Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section C / C12 / Mixing by effusion

Let L be the vessel initially containing argon and R the vessel initially containing helium. Both have volume $V$ and fixed temperature $T_0$, and the aperture has area $A_h$. The incoming and outgoing rates of each species must be considered separately. For species $s$,Thus its density difference relaxes exponentially with time constantThe two-vessel effusive mixing solution for the four densities isUsing atomic masses of approximately $40$ and $4$ atomic mass units gives ${\tau }_{\mathrm{He}}/{\tau }_{\mathrm{Ar}}\simeq 1/\sqrt{10}\simeq 0.32$. Helium therefore transfers faster. Each species eventually reaches $n_0/2$ in each vessel, but its transient is different. In particular, the total density temporarily rises in the initially argon vessel and falls in the initially helium vessel, before both return to $n_0$.

For unequal maintained temperatures $T_L,T_R$, replace the rate equation byWith each species' total density sum still $n_0$, its final densities and relaxation time areEvery species has the same final density ratio $\sqrt{T_R/T_L}$ and partial-pressure ratio $\sqrt{T_L/T_R}$. Thus there is no permanent composition separation in this ideal steady state, although the rates remain mass dependent. A maintained hot side has lower density and higher pressure. Equal particle rates do not imply thermal equilibrium: atoms arriving from the hotter side carry more energy, so external heating and cooling are required to hold different temperatures. The model assumes thin-hole effusion, local equilibrium within each vessel and negligible aperture volume.