# Confusion on work done by an ideal gas

When an ideal gas,in a piston kind of system and whose equilibrium state is mentioned, is allowed to expand (piston is allowed to move and not gas leaking )against a constant external pressure very quickly, then, is the work done by gas zero or not zero ?

The arguement for work being zero is that "the constituents take sometime to move towards the piston, as the piston moves away, and hence there's no pressure one the piston due to one side (inside ) and hence zero " I feel this argument is wrong because the pressure is 0 from the inside then the piston would be compressed by the external pressure.

Thank you in advance.

"the constituents take sometime to move towards the piston, as the piston moves away, and hence there's no pressure one the piston due to one side (inside ) and hence zero

Yes, the argument is wrong.

The piston would keep moving one way or the other till the pressure inside the cylinder(let, the shape of the container be cylinder) and outside the cylinder is equal.

And remember that you are in macroscopic world. There would not be few particles there would be particles millions of millions of particles, in order of 23. So, when some particles are away from piston, then there are others at work.

is the work done by gas zero or not zero ?

If the piston moves then work done by the gas is not zero done. You can apply the classical mechanics, where work done is equal to force x displacement.

• Just to clarify, W in the first law of thermodynamics is external pressure times the change in volume here right ? – Anton Aug 27 '18 at 10:20
• @allmysolvere Yes, W in the Ist law is external pressure times the change in volume. – TontyTon Aug 28 '18 at 9:32

When you release the piston, since the gas has inertia, a dilatation region begins to form in the gas immediately adjacent to the piston face. Within this region, assuming a massless piston, the pressure is virtually equal to the outside pressure (since the piston has no mass, it can accelerate at finite speed with even the slightest differential in pressure between inside and outside). As time progresses, the size of the dilatation region gets larger, and its leading edge travels backwards at the speed of sound within the gas (along the cylinder axis). So the phenomenon is similar to that of a sound wave.

In addition to this, there are viscous fluid stresses that develop within the gas that also contribute to the force per unit area exerted by the gas on the inside face of the piston. But, again, for a massless piston, Newton's third law tells us that the force exerted by the gas on the inside face of the piston must equal the constant external force exerted on the outside face of the piston.

I think that you can use, in your mind, the formula: $dW = p\cdot dV$

The external pressure is known and is constant as you say. So p = pressure is not zero. The gas expands its volume so dV is not zero aswell. Finally, dW is not zero.

I think a more straightforward answer would be "the piston only moves because the gas particles are colliding with it". So there is no region where the gas is of lower density. If the pressure on the inside and the outside is not the same, the gas particles push the piston, doing work on it and expanding the region until the pressure drops far enough for equilibrium to be reestablished.

My answer is idealized in some way, but consistent with the model of an ideal gas. If there ever was such an underdensity in the gas, it would essentially not be ideal any longer.