# Why no need to consider collision in kinetic theory?

In standard elementary thermodynamics textbook, in the kinetic theory it is assumed that the gas particles have no interactions except during collisions. However, in the subsequent derivation of $$pV=\frac{2}{3}U$$ only the collisions of the gas particles with the container walls are considered, but not the collisions between gas particles. Why we can ignore it?

We can ignore the collisions between particles when deriving that equation because we are assuming that we are dealing with a rarified gas, i.e. the average distance between a two molecules is large compared to their size.

When calculating pressure, we are mainly concerned with the interaction with the walls, so this approximation is quite good; actually, an ideal gas is formally made of point particles, so the probability of a particle-particle collision would be rigorously zero in the ideal gas approximation.

But of course, to compute other quantities such as the mean free path we must consider the particles to be of finite size, i.e. we have to give up the ideal gas approximation. However, in such calculations, the ideal gas law (IGL) is used: this appearently makes no sense, because the IGL is derived under the assumption that we are dealing with point particles!

The catch is that, even if the IGL is only rigorously valid for a gas of point particles (which by definition cannot collide with each other), it is a quite good approximation also for a gas of finite-size, but small particles: this is why we can use the IGL in the derivation of quantities such as the mean free path. And this is also why you can find in many books (or on Wikipedia) the somewhat misleading definition "An ideal gas is a theoretical gas composed of point particles that do not interact except when they collide elastically".

There is a more fundamental problem about the usual textbook derivation of $pV = \frac 23 U$.

The mass of the particles appears in the derivation, but not in the end, it cancels out exactly - if all energies are equal. Since they are not equal, the derivation just inserts "mean energy" instead of "energy" and is done.

Now it is obvious, that a gas of identical particles will have a mean energy, and since the contribution to $p$ is proportional to the energy, the derivation is perfectly accurate, see below.

What is not obvious is, why the mean energies of sets of different particles should coincide. If they don't, then the whole derivation gets much more complicated (the collisions with the walls and between particles are not "elastic" any more), and collisions matter.

But an elastic collision between identical particles is very simple: they swap their momenta. In other words, the particle just gets exchanged, and continues as if there had been no collision.

• Originally I also thought they just swap their momenta. But think more carefully, it isn't true. Say, in the CM frame, they move directly towards each other along a certain line with equal speed $v$. But after the collision, by conservation of momentum and KE, they must have the same speed $v$, true, but can be moving directly opposite to each other along a line making an arbitrary angle with the original line. – velut luna Nov 17 '16 at 13:02
• Oh no! you are right, how could I mistake this :( it's only true in 1 dimension. Well, you are right, the direction after the collision can be arbitrary. But I still disagree, that the collisions have an influence, i just cant find a "proof" – Ilja Nov 17 '16 at 14:08