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JeanetteBaker

prove that d(A/T)/dT = -E/(T^2)

  • 11 months ago
  • 11 months ago

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

    • 8 months ago
  3. Frostbite
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    True most of the way except that: \[A=U-TS\] \[\left( \frac{ \partial A }{ \partial T } \right)_{V}=-S\] S is the entropy.

    • 8 months ago
  4. Frostbite
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    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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