Thermal equilibrium without transfer of heat

Question

Suppose one has a box divided into two halves by a partition, with (possibly different) gases on either side. Both the walls of the box and the partition are perfect insulators so that no heat transfer between the two halves occurs. However the partition can slide in response to a difference in pressure between the two halves. Thus only transfer of volume occurs.

You would expect in equilibrium the pressures in the two halves $P_1 = P_2$ to be equalised, though since there is no heat flow we would not necessarily expect an equality of temperature. I would like to derive this.

Let $U_1 + U_2 = U$ be the total energy of the two halves, $V_1 + V_2 = V$ the total volume. Then by the first law $$dU_1 = T_1 dS_1 - P_1 dV_1$$ $$-dU_1 = T_2 dS_2 + P_2 dV_1$$ Then $$dS = dS_1 + d S_2 = dU_1 \left( \frac{1}{T_1} - \frac{1}{T_2} \right)+dV_1\left( \frac{P_1}{T_1} - \frac{P_2}{T_2} \right) $$ Equilibrium is determined by $dS= 0$. What confuses me is several things:

1. I want to assume no heat flow. But I am not sure how to put this assumption in. The equalities $dQ_1 = T_1 dS_1$, $dQ_2 = T_2 dS_2$ are only true for reversible processes and the establishing of equilibrium of pressures can't happen reversibly. All you can say is $dQ_i \leq T_i dS_i$. It seems obviously meaningful to insist on no heat transfer (imagine a polystyrene partition), but since heat is not expressible in terms of state functions the assumption seems to dangle without doing anything.
2. If I do pretend that this process could be done reversibly then it immediately follows that $dS_1 = dS_2 = 0$ and so no equilibrium condition could be derived.
3. It seems, though I am guessing, that in an irreversible process such as this there will not be any fixed relationship between $dU_1$ and $dV_1$ and so these differentials could be treated as independent. It would then follow from $dS=0$ that $$T_1 = T_2$$ $$P_1 = P_2$$ Thus while I have assumed no heat transfer, the work done by the sliding of the partition is enough to generate equality of temperatures (?!).

Answer 1

I disagree that irreversibility implies a lack of relationship between $V$ and $U$ (although that relationship will vary depending on how you impose this irreversibility).

The outcome for the reversible case is perpetual oscillation of the adiabatic partition, which is why you couldn’t identify a (static) equilibrium. Even if you dampen this motion to a halt by adding a certain amount of friction on each side, there will still be a fixed relationship between $V$ and $U$ for each side, depending on the details of the friction. These details will determine the individual temperatures on each side once the partition comes to a stop.

(Assuming as you did that $V$ and $U$ become uncoupled is actually an admission of heat transfer through the partition, which violates an earlier assumption.)

In other words, to consider the irreversible case, you have to revise your energy terms to incorporate how much heat is dumped on each side from braking the partition.