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LolWolfBest ResponseYou've already chosen the best response.3
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: \[ E\cdot\frac{1}{T^2}=\frac{E}{T^2} \]
 11 months ago

FrostbiteBest ResponseYou've already chosen the best response.0
A is the Helmholtz energy, A for Arbeit in German, E or mostly written U is the internal energy.
 8 months ago

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

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