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Thanks to the help of @ChetMiller, the following fact is essentially concluded in this thread. Consider a rigid, thermally isolated container divided by a massless barrier parallel to its base into two parts, left and right, each containing $n$ moles of the same (ideal) gas, with the barrier preventing the exchange of particles and heat between the two parts. If at a certain instant the moving wall is left free to flow without friction and the system, after a rapid transformation, reaches a new equilibrium condition, then this transformation must necessarily be irreversible.

Now, consider the same physical situation, except that between the moving wall, of small but non-zero mass (unlike before) so that the weight rests on the edges of the box, and the inner edges of the box, there is an initially infinite friction coefficient $k$, which we then relax very slowly until $k=0$.

Haven't we essentially brought the system from the initial state to the final state, passing only by quasi-static transformations? If not, I must be missing something; if so, is there a formal explanation for this?

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Haven't we essentially brought the system from the initial state to the final state, passing only by quasistatic transformations? If not, I must be missing something;

What you are missing is quasi-static is a necessary but not sufficient condition for a process to be reversible. It must also not involve friction.

Hope this helps

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  • $\begingroup$ Thanks. As far as I understood from your answer, quasi-static does not always correspond to reversible. But I would like clarification on one thing: with your answer are you also saying that the quasi-static nature of the process cannot be guaranteed if you start with an infinite coefficient of friction and gradually release it to zero? Or, even though there is no reversibility, with this arrangement of the friction coefficient you still only go through quasi-static transformations? $\endgroup$
    – Bml
    Commented Dec 18, 2023 at 15:16
  • $\begingroup$ @Bml You can have a quasi-static process with friction but it will not be a reversible process. $\endgroup$
    – Bob D
    Commented Dec 18, 2023 at 15:25
  • $\begingroup$ OK. I formulate my request differently. Regarding the physical situation I described, why might it not be a quasi-static process? From my point of view, even if not reversible, it should necessarily be quasi-static, no? I have also modified my initial request to make it more in line with what I would like to understand. Thank you. $\endgroup$
    – Bml
    Commented Dec 18, 2023 at 15:29
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    $\begingroup$ @Bml I've already told you it it can be quasi-static. Just as long as it happens slow enough that the difference in pressures between the sides is kept infinitesimal and there are no temperature gradients within each gas. $\endgroup$
    – Bob D
    Commented Dec 18, 2023 at 15:35

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