# Ideal gas in a vessel: kinetic energy of particles hitting the vessel's wall

Reading Landau's Statistical Physics Part (3rd Edition), I am trying to calculate the answer to Chapter 39, Problem 3.

You are supposed to calculate the total kinetic energy of the particles in an ideal gas hitting the wall of a vessel containing said gas.

The number of collisions per unit area (of the vessel) per unit time is easily calculated from the Maxwellian distribution of the number of particles with a given velocity $\vec{v}$ (we define a coordinate system with the z-axis perpendicular to a surface element of the vessel's wall; more on that in the above mentioned book): $$\mathrm{d}\nu_v = \mathrm{d}N_v \cdot v_z = \frac{N}{V}\left(\frac{m}{2\pi T}\right)^{3/2} \exp\left[-m(v_x^2 + v_y^2 + v_z^2)/2T \right] \cdot v_z \mathrm{d}v_x \mathrm{d}v_y \mathrm{d}v_z$$

Integration of the velocity components in $x$ and $y$ direction from $-\infty$ to $\infty$, and of the $z$ component from $0$ to $\infty$ (because for $v_z<0$ a particle would move away from the vessel wall) gives for the total number of collisions with the wall per unit area per unit time: $$\nu = \frac{N}{V} \sqrt{\frac{T}{2\pi m}}$$

Now it gets interesting: I want to calculate the total kinetic energy of all particles hitting the wall, per unit area per unit time. I thought, this would just be: $$E_{\text{tot}} = \overline{E} \cdot \nu = \frac{1}{2} m \overline{v^2} \cdot \nu$$ The solution in Landau is given as: $$E = \nu \cdot 2T$$

That would mean that for the mean-square velocity of my particles I would need a result like: $$\overline{v^2} = 4\frac{T}{m}$$ Now, I consider that for the distribution of $v_x$ and $v_z$ nothing has changed and I can still use a Maxwellian distribution. That would just give me a contribution of $\frac{T}{m}$ each. That leaves me with $2\frac{T}{m}$, which I have to obtain for the $v_z$, but this is where my trouble starts:

How do I calculate the correct velocity distribution of $v_z^2$?

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The reason your calculation is not right is because the mean energy of the molecules hitting the wall is not the mean number times the mean energy per molecule, because the fast molecules hit the walls more frequently than the slow ones.

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Okay, but how do you calculate those particles' velocity distribution? –  mSSM Apr 30 '12 at 10:40
It's the maxwell distribution, you just need to count the mean energy deposited--- number of molecules hitting per unit time with velocity v, times mv^2/2. It's the same integral you did, with an extra "v^2" inserted, and the answer is dimensionally the average energy times the number of particles, but the dimensionless factor changes. To do these, you need the moment of a gaussian integral and a half-gaussian integral (for odd moments), both of which are good exercizes and are found in many places. –  Ron Maimon Apr 30 '12 at 13:05
Sweet, thank you. I have to ponder a bit over this, but I think it makes perfect sense. It seems, at some point I had exactly the same idea, but I dismissed it because I made an error in a calculation so things didn't add up... –  mSSM Apr 30 '12 at 14:29

The following calculation gives the correct answer: $$Z\int_0^{\pi/2}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT}\; v \cos\theta\; \frac{1}{2}mv^2,$$ where $Z$ is such that $$Z\int_0^{\pi}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT} = n,$$ where $n$ is the particle number density.

The correct answer is $$\left(\frac{2kT}{\pi m}\right)^{1/2}\; nkT = \left(\frac{2kT}{\pi m}\right)^{1/2}\; p.$$

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