I have tried multiple sources and methods, but my attempts at a proof of the number of particles leaving a gas using statistical mechanics keep finding the same wrong result. I have tried to read other answers on this site but none of them had exactly what I wanted or just jumped to conclusions.

Let $Φ$ be the number of particles per unit of time and area leaving a box of gas by means of a small hole.

We know from the kinetic theory of gases that the average $Vx$ (modulus of velocity in the x direction) is $<Vx> = <Vel>/4$, where $Vel$ is the velocity of particles.

Let us calculate the number of particles with x-direction velocity equal to $Vx$ that escapes through the hole of area $S$ during time $dt$. The particles come from a cylinder of height $Vx.dt$ and area of the base $S$. So the particles come from the region of volume $S.Vx.dt$.

If the gas has a number $n(Vx)$ of particles with x-velocity equal to $Vx$ per unit of volume, then from the cylinder will come, in time $dt$, a number $Nxc = n(Vx).S.Vx.dt$ of particles.

If we sum all the $Nxc$ for different x-velocities we find the number $Nc$ of particles that escaped. Then

$Nc = S.dt.ΣVx.n(Vx)/2$, where $/2$ appeared because only half of the particles have $Vx > 0$, i.e., are going in the direction of the hole.

If the gas has a total volume $V$, we can call $n(Vx) = N(Vx).V$ where $N(Vx)$ is the total number of particles with velocity $Vx$ in the whole gas. Then:

$$2Nc = S.dt.ΣN(Vx).Vx/V$$

We multiply the right side by $N/N$, where $N$ is the total number of particles in the gas:

$2Nc = S.dt.(N/V).[ΣN(Vx).Vx/N]$

Well, $ΣVx.N(Vx)/N$ is just $<Vx>$, right? So:

$$2Nc = S.dt.(N/V).<Vx>$$

But we can write $Nc/(S.dt) = Φ$ and $N/V = n$:

$$2.Φ = n.<Vx>$$

But $<Vx> = <Vel>/4$

$$Φ = n.<Vel>/8 (!!!)$$

This is wrong. It is well known that $Φ = n.<Vel>/4$, not $/8$. I can't seem to find the mistake.

I asked a person and they told be I should've used $<Vx> = <Vel>/2$ in this case for some reason, but I can't see why and this doesn't agree with Maxwell-Boltzmann distribution.

Where is my mistake? I can't seem to find anything, really. Even if I increase the mathematical rigor of $<Vx>,<|Vx|>$ etc, I can't find another answer. Where am I wrong?

It appears to me that some sort of double integral is used to find the apparent result of $<Vx>= <Vel>/2$ (?). I cannot, however, understand double integrals: only simple integration.

Anyway, I have tried multiple sources but most proofs seem to jump some or the steps I did. Where is my mistake and how can I solve it?

  • $\begingroup$ Please use LaTeX. $\endgroup$ – Pieter Oct 18 '18 at 22:35
  • $\begingroup$ I think this issue may be to do with the fact you are only allowing the particles to move in the x direction, but it would be much easier to follow if this were in LaTeX. $\endgroup$ – jacob1729 Oct 18 '18 at 22:38
  • $\begingroup$ I am sorry. I have fixed my post so it is now using LaTeX. $\endgroup$ – João Vítor G. Lima Oct 18 '18 at 23:14
  • $\begingroup$ @jacob1729 Given that integrating over one half-period of $\cos(\theta)$ gives you precisely a factor of 2, I wouldn't be surprised if that was the case. $\endgroup$ – probably_someone Oct 18 '18 at 23:46
  • $\begingroup$ But shouldn't an average be divided by the period, in this case? $\endgroup$ – João Vítor G. Lima Oct 19 '18 at 0:44

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