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According to Wikipedia:

"In thermodynamics, a quasistatic process is a thermodynamic process that happens slowly enough for the system to remain in internal equilibrium."

But if we look things more closely, I don't think that the above statement is achievable. Suppose that we have the classic thermodynamic system (a container with gas inside) thermally isolated and at equilibrium $ (P,V,T) $. We assume that there is no gravity and there is no friction between the piston shown and the walls. A schematic of the system can be seen in the following figure.

[1]: https://i.stack.imgur.com/pScTS.png

Where the red rectangular shape represents the piston. At this stage, the pressure at the outer side of the piston equals $ P $ to keep the piston at equilibrium. Suppose now we want to change the state of the system to $ (P',V',T') $ with $V' < V$ via a quasi - static process. We do that by exerting a force $\vec{F} $ with $ |\vec{F}| \rightarrow 0 \ $as shown:

enter image description here

And thus, in the beginning the piston accelerates a little$^{1}$ to gain some velocity which is then held constant (and small). During this phenomenon, the pressure changes near the piston and the new pressure gets propagated through the entire gas$^{2}$ (This happens until the process stops).

  • $^{1}$My first point is that, because of that initial acceleration that needs to happen to get the piston going there is a point in time where the function of pressure $P(\vec{r},t)$ is not homogeneous.

  • $^{2}$ Even when the velocity of the piston is constant, the new pressure are propagated throughout the gas and thus there are times (after the first step of accelerating the piston) where $P(\vec{r},t)$ is again not homogeneous.

In conclusion, however slow a process might be, (I think) its not capable of preserving the system at internal equilibrium at all times. The time intervals of pressure propagation might be really small, but they are not zero. Is then the concept of a quasistatic process an idealization ? if not, where am I wrong ?

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    $\begingroup$ You're correct that it's an idealization the is exact only in the limit. The same goes for perfect geometric objects like squares and circles. However, many processes in real life come pretty close to being reversible. $\endgroup$ – Chet Miller Mar 14 at 2:00
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The important point to note is that if you change the pressure $\delta P$ , the entropy created is second-order in $\delta P$.

Thus, by increasing the number of steps, you can make as small as you want the created entropy.

For a Joule expansion (in the vacuum) it is not the case: the variation of entropy is of the first order compared to the change of volume. In this case, a succession of a large number of small expansion remains irreversible.

Sorry for my poor english !

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