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Everyone knows: imagine the isolated system consisting of a reservoir and a box with a gas and one of its sides is a piston. If we (inside the system) compress the piston quasi-statically, the entropy increase of the isolated system is zero (or negligible if we do it slowly enough). If we on the other hand charge and bump into the piston and compress it all at once, the entropy change is considerable.

Now I was wondering how I could imagine this quantity S in this situation and more specifically: try to explain it (intuitively).

So I was thinking of doing it with classical thermodynamics with the use of Clausius' relation, but then your argument depends on you classifying it as a reversible or irreversible, which is kind of what you are trying to explain.

If we look at Statistical Mechanics, I see no connection at all with (ir)reversibility. If we denote the entropy of the isolated system as S, then S = k log(omega). Now is there anything in the theory of S.M. that predicts (and maybe even explains...) that in the first scenario, omega is constant, and in the latter, omega grows?

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# Explaining (ir)reversibility with S.M.

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