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Let's say I have an isolated, adiabatic chamber which is divided into two parts with a massless, frictionless piston. One part has vacuum, while the other has a gas. In the beginning, they are at equilibrium. I release the piston, and the gas expands, thereby pushing the piston towards the end of the box. Since the process is a free expansion, there is no work done by the gas. One could thus say, that the gas applied no force whatsoever, on the piston. Then why did the piston move in the first place?

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    $\begingroup$ You can't do any work on a massless piston - it has no inertia, therefore no energy is expended by accelerating it. $\endgroup$
    – Brionius
    Jan 14, 2015 at 15:39
  • $\begingroup$ Then are you saying that this particular example to illustrate free expansion is flawed? $\endgroup$
    – Lexicon
    Jan 14, 2015 at 15:41
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    $\begingroup$ it can be seen as an unphysical limit $m\to0$ showing that the work done tends to 0 as well, hence in free expansion, where the separation between the two chambers is removed, there is no work done. $\endgroup$
    – Phoenix87
    Jan 14, 2015 at 15:43
  • $\begingroup$ Okay, so in contrast, if i had a piston with mass, would there still be no work done, as essentially, there is no opposing force? $\endgroup$
    – Lexicon
    Jan 14, 2015 at 15:46
  • $\begingroup$ you need a force to accelerate a mass and when the force is causing a displacement (which is not orthogonal to the force) there is work involved $\endgroup$
    – Phoenix87
    Jan 14, 2015 at 15:47

2 Answers 2

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Let the piston have a bit of mass, and just think it through. The piston will accelerate and this involves energy moving from the gas to the piston. The piston then hits the end of the rigid chamber and comes to a stop giving up its kinetic energy. Where does that energy go? It goes to raising the temperature of gas and piston. But the gas has, at this stage, cooled a little. So the temperature of the gas first falls a bit while the piston accelerates, and then rises back up as the piston gives up its energy. Eventually the system settles down to an equilibrium in which no energy has left the chamber (if we assume this piston is internal, so did no work on any external body).

In the case of an ideal gas, in this final equilibrium, the final temperature is the same as the initial temperature.

The point is that the gas does no work on anything outside the chamber, and nothing inside the chamber acquired any energy (in the end) from the gas if we assume the heat capacity of the piston is negligible.

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Work:

$$W=Fd$$

$F$ is force done to do displacement $d$.

Think of it intuitively like this:

If you push something very hard (large force), but it doesn't move at all - like pushing on a building - you don't do any work of any use. If you push something very far, but it didn't require much force - like pushing a balloon - then you also really didn't do much work (it wasn't much of an effort).

If the piston is a balloon that you hold in place in front of the opening to the gas chamber, then when you let go, the gas pushes this very light balloon out of the way. That doesn't require much force from the gas, so not much work is done.

Remember that to make something move, you must apply a force $F=ma$. If this mass $m$ is very, very small, then the force needed is also very, very small.

A massless piston is a idealization - to move zero mass requires zero work.

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    $\begingroup$ Aah, i see. As phoenix87 said, as m---->0, the work also tends to 0. If the mass was exceedingly small, the work would also be exceedingly small, but non zero nonetheless. Thanks! $\endgroup$
    – Lexicon
    Jan 14, 2015 at 15:58
  • $\begingroup$ What do you mean by "just wasted your energy"? Where did the energy go? Do you pull the chemical processes in the muscles here too? $\endgroup$
    – Ruslan
    Jan 14, 2015 at 16:51
  • $\begingroup$ @Ruslan a human being dissipates energy within the muscles to maintain a constant pushing force on an object. This is converted into heat in the muscles in the body. But you are right that energy should not be mentioned in this context - it is confusing matters. I have removed it from the answer. $\endgroup$
    – Steeven
    Jan 14, 2015 at 18:45

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