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I'm trying to justify that at constant volume, thermal equilibrium is achieved at a minimum of the Helmholtz function.

For any thermodynamic process,

$$dF = dU - TdS - SdT$$

$$\text{if $dF = 0$}$$

$$\implies dU - TdS = SdT$$

$$ \implies TdS - PdV - TdS = SdT$$

$$\implies SdT = - PdV$$

$$\text{constant volume, so implies $dV = 0$}$$

$$\implies SdT = 0$$

This implies an isobaric process where $dF = 0$ implies an isothermal process.

I'm not sure if I can get away with saying this, as the simplicity almost feels like cheating, but I tried to take this logic to justify that the Gibbs free energy is minimum for an isobaric and isothermal process at chemical equilibrium.

$$dH = dU - PdV - VdP$$ $$dG = dH - TdS - SdT$$ $$\text{if $dG = 0, \implies dH =TdS + SdT$}$$ $$\implies TdS + SdT = dU - PdV - VdP$$

$$\text{isobaric and isothermal, so $dV = dT = 0$}$$

$$\implies TdS = dU - VdP$$ $$\implies TdS = TdS - VdP$$

$$\implies VdP = 0$$

This seems to imply mechanical equilibrium occurs at a minimum for Gibbs free energy. Where is my thinking going wrong?

Also, I may have asymmetrical logic between the first and second 'proofs'. I didn't a priori assume $dT=0$ and that would bring me to $ 0 = 0$.

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    $\begingroup$ I'm pretty sure you have to consider your system as being in contact with a larger reservoir. Then you show that maximizing entropy of the entire system +envrironment is the same as minimizing just the free energy of your system. $\endgroup$ – Aaron Stevens May 2 at 12:11
  • $\begingroup$ You have it backwards. dF = 0 for isochoric and dG = 0 for isobaric. $\endgroup$ – Chet Miller May 2 at 13:31
  • $\begingroup$ Oh, thank you for spotting. I’ll rectify this. $\endgroup$ – sangstar May 2 at 13:40

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