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LolWolf
 one year ago
Best ResponseYou've already chosen the best response.3Note 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: \[ E\cdot\frac{1}{T^2}=\frac{E}{T^2} \]

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0A is the Helmholtz energy, A for Arbeit in German, E or mostly written U is the internal energy.

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0True most of the way except that: \[A=UTS\] \[\left( \frac{ \partial A }{ \partial T } \right)_{V}=S\] S is the entropy.

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0But if you substitute the equations back and forth you come to the conclusion you did so well done by lolwolf.
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