# Statistical Relationship Between Helmoltz Free Energy and Entropy

1. Jul 16, 2012

### Aero51

1. The problem statement, all variables and given/known data
From Hill's "Introduction to Statistical Thermodynamics", question 3-4 reads:

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

2. Relevant 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 = $- (\partial A / \partial E) = kT ( \partial ln(Q) / \partial T )+ k ln(Q)$

3. 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: Jul 16, 2012
2. Jul 17, 2012