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I have read that no work is done during free expansion and temperature of an ideal gas undergoing free expansion doesn't change.

Let us assume an ideal system of piston and gas which has vacuum around it. The gas will expand but we say there is no work done by the gas. Why so ? Gas applies a pressure(& hence a force) on the piston and the piston gets displaced. So acc to F.s = work , there is work done by the gas !

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    $\begingroup$ The words "free" and "piston" contradict each other. $\endgroup$ Oct 14 '15 at 9:22
  • $\begingroup$ @ivan I have been taught in school that if we assume a weightless piston and vacuum around, we can regard it as free expansion. Is that incorrect ? $\endgroup$
    – biogirl
    Oct 14 '15 at 9:25
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    $\begingroup$ Weightless piston is another story. It is as good as no piston at all. $\endgroup$ Oct 14 '15 at 9:34
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    $\begingroup$ I think what biogirl is referring to is a massless piston. If you do a force balance on the piston (using Newton's 2nd law), the force exerted by the gas on one side of the piston is equal to the force exerted by the vacuum on the other side of the piston (i.e., zero force). So the force that the gas exerts on the massless piston is zero, and the gas does no work on its surroundings. $\endgroup$ Oct 14 '15 at 17:44
  • $\begingroup$ @ChesterMiller I know this is a silly doubt but can the pistin move freely inside the container? $\endgroup$
    – Scáthach
    Mar 28 '18 at 7:29
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Little bit tricky question. As you say, the work is force acting along the path. But there is no force acting, as there is weightless piston with vacuum on the opposite side.

To visualize the process better, imagine an inflatable balloon in outer space, when you puncture it. All gas molecules continue to fly in the direction and speed they had in the moment of accident (therefore the temperature is the same) and they exert no work (no collision thereafter).

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  • $\begingroup$ Is there a reason for which we considered a weightless piston $\endgroup$
    – Scáthach
    Mar 28 '18 at 7:36
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I have seen some answers on this similar question they all point to same thing that pressure outside is zero so from $W = p.\Delta V$, work done will be zero or as said in the answer from @ssavec that there is no force acting on the piston, the point is that you cannot have a balloon in space, it will burst due to inner pressure.

My physics teacher gave another perspective for this situation which in my opinion is much more relevant. It goes like this:

Since the gas is in a vessel with massless piston, then it will definitely have some pressure in normal conditions and even in vacuum, but the twist here is that, this pressure once pushes the piston and after that because piston had no resistance, it went out with some constant velocity. There was no direct force and displacement relation but just a sudden push from gas which made piston leave immediately.

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