3
$\begingroup$

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?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ 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 '14 at 16:27
  • $\begingroup$ I don't know why but the entropy is maximal in equilibrium. $\endgroup$ – tonydo Mar 27 '14 at 16:38
  • 1
    $\begingroup$ 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 '14 at 19:15
4
$\begingroup$

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 )

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ ok, so mechanical, chemical, etc. equilibrium implies thermal equilibrium. And there can't be a mechanism compensating the energy loss/gain at both systems? $\endgroup$ – tonydo Mar 27 '14 at 21:58
  • $\begingroup$ ok, the point is heat must be allowed to change. Not just energy. $\endgroup$ – tonydo Mar 28 '14 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 '14 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 '14 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. $\endgroup$ – tonydo Mar 28 '14 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.