmholtz Free Energy (A) and entropy (S) using the most probable distribution from Hill's "Introduction to Statistical Thermodynamics". The key equation derived is A = -NkT ln(∑ e^{β ε_j}), where the most probable distribution is represented as C^{str}. The participant initially computed C^{str} and the number of accessible quantum states, but encountered difficulties in simplifying the logarithmic expressions correctly. The oversight regarding the N factor in the equation for C was identified as a potential source of error.

PREREQUISITES
  • Understanding of statistical thermodynamics concepts, specifically Helmholtz Free Energy and entropy.
  • Familiarity with the most probable distribution for independent indistinguishable molecules.
  • Knowledge of quantum states and the formula for accessible states, Ω(C).
  • Proficiency in manipulating logarithmic expressions and derivatives in thermodynamic equations.
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  • Study the implications of the most probable distribution in thermodynamic systems.
  • Learn about the partition function Q and its relationship to entropy and free energy.
  • Explore advanced topics in statistical thermodynamics, including approximations in quantum state calculations.
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Homework Statement


From Hill's "Introduction to Statistical Thermodynamics", question 3-4 reads:

Show that the substitution of of the most probable distribution in
S=k*ln(\Omega(C^{str})) leads to
A=-NkT*ln(\sum e^{\beta \epsilon_j})

(note "STR" denotes the case of most probable distribution and should be read as C*)

Homework Equations


The most probable distribution for a system of independent indistinguishable molecules:
\eta_i = \bar{C_j} /N =C^{str}/N = e^{-\beta \epsilon_j} / \sum_i e^{-\beta \epsilon_i}

Formula for accessible quantum states for an ensemble of independent indistinguishable molecules:
\Omega(C) = N! /( \prod_j C_j) !

Relationship between S and A
S = - (\partial A / \partial E) = kT ( \partial ln(Q) / \partial T )+ k ln(Q)


The Attempt at a Solution



First, I computed C in terms of the given equations. This is pretty straight forward:

C^{str}_j = Ne^{-\beta \epsilon_j} /{ \sum_i e^{-\beta \epsilon} }

Then, I computed the number of accessible quantum states with the above expression for C
\Omega (C^{str}) = { N! }/{\prod C^{str} }

Skipping a lot of algebra I arrived as this expression for ln(Ω(C)):
ln(C^{str}) = ln(N!)- \sum(ln(e^{-\beta \epsilon}) )- ln(\sum(e^{-\beta \epsilon})

Which can be reduced to:
ln(C^{str}) = ln(N!) + \sum(\beta \epsilon) - ln(Q) = ln(N!) + E \beta -ln(Q)

I am not sure exactly where I went wrong, as the formula I derived is very similar to the relationship between A and S. I think an approximation may have to be applied somewhere or perhaps the natural logs should not have been broken down into their operations. I am not sure, hence why I am posting here!

Thanks for the help.

PS, if needed I can post a small nomenclature of the variables.

EDIT:
Just realized I did not take into account the N factor in the equation for C, I believe this makes N! become (N-1)! in the following equations. Also, I did not show the additional formula if one were to expand \partial ln(Q) / \partial T
 
Last edited:
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