Past exam of the physics B course of the University of Cambridge / 2016 / Paper 2 / Section C / C12

For the two unlabelled follow-up requests, distinguish re-equilibrating effusion from collisionless effusion. The calculations below show that escaping atoms carry mean kinetic energy $2k_{B}T$, whereas a bulk atom has mean kinetic energy $3k_{B}T/2$. If the aperture is small and interatomic collisions maintain near-equilibrium conditions, faster atoms escape preferentially, and the remaining gas relaxes toward a lower-temperature Maxwell-Boltzmann distribution. No heat supplied by the walls is available to replace the lost energy.

For a monatomic gas at fixed container volume, conservation of energy gives a quantitative check on this effusive cooling picture:Since $dN<0$, the temperature falls. Under these idealizations $T\propto N^{1/3}$; the gas is not left with a permanently truncated velocity tail when collisions rapidly restore equilibrium.

If interatomic collisions are absent, the individual surviving atoms do not rethermalize to a new common temperature. Their speed distribution becomes selectively depleted, generally favouring slower survivors, and their angular and spatial distribution can also develop structure determined by which trajectories reach the hole. A single lower temperature does not describe the full nonequilibrium distribution. With a direction-randomizing but energy-preserving wall model, an illustrative survival law is $f(v,t)\propto f(v,0)e^{-\kappa vt}$; specular walls can give more detailed directional selection and even persistent nonescaping trajectories. The question supplies no wall-scattering or container-geometry model from which a unique collisionless loss law could be inferred.

For the labelled wall integrals, choose the polar axis toward the wall so that incoming atoms have $0\le \theta \le \pi /2$, and putThe needed radial integrals are $I_3=1/(2{\alpha }^2)$, $I_4=3\sqrt{\pi }/(8{\alpha }^{5/2})$ and $I_5=1/{\alpha }^3$. They are related to the supplied integrals by rescaling the speed.

The collision flux contains an extra normal velocity $v\cos \theta$: a faster atom sweeps out a longer approach cylinder per unit time. Integrating the Maxwell-Boltzmann distribution over the incoming hemisphere givesThereforeHere $J$ is the number striking unit area per unit time and $\overline{v}=\sqrt{8k_{B}T/(\pi m)}$ is the bulk mean speed. A simple factor of one half would count the incoming hemisphere but miss the projection factor and velocity weighting.

An elastic reflection transfers normal momentum $2mv\cos \theta$ to the wall. Multiplying this impulse by the collision flux and integrating givesUsing the radial integral and $\alpha =m/(2k_{B}T)$ yieldsThis agrees with the ideal-gas equation of state. The calculation assumes a stationary wall and elastic normal momentum reversal, appropriate to a gas in equilibrium with it.

The energy average must be taken over the atoms that actually arrive, not over the bulk population. Weighting the collision flux by $m{v}^2/2$ givesThe wall-impact speed distribution is proportional to $v^3{e}^{-\alpha v^2}$ rather than the bulk $v^2{e}^{-\alpha v^2}$. Faster atoms are overrepresented among arrivals. Equivalently, the mean normal kinetic energy of the arriving flux is $k_{B}T$, while each tangential component still contributes $k_{B}T/2$.