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In the book of Kondepudi & Prigogine, Modern Theormodynamics, at page 114, it is stated that

In a reversible expansion of a gas, the pressure of the gas and that on the piston are assumed to be the same. If we consider an isothermal expansion of a gas that has a constant temperature T by virtue of its contact with a heat reservoir, the change in entropy of the gas deS = dQ/T, in which dQ is the heat flow from the reservoir to the gas that is necessary to maintain the temperature constant. This is an ideal situation. In any real expansion of a gas that takes place in a finite time, the pressure of the gas is greater than that on the piston.

and they go on showing what happened when they are not equal.

However, I'm having trouble understanding how can such a thing even possible ?

I mean, by definition, the pressure on the piston is the pressure applied on the piston by the gas (it is not like the piston has its own internal pressure or something), so how can a gas apply a lower pressure than its own pressure to an external object. I mean, as far as I know, the way how we measure the pressure of a gas is by means of such an external object, so it really does not make any sense to me.

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  • $\begingroup$ Which “thing” are you having trouble understanding, the quoted statement or what happens when the pressures aren’t equal? $\endgroup$
    – Bob D
    Commented Sep 2, 2019 at 11:15
  • $\begingroup$ @BobD I'm having trouble in understanding how can those two values be different. $\endgroup$
    – Our
    Commented Sep 2, 2019 at 11:21
  • $\begingroup$ In that case you need to tell us what the author went on to say so that we can see the context $\endgroup$
    – Bob D
    Commented Sep 2, 2019 at 11:30
  • $\begingroup$ Suppose there are reservoirs of gas on both sides of the piston, but at very different pressures. $\endgroup$
    – gandalf61
    Commented Sep 2, 2019 at 11:30
  • $\begingroup$ Are you familiar with Newton's 3rd law of action-reaction which says that the force exerted by B on A is equal in magnitude and opposite in direction to the force which A exerts on B. $\endgroup$
    – Chet Miller
    Commented Sep 2, 2019 at 11:50

3 Answers 3

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When the text refers to the pressure "on the piston" it clearly means the pressure applied to the other face of the piston. If the pressure applied to the other face of the piston were always exactly the same as the pressure of the gas then the piston would not move and there would be no expansion or compression of the gas. On the other hand, if we have a vacuum on the other face of the piston (and assume a light piston with no friction) then the gas will expand very rapidly, and this process is irreversible.

A reversible expansion is an ideal situation where the pressure on the piston is infinitesimally less than the pressure of the gas, so the piston moves at an infinitesimal speed. See https://en.wikipedia.org/wiki/Reversible_process_(thermodynamics).

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Your edit changes things considerably. Now, instead of looking at a reversible expansion, you are looking at an irreversible expansion. But, in the sense that you originally indicated, the authors are incorrect in saying that the gas pressure is higher than that exerted by the piston. If, by pressure, they mean the force per unit area (normal stress) exerted by the gas on the piston face, then this is always equal to the force per unit area that the piston exerts on the gas. But, if they mean the pressure calculated from the ideal gas law $nRT/V$, this value will be higher than the force per unit area exerted by the gas on the piston and the piston on the gas. The difference between these two views is related to the irreversible fluid dynamics behavior that is occurring. In particular, it is related to the viscous behavior of the gas as it is deforming. I can get into this in more detail if you would like, but I'm not sure you would be interested.

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If the piston in a cylinder is moving rapidly away from the gas, then the momentum transferred to the piston from each colliding gas molecule will be less, thus reducing the effective pressure.

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  • $\begingroup$ "then the momentum transferred to the piston from each colliding gas molecule will be less": how ? $\endgroup$
    – Our
    Commented Sep 2, 2019 at 19:44

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