@ChetMiller I deduce from your answer that this is wrong. Could you explain why?

You first. You made a statement. Please explain your rationale.

Isn't it well known that the velocity of the barrier would equate or exceed the speed of sound in the ideal gas at initial time, when the process is irreversible? Furthermore, this statement was also made by @JohnRennie, but I don't know how to quote the source. (To be honest, I don't even understand why the downvote, since this question could be useful and I don't think it is obvious...)

Even with a massless barrier, the barrier would not travel with the speed of sound. But compression- and expansion waves adjacent to the barrier would travel at local sonic velocity. The important thing is that, within these expansion and compression zones, the gas is deforming very rapidly, and viscous dissipation of mechanical energy into internal energy is occurring. Even in the case of a piston with significant mass, the process would not be reversible since, eventually the kinetic energy of the piston would be dissipated by viscous effects.

@ChetMiller The OP and I have been discussing this in a chat room. My comment about the speed of sound was that if the barrier moves at significantly less than the speed of sound, sat 10% or less, the pressure in the chambers would have time to equilibrate and the process would be sufficiently close to reversible that the adiabatic equation $PV^\gamma=\textrm{const}$ could be used to describe it.

@JohnRennie If the barrier is moving at even on the order of 10% of the speed of sound, there will still be a siighnificant rate of deformation of the gas. on both sides of the barrier, and significant viscous dissipation within the gas on both sides. see Transport Phenomena by Bird et al, Homework problem 11.D.1 Entopy generation balance. In addition, if you assume that the process is adiabatic and reversible on both sides of the barrier in this particular problem, you will not be able to satisfy the condition that the internal energy of the combined system does not change,

@ChetMiller Shouldn't there be no change in internal energy regardless of whether entropy is generated since the system is isolated from the surroundings? What am I missing?

Yes, but if you use those adiabatic reversible equations for both gases, you will not be able to satisfy the aero change in internal energy condition. Tr it and see what you get.

@ChetMiller So you are saying it is impossible for one gas to undergo a reversible adiabatic compression at the same time as the other gas undergoes a reversible adiabatic expansion, correct? I will try it, but a physical explanation would be appreciated.

@BobD Both gases undergo irreversible changes. If only one undergoes reversible, the is no basis for deciding which one.

@ChetMiller What I meant is, is it theoretically impossible to internally control the partition in such a way that the left side slowly expands while the right slowly compresses so that both the expansion and compression are reversible? I'm thinking the problem may be that they cannot both end up with the same final pressure without a change in internal energy.

@BobD You can, of course, do it by doing work on the partition. But you can't if it is free to move.

@ChetMiller Could you explain to me why this process cannot be even close to being reversible (e.g. assuming the barrier is moving at a speed with a low relative percentage to the speed of sound)?

The only way to make the barrier move such that the process is reversible is if one exerts external force to control the motion of the barrier. If the barrier is free to move, but comes to rest in the final state, there will be generation of entropy with each of the two compartments, and the reversible equations will not apply.

@ChetMiller Sorry to belabor the point, but what I'm trying to say is IF, as a thought experiment, the barrier could be controlled internally, then the processes could be reversible with no contradiction to the need for internal energy to remain the same.

@BobD What you are describing would really be equivalent to controlling it extremely.

@ChetMiller Sorry again, but I am not clear on a few points. We are only told that a rapid transient takes place, but how do we know what kind of transformation takes place during? Under conditions of final equilibrium, don't we have that the two final pressures are equal and stop? Being ideal gases, isn't the only thing we can say is that gases obey the perfect gas law, i.e. $pV = nRT$ and that's it? Or is there some more subtle reason? I fail to see how talk of viscosity can fit in if during the transient we know nothing about the type of transformation and if the gas is ideal...

Do you think that the ideal gas law is valid during rapid irreversible expansions or compressions, or is it only valid at thermodynamic equilibrium??

@ChetMiller At thermodynamic equilibrium. I cannot understand how one can say that the transformation is irreversible or reversible if no information is given... All that is said is that there is a rapid transient....

See Transport Phenomena, Bird et Al, homework problem 11.D.1

@ChetMiller I watched it, but unfortunately it is not done and I did not understand much. I would like to understand one thing: am I wrong in saying that it is not possible to determine the nature of the transformation (reversible or irreversible) at thermodynamic equilibrium, since there is only a rapid transformation? If I am wrong, could you clarify this point? I would be grateful if you could provide further explanation.

If the initial and final states are thermodynamic equilibrium states, that is not enough. For a process to be reversible, the processs must be at thermodynamic equilibrium at all intermediate states. In other words, the process must consist of a continuous sequence of thermodynamic equilibrium states (over the entire process path). No spontaneous changes, involving non-equilibrium states, are allowed.

@ChetMiller OK, I understand. What I also want to say is: if I wanted to analyse the situation at thermodynamic equilibrium, first calculating the two final pressures (equal to each other), then the temperatures and volumes, it would really be necessary to know the nature of the transformation (whether reversible or irreversible)? If it is an ideal gas, one cannot impose that it satisfies the ideal gas law (being just such) to calculate the state variables $p, V, T$?

@ChetMiller Comments have been moved to the chatroom. Could you please clarify the previous question?

How about we solve this assuming that the gas undergoes adiabatic reversible expansion in each chamber and see where it leads. If you want to do this, go to PhysicsForums.com, register, and start a conversation with user chestermiller.