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Effusion is molecular passage through a small opening whose dimensions, including its thickness, are much smaller than the intermolecular mean free path. For an equilibrium ideal gas, the one-way particle rate through a thin hole of area $A$ is $\dot{N}=Ap/\sqrt{2\pi m{k}_{B}T}$. This weights molecules by their wall-normal velocity, rather than just counting all molecules in the reservoir. ^[ref]

Which speed distribution does an arrival counter measure? Suppose an equilibrium, nonrelativistic ideal gas escapes by effusion through a thin opening into vacuum. The opening is small compared with the mean free path; the reservoir remains at temperature $T$. Let $f(v)$ denote its normalized Maxwell speed distribution, with mass $m$ and $a=m/(2k_{B}T)$:At a given direction toward the opening, the number crossing per unit time is proportional to the normal component $v\cos \theta$. Integrating over a fixed accepted cone contributes an angular factor independent of $v$. Thus the normalized probability density function for speeds among arriving particles isIts integral is one because ${\int }_0^{\infty }{v}^3{e}^{-a{v}^2}dv=1/(2a^2)$. The reservoir mode is $v=\sqrt{2k_{B}T/m}$; the arrival mode is $\sqrt{3k_{B}T/m}$. Faster particles get more opportunities to cross the opening. This is a selection effect, not heating after emission.

Specify the observable before applying the formula. An instantaneous count in a fixed segment of the beam weights the arrival flux by the time spent there, proportional to $1/v$; in this ideal model its normalized speed law returns to $f(v)$. A detector's speed-dependent efficiency, source collisions or a velocity selector changes the arriving law again. Counting arrivals and counting particles currently present are different measurements.

If a classical monatomic gas re-equilibrates by collisions while losing atoms by effusion, the escaping atoms carry mean kinetic energy $2k_{B}T$. Without wall heating, conservation of energy gives $d(3N{k}_{B}T/2)=2k_{B}TdN$ and therefore $T\propto N^{1/3}$. This assumes a fixed container volume and a small aperture in the molecular-flow regime.

Immanuel Estermann, O. C. Simpson and Otto Stern used gravitational deflection to measure atomic speeds. Stern describes scanning a hot-wire surface ionization detector through a cesium beam and recording ion electric current.^[ref] Their 1947 paper reports potassium and cesium beams; a deficit of slow arrivals increased with oven pressure, supporting a source-slit collision explanation.^[ref]

First consider an original point-launch model, with horizontal speed $v>0$, zero initial vertical velocity, flight distance $L$, and uniform gravitational acceleration $g$. Downward displacement $y$ follows from flight time $t=L/v$:Slow particles fall farther. However, a detector scan counts arrivals per interval of height, not per interval of speed. Change of variables for probability densities supplies the Jacobian determinant:Here $g(v)$ denotes the effusive beam speed distribution; the unadorned constant $g$ in $C$ denotes gravitational acceleration. The integral of $p_y$ is one: substituting $u=b/y$ reduces it to ${\int }_0^{\infty }u{e}^{-u}du=1$. Its mode is $y=b/3$. Simply mapping the most probable arrival speed into a fall distance instead gives $2b/3$. The peaks differ because equal height bins cover unequal speed intervals.

For those chosen parameters, $b\simeq 176.54\mu \mathrm{m}$, so the detector-height probability density peaks near $58.85\mu \mathrm{m}$. A detector covering only $0<y<Y$ accepts the fractionAt $Y=0.80\mathrm{mm}$ this is about $97.9\%$, even before allowing for imperfect detection. Discarding the remainder changes the observed normalization.

Real slits transmit a range of launch heights and vertical velocities. Their geometrical response must be combined with the gravitational deflection and detector width; the point-launch formula is not an exact reduction of the historical apparatus. In the point-launch model, an initial vertical velocity $w$ would add $wL/v$ to $y$. Source collisions, detector efficiency and alignment can also distort a profile. Varying oven pressure helps distinguish such source losses from an incorrect thermal distribution; inferring a source temperature from the peak alone ignores both selection and the detector's coordinate transformation.

For speed distribution $f(v)$ normalized to one and isotropic directions, the incident number flux isThe positive normal velocity weights the population reaching the wall. Consequently its mean translational energy isThe bulk value $3k_{B}T/2$ would omit the arrival weighting.

For the two vessels, let $r=\sqrt{T_1/T_0}$, with subscripts h and c for hot and cold. Equal opposing effusion rates give $p_h/\sqrt{T_1}=p_c/\sqrt{T_0}$. Number conservation in equal, fixed volumes adds $p_h/T_1+p_c/T_0=2p_0/T_0$. HenceBoth final pressures must be determined; holding the cold temperature fixed does not hold its particle number fixed.

For aperture area $A_h=\pi d^2/4$, the common one-way flux is $J=p_c/\sqrt{2\pi m{k}_B{T}_0}$. The translational heat transfer from hot to cold is thereforeThis is the total heat transfer for a monatomic gas with no active internal modes, neglecting radiation and conduction through vessel walls. For a molecular gas, add $A_{h}J[u_{\mathrm{int}}(T_1)-u_{\mathrm{int}}(T_0)]$ for its internal energy per molecule. The maintained unequal temperatures describe a stationary nonequilibrium state even though the net particle current is zero.

For isotropic planar motion, the planar effusion flux gives an escape rate through a gap of width $w$ is $\Gamma =w(N/\mathcal{A})⟨v⟩/\pi$, where $\mathcal{A}$ is tray area. At fixed equilibrium kinetic energy, or fixed shape of the scaled speed distribution, $v\propto m^{-1/2}$. Hence halving the mass increases the rate by $\sqrt{2}$. This planar effusion result assumes the same spatial density and an unrestricted small opening. Strictly, average energy alone fixes $⟨v^2⟩$, not $⟨v⟩$; an arbitrary change of distribution shape would need more information.

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.

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.

Choose a flat wall and let $\theta$ measure the angle from the direction towards it. Particles of speed $v$ in solid angle $d\Omega$ occupy fraction $f(v)dvd\Omega /(4\pi )$ of an isotropic gas. In time $dt$, those within normal distance $v\cos \theta dt$ reach the wall. Their number per area per time is therefore $nf(v)dvv\cos \theta d\Omega /(4\pi )$.

Only the incoming hemisphere contributes. Integrating gives the one-way collision rate per unit area,Its units are ${\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. This is not $n\overline{v}/2$: selecting the incoming half of the particles must also account for their mean normal velocity. For effusion through a thin hole whose diameter is much less than the mean free path, the same one-way rate applies because collisions within the opening can be neglected.

Thermal transpiration is gas transport driven by an imposed temperature difference in a rarefied connection. Gas can move from cold to hot even when the two end pressures start equal. If the reservoirs are closed, redistribution builds a counteracting pressure difference until the net transfer stops. This final stationary state is not global thermal equilibrium: the two temperatures remain different. It explains why a heated pressure sensor can read a different pressure from the vessel to which it is connected. The correction depends on gas, connection geometry and molecular interaction with its surfaces. ^[ref]

The simple molecular limit is beautiful: equal opposing particle rates need not mean equal pressures. Consider effusion through a thin hole connecting two locally equilibrated reservoirs of the same gas, at $T_h>T_c$. Equating their one-way rates givesAt equal pressure, the hotter reservoir has lower number density; its faster molecules do not fully compensate for that reduction in outgoing particle rate. With $T_c=300\mathrm{K}$ and $T_h=351\mathrm{K}$, the ideal zero-transfer ratio is $\sqrt{351/300}\simeq 1.0817$. The hydrodynamic, zero-flow limit instead has equal pressures when other forces are absent. These are limiting models, not an interpolation formula for a glass capillary. ^[ref] In a real tube, incomplete exchange of molecular energy with the wall can alter the limiting ratio; a Knudsen number does not specify that exchange. ^[ref]

For a species $s$ in two well-mixed vessels of equal volume $V$ joined by a thin opening of area $A_h$, the effusion law gives ${\dot{n}}_{s,L}=-(A_h/4V)({\overline{v}}_{s,L}{n}_{s,L}-{\overline{v}}_{s,R}{n}_{s,R})$. Each species' total inventory is conserved. With fixed temperatures and total density sum $n_{s,L}+n_{s,R}=n_{s,\Sigma }$, the left density relaxes exponentially with time constant ${\tau }_s=4V/[A_h({\overline{v}}_{s,L}+{\overline{v}}_{s,R})]$ to $n_{s,L}^{\infty }=n_{s,\Sigma }\sqrt{T_R}/(\sqrt{T_L}+\sqrt{T_R})$.

At equal temperatures, ${\tau }_s=2V/(A_h{\overline{v}}_s)\propto \sqrt{m_s}$, and initially separated species mix at different rates. At unequal temperatures, the final density ratio is $n_{s,L}/n_{s,R}=\sqrt{T_R/T_L}$ for every species, while the partial-pressure ratio is $p_{s,L}/p_{s,R}=\sqrt{T_L/T_R}$. The mass-independent final ratios do not imply mass-independent transient rates. Maintaining different temperatures requires external heat exchange even after net particle transfer vanishes.

Let each drive region have length $\ell$ and the drive-free gap have length $L$, measured edge to edge. A particle of constant speed $v>0$ experiences $\tau (v)=\ell /v$ and $T(v)=L/v$. For a normalized detected-beam speed probability density function $f(v)$, sum the transition probabilities, not the probability amplitudes of different particles:The displayed $f$ is an ideal effusion flux distribution, with speed scale $u$, uniform transmission and speed-independent detection efficiency. It is not a claim about any particular clock's selected beam. The substitution $x=(v/u{)}^2$ gives $f(v)dv=x{e}^{-x}dx$, proving normalization. Actual apertures, state selection and readout can change the effective weights.

For an original example choose $\ell =0.10\mathrm{m}$, $L=1.00\mathrm{m}$, $u=v_{\ast }=200\mathrm{m}{\mathrm{s}}^{-1}$ and $\Omega =2\pi \times 500\mathrm{rad}{\mathrm{s}}^{-1}$. The reference particle has ${\tau }_{\ast }=0.500\mathrm{ms}$, $T_{\ast }=5.00\mathrm{ms}$ and resonant $\pi /2$ pulses. The instantaneous-pulse width would be $100\mathrm{Hz}$. The exact finite-pulse width for that single speed is about $88.7\mathrm{Hz}$; the chosen distribution also reduces the on-resonance average to approximately $0.738$ because not every particle receives a $\pi /2$ pulse.

Different velocities smear side fringes because $\delta L/v$ varies across the beam. Nevertheless every term is even in $\delta$, so this symmetric ideal-model average has no odd line asymmetry that could displace the central lock point. Reduced contrast or altered width alone is not evidence of a shifted line center. A speed-dependent transition shift or unequal drive quantum phases changes that conclusion; atomic clock frequency shifts explains why these distinctions matter experimentally.