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The Second Law of Thermodynamics says, that in equilibrium the Entropy S of an isolated system is maximized.

The First Law of Thermodynamics says that an isolated system has fixed internal energy U. But it doesn't say that a system with fixed U is isolated.

Therefore I don't think it's valid to replace "isolated system" in the Second Law with "system with fixed U".

However, all deductions of the so called "Minimum energy principle" start by claiming that "In equilibrium, S is maximized in systems with fixed U". They don't demand isolation, only fixed U, and continue from there with flawless math about convexity and monotonicity.

Unfortunately as I said before, I find their premise is simply false and therefore the "Minimum energy principle" should be false.

Can someone show why is the "Minimum energy principle" considered valid?

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  • $\begingroup$ "But it doesn't say that a system with fixed U is isolated". It doesn't say that because it isn't necessarily true. The internal energy of an ideal gas system undergoing a reversible isothermal expansion or compression is constant yet the system isn't isolated (energy transfer by work and heat is occurring with the surroundings). $\endgroup$
    – Bob D
    Aug 2, 2022 at 14:00

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You're taking some shortcuts that indeed make the conclusion dubious.

Saying that energy is minized at equilibrium due to the second law is correct, but:

  • not because the system is isolated (that would make the result mostly useless in practice)
  • the energy in question doesn't have to be the internal energy.

In a nutshell, depending on the type of transformation, a different form of energy is selected and gets minized at equilibrium due to the second law.

For example, for systems with no chemical reactions:

  • If the transformation is with $V$ and $S$ constant, then $U$ (internal energy) is minimized.
  • If the transformation is with $P$ and $S$ constant, then $H$ (enthalpy) is minimized.
  • If the transformation is with $P$ and $T$ constant, then $G$ (Gibbs's free energy) is minimized.

And so on. In every case, the proof in non-trivial and does not assume that the system is isolated.

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  • $\begingroup$ Oh... So the Entropy of the global, isolated system is maximized. But its Energy is not minimized. It is only the Energy of one of its non-isolated subsystems that is minimized (depending on which of its parameters is held constant). Is this correct? $\endgroup$
    – Juan Perez
    Aug 2, 2022 at 16:12
  • $\begingroup$ IF the system is isolated and IF the energy you're talking about is internal energy, then the latter is trivially constant. Again, you have to specify which energy you're talking about. $\endgroup$
    – Miyase
    Aug 2, 2022 at 16:16
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You are trusting an incomplete description in the Wikipedia page you cite. Wikipedia is a great resource, but it should not be a substitute for a good textbook if you want to understand a subject of Physics.

In the link you cite, the person who wrote the page omitted to specify that partial derivatives of $S$ with respect to the constrain variable $X$ at constant $U$ should also be intended at constant volume and number of particles. I would agree that constant $U$ alone is not enough to consider the system isolated, but adding constant volume and number of particles is really equivalent to isolation.

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