Consider the experiment of Gay-Lussac: A cylinder is split into two compartments $A$ and $B$ by a barrier. Compartment $A$ initially contains some gas and compartment $B$ is initially empty. The walls of the cylinder as well as the barrier are assumed to be adiabatic.

Experiment shows that if a small opening is made in the barrier, allowing the gas to pass between the two compartments, then after some time the gas will fill the whole volume of the cylinder. Indeed it's a well-known property of gases that they fill the entirety of the volume of the container in which they are confined.

During this transformation, no heat is added to the gas, and no work is done on it either. Also the net external force on the gas is zero. Initially, the center of mass of the gas is at rest, located at some point in the interior of compartment $A$. But since the gas fills the whole volume of the cylinder at equilibrium, the center of mass must move.

But if the net external force on the gas is zero, the center of mass should always remain in the same place, no?

  • $\begingroup$ What if the glass was really half-empty? $\endgroup$
    – rob
    May 27, 2017 at 12:57
  • $\begingroup$ If you are in outer space the container will move so the position of the center of mass stays at the same place. On earth, friction or other forces holding the container will prevent that. in any case, think about a bunch of moving frictionless billiard balls instead of a gas. do you think they will not eventually move around all available space? $\endgroup$
    – user126422
    May 27, 2017 at 14:28
  • $\begingroup$ If I look at the fbd diagram of the gas before the partition is punctured then the forces on it from the jar wall and the partition and the jar wall would all add up to zero.After the puncture the force would be unbalanced in a direction pointing to the emty half.If I look at the forces on the jar and partition before and after a net force acts On the jar wall opposite the partition which has been punctured. $\endgroup$
    – Chappy
    May 28, 2017 at 0:40
  • $\begingroup$ By the act of pumping in the gas into the jar we have built up internal stress which gets released when the partition is punctured. $\endgroup$
    – Chappy
    May 28, 2017 at 1:17

2 Answers 2


You've proved that such a container, in the absence of any external forces or constraints, must shift somewhat in the opposite direction, which is the correct answer. In this way, the center of mass stays motionless.

Forcewise, the container is accelerated because the pressure of the gas is unbalanced after the opening is made. It is then decelerated as the gas molecules hit the opposite wall in the process of establishing a new equilibrium.

  • $\begingroup$ This is principle of jet propulsion. The gas leaving container A is the jet. The unbalanced force on A's wall opposite the hole through which the jet leaves provides the propulsion. $\endgroup$ May 27, 2017 at 12:38
  • $\begingroup$ @Chemomechanics: are you saying the whole container would move if I placed it on a slippery surface, even though there are no net external forces acting on it, and no gas is leaving the container (remember, we only have that the gas passes in between compartment $A$ and compartment $B$)? $\endgroup$
    – math_lover
    May 27, 2017 at 12:52
  • $\begingroup$ Depends what you mean by an "external" force. If you are a chunk of the container, how do you know which is your inside and which is your outside? $\endgroup$
    – Philip Roe
    May 27, 2017 at 18:06
  • $\begingroup$ @JoshuaBenabou Yes. $\endgroup$ May 28, 2017 at 11:33

Gas molecules have temperature, which means speed. There are trillions of them, shooting around, bouncing off each other and the walls. If you remove the barrier, the molecules that had been bouncing off it keep going into the empty part. And the ones that would have bounced off those also keep going. Basically, the randomness means that the probability of not filling the empty space, while not zero, is maybe 10^-kajillion.

And yes, the center of mass will remain in the same place, meaning that the entire container will move a certain distance.

How? you might ask. Because in the interval of time before the empty part filled up, the pressure inside was unbalanced, making it move. And when the gas sloshed up against the opposing wall, the motion stopped.

That's how rockets work. There is no opposing wall.

There are (yes there are) people who think rockets could not work in vacuum because there's no air to push against. They would just "fizzle" :)

  • $\begingroup$ I don't understand... This is not a rocket. The container (compartment A+compartment B) is closed, i.e there is no gas leaving the container. You are telling me, if I place the container on a frictionless surface, the net external force being zero and the system initially at rest, then the container is capable of moving as a result of stuff happening inside of it? $\endgroup$
    – math_lover
    May 27, 2017 at 13:31
  • $\begingroup$ @JoshuaBenabou Yes, of course. Look up 'jumping bean' for a macroscopic example of such a thing. $\endgroup$
    – user107153
    May 27, 2017 at 13:41
  • $\begingroup$ @JoshuaBenabou: It won't move far, but it will move and then stop. It all comes down to conservation of momentum. In this case, the gas moves one way and stops, so the container has to move the other way and stop, to keep the center of mass of both together having zero velocity. In a rocket the gas keeps going, and so does the container, still keeping the center of mass of both together having zero velocity. $\endgroup$ May 28, 2017 at 12:16

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