How Do Temperatures Equalize in a Partially Adiabatic Cylindrical Gas System?

[Gianmarco]

Homework Statement

A cylindrical container with rigid walls is divided in two sections by a barrier with negligible mass that is free to move without friction along the axis of the cylinder. The barrier is adiabatic and so is half of the cylinder while the other half let's heat through. The cylinder contains ##n = 2\:mol## of perfect monatomic gas distributed in the two sections. Initially the container is in contact with the environment at ##T_0=300\:K## and the barrier is at equilibrium in the center. The cylinder is then put in contact with a reservoir at ##T_f = 280\:K## and the barrier is at equilibrium in a place such that the ratio of the volumes of the two sections is ##R = \frac{10}{9}##. Calculate the initial temperatures in the two sections of the cylinder

Homework Equations

PV = nRT

The Attempt at a Solution

So we are not given the amount of moles in each section, but we know that since we are at equilibrium in both initial and final states, then ##P_{1i} = P_{2i}## and ##P_{1f} = P_{2f}##. So, since at the beginning the volumes are the same because the barrier is in the middle: ##P_{1i}V_{1i} =P_{2i}V_{2i} \rightarrow n_1T_{1i}=n_2T_{2i}\\ T_{1i}=\frac{n_2}{n_1}T_{2i}##
The final state is what confuses me. I wasn't able to find out the respective values of ##n_1,\:n_2## so I looked through the solutions and it says that ##T_{1f} = T_{2f}=280\:K## when the system reaches equilibrium in contact with the reservoir. How is this possible? Shouldn't only the non-adiabatic section be at T = 280 K?

 

[Brian T]

Well if its the same gas on both sides, and you know it is at equilibrium, and you know the volume ratio, you should be able to determine the mole ratio.

Also, just because the barrier is adiabatic, this does not imply that the temperature on the adiabatic side of the cylinder can't change. The problem says the barrier can move, so the internal energy can change by work being done (the barrier sliding means compression/expansion) even though no heat can flow into that region.

 

[Gianmarco]

Yes I have the volume ratio. But if I don't know that ##T_{1f} = T_{2f}## I can't determine the mole ratio. It is at equilibrium so ##P_{1f}V_{1f}=n_1RT_{1f}## and ##P_{2f}V_{2f}=n_2RT_{2f}##. By taking the ratio of these two equations I get ##\frac{10}{9}=\frac{n_1T_{1f}}{n_2T_{2f}}##. There's nothing I can do here unless I know that ##T_{1f} = T_{2f}##

 

[Gianmarco]

Also, just because the barrier is adiabatic, this does not imply that the temperature on the adiabatic side of the cylinder can't change. The problem says the barrier can move, so the internal energy can change by work being done (the barrier sliding means compression/expansion) even though no heat can flow into that region.

I see what you are saying and I agree. The problem is that the solution explicitly says that the temperature is the same in both sections because all the gas is in contact with the reservoir. I don't understand how the gas in the adiabatic section could be in contact with the reservoir, it makes no sense to me. Also the transformation is supposed to be irreversible so I can't even use the equations of adiabatic transformations to find the temperature