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Now, by this I don't mean the effect when pressure is too high such that the molecules are taking up a significant amount of space, however, I think it might be a different angle to approach the pressure problem from.

Pressure exerted by a gas on the wall of a container can only be through the elastic collision between the gas molecule and the wall. If there was just a low amount of particles it would make sense that $P \propto \frac{n}{V}$ because every single molecule would bounce against the wall exerting pressure.

But this doesn't feel like it should work if we had a lot of particles. For example, there may be one unlucky molecule that keeps bouncing against other molecules molecules but itself never gets to hit the wall. This would mean that this particle never exerts pressure on the wall. Thus the proportionality is no longer true.

My question is, is this a real effect and if so why doesn't it affect the equation? Is this just another reason why high pressure doesn't work for the ideal gas assumption?

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Indeed, the ideal gas must be sufficiently sparse that we can neglect the collision between the molecules, apart from the fact that these collisions lead to the establishment of thermal equilibrium. When the density of molecules is too high, we deal with a non-ideal gas or even a liquid. The first approximation for non-ideal gases is van der Waals equation $$\left(P + a\frac{n^2}{V^2}\right)\left(V - nb\right) = nRT,$$ as an alternative to the ideal gas equation $$PV = nRT.$$ Here $a$ and $b$ are phenomenological coefficients describing the intermolecular attraction and the finite volume occupied by the molecules. In particular, the pressure of this gas on the walls of a container would be different that that of an ideal gas (for the same $V,T$).

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  • $\begingroup$ so basically in the ideal gas law we assume that all the particles will get a chance to hit the wall within a reasonable time frame? But I don't see this correction in that formula - it only seems to correct for intermolecular attraction and the volume of the molecules not for the probability that particles may never hit the wall $\endgroup$
    – John Hon
    Apr 21, 2020 at 23:57
  • $\begingroup$ If you express pressure, it is reduced by the term proportional to the strength of the intermolecular interaction and molecular density: $-an^2/V^2$. It is supposed to take account for your scenario and other effects reducing pressure. You probably want to see a direct calculation of pressure as the rate at which molecules hit the wall - it is hard to do for non-ideal gas. In practice the pressure for interacting systems is calculated from the thermodynamic identitirs. $\endgroup$
    – Roger V.
    Apr 22, 2020 at 2:59
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Why would there be a single "unlucky" molecule? How would that be singled out?

One of the assumptions behind an ideal gas (and statistical mechanics in general) is that there are many (many many) constituents that are essentialy equivalent. Hence, they all have the same probability of hitting a wall in a given time, i.e. a hitting rate.

Note that properties such as pressure etc. are averaged over many molecules and many collisions in the gas and with the walls, so even if a given molecule stays in the interior of the gas for some time, it will be near the walls at some other time.

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  • $\begingroup$ Well it wouldn't have to be a single unlucky molecule. It's just a thought experiment and the idea seems highly plausible. One molecule can be thought to be constantly bouncing against other gas molecules instead of the wall itself. This would mean it never exerts pressure $\endgroup$
    – John Hon
    Apr 21, 2020 at 23:44
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    $\begingroup$ @JohnHon well, (a) it would exert pressure indirectly through the other molecules, but, more importantly, (b) all the molecules are statistically equivalent, so they all have the same probability of hitting the wall, and, in the long run, the same rate. So its not plausible (it has probability zero) that one (or a subset) of molecules is special in this sense. $\endgroup$
    – Toffomat
    Apr 22, 2020 at 5:54
  • $\begingroup$ Ohh! That makes a lot of sense - the inner particle could translate it's momentum to the outer molecules and thus still exert pressure! And yes I understand what you mean by b) $\endgroup$
    – John Hon
    Apr 22, 2020 at 13:11

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