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In free expansion W=0, even though volume changes and if q=0, then temperature and internal energy do not change. But what about pressure? I will present some arguments, please tell me which one(if any) is right, and also give your own arguments.

Consider ideal gas in an adiabatic container separated from vacuum side by a movable partition. Following are the arguments:

  1. Pressure of a gas is its momentum transfer to a wall/boundary, and momentum is mass times velocity. But R. M. S velocity of an ideal gas is a function of its temperature and since temperature does not change in free expansion, therefore velocity should remain same, and therefore pressure also remains same.

  2. Since temperature and amount of substance remain constant, therefore according to boyle's law, pressure should decrease.

  3. Since volume increases, therefore the distance travelled by a molecule to hit the side wall also increases, decreasing the velocity, thus decreasing the pressure. But here the counter argument could be that the number of collisions between molecules, will also decrease as volume increases, which may neutralise the effect of increased volume on velocity.

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Pressure doesn't remain the same because the number of collisions per unit area decreases when the volume increases. The velocity distribution of the molecules doesn't change because the temperature doesn't change.

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2 is correct, as is something along the lines of the first part of 3. The temperature and amount of gas have remain constant and the volume has increased, so the ideal gas law says that the pressure must have decreased. For a microscopic picture of why this has happened, the simplest argument is that the density of the gas has fallen, so the rate of collisions with a given part of the wall has also fallen.

The problem with argument 1 is that it does not consider the rate of collisions, only the mean momentum transfer with each collision.

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  • $\begingroup$ This is what i have inferred from the answers posted by Chester Miller and you: If we were to measure the pressure exerted on side wall before and after expansion for a duration of 1 minute, then the pressure after expansion would be less, because of fewer collisions per second. However, the momentum transfer to wall by 1 molecule per collision would remain the same in both situations i.e before and after volume change. Am i right to think this way? $\endgroup$ – Mohammad Nayef Jul 11 '18 at 12:37
  • $\begingroup$ essentially yes (although at around room temperature and pressure, measuring the the pressure over a fraction of a second would be enough to get a reliable average) $\endgroup$ – By Symmetry Jul 11 '18 at 14:09

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