Consider two ideal gases of differing heat capacities. They occupy separate compartments of the same total volume. Assume their initial temperatures and that the amount of substance for both are the same. Assume that there is a pressure/volume imbalance such that the system is not in equilibrium, and that the two compartments are separated by an adiabatic and impermeable wall that allows for movement.

Now, at least according to the solutions to my problem set, the final temperatures for both gases are equal and also equal to the initial temperature. But if the gases are expanding and contracting, wouldn't this imply a change in said temperatures?

Is it simply that the process evolves such that $PV$ is always constant and (since $nR$ is constant, too), $T$ therefore remains fixed? If that's the case, how can we infer this from the setup? Is it because the system is initially in thermal equilibrium and therefore is not going to evolve in such a way that "negates" that?

  • $\begingroup$ More discussion of the challenging analysis of the adiabatic partition: 1, 2, 3. $\endgroup$ May 27 at 16:11
  • $\begingroup$ Can you share the exact text of the justification that both final temperatures are equal? Chet raises an interesting point below. $\endgroup$ May 27 at 17:06
  • $\begingroup$ @Chemomechanics There is no justification given in the solutions, it is just stated that the final temperature is equal to the initial temperature (and that it is equal between the compartments). $\endgroup$
    – agaminon
    May 27 at 17:08

1 Answer 1


It is a consequence of the first law of thermodynamics applied to the whole system. Since the whole system is isolated $dU=dU_1+dU_2=0$, also at equilibrium the final temperature of the two compartments is the same, combining this two conditions you can obtain that $T_f=T_0$.

Also, the final pressure must be the same to ensure mechanical equilibrium.

(Remember that for an ideal gas $dU=C_vdT$)

  • $\begingroup$ Oh, I suppose that you can deduce it directly from the fact that $dU=0$ implies that the net temperature change of the system is also 0, and through equilibrium Tf=T0. That's what you mean, right? $\endgroup$
    – agaminon
    May 27 at 2:14
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    $\begingroup$ One also needs to assume that $U$ depends only on $T$ (a special characteristic of the ideal gas). The final temperature could easily differ from the initial temperature for any real material, which is one of the reasons why the solution might seem counterintuitive. $\endgroup$ May 27 at 2:16
  • $\begingroup$ Yeah, $dU=0$ and the fact that the initial temperatures are the same implies that the temperature doesn't change for any of the compartments. $\endgroup$ May 27 at 2:24
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    $\begingroup$ With an adiabatic partition, there is no reason why the final temperatures should be equal. In particular, the initially higher pressure gas does net work on the initially lower pressure gas. $\endgroup$ May 27 at 10:18
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    $\begingroup$ @ChetMiller This paper discusses how temperatures tend to equilibrate on either side of even an adiabatic movable piston via momentum transfer due to pressure fluctuations. However, the time scale is very long for macroscale systems, and this subtle behavior is typically not discussed in undergraduate- or even graduate-level treatment of thermodynamics. I'm curious to see the exact text of agaminon's solution sets to see how this point is addressed. $\endgroup$ May 27 at 16:57

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