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Consider a closed composite system with an internal movable adiabatic wall. If we release the piston, thermodynamics cannot determine the final equilibrium state (the temperature cannot be determined).

(About the previous example, you can refer to Callen's Thermodynamics and an Introduction to Thermostatistics 2ed. P53 Problem 2.7-3 and P100 Problem 4.3-1: enter image description here )

How to understand this? Consider a quasi-static process. As the piston moves, the temperature and the pressure of both chambers will change until the pressures coincide. Then there is a well-defined unique final equilibrium state. What is wrong with this argument?

Does it mean that thermodynamics is incomplete?

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  • $\begingroup$ I don't understand this line: "How to understand this? Consider a quasi-static process intuitively, there must be an end to any process." And I believe this is the main crux of your query. Care to clarify? $\endgroup$ – Gummy bears Jun 30 '14 at 6:01
  • $\begingroup$ I have improved the question. $\endgroup$ – Andy Jun 30 '14 at 6:27
  • $\begingroup$ "If we release the piston, thermodynamics cannot determine the final equilibrium state (the temperature cannot be determined)." Why do you think so? If the state is thermodynamic equilibrium, it should be possible to find it from the condition of maximum entropy. $\endgroup$ – Ján Lalinský Jun 30 '14 at 8:46
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    $\begingroup$ « As the piston moves, the temperature and the pressure of both chambers will change until the pressures coincide. » No: during the move, the piston acquires some kinetic energy. Callen states that this kinetic energy must be somehow gotten rid of, otherwise the piston will restore it to the gases, leading to perpetual oscillations. If the gases are viscous, some of the energy will be converted into heat, warming up the gases. Depending on how this energy will be distributed among the gases, the final states will be different and determined by the dynamics of evolution. $\endgroup$ – Tom-Tom Jun 30 '14 at 12:13
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    $\begingroup$ I asked a similar question some time ago - physics.stackexchange.com/questions/257815/… $\endgroup$ – Alexander Feb 8 '17 at 23:29
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  1. I would like to repeat the comment by Rossetto [user:35964, Tom-Tom]:

    During the move, the piston acquires some kinetic energy. ... If the gases are viscous, some of the energy will be converted into heat, warming up the gases. Depending on how this energy will be distributed among the gases, the final states will be different and determined by the dynamics of evolution.

  2. You must remind yourself that the motion of the piston and of the gases is not slow or isentropic, as it is in so many textbook problems.

  3. To define the state of a homogeneous material whose equation of state is known, we need two quantities (functions of state) - for example, temperature and pressure. There are two separate fluids, so we have 4 unknowns in all. How many equations do we have?
    a. $V_1+V_2=constant$
    b. Energy conservation. (The fluids and the piston all end at rest. The only energy remaining is thermal energy in the fluids, which must equal its initial value.)
    c. Mechanical equilibrium: $p_1=p_2$

There are not enough equations to determine the final state.

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The cited problem shows, that the first and second law of thermodynamics are not necessarily enough to fully determine all parameters of the system. Both laws give you restrictions on possible solutions, i.e. they sort solutions into 'allowed' and 'forbidden'. However, you can have more than one state in the 'allowed' basket. In that case you need additional information to decide which of the allowed states will be realized. In the problem you cited, the two laws only give you a restriction on the pressure. The system setup gives you a relation between the volumes of the subsystems, but due to the adiabatic wall, the temperatures are only related to each other since in both subsystems volume, pressure and temperature are connected via an equation of state. This leads to a situation, where you can have multiple solutions, that all fulfil both laws of thermodynamics. Which of these allowed solutions is actually realized depends on other parameters, here on the viscosities in the subsystems.

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  • $\begingroup$ I think you are right. But as I mentioned in the problem, it seems that as the piston moves, the temperature and the pressure of both chambers will change until the pressures coincide. Then there is a well-defined unique final equilibrium state? $\endgroup$ – Andy Jul 1 '14 at 3:05
  • $\begingroup$ Yes, there is a well defined final state. But you can't determine this state from the first and second law of thermodynamics alone. You need the viscosities as well. The second law doesn't claim there's only one state with maximum entropy. $\endgroup$ – taupunkt Jul 1 '14 at 9:51

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