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There are various things I am not sure I fully understand regarding reversible and quasi-static processes. I have tried to find similar questions on this forum, but so far none have fully answered my questions here:

  1. When we say a "process" in this case, then do we mean a transition from an initial well defined equilibrium state between the system and it's surroundings (i.e. the pressure and temperature of the system and surroundings are well defined and equal) to a final one? If the answer is yes, then I think it clears up some of the other questions below.

  2. Why are all reversible processes quasi-static? If the answer to the above question is yes, then I guess if the process is not quasi-static (I.e. at some point the system is not in an internal equilibrium state), then an entropy increase is required for the system to go back to equilibrium, right?

  3. Why is the work given by dW=-pdV during a reversible process, and why is the work done on the system greater than this in any irreversible process? I understand that if the external pressure (I.e. the pressure from the surroundings) is the same as the internal pressure, then the work is indeed equal to dW=-pdV, but I see no reason why this can't be true for irreversible processes as well, especially not if we have a quasi-static irreversible process.

  4. Why do we require that there is always thermodynamic equilibrium between the system and surroundings during a reversible process? If the answer to question 1 is yes, then much like question 2, I guess we again have that an entropy increase is required to go back to equilibrium again if at some point there is not equilibrium during the process?

Thanks for any answers in advance!

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  • $\begingroup$ Regarding item 3, do you think that the ideal gas law (or other equation of state) is satisfied within the system during an irreversible rapid-deformation process? At the interface between the system and surroundings, is the force per unit area exerted by the system on the surroundings equal to to the force per unit area exerted by the surroundings on the system (i.e., Newton's 3rd law)? $\endgroup$ Dec 28, 2020 at 13:00

2 Answers 2

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  1. A thermodynamic process is a passage of a thermodynamic system from an initial to a final state of thermodynamic equilibrium. The initial and final states are the defining elements of the process. So Yes!
  2. Any reversible process is a quasi-static one. However, quasi-static processes involving entropy production are not reversible. More detail here.
  3. The equation $$\delta W=-pdV$$ is only strictly true for a reversible change. The idea is that if the the piston is not frictionless, or if you move the piston too suddenly and generate shock waves, you will need to do more work to compress the gas because more heat is dissipated in the process.
  4. The reversible process has great importance for the engineer who designs engines, in which you want to waste as little heat as possible to make your engine as efficient as possible.

Recall Carnot's theorem for instance

Of all the heat engines working between two given temperatures, none is more efficient than a Carnot engine.

Carnot engine is a reversible one. This alone is not important for the reversible process.

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  • $\begingroup$ "The initial and final states are the defining elements of the process". This may be misconstrued to mean the initial and final states define the process, which of course they do not. $\endgroup$
    – Bob D
    Oct 3, 2021 at 15:34
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  1. Yes. But there may be many potential paths (combination of processes) that connect the same two equilibrium states.

  2. All reversible processes are quasi-static, but not all quasi-static processes are reversible. If the process is not reversible then entropy will be generated. But the change in entropy of the system between the same two equilibrium states will be the same regardless of whether the process is quasi-static because entropy is a state function and is independent of the path. For this to be the case, in general any entropy generated in the system must be transferred to the surroundings in the form of heat (an exception being an irreversible adiabatic process).

Why is the work given by dW=-pdV during a reversible process,...I see no reason why this can't be true for irreversible processes as well, especially not if we have a quasi-static irreversible process.

It applies to both a reversible and irreversible process if $p$ only represents the external pressure and the pressure at the boundary between the system and surroundings. For an irreversible process the pressure of the system varies spatially internally (i.e., there is no single defined pressure). For a reversible process, $p$ is the equilibrium pressure for both the system and surroundings.

For an irreversible process the equation is sometimes written as $dW=-p_{ext}dV$ to denote external pressure only. (As an aside, $dW=-pdV$ applies to the $\Delta U=Q+W$ version of the first law, commonly used in chemistry, whereas $dW=+pdV$ for the $\Delta U=Q-W$ version commonly used in physics).

...and why is the work done on the system greater than this in any irreversible process?

As noted in my answer to (2), in general for an irreversible process generated entropy is transferred to the surroundings in the form of heat. That leaves less heat available to perform work than for a reversible process.

  1. Why do we require that there is always thermodynamic equilibrium between the system and surroundings during a reversible process?

If a process is reversible both the system and surroundings can be returned to their original state, so that the total entropy change is zero. For an irreversible process the system may be returned to its original equilibrium state but the entropy of the surroundings will have increased, so the total entropy change (system + surroundings) is greater than zero.

Hope this helps.

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