Does air filling a vacuum produce a force?

Question

Imagine you have a vacuum chamber set at a complete vacuum, and there is a valve to let air in.

If you open this valve, the air molecules will start moving into this empty area, eventually filling it.

If you were to put this vacuum chamber on a frictionless surface, would it start to move?

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In other words, would this make ‘thrust’?

I know every action has an equal and opposite reaction, but in this case the vacuum chamber isn’t doing an action. The air is filling the void on its own, the vacuum chamber isn’t doing anything.

Because of that, I don’t think it would produce a force.

Am I missing something, or am I correct? (correct meaning that the situation I described wouldn’t make any force)

Answer 1

If no external forces are applied to your vacuum chamber the center of mass’ velocity will remain constant. So in case we do assume the chamber to be at rest initially it will also be at rest once the gas spread into the empty area and the center of mass will be constant at all times. However the walls of the chamber might be very will moving during the spreading, because the center of mass will be initially the center of the gas filled area, while at the end it will be the center of the whole area, but it has to be fixed in the inertial system, which has it at rest initially.

One way to think of it is, that the gas pressure on the one sight of the box with the full wall will exert more force on the wall then on the other side with the hole in the wall. The gas streaming out of the hole will have some momentum and there fore exert a larger force to the solid wall of the new area initially, which will stop the boxes movement again, before eventually equlibrating to exerting the same force on both side again.

Another way to picture it would be to think of the box half with the hole as a rocket, pressing gas out one side and therefore accelerating to the other, while the other half could be thought of as kind of an inverse rocket letting gas in on one side and absorbing its momentum. Both effects will cancel each other out eventually, however, because the accelerating starts a bit before the particles reach the opposite side of the new area, the box (meaning its hull not its center of mass) will move a bit before sitting at rest again.

Answer 2

Yes, the vacuum chamber is doing work.

Since the principles should be the same whatever the air pressure, let’s consider the case where the air pressure is so low that there is just one air molecule.

The molecule happens to be moving in a direction which takes it into the chamber. At this point nothing has happened. No force has been exerted by anything on anything. No work has been done.

The molecule reaches the far wall of the chamber and bounces off it. Thus its momentum has changed from $-mv$ to $+mv$ - the wall has done work on it - and accordingly the momentum of the chamber itself has changed by $-2mv$.

Some time later, the bouncing particle reaches the near wall off the chamber and bounces off it. This puts everything back to how it was at the beginning. The work done by the chamber reverts to $0$.

This reflection process can go on indefinitely. I have of course oversimplified, assuming a 1-dimensional chamber; but scaling up to 3 dimensions with possibly oblique reflections does not change the principle.

Now you can scale this up to more air molecules. At any given time, half of them are in the “having been reflected an odd number of times” state (thus, having had work done on them) and half are in the “reflected an even number of times” state, thus having had no work done on them.

In total, therefore, the incoming air has had work done on it by the vacuum chamber.

It is possible to approach this question by considering the momentum of the flowing air rather than individual molecules, and that is probably easier for quantitative calculation. But the fundamental principle is made more obvious by considering the molecular level.