Average kinetic energy of escaping molecule from a container

Question

I run into a question which haunts me for a while. Can anyone help me out?

The problem is:

Consider the effusion of molecules through an opening of diameter d in the walls of a container with volume V. Show that, while the average kinetic energy of the molecules in the container is 3/2kT, the average kinetic energy of the effusing molecules is 2kT, where T is the quasi-static temperature of the gas in the container.

I understand how people get 3/2kT by using Maxwell-Boltzmann distribution; I don't know how the 2kT appear by using Maxwell-Boltzmann distribution. Can't do the integral right.

Answer 1

Picking up where @CR Drost leaves off (thank you Dorst getting this started. For completeness sake, I'll complete the question and address the normalization constant you mentioned), we have the volume element of velocity space that is exiting the container. $$v_z f(v) \theta(v_z) dv d\theta d\phi$$ We can integrate over velocity space to determine the total number of particles that are exiting. $$\left< N \right> = \int d^3v \, n v_z f(v) \theta(v_z) $$

Note: (1) $\theta(v_z)$ is the step function because (as Drost notes) only positive velocity will travel through the hole. (2) $d^3v= dv d\theta d\phi$ and $n$ is the multiplicity factor for number flux. (3) The number flux will also now serve as our normalization constant (just like in $\bar{x} = \sum x_i p_i = \frac{1}{N} \sum x_i n_i $)

So to calculate the expectation value of an exiting particle's kinetic energy we compute compute the average kinetic energy over all the escaping particles (with appropriate normalization): $$\left< \frac{mv^2}{2} \right> = \int d^3v \, \left( \frac{mv^2}{2} \right) n v_z f(v) \theta(v_z) / \int d^3v n v_z f(v) \theta(v_z) $$ For a classical gas we have the Boltzman distribution $f(v) = exp(-\beta m v^2 /2)$, where $\beta=1/T$, the inverse temp. So simplifying with $a = \frac{\beta m}{2}$ and $d^3v= 4 \pi v^2 dv$ then we arrive at, $$\left< \frac{mv^2}{2} \right> = \frac{m}{2} \frac{\int^{\infty}_0 dv \, v^5 e^{-av^2}}{\int^{\infty}_0 dv \, v^3 e^{-av^2}}$$ for which we can evaluate the integral (https://www.wolframalpha.com/input/?i=Integral%5Bx%5EnExp%5B-ax%5E2%5D%2C%7Bx%2C0%2Cinfinity%7D%5D) to arrive at $$\left< \frac{mv^2}{2} \right> = \frac{m}{2} a^{-1} \frac{\Gamma(3)}{\Gamma(2)} = 2T $$ Note: Factors of $K_b$ are dropped throughout entire derivation

Answer 2

Reading Landau and Lifschitz I think this is the issue:

Imagine that we fix a hemisphere at distance $r$ from the hole. In some time $dt$ some particles with speed $v$ from this hemisphere have passed through the hole, and we want to get a grip on the velocity distribution of the particles that pass through. I will use spherical coordinates with $\hat z$ pointing normal to the surface into the gas, where the hemisphere is $0\lt \theta \lt \pi/2$ and $0\lt \phi \lt 2\pi$.

At some point $(r, \theta, \phi)$ on the hemisphere there is a small volume $v~dt~r^2\sin\theta~d\theta~d\phi$ of particles. Now get into the psychology of one of these particles: their direction is uniformly distributed about the spherical "sky" around them as they see it; and the area $A$ of the hole only takes up a proportion $A \cos\theta/(4\pi~r^2)$ of this sphere because it's seen partly in profile depending on $\theta.$ Combining these we can see that the flux of particles with velocities in $(v, v+dv)$ which make it to the hole must look like $$f(v)~dv~\cdot v~\sin\theta~\cos\theta~d\theta~d\phi.$$ There is a normalization constant on this and it's something like $NA/(4\pi V),$ but there is probably a better way to work it out explicitly by demanding that the resulting probability distribution sum to 1. Adding the $v~\cos\theta$ term can be interpreted as adding a $v_z$ term and normalizing basically means that we construct the expectation value $\langle U \rangle_\text{walls}$ for any dynamical variable $U$ is the expectation $\langle U~v_z\rangle/\langle v_z\rangle$ in terms of the Maxwell-Boltzmann distribution. This would also presumably apply to the kinetic energy $\frac12 m v^2.$