If a gas expand in vacuum, $P_{ex.} =0$. Then why is the work done 0? Since the gas is expanding, it is doing some work on the piston so the work done must have some negative value.

  • $\begingroup$ How about a balloon bursting with the gas expanding into vacuum? Or opening a valve (you would do some work opening the valve, but the gas would not be involved)? There are many situations where the process described is an "ideal" one that can only be approximated in reality. $\endgroup$ – Karsten Theis Dec 5 '19 at 16:07

There is no work done on moving a piston against zero external pressure (not if it is ideal, ie lacks mass and there is no friction in its movement). You might consider the definition of $pV$ work $$w = -\int_{V_\mathrm{ini}}^{V_\mathrm{fin}}p_\mathrm{ext} \mathrm{d}V$$ Obviously if the external pressure $p_\mathrm{ext}=0$ then $w=0$.

Conceptually, the point is that a collision by a single gas molecule would suffice to drive the piston to its extended position. In addition, that gas molecule would expend no energy, it would not slow down or change course in pushing the piston, it wouldn't feel it at all (this can get complicated, as explained further below). As absurd as that sounds, this is accurate within the constraints imposed by this conceptual question.

The following schematic contrasts the effect of a gas molecule impinging upon an ideal versus a slightly more realistic piston.

enter image description here

In the real case forces during the collision transfer momentum to the piston and reduce or reverse the forward momentum of the gas molecule. In the case of the ideal piston the collision is best modeled as a perfectly inelastic collision between the massless piston and the gas molecule, the resulting united body having the same momentum (in the direction of the piston axis) as the original momentum of the gas molecule. There are no forces however and therefore no work is done.

† Components of the molecule's momentum along the orthogonal direction are conserved as well, and the molecule must be reflected once it reaches the chamber wall.

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