# Difficulty in calculating effusion rate using the kinetic theory of gases and statistical mechanics

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 $$ = /4$$, 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 $$$$, right? So:

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

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

$$2.Φ = n.$$

But $$ = /4$$

$$Φ = n./8 (!!!)$$

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

I asked a person and they told be I should've used $$ = /2$$ 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 $$,<|Vx|>$$ 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 $$= /2$$ (?). 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?

• Please use LaTeX. – Pieter Oct 18 '18 at 22:35
• 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. – jacob1729 Oct 18 '18 at 22:38
• I am sorry. I have fixed my post so it is now using LaTeX. – João Vítor G. Lima Oct 18 '18 at 23:14
• @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. – probably_someone Oct 18 '18 at 23:46
• But shouldn't an average be divided by the period, in this case? – João Vítor G. Lima Oct 19 '18 at 0:44