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(The equation of ideal gas is $pV=NRT$. If p = 1 atm, N = 1 mole, T = 0°K, and R = gas constant, then volume = V = zero. Hence, the volume of an individual molecule of ideal gas is zero).

An individual molecule of ideal gas is assumed to have zero volume. The molecules of ideal gas are assumed to be dimensionless points. Then how do the dimensionless points collide with each other in accordance with the kinetic theory of gases?

I assume that the individual molecules of the gas have non-zero volume such that they are able to collide with other molecules or the wall of the container. If a molecule has zero volume (i.e. a dimensionless point), then how can it collide with other molecules (how can points collide with each other)?

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    $\begingroup$ They don't collide for the reason you have given, they are point particles. You don't need to have interatomic collisions to derive the ideal gas law. The wall is modelled as a continuous surface, even point particles can collide with that. $\endgroup$
    – Ian Bush
    Aug 30, 2022 at 10:32
  • $\begingroup$ @IanBush But they have to mutually collide to achieve thermodynamical equilibrium and Maxwell-Boltzmann speed distribution. // The idea is zero-like as negligible volume compressibility-wise, but non-zero volume collision-wise. BTW 0 K is not achievable. $\endgroup$
    – Poutnik
    Aug 30, 2022 at 10:54
  • $\begingroup$ No they don't, not in the ideal gas model. Thermodynamics says what the final state will be. It says nothing about how that state is reached. That is kinetics. c.f. forbidden transitions in Quantum mechanics. $\endgroup$
    – Ian Bush
    Aug 30, 2022 at 11:01
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    $\begingroup$ @Poutnik The speed distribution could work via the wall. $\endgroup$
    – Karsten ♦
    Aug 30, 2022 at 12:25
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    $\begingroup$ @Karsten The Boltzmann distribution doesn't require collisions. which is what the speed distribution really is. It is just how classical particles occupy the available energy levels at a given temperature. The ideal gas model has no interparticle collisions. $\endgroup$
    – Ian Bush
    Aug 30, 2022 at 13:02

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The ideal gas is a 'model' of how an actual gas behaves. It is still used because it agrees with many different types of experiments on gasses, but its limitations are also recognised.

The initial assumptions are that molecules are represented by perfect and elastic spheres of infinitesimal size, these particles travel in straight lines and strike the walls of any vessel in which they are contained. By being elastic means should they collide with one another no energy is lost, i.e. no energy is retained within a particle so that the total kinetic energy after collision is the same as before collision, but their direction and hence momentum changes.

On collision ideal gas particles impart momentum to any wall thus producing pressure. The collision with a wall is elastic so no energy is lost. The incident and reflected angles are the same. To calculate the pressure ($p$) we do not need particle to particle collisions and all molecules can travel at the same speed then

$$p=\frac{2\cdot \mathrm{kinetic\; energy}}{3\cdot V}$$

where $V$ is volume. Even if there were collisions this would not affect the pressure as this is caused by collisions with the walls.

As kinetic energies are additive so are pressures, this means that the pressure of mixtures is the same as that of each gas taken separately which is Dalton's Law. If the volume changes it does so in inverse to the pressure which is Boyle's law.

If two gasses at different temperature are allowed to mix, say by removing a partition between them, they will diffuse into one another and at the same time the temperature will eventually become uniform throughout the gas. This can only happen if collisions between particles occur, so this is an essential assumption of the ideal gas model, unrealistic though it may seem for particles of an infinitesimal size. Once the absolute temperature is introduced the kinetic energy can be shown to be $3RT/2$ which produces the ideal gas law via the previous equation.

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    $\begingroup$ The zeroth law says that two bodies in thermal equilibrium with a third will be at the same temperature. I don't see the requirement for direct collisions when you mix gases. $\endgroup$
    – Buck Thorn ♦
    Aug 30, 2022 at 22:21
  • $\begingroup$ @Buch Thorn Suppose that I have a vast volume of gas so large that the walls are unimportant and I excite just the centre part with a focussed laser, without collisions I will have the gas at two different temperatures persisting for ever. Does this happen in practice? Or I could have two gas clouds in space at different temperatures that collide but retain their individual temperatures. Does this happen? $\endgroup$
    – porphyrin
    Aug 31, 2022 at 14:21
  • $\begingroup$ Yes, I see what you mean. Two mixed gases at different T would be strange indeed. But I suppose something like it could be generated. The excited focal point would dissipate, but it need not do so into a uniform gas, since there is no wall to ensure mixing. The universe is a good example. It is not uniform and the absence of a boundary to keep things mixed is probably why (ok, I actually assume this, but I would like to hear the counterargument). $\endgroup$
    – Buck Thorn ♦
    Aug 31, 2022 at 14:30
  • $\begingroup$ Gas clouds in space? they would be at a T low enough to prevent dispersion. A heated gas mass expands radially to equalize pressure or gravitational energy. To see what happens compare the planets in the solar system in relation to the sun. The collisions with the wall are more complicated ; the wall must also have a 2mv change in momentum. The wall is superfluous; all one needs is a bunch of molecules with the same average KE; when they hit they bounce. $\endgroup$
    – jimchmst
    Aug 31, 2022 at 20:55
  • $\begingroup$ @jimchmst yes, but you are bringing gravity into the story, and there is no comparable force at the atomic scale (other than electromagnetic, but we assume an ideal gas). The point of the wall is that it leads to exchange of energy (in hindsight mixing is not the key point, but is important too given the counterargument that started the discussion). $\endgroup$
    – Buck Thorn ♦
    Sep 1, 2022 at 19:09

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