2
$\begingroup$

This is a conceptual question more than anything else.

Assume you have a container with two ideal gases separated by a partition. The container is somewhere out in empty space and is not influenced by external forces or friction.

The two gases occupy the same volume $V$, have the same temperature $T$ and but have different total mass and Pressure $M_1$, $M_2$ and $P_1$, $P_2$ respectively with $M_1> M_2$. One can assume that the container and partition are massless compared to the masses of the gases.

Initially, before the partition is removed, the entire system is in equilibrium since the pressure of each individual gas is applying exactly the same force in all 6 walls that surround it. The net force on the container is zero.

At some point, the partition vanishes instantaneously. From entropic arguments one expects the gases to mix perfectly and reach some equilibrium values for their thermodynamic state parameters. When the mixing is complete the center of mass of the system lies in the center of the container. However in the initial configuration since the gases had the same volume but different masses, the center of mass was off-center and closer to the side which the heavier gas occupied. How is it possible that the center of mass of the container moves with respect to the container when there is no external force? How can that be consistent with Newton's laws?

The question above was the original form of the question I was trying to answer. I believe I have the correct answer for the above question, but I am struggling to figure out the answer to a variation of the above setup. So here is my answer to the question above:

This is really a relativity question more than a thermodynamics question. One is wrong to assume that the container itself is an inertial frame of reference. There is nothing special about the box. Instead the inertial frame of reference is by definition the one for which Newton's laws hold. In that sense the center of mass is always stationary and at the same location both before and after the gases mix. Yet however the center of mass moves from off-center to the center of the box. That is because the box moves backwards to accommodate the mixing of the gases in a way that will leave the center of mass unchanged. A coordinate system fixed on the container would simply be non-inertial since we need to violate Newton's law to account for the change of the center of mass. This is a reasonable expectation based on what will happen microscopically when the partition vanishes. Immediately after the partition vanishes and before the gases mix there is a net force on one side of the container as opposed to the other because of the concentration of the heavier gas (and also with the higher pressure) on that side. This implies that the container will accelerate towards the side with the heavier gas. Eventually after the gases mix conservation of momentum will force the container to be decelerated and stop since it is moving with respect to the net zero momentum of the gas particles. The end result is a movement of the box towards the side of the heavier gas in a way that makes the center of mass be at the center of the box.

All this is clear to me but I am struggling to understand the microscopic interpretation of this problem when instead of the gases having different pressures, they have different temperatures and the same pressure! The setup I described is exactly the same, you just compensate for the difference in mass, by assuming a difference in initial temperatures of the two gases in the initial configuration. It is still obvious to me that the center of mass remains stationary and the massless box moves backwards to accommodate the mixing of the gases but in that case, what causes the internal transient force on the container since we do not have a pressure differential?

$\endgroup$

1 Answer 1

1
$\begingroup$

The answer to the general question, as you know, is that the centre of gravity of the box and its contents never moves. If it moves relative to the box, that's because the box is moving.

Now let's take your more specific scenario in which the two parts of the box contain different masses of gas which are at different temperatures but the same pressure. When you remove the partition, you can no longer maintain the temperature differential- the temperature of the hotter gas will go down, while the temperature of the cooler gas will rise. That means the pressure exerted by the hotter gas will fall, while the pressure exerted by the cooler gas will rise. Since the cooler gas is the more dense, the temperature drop experience by the hotter gas will be greater than the temperature rise experienced by the cooler gas. The result is that the gas in the hotter side of the box will experience the greater drop in pressure when the partition is first removed, which will have the effect of causing an imbalance of forces on the hot and cool ends of the box. The end result is that the cool end of the box will recoil as the heavier cool gas disperses toward the hot end of the box.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.