From Statistical Physics, 2nd Edition by F. Mandl:
Two vessels contain the same number $N$ molecules of the same perfect gas. Initially the two vessels are isolated from each other, the gases being at the same temperature $T$ but at different pressures $P_1$ and $P_2$. The partition separating the two gases is removed. Find the change in entropy of the system when equilibrium has been re-established, in terms of the initial pressures $P_1$ and $P_2$. Show that this entropy change is non-negative.
I'm a little confused about a few things.
- Is there a temperature change in this process? Intuitively, I would say no because $(T+T)/2=T$. My other guess would be that the temperature must change because we now have a third pressure, $P_3$, that is different from the pressure of the other two and also because we have increased the volume.
- I believe this is an irreversible process, correct? Because you can't realistically separate the gasses into that which came from vessel A and that which came from vessel B.
Can the change in volume simply be called $V_A+V_B$? I thought it would be that simple but thinking more about it I feel as though the change in pressure and possible change in temperature might change things.
When the partition is removed, is there a heat exchange between the 2 gases? My intuition says no because heat can only flow when there is a temperature difference and in this case both vessels are at temperature $T$.
My attempt:
So all in all I need to solve $\Delta S= \int \frac{dQ}{T}$
$$\Delta S= \int \frac{dQ}{T}$$ $$=\int \frac{dE+dW_{by}}{T}$$ We know that $dE=0$ and that $dW=PdV$
$$=\frac{1}{T}\int PdV$$ $$=\frac{P\Delta V}{T}$$
This is where I'm stuck - I don't think there is a valid thing to put in for $\Delta V$ because there were 2 systems that formed into 1 bigger system. If the final system is $V_1+V_2$, then what was its previous size? $V_1$ or $V_2$? Or can I say that it was $\frac{V_1+V_2}{2}$?