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I've tried to derive the relation between free energy, $F$, and equilibrium of a closed system. From the derivation i've written below it seems that if $T$ is constant and the system can't exchange work with the environment then, $F$ is at a minimum. If my reasoning is correct, can you give me an example of such a system that is not just an isolated one?

derivation

$$dE-\delta W =\delta Q \\ dS \ge \delta Q/T $$
substituting $$dE-TdS \leq \delta W $$

using the definition of the free energy

$$F=E-TS \\ dF=dE-TdS-SdT $$ considering an isothermal transformation $$dF=dE-TdS$$

substituting again $$dF \leq \delta W=0 \\$$

It follow that if the system can spontaneously evolve it will reach a state with less free energy. So, if there aren't available states with less free energy the system can't evolve, namely it is in equilibrium.

SimoBartz
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  • T is not the temperature of the system but the temperature of the heat source with which the system can exchange entropy and thermal energy, (fixed temperature reservoir). Only at the end of the process when F is minimum the system will be in equilibrium with the reservoir. – hyportnex May 18 '22 at 12:06
  • I think you are are thinking about the statistical approach. This is about thermodynamics and thermodynamics deals only with equilibrium systems – SimoBartz May 18 '22 at 12:42
  • i'm not saying that you are wrong, just that we are speaking of different things – SimoBartz May 18 '22 at 12:43
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    Does this answer your question? Why are thermodynamic potentials minimised? – Chemomechanics May 18 '22 at 17:29
  • You are asking for an example that is "not isolated" but the minimization of $F$ has nothing to do with isolated systems. It applies to an isothermal system, which means it exchanges heat with the surroundings. – Themis Aug 05 '22 at 10:03

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