Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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).

share|improve this question
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 $\int_0^{\pi / 2} d \theta \sin \theta~ \cos \theta$ –  Trimok Oct 7 '13 at 18:18
@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? –  Satwik Pasani Oct 8 '13 at 13:40

1 Answer 1

up vote 1 down vote accepted

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 $I = \large \frac{nA v_{avg}}{4}$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.