According to the kinetic molecular theory obeying Maxwell-Boltzmann distribution of speeds, the rate of effusion through a pinhole of area $A$ is
$$R=\frac{PA}{\sqrt{2\pi M R T}}$$ where $M$ is the molecular weight, $R$ is the gas constant and $T$ the absolute temperature.

To derive this, I consider the collision frequency on any small area ($A$):-
using $$v_{avg}=\sqrt{\frac{8RT}{\pi M}}$$ I get the result, that the atoms in volume $v_{avg}A$ can hit the area. The number of particles in this volume is $nv_{avg}A$ ($n$=number density). But the derivation includes a factor of $\frac14$ before this term to find the actual number of atoms in this volume hitting the wall. I want to know how that factor of $\frac14$ came into picture to make the collision frequency per unit area as $\frac14nv_{avg}=\frac{P}{\sqrt{2\pi M R T}}$.

I know the origins of a factor of $\frac12$ before the pressure term while calculating it, considering change in momentum of a molecule on collision with the wall, which is to account for the fact that $<v_x^2>$ includes both the $v_x$ terms, going towards and going away from the wall(positive and negative directions) but the ones colliding are just the half of these $\frac12<v_x^2>$(going in any one direction).

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    $\begingroup$ In considering the flow passing by the area $A$, you have a $\frac{1}{2}$ factor coming by selecting one way, and I think that the other $\frac{1}{2}$ comes because you have to consider the projection of the speed on the normal to the surface $A$ , see also this answer. The problem is different, but just look at the difference between integrals in $\theta$ in equations $(2)$ and $(4)$, and you get your $\frac{1}{4}$ factor.This is the difference between $\int_0^\pi \sin \theta ~d \theta$ and $\int_0^{\pi / 2} d \theta \sin \theta~ \cos \theta$ $\endgroup$
    – Trimok
    Commented Oct 7, 2013 at 18:18
  • $\begingroup$ @Trimok For my problem, i calculated the integral to be:- $n\int_{0}^{\pi/2}\frac12 v_{avg}At\cos\theta d\theta$ which does not yield the given answer and hence must be wrong. Can you tell me what the integral for my problem should be? $\endgroup$ Commented Oct 8, 2013 at 13:40

1 Answer 1


If your Maxwell-Boltzmann distribution is $\mu(\vec v) = \mu(v) = (\frac{m}{2 \pi k T})^{3/2} e^{- \frac{m v^2}{2 k T}}$, then, if I am not mistaken, you should have to perform an integral with $\theta$ limited between $0$ and $\pi/2$ of kind $ I = nA \int_{0 \le \theta \le \pi/2} d^3 \vec v \mu(\vec v) (\vec v.\vec n)$, where $\vec n$ is the unit normal to $A$, and we choose $\theta$ such as $\vec v.\vec n = v \cos \theta$,so you would have : $ I = nA\int_0^{2 \pi} d\phi \int_0^{\pi/2} d\theta \sin \theta \cos \theta \int_0^{+\infty} dv ~v^3 ~\mu(v)$. Note that $v_{avg} = \int_0^{2 \pi} d\phi \int_0^{\pi} d\theta \sin \theta \int_0^{+\infty} dv ~v^3 ~\mu(v)$. Your result should be $I = \large \frac{nA v_{avg}}{4}$


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