Statistical Relationship Between Helmoltz Free Energy and Entropy

In summary, the conversation discusses question 3-4 from Hill's "Introduction to Statistical Thermodynamics" and the relationship between A and S. The most probable distribution for a system of independent indistinguishable molecules is computed using the equations provided, and the number of accessible quantum states is also calculated. An attempt is made to derive an expression for ln(Ω(C)), but it is not clear where the error lies. The question poster realizes they did not take into account the N factor in the equation for C and wonders if an approximation needs to be applied or if the natural logs should not have been broken down. They also mention a possible error in not considering the additional formula for expanding \partial ln(Q) / \partial T.
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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
[itex]S=k*ln(\Omega(C^{str}))[/itex] leads to
[itex]A=-NkT*ln(\sum e^{\beta \epsilon_j})[/itex]

(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:
[itex]\eta_i = \bar{C_j} /N =C^{str}/N = e^{-\beta \epsilon_j} / \sum_i e^{-\beta \epsilon_i}[/itex]

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

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


The Attempt at a Solution



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

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

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

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

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

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 [itex] \partial ln(Q) / \partial T[/itex]
 
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FAQ: Statistical Relationship Between Helmoltz Free Energy and Entropy

What is the statistical relationship between Helmoltz free energy and entropy?

The statistical relationship between Helmoltz free energy and entropy is described by the equation: F = U - TS, where F is the Helmoltz free energy, U is the internal energy, T is the temperature, and S is the entropy. This equation shows that as the temperature increases, the free energy decreases, and the entropy increases.

2. How does the Helmoltz free energy relate to the spontaneity of a reaction?

The Helmoltz free energy is a measure of the available energy in a system that can be used to do work. If the free energy decreases during a reaction, it indicates that the reaction is spontaneous and releases energy. In contrast, if the free energy increases, the reaction is non-spontaneous and requires an input of energy to occur.

3. Can the statistical relationship between Helmoltz free energy and entropy be applied to all chemical reactions?

Yes, the statistical relationship between Helmoltz free energy and entropy can be applied to all chemical reactions. However, it is most commonly used in the study of thermal equilibrium and thermodynamic processes.

4. How does temperature affect the statistical relationship between Helmoltz free energy and entropy?

Temperature has a direct effect on the statistical relationship between Helmoltz free energy and entropy. As the temperature increases, the free energy decreases, and the entropy increases. This is because at higher temperatures, there is more thermal energy available for molecules to move and rearrange, leading to an increase in disorder (entropy).

5. What is the significance of the statistical relationship between Helmoltz free energy and entropy in thermodynamics?

The statistical relationship between Helmoltz free energy and entropy is significant in thermodynamics because it helps to predict the spontaneity and direction of a chemical reaction. It also provides insight into the energy changes and equilibrium of a system. Additionally, the relationship between free energy and entropy is a fundamental concept in the study of thermodynamics and has many practical applications in fields such as chemistry, physics, and engineering.

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