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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.

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.

Wall arrivals are not an unbiased sample of the gas: faster atoms arrive more frequently. The normalized speed distribution among incident atoms isThe angular factor in the arrival rate cancels when calculating the mean kinetic energy. ThusUsing $I_5=2I_3/\alpha$ givesBy contrast, an atom chosen randomly from the bulk has mean kinetic energy $3k_{B}T/2$. The arrival weighting raises the normal-direction contribution to $k_{B}T$ while each tangential direction still contributes $k_{B}T/2$. This is also the mean energy carried by atoms effusing through a thin aperture from an equilibrated reservoir.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

For an isotropic two-dimensional velocity distribution, a straight opening of width $w$ in a region of number per area $n_2$ loses particles at the displayed rate. The angular integral is ${\int }_{-\pi /2}^{\pi /2}\cos \theta d\theta /(2\pi )=1/\pi$. At fixed equilibrium kinetic energy the speed distribution scales as $m^{-1/2}$, so the rate has the same scaling. A fixed second moment alone does not determine the first moment without a distribution assumption.

A speed distribution is the probability distribution of the magnitude of velocity. When it has a probability density function $f$, the fraction of particles with speeds between $v$ and $v+dv$ is $f(v)dv$, and ${\int }_0^{\infty }f(v)dv=1$. The Maxwell speed distribution describes an equilibrium ideal gas; the effusive beam speed distribution weights that population by its rate of arrival.

The speed distribution of atoms striking a wall is weighted by the normal component of velocity, relative to the distribution of atoms present in the bulk. For an isotropic classical monatomic gas, the impact speed density is proportional to $v^3{e}^{-m{v}^2/(2k_{B}T)}$, and the mean impact kinetic energy is $2k_{B}T$, rather than $3k_{B}T/2$.