Thermodynamics: A piston can freely move inside a horizontal cylinder which is closed from both ends

Question

Is it safe to say that regardless of where the piston is, the final pressure on each side of it would remain the same. And if yes, then why?

Answer 1

I assume that the two subvolumes separated by the piston are a single closed, isolated system. The answer is affirmative. However, it requires different justifications according to the nature of the piston (is it thermal conducting or not?).

Suppose the piston is a diathermal wall (i.e., it allows the exchange of energy between the two subregions it divides the volume). In that case, the conditions for thermodynamic equilibrium of the global isolated system (at fixed total energy $E=E_1+E_2$, and total volume $V=V_1+V_2$) correspond to maximizing entropy with respect to the energy $E_1$, and volume $V_1$. This maximum implies $$ \frac{\partial S_1}{\partial E_1} = \frac{\partial S_2}{\partial E_2}, $$ i.e. the temperatures of the two subvolumes must be equal: $$ \frac{1}{T_1}=\frac{1}{T_2}.\tag{1} $$ Moreover, we have $$ \frac{\partial S_1}{\partial V_1} = \frac{\partial S_2}{\partial V_2} ~~~~~~\Rightarrow~~~~~~~\frac{P_1}{T_1}=\frac{P_2}{T_2}. $$ Using the condition on the equality of temperatures, we get the equality of pressures.

If the piston is an adiabatic wall, the situation requires a more careful analysis. According to Callen (Thermodynamics and an Introduction to Thermostatistics), in the presence of an adiabatic piston, the problem of finding the equilibrium state is indeterminate. In the absence of viscous damping, the piston would oscillate forever. In contrast, with the addition of viscous damping, eventually, the piston will come to rest. In such a case, the pressure on both sides must be equal. The final temperatures of the two subvolumes would be undetermined. However, more recent work has shown that this last conclusion is too pessimistic. In any case, there is agreement about the conclusion of a final pressure.