I This gas work violates 2nd law of thermodynamics?

[Chestermiller]

I've tried to simplify the problem, hopefully retaining its essence, and do an analysis, but I ran into some difficulties. Maybe someone can tell me how to proceed.

If I have a closed cylinder as in the sketch but only 1 piston between gas A and gas B, so no vacuum, then I don't need to think about collisions between pistons.
Initially I have the following situation:
Gas A occupies 1/3 of the cylinder on the left, gas B occupies 2/3 on the right and they are separated by a piston. The number of moles (n) is the same for both and remains constant.

For the starting values of volume, pressure and temperature we have:
V_B = 2 V_A
P_A = 2 P_B
T_A = T_B = T

There is no heat transfer between a gas and anything else (piston, cylinder or the other gas).
The piston is extremely heavy so that expansion and compression will be quasi-static.

If I release the piston gas A will expand very slowly and gas B will be compressed. The sum of volumes is constant.

This process should stop when the pressures of gas A and B are the same.
Gas A will have cooled off and gas B will have heated up.

Here's my problem:
If I treat the expansion of A and the compression of B as reversible adiabatic, then the whole process will be reversible since nothing happens in the surroundings.
(I checked that analytically, just to be sure, and if I use expressions for reversible adiabatic processes ( P V^γ = constant, for example), I get ΔS = 0 for both gases (with some rounding errors).

However, by just looking at it, I would swear the process must be irreversible.

How would you calculate the final temperatures and volumes of the gases and the entropy-changes?

There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.

 

[aliinuur]

There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.

There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.

I think gas A and gas B volumes oscillate between 1/3V and 2/3V, so yeah, this is reversible process.

 

[Chestermiller]

This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.

 

[Herman Trivilino]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

There is no transfer of heat in this scenario. There is work being done, and temperatures are changing, but neither of those things need be associated with a transfer of heat.

Heat is a transfer of energy due to a temperature difference. Just because there's a temperature change you can't assume there's heat energy being transferred. There could instead be (and indeed there is) work done. That is the essence of the 1st Law of Thermo.

 

[Philip Koeck]

This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.

I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?

 

[Chestermiller]

I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?

Definitely. See chapter 1, Transport Phenomena by Bird el al.