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How would I prove that the number of collisions by gas molecules on a wall of unit area per unit time is $r = \frac 1 4 n \left< v \right>$ using the Maxwell-Boltzmann speed distribution function? $r$ is the rate of collisions, $n$ is the number of molecules per unit volume, and $\left< v \right>$ is the average speed of the molecules.

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  • $\begingroup$ You don't need to actually know the distribution function for this, just that its isotropic. It might help if you add what $r$ is to your question (presumably rate of particle effusion through a hole in some container). $\endgroup$ – jacob1729 Jun 8 '18 at 18:28
  • $\begingroup$ What quantity is represented by $r$? (To my knowledge, $r$ is not a common sign for a thermodynamic quantity.) $\endgroup$ – Sebastian Riese Jun 8 '18 at 18:28
  • $\begingroup$ I think $r$ is the rate of collisions of the gas molecules. $\endgroup$ – Alex Wang Jun 8 '18 at 18:32
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So imagine the entire space $z > 0$ is filled with this gas and you are looking at collisions on the plane $z=0$ in some time $t$. Considering the infinite space gives us translational invariance, so look at a small circle of area $dA$ centered at $(x, y) = (0, 0)$. Use spherical coordinates with $\phi$ being a polar angle down from $x=0, y=0, z>0$ and $\theta$ being an azimuthal angle counterclockwise from $y=0, x > 0,$ both measured in radians.

Somewhere out in space we have a small volume $dV = r^2\sin\phi~dr~d\phi~d\theta$ and it has a number of particles $dn = \rho~dV.$ What fraction of those pass through your square $dA$?

Well there's two concerns. One is the amount of solid angle occupied by the area $dA$. The particles at $(x, y, z) = (1, 0, \epsilon)$ for small $\epsilon$ see this area nearly edge-on and have a lot more trouble getting into it than the particles at $(x, y, z) = (0, 0, 1)$, which see it perpendicular. This boils down to a factor of $\cos \phi$, if you work it out.

In addition there's the more obvious concern that even if it were parallel, the chance of the particle having the right direction goes like $dA / 4\pi r^2.$

Finally it needs to have a velocity sufficient to traverse the distance in the given time. The easiest way to do this is with a cumulative distribution function: $F(s)$ is the probability that the speed $S$ is less than or equal to some value $s$, with $F(0) = 0$ and $F(\infty) \to 1.$ So you've got an expression for the rate $\eta$ of collisions, $$\frac{\eta}{dA} = \int_0^\infty dr~\int_0^{\pi/2}r~d\phi~\int_0^{2\pi}r~\sin\phi~d\theta~~\frac{\cos\phi}{4\pi r^2}~[1 -F(r/t)].$$You can immediately complete the angular integrals:$$\frac{\eta}{dA} = \frac{1}{2}~2\pi~\frac{1}{4\pi}~\int_0^\infty dr ~[1 - F(r/t)].$$So there's your mysterious $1/4$. As for the rest, you may first substitute $s = r/t, dr = t~ds $ to find $$\frac{\eta}{t~dA} = \frac14 \int_0^\infty ds~[1 - F(s)],$$ then integrate by parts, raising $ds \to s$ and lowering $[1 - F(s)] \to -f(s)~ds$ to find an expression $\int s~f(s)~ds = \langle S \rangle,$ and you just check that the boundary terms are both zero.

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