# Slow expansion: Work done by gas vs. Work done by environment

Suppose an ideal monoatomic gas expands and fills a container of volume $$V,$$ where one wall is a movable piston.

The text-book scenario of "reversible adiabatic expansion" 1 is that there is sand on the piston, and as one removes sand grain by grain, the piston moves slowly as the gas expands and pushes against the piston. This leads to the gas doing work on its environment $$dW = -p dV$$.

If, on the other hand, the piston is rigid, and we move it slowly by hand, the process is still adiabatic as it is moved slowly enough ("quasistatically"). However, in this case, the physical origin of the expansion is not the gas pushing against the piston, but an external force moving it. My intuitive understanding is that, according to the argument above, the gas does not do any work on the piston. Is this true, i.e., will there still be work done by the gas?

• In force–displacement work, gas molecules lose energy when bouncing off a retreating wall (and gain energy when bouncing off an approaching wall). This clarifies that it doesn't matter for the work calculation if the wall is moving predominantly because the gas is pushing it or predominantly because something else is pushing it. Commented Jul 28 at 17:07

In both cases, the force exerted by the gas on the piston is the same as the force exerted by the piston on the gas up to the sign (Newton’s third law). Furthermore the displacement $$\Delta z$$ used to compute the work is the same in both cases. As a consequence, the work of the gas on the piston and the one of the piston on the gas are always identical up to the sign. It does not matter if there is some other force on the piston…