# Piston cylinder assembly with separate compartments [closed]

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?

## closed as off-topic by user191954, stafusa, John Rennie, ZeroTheHero, Jon CusterSep 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 $$W=P_{1A}\Delta V$$
• 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. – Drew Sep 15 '18 at 14:55