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Piston cylinder assembly with 3 compartments

Consider an ideal gas in a piston cylinder assembly, initially divided into 3 compartments by impermeable diathermal membranes. The compartments initially have the same mass and temperatures, but different pressures. The membranes are punctured, and the system settles to the final state with a uniform pressure, temperature, and volume.

The goal is to find the final temperature $T_2$ in terms of the initial pressures $P_{1A}$, $P_{1B}$, $P_{1C}$, initial temperature $T_1$, and ideal gas properties.

The first law of thermodynamics yields

$\Delta E = Q-W=\frac{m}{3}c_v\Delta T + \frac{m}{3}c_v\Delta T + \frac{m}{3}c_v\Delta T$

But $Q=0$ so we have

$-W=mc_v\Delta T$

Now, this is easy if the process is isobaric, so that

$W=P_2(V_2-V_1)$

but is this a valid assumption? I'm going to assume that it is, and that the final pressure $P_2$ is the average of all the initial pressures, $P_2=\bar{P_1}=\frac{P_{1A}+P_{1B}+P_{1C}}{3}$.

So we have

$W=\bar{P_1}(V_2-V_1) = \bar{P_1}(\frac{mRT_2}{\bar{P_1}} - \frac{mRT_1}{3P_{1A}} - \frac{mRT_1}{3P_{1B}} - \frac{mRT_1}{3P_{1C}})$

Substituting this back into the first law will allow us to solve for $T_2$ in terms of $P_{1A}$, $P_{1B}$, $P_{1C}, T_1$ and gas properties.

My solution, however, is contingent assuming that $P_2=\bar{P_1}=\frac{P_{1A}+P_{1B}+P_{1C}}{3}$. I made this assumption because intuitively I think the pressures need to equilibrate to some value, and that value must be an average of all the pressures if the masses in each compartment are the same. Is this a good assumption? Is there a better way to solve this problem?

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closed as off-topic by user191954, stafusa, John Rennie, ZeroTheHero, Jon Custer Sep 16 '18 at 3:38

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Actually, in the end, the work done by the overall combined system is just that required to raise the weight of the piston to its new elevation and to push back the outside atmosphere. For the initial state, $$P_{1A}A=Mg+P_{atm}A\tag{1}$$where M is the mass of the piston, and the expansion work done by the system is $$W=\left(\frac{Mg}{A}+P_{atm}\right)\Delta V\tag{2}$$So, combining these equations, the work done by the system is just $$W=P_{1A}\Delta V$$

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  • $\begingroup$ This makes sense mathematically but I would have never expected this. I even thought the isobaric assumption was strange and I only made it because it made things easy. Is there any intuitive reason why the pressure remains constant? It's even less intuitive that the bottom compartment pressures don't affect the top compartment pressure. $\endgroup$ – Drew Sep 15 '18 at 14:12
  • $\begingroup$ Well, what you are really integrating is the external pressure on the inner face of the piston, which, over the time for deformation plus equilibration starts and ends with zero kinetic energy for the piston, and so the work the gas does is just the work to raise the piston plus push back the atmosphere. Because of the initial equilibrium, P3A is just equal to the pressure of the piston weight plus the outside atmosphere. $\endgroup$ – Chet Miller Sep 15 '18 at 14:47
  • $\begingroup$ I see why it's constant pressure now; the piston weight and atmospheric pressure are always the same. But did you mean that $P_{1A}=P_{piston} + P_{atm}$? I don't see why $P_{3A}$ has any constraints, it could be anything. $\endgroup$ – Drew Sep 15 '18 at 14:55
  • $\begingroup$ Oops. Yes. I meant P1A. sorry. $\endgroup$ – Chet Miller Sep 15 '18 at 22:48

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