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I have the following definitions:

  • A transformation is said to be quasistatic if it passes only through equilibrium states (that is, the thermodynamical coordinates of the system are defined at every time instant during the transformation).
  • A transformation is said to be physically possible if it does not violate the conservation of energy and the second law of thermodynamics expressed by the Kelvin-Planck statement, or equivalently, by the Clausius statement.
  • A transformation F is said to be reversible if there exists a physically possible transformation G such that 1) the initial state of G equals the final state of F; 2) the final state of G equals the initial state of F; 3) the heat $Q_F$ exchanged by the system during F equals the opposite of the heat $Q_G$ exchanged during G (that is $Q_F = -Q_C$); and 4) the work $W_F$ exchanged by the system during F equals the opposite of the work $W_G$ exchanged during G.

My question is: is my definition of reversible transformation well-posed? If it is, then how can I logically and rigorously prove that if a transformation is reversible then it has to be quasistatic without using the concept of entropy, but using only the Kelvin-Planck and Clausius versions of the second law of thermodynamics?

I know how to prove that not all quasistatic transformations are reversible with the previous definitions, so I am asking for a logical and rigorous proof for the opposite: all reversible transformations are quasistatic.

If my definition of reversibility is wrong, then what is a definition of reversible transformations that does not refer to "quasistaticity" and does not use the concept of entropy such that I can prove that all reversible transformations are quasistatic, again, without using entropy?

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3 Answers 3

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You cannot prove that a process is reversible from being quasi-static because it is not true. The most obvious example is B/H ferromagnetism or stress/strain beyond elasticity. It does not matter how slow the process is you will have a hysteresis whose size is not rate dependent.

The simplest way to formulate reversibility mathematically is having all relationships among the thermodynamic variables be piecewise differentiable functions, no rates, no gradient, "no nothing", only simple variables and their relationships as functions. Entropy is much easier concept than "heat" is, and you could get to the 2nd law without the 1st law, hence entropy is more fundamental.

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  • $\begingroup$ My question asked for the opposite: I wrote that I know that if a process is quasistatic it does not necessarily mean that it is reversible. But it is known to be true that if a process is reversible (see the first answer in this post physics.stackexchange.com/questions/328168/…), then it is quasistatic: well, how do we prove it? $\endgroup$ Nov 26, 2022 at 0:22
  • $\begingroup$ as I wrote above, if reversible you have equations without time and rates (time derivatives) that is static, and the quasi is because we move from one static equilibrium state to another without any time or rate dependence. $\endgroup$
    – hyportnex
    Nov 26, 2022 at 0:46
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You cannot prove it, because not all the reversible processes are quasi-static.

As an example,

  • many elastic processes are reversible,
  • many processes involving the fluid, with no shock, negligible viscosity and heat conductivity are approximately reversible as well.
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This is not a matter of proof but one of definition. By quasi-static we mean that the system is always at true equilibrium except for infinitesimal gradients in $T$, $P$ and $\mu$ in order to nudge the system in the desired direction. Under this idealization the system is always at equilibrium, therefore $S_\text{gen} = 0$ for such process.

A practical interpretation of quasi-static is "do things slowly". This, however, is not a guarantee of reversibility, as others have commented here. Doing things slowly is often the key to minimizing gradients but should be unconditionally treated as proof of reversibility.

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