## anonymous 3 years ago prove that d(A/T)/dT = -E/(T^2)

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1. anonymous

Note that this is: $\frac{d}{dT}\frac{A}{T}=\frac{dA}{dT}\cdot\frac{d}{dT}\left(\frac{1}{T}\right)$This becomes: $\frac{dA}{dT}\cdot\frac{-1}{T^2}$And, I can take a guess, although I don't quite know what $$A$$ is in this context, we have: $\frac{dA}{dT}=E$So: $E\cdot\frac{-1}{T^2}=-\frac{E}{T^2}$

2. Frostbite

A is the Helmholtz energy, A for Arbeit in German, E or mostly written U is the internal energy.

3. Frostbite

True most of the way except that: $A=U-TS$ $\left( \frac{ \partial A }{ \partial T } \right)_{V}=-S$ S is the entropy.

4. Frostbite

But if you substitute the equations back and forth you come to the conclusion you did so well done by lolwolf.