# An attempt at justifying effects of minima for Gibbs free energy and Helmholtz free energy

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 $$0 = 0$$.

• 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. – Aaron Stevens May 2 at 12:11
• You have it backwards. dF = 0 for isochoric and dG = 0 for isobaric. – Chet Miller May 2 at 13:31
• Oh, thank you for spotting. I’ll rectify this. – sangstar May 2 at 13:40