I am trying to prove that the average kinetic energy of gas molecules hitting a containers surface is $2k_{B}T$ instead of the average for the entire gas, which is $\frac{3}{2}k_{B}T$, where $k_{B}$ is the Boltzmann constant and $T$ is the thermodynamic temperature.

I have an expression for the flux of molecules hitting the wall with speed $v$ and angle $\theta$, given by:


Where $\widetilde{f}(v)$ is the Maxwellian speed distribution. Thus, if we want the average energy $\langle E \rangle$ we get:

$$\langle E \rangle = \int_{0}^{\pi}\int_{0}^{\infty}\frac{mv^{3}}{4}\widetilde{f}(v)\sin(\theta)\cos^{2}(\theta)\:\mathrm{d}v\:\mathrm{d}\theta$$


$$\widetilde{f}(v)=4\pi v^{2}\left[\frac{m}{2\pi k_{B} T}\right]^{\frac{3}{2}}\exp\left(-\frac{mv^{2}}{2k_{B}T}\right)$$

Evaluating the integral, I get:

$$\langle E \rangle = \frac{4\sqrt{2} (k_{B}T)^{3/2}}{3\sqrt{m}}\neq 2k_{B}T$$

I've checked the integral using Mathematica and that's not the problem, so one of my expressions is incorrect, but I cannot see which one it could be?

  • $\begingroup$ Can you explain a bit the formula of $d\Phi $ ? Where from comes 1/2, and who is $n$ ? I understand that $v \ sin(\theta) \ d \theta \ dv$ comes from the fact that you took an element of volume in the velocity space. But then, there should be also a $2 \pi$ because of the integration over $\phi$, and $v^2$ instead of just $v$. $\endgroup$ – Sofia Jan 12 '15 at 12:39
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    $\begingroup$ Is this a fact you were given, or are you theorizing? And are you sure you're calculating K.E. and not delta velocities? There's no variation in K.E. with angle in an isotropic gas, for example. $\endgroup$ – Carl Witthoft Jan 12 '15 at 12:42
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    $\begingroup$ Indeed, you will have a very hard time proving something that is not true. $\endgroup$ – Jon Custer Jan 12 '15 at 14:49
  • $\begingroup$ @CarlWitthoft It is a fact that I am given, the question as is identically: "Using the results of Q3.3, show that for a gas obeying the Maxwellian distribution, the average energy of all the molecules is $(3/2)k_{B}T$, but the average energy of those hitting the surface is $2k_{B}T$" Where in Q3.3 I showed that the speed distribution was defined as: $$\widetilde{f}(v)\:\mathrm{d}v = \frac{4v^{2}}{\sqrt{\pi}v_{th}}\exp\left(-\frac{v^{2}}{v_{th}^{2}}\right)\: \mathrm{d}v$$ And calculated that generally: $$\langle v^{n} \rangle = \frac{2v_{th}^{n}\Gamma(\frac{3+n}{2})}{\sqrt{\pi}}$$ $\endgroup$ – Thomas Russell Jan 13 '15 at 15:29
  • $\begingroup$ @JonCuster It's something I've seen in both Thermodynamics textbooks and in my lecturers notes; would you mind proving its invalidity? $\endgroup$ – Thomas Russell Jan 13 '15 at 16:33

You want the average energy of particles hitting the wall, so you need to normalise with respect to the number of particles hitting the wall (you need to divide your result by the number of particles hitting the wall). Using unnormalised distribution (the normalisation cancels), gives you $$\frac{\int_0^\infty m v^3 f(v) dv}{2\int_0^\infty v f(v) dv}\,,$$ where the integral over $\theta$ and all other constants from the flux (number of particles hitting a wall) cancelled. This gives you $2 k T$ when the integrals are evaluated.


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