Confused about Reversible and Irreversible processes

Question

Let us look at this situation.

Consider a container having a gas trapped inside, facing pressure from masses M1 and M2, assume that this system is at equilibrium. Here point A represents the initial position of the piston before mass M2 is removed.

After removing mass M2 the pressure decreases and the volume of the gas increases forcing the piston to rise up. Here point B represents the final position of the piston after the system has attained equilibrium again.

"The aforementioned situation is an irreversible process." , said my teacher.

My question is -

Why and how will the parameters(state functions) change from their initial values while going back from B to A resulting in the journey from A to B to become irreversible?

Answer 1

Why and how will the parameters(state functions) change from their initial values while going back from B to A resulting in the journey from A to B to become irreversible?

There will be no change in the state functions (S, U, P, T, V, etc.) of the system if, by any means, it can be brought back to its original state A. That is because state functions of the system do not depend on the path between equilibrium states. But because the expansion processes is irreversible (due to the sudden removal of $M_2$) there will be a change (increase) in the entropy of the surroundings if the system is brought back to its original state.

In order to calculate the change in entropy of the surroundings, you would need to have more information on the system and its surroundings. Are the cylinder and piston thermally conductive or insulated? If conductive, are the surroundings considered to be a thermal (constant temperature) reservoir?

Hope this helps.

Answer 2

For the two-step process you describe, the amount of work done by the gas in expansing is less than the amount of work done on the gas in compressing. This is because, in expanding, the gas is only lifting M1 while, in compression, the gas is being compressed by M1+M2. So the net amount of work done by the masses on the gas is positive. If the initial and final temperatures of the gas is the same (say by allowing heat to exchange between the gas and a constant temperature reservoir), the amount of heat that has to be removed from the gas in the process will be positive and equal to the net of work done on the gas by the masses. This means that, even though the gas returns to its initial state after the two-step process, heat has been transferred from the gas to the surroundings (the reservoir), and the reservoir is no longer in its initial state. So, by definition, the process is not reversible.

Answer 3

Irreversibility occurs whenever a difference in an intensive property (temperature, pressure, surface tension, voltage, magnetic field, mass concentration) leads to action.

“But that’s every process.”

Yes, every real process is irreversible. Note, however, that we could approach reversibility by conducting processes nearly in balance, with very low friction.

“If you reduced the driving force for a process toward zero, it would slow down to the point of unusability.”

That’s true. Nevertheless, this idealized case is still useful for analysis (in technical terms, it’s the only case that doesn’t generate entropy).

“How does this apply to this problem?”

When the mass is suddenly removed, the gas—now out of equilibrium—accelerates the piston upward. The forces are unbalanced during this process, so it’s irreversible.

“But I could return the gas to its original state. Doesn’t this connote reversibility?”

Not in a global thermodynamic sense, which is usually the context of thermodynamic (ir)reversibility discussions. The braking of the piston heats the universe (and the larger the removed weight is, the greater the imbalance, the greater the acceleration, the greater the required braking, and the greater the resultant heating and entropy generation), and this heating cannot be removed. That constitutes irreversibility.