Internal Energy/Helmholtz Free Energy Proof

mobwars
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Homework Statement



Show that (∂(βA)/∂β)N,V = E, where A = E - TS is the Helmholtz Free Energy and E is the Internal Energy.

Homework Equations


A = E - TS
dE = TdS - pdV + ΣUidni
β = 1 / (kBT)

The Attempt at a Solution


(∂(βA)/∂β)N,V = (∂/∂β) * (βE - βTS)
(∂(βA)/∂β)N,V = (∂/∂β) * (βE - TS/(kBT))
(∂(βA)/∂β)N,V = (∂/∂β) * (βE - S/(kB))
(∂(βA)/∂β)N,V = (∂(βE)/∂β) [S/(kB) goes away because S is constant for Helmholtz Free Energy]
(∂(βA)/∂β)N,V = E

This solution just feels entirely too easy and simplified. I think you're supposed to do something with the fact that T and S are actual variables in the equation and some chain rule is needed, but that didn't seem to get me anywhere either. Anyone know what's really going on here?
 
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mobwars said:
(∂(βA)/∂β)N,V = (∂(βE)/∂β) [S/(kB) goes away because S is constant for Helmholtz Free Energy]
(∂(βA)/∂β)N,V = E

The two lines above are incorrect. As you suspected, S is not generally constant for a process in which N and V are kept constant.
Also, ∂(βE)/∂β ≠ E because E is not generally constant either.

It might help to make use of the first law as written in your second relevant equation.
 
Start with your first two relevant equations by taking dA and then eliminating dE.

Chet
 
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