Kinetic-theory model of pressure

The model is a deliberate caricature: N identical molecules of mass m in a volume V, with N large enough that averages are steady; each molecule tiny compared with the spacing between molecules, so the volume they occupy is negligible beside V; no forces between them except during brief collisions, so between collisions they move in straight lines under Newton's laws; and collisions with the walls elastic, so no kinetic energy is lost to the container. None of this is exactly true, and that is the point — each clause is a claim you can later test. Room air holds about 2.5 × 10²⁵ molecules per cubic metre, which is a mean spacing of 3.4 nm against a molecular diameter near 0.3 nm, so the molecules themselves take up about 0.04% of the box. At that density the caricature is very nearly the truth.

Take a wall perpendicular to x. Smooth means it exerts no tangential force, so v_y and v_z are untouched; elastic means the speed is unchanged, so vₓ simply reverses to −vₓ. The molecule's momentum changes by (−mvₓ) − (mvₓ) = −2mvₓ, and by Newton's third law the wall receives +2mvₓ. For a nitrogen molecule, m = 4.65 × 10⁻²⁶ kg, approaching at vₓ = 480 m s⁻¹, that is 4.46 × 10⁻²³ kg m s⁻¹ delivered in one hit. Only the normal component counts: a molecule at the same speed grazing in at 80° to the normal carries vₓ = v cos 80° and delivers just 17% as much.

A gauge does not feel individual hits, it feels their time average, so the rate matters as much as the size. Between two hits on the same face the molecule crosses the box and comes back, a distance 2L at unchanged vₓ, so Δt = 2L/vₓ and the hit rate is vₓ/(2L). The time-averaged force from one molecule is F₁ = Δp/Δt = 2mvₓ × vₓ/(2L) = mvₓ²/L. For the nitrogen molecule above in a 0.250 m cube: 960 hits per second, F₁ = 4.29 × 10⁻²⁰ N, a pressure of 6.9 × 10⁻¹⁹ Pa on that face. The obvious objection is that molecules collide with one another and never make clean round trips. In a steady state that does not matter: a molecule knocked off its path is replaced by another, and the wall sees the population average, which this bookkeeping reproduces. Counting the flux of molecules arriving at the wall gives the same answer without ever mentioning a round trip.

Define the mean square of the x-components, ⟨vₓ²⟩ = (1/N) Σ vₓᵢ². Summing F₁ over all N molecules gives F = (m/L) Σ vₓᵢ² = (N m/L) ⟨vₓ²⟩, and dividing by the face area L² gives p = N m ⟨vₓ²⟩ / L³ = (N/V) m ⟨vₓ²⟩. Now bring in the other two directions. For each molecule v² = vₓ² + v_y² + v_z², so averaging gives ⟨v²⟩ = ⟨vₓ²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩, and in random motion no axis is special, so the three are equal and each is ⅓⟨v²⟩. Hence p = ⅓ (N/V) m ⟨v²⟩, or pV = ⅓ N m ⟨v²⟩. Notice that L has vanished: the cube was scaffolding, and the result holds for a container of any shape and on every wall of it.

Two bulk quantities carry the microscopic content. The number density is n = N/V in m⁻³, and the mass density is ρ = n m in kg m⁻³, so the pressure can be written p = ⅓ n m ⟨v²⟩ or p = ⅓ ρ ⟨v²⟩. The second form is the useful one, because ρ is weighable. Inverting it defines the root-mean-square speed, vᵣₘₛ = √⟨v²⟩ = √(3p/ρ). Room air at 1.013 × 10⁵ Pa with ρ = 1.20 kg m⁻³ gives vᵣₘₛ = 503 m s⁻¹; helium in that same room, at the same pressure and temperature, has ρ = 0.166 kg m⁻³ and so vᵣₘₛ = 1.35 × 10³ m s⁻¹, because a lighter molecule must move faster to press as hard. The 503 m s⁻¹ is checkable: sound in that air travels at 343 m s⁻¹, which is √(γ/3) = 0.683 of vᵣₘₛ for γ = 1.40, since sound is carried by the same molecular motion.

The assumptions fail one at a time as a gas is compressed, and for air near room temperature the attraction gives way first: a molecule on its way to the wall is pulled back by the ones it is leaving behind, so it arrives with slightly less momentum, the hit is softened, and the measured pressure sits below the ideal value. Squeeze harder and the molecules' own volume takes over — the space left for free flight is less than V, the wall is struck more often than the bookkeeping assumes, and the pressure runs high instead. Those two corrections are precisely the a and the b of the van der Waals equation, and warming the gas past a few hundred kelvin reverses which of them leads. In the opposite direction, if the gas is so rarefied that molecules cross the container without meeting one another, there is no local equilibrium to average over and the walls, not the gas, set the behaviour. Two things do not break the result: rotation and vibration of the molecules, since only translational motion carries momentum to the wall, and the discreteness of the hits, whose relative fluctuation is 1/√N and so 10⁻¹⁰ for N = 10²⁰.