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Let's say there are two systems which can interact by a moving wall but cannot exchange heat. Then the system will be in mechanical, but not necessarily in thermal equilibrium.

The maximality of entropy in mechanical equilibrium requires only the ratio p/T to be equal. So there is a possible equilibrium state where one system has double pressure and double temperature...

Where is my mistake?

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  • $\begingroup$ Why would a mechanical system, i.e. one whose dynamics are governed by Newton's laws, obey an entropy maximum principle? $\endgroup$
    – webb
    Mar 27, 2014 at 16:27
  • $\begingroup$ I don't know why but the entropy is maximal in equilibrium. $\endgroup$
    – dan-ros
    Mar 27, 2014 at 16:38
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    $\begingroup$ webb, it would because the mechanical system is also a thermodynamical system and mechanical equilibrium is necessary condition for thermodynamic equilibrium. $\endgroup$
    – Ján Lalinský
    Mar 27, 2014 at 19:15
  • $\begingroup$ SImilar question with great answers: physics.stackexchange.com/a/258592/123183. $\endgroup$
    – Juan Perez
    Jul 31 at 16:16

2 Answers 2

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This is a classic conundrum and it is called the "problem of the adiabatic piston". You can find it discussed in books on thermodynamics by Landau & Lifhsitz and by Callen. Another very thorough analysis is by Gruber "Thermodynamics of systems with internal adiabatic constraints: time evolution of the adiabatic piston". (You can find Gruber's article here free http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.55.995 )

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The maximality of entropy in mechanical equilibrium requires only the ratio p/T to be equal. So there is a possible equilibrium state where one system has double pressure and double temperature... Where is my mistake?

If the systems are allowed to change their volumes, they are allowed to change their internal energies as well. When the wall moves, transfer of energy from one system to another occurs in the form of work. The maximum entropy principle then implies that for equilibrium, both temperature $T$ and the ratio $P/T$ have to be the same for both systems, hence $P$ has to be the same.

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  • $\begingroup$ ok, so mechanical, chemical, etc. equilibrium implies thermal equilibrium. And there can't be a mechanism compensating the energy loss/gain at both systems? $\endgroup$
    – dan-ros
    Mar 27, 2014 at 21:58
  • $\begingroup$ ok, the point is heat must be allowed to change. Not just energy. $\endgroup$
    – dan-ros
    Mar 28, 2014 at 9:25
  • $\begingroup$ No, it is the other way around. Thermodynamic equilibrium implies mechanical and chemical equilibrium, not the other way around. $\endgroup$
    – Ján Lalinský
    Mar 28, 2014 at 18:12
  • $\begingroup$ What do you mean, heat must be allowed to change? Heat is not a state quantity. $\endgroup$
    – Ján Lalinský
    Mar 28, 2014 at 18:13
  • $\begingroup$ forget my first comment above. if the moving wall is not conducting heat, then both system can be at different temperatures. Only the pressure needs to be the same. $\endgroup$
    – dan-ros
    Mar 28, 2014 at 23:55

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