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