If the gas alone makes up the system (making piston and outside the surroundings), would it be correct to say that $P_{ext}\Delta V$ is the work done by the surroundings on the gas and $P_{int}\Delta V$ is the work done by the gas on the surroundings (where $P_{int}>P_{ext}$ if the gas expands)?

@cloud No, because the surroundings (including the piston) acts on gas with pressure $P_{int}$. So the gas does work $P_{int}\Delta V$ on the piston, and the piston does work $-P_{int}\Delta V$ on the gas.

But if the pressures the gas exerts on the outside and the outside exerts on the gas are equal, how can the gas expand in the first place? Wouldn't they be in machanical equilibrium? Also, in your answer, shouldn't $P_{ext}\Delta V$ (where $ext$ refers to only atmosphere and not piston) be the work the atmosphere does on the piston, not the other way around?

No. Sum zero of forces does not imply zero motion. It thus does not imply mechanical equilibrium. It's the other way around: mechanical equilibrium implies zero motion, thus zero acceleration, thus zero net force.

$P_{ext}\Delta V$ is work of piston on the atmosphere with pressure $P_{ext}$. The work of this atmosphere on the piston is $-P_{ext}\Delta V$.

Some people may define mechanical equilibrium as net force on body being zero, and thus include a uniformly moving wall into states that are in mechanical equilibrium. But the important point is the two reaction pair forces having equal magnitude does not imply any equilibrium - it's just Newton's third law, the magnitudes are always equal, even in non-equilibrium, when the wall accelerates.

What if the pistol is massless and frictionless?

@ChetMiller then the piston has infinite acceleration. In reality it won't be possible to have difference of pressure without some additional force stabilizing the piston.

The piston doesn't have infinite acceleration. Instead, the force per unit area exerted by the gas on the inside face of the piston will be equal to the external pressure exerted on the outside face of the piston. But, since the system is not at thermodynamic equilibrium, the force per unit area on the inside face of the piston is not equal to the value calculated from the ideal gas law. The ideal gas law applies only at thermodynamc equilibrium.l. Are you familiar with the rheological equation for a Newtonian fluid?

> The piston doesn't have infinite acceleration. The premise of the question is that there is pressure difference. You added another assumption, massless piston. Then there is net force on massless object, thus infinite acceleration. It's a thought experiment about idealized concepts. The realistic version is that there is no such thing as massless piston, it has some mass, and there is also some friction, or other force maintaining the pressure differential.

> But, since the system is not at thermodynamic equilibrium, the force per unit area on the inside face of the piston is not equal to the value calculated from the ideal gas law. How does this connect to the original question? These are additional assumptions that only complicate the discussion.

By your rationale, it is useless to study plane geometry since there is no such think as a perfect circle, a perfect square, a perfect rectangle, etc; boring with these involves additional assumptions that only complicate the discussion (of physics, for example). I have only one more question for you. In a rapid irreversible expansion or compression, is the "internal pressure' of the gas described by the ideal gas law, even though the gas is not at thermodynamic equilibrium?

@ChetMiller No, you're making further assumptions. My point is that I don't see why these things you invoke are relevant. The question does not ask about massless piston, or ideal gas law being valid during rapid irreversible expansion. It's based on a lack of understanding of the concept of work and what work of the gas vs work of the piston/wall is.

What you can surely say that between two equilibrium states of pressures, initial $p_1$ and final $p_2$, if the volume change has been $\Delta V$ then the useful mechanical work (exergy) loss is $(p_1-p_2)\Delta V$. If this has been done reversibly then there is an equal amount of work gained by another process or processes, say, gravitational or electric or thermal, etc. Otherwise, some or all of it would have been lost as heat: some of it if there is another work process and all of it if there is none.

@hyportnex I don’t follow this at all. If a gas goes between these two equilibrium states, the work done by the gas must be $\int{pdV}$.

@ChetMiller $\int_1^2 pdV$ is one kind of work, another kind is $(p_1-p_2)\Delta V$ is the lost useful work (exergy, or available work). The beauty of this quantity, availability/exergy, is that it also informs us how things get lost in dissipation and in the environment. It is a very intuitive and primal quantity, not as abstract as $pdV$ is.

I don’t see how we ca say this is the net work done by the gas. It’s the net work done on the container due to the sum of the work done by the gas and the work done by the external (to the container) agent.

@JánLalinský In thermodynamics, we divide the universe into two parts: the system and the surroundings. Do. you regard the piston as part of your system or part of its surroundings?

@ChetMiller Both can be valid. It depends on the question, there isn't a need to have a generally valid assignment of the piston. If the question is about work on the atmosphere, then piston does the work on the atmosphere, not the gas, thus the piston may be considered part of the system that does the work on the atmosphere. If the question is about work on the gas, then the gas is the system being acted upon. If we have both kind of questions, then we may consider the piston to be a third system.

Suppose you choose the gas as your system. Then $P_{ext}$ is the magnitude of the force per unit area exerted by the inside face of the piston on the gas, and $P_{int}$ is the magnitude of the force per unit area exerted by the gas on the inside face of the piston. How are $P_{ext}$ and $P_{int}$ related?

@ChetMiller they have the same value. But where's the point of this?