Kinetic model of gas pressure

The kinetic model is a list of assumptions, and every later step leans on one of them. The gas is N molecules in a volume V with N enormous, so averages are sharp. Each molecule is a point: its own volume is negligible against the space it roams. At 0 °C and 101 kPa the number density is 2.69 × 10²⁵ m⁻³, so the mean spacing n^(−1/3) is 3.3 nm against a molecular diameter near 0.37 nm for N₂, and the molecules fill under 0.1% of the volume. Between collisions no forces act, so paths are straight lines at constant speed. Collisions with the wall are elastic, so no kinetic energy is lost there. Velocities point every way with no preferred axis. And the gas is in a steady state, so the wall returns on average exactly the momentum flux a perfectly specular bounce would, whatever its roughness. Remove any one of these and the derivation must change — which is how the model tells you where it stops.

Put one molecule of mass m in a cube of side L and watch the wall at x = L. When the molecule strikes it with normal velocity component vₓ the collision reverses that component and leaves v_y and v_z alone, so the wall receives momentum 2m vₓ. The molecule then crosses the box, rebounds from the far wall and returns after a time 2L/vₓ. Averaged over many round trips the force on the wall is impulse divided by interval: F̄ = 2m vₓ ÷ (2L/vₓ) = m vₓ²/L. Notice where vₓ appeared: once in the impulse and once in the rate, and it is that second appearance that will make a mean of squares unavoidable. For a nitrogen molecule (m = 4.65 × 10⁻²⁶ kg) with vₓ = 300 m s⁻¹ in a 0.10 m box, each hit delivers 2.79 × 10⁻²³ kg m s⁻¹, hits arrive 1500 times a second, and the mean force is 4.2 × 10⁻²⁰ N. Collisions with other molecules do not spoil the bookkeeping: they conserve momentum, so what one molecule fails to deliver another delivers instead, and the population sum that follows is unchanged.

Add the contributions of all N molecules: F = (m/L) Σ vₓ² = (N m/L) ⟨vₓ²⟩, where ⟨vₓ²⟩ is the population mean of the squared x-component. Dividing by the wall area L² and writing n = N/V for the number density gives p = n m ⟨vₓ²⟩. The average sits over the square, and no rearrangement will move it: pressure depends on the mean of vₓ², not on the square of a mean speed. Try three molecules with |vₓ| of 200, 400 and 600 m s⁻¹. Their mean is 400 m s⁻¹, so the square of the mean is 1.60 × 10⁵ m² s⁻², but the mean of the squares is (0.40 + 1.60 + 3.60) × 10⁵ ÷ 3 = 1.87 × 10⁵ m² s⁻², larger by 17%. The gap is the variance: ⟨vₓ²⟩ = ⟨vₓ⟩² + Var(vₓ), always. For a gas at rest the signed component averages to zero, ⟨vₓ⟩ = 0, so the square of the mean would predict no pressure at all — the whole of p is the variance of the molecular motion.

The box is scaffolding, and a second-year derivation removes it. Let f(vₓ) dvₓ be the fraction of molecules with x-velocity between vₓ and vₓ + dvₓ, normalised so that ∫ f dvₓ = 1 over all vₓ. That class numbers n f(vₓ) dvₓ per unit volume, and the members that reach a wall of area A in a time Δt are those inside a slab of thickness vₓ Δt heading toward it: n f(vₓ) dvₓ vₓ A Δt of them, each delivering 2m vₓ. Summing the momentum over every approaching class (vₓ > 0) and dividing by A Δt gives p = ∫₀^∞ 2m vₓ² n f(vₓ) dvₓ. For a gas at rest f is even, f(−vₓ) = f(vₓ), so the half-range integral with its factor 2 equals the full-range integral without it: p = n m ∫ vₓ² f(vₓ) dvₓ = n m ⟨vₓ²⟩. The integrand vₓ² f(vₓ) is the solid curve in the figure: it vanishes at vₓ = 0, where the count of molecules peaks, and for the bell-shaped f drawn there, with rms width σₓ = √⟨vₓ²⟩, it peaks at ±√2 σₓ, where molecules are rarer but each arrives sooner and hits harder. The container has left the argument, so p is a property of the gas itself — the same momentum flux crosses any imaginary plane drawn inside it.

Nothing distinguishes x from y or z in a gas with no bulk flow, so ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩. Since v² = vₓ² + v_y² + v_z² for every molecule, averaging gives ⟨v²⟩ = 3⟨vₓ²⟩ and the pressure becomes p = ⅓ n m ⟨v²⟩ = ⅓ ρ ⟨v²⟩, with ρ = n m the mass density. Check the dimensions: kg m⁻³ × m² s⁻² = kg m⁻¹ s⁻² = N m⁻², a pressure. Check the limits: empty the box, n → 0, and p → 0; freeze the motion, ⟨v²⟩ → 0, and p → 0; halve the volume at the same speeds, so n doubles, and p doubles, which is Boyle's law read molecularly. Rewrite it as p = ⅔ n ⟨½ m v²⟩ and a third reading appears: pressure is two-thirds of the translational kinetic energy per unit volume, so air at 101.3 kPa stores 3p/2 = 152 kJ of translational kinetic energy in every cubic metre. Multiplied by V, pV = ⅓ N m ⟨v²⟩ now stands ready against the measured pV = NkT; that comparison, and the meaning of temperature it yields, is the next topic.

Invert the result: vᵣₘₛ = √⟨v²⟩ = √(3p/ρ), and two bench readings fix a molecular speed. Nitrogen at 0 °C and 101.3 kPa has density 1.25 kg m⁻³, so vᵣₘₛ = √(3 × 101 300 ÷ 1.25) = √(2.43 × 10⁵ m² s⁻²) = 493 m s⁻¹. Hydrogen under the same conditions has ρ = 0.0899 kg m⁻³ and vᵣₘₛ = 1.84 × 10³ m s⁻¹: the same pressure from a fourteenth of the mass, so √14 ≈ 3.7 times the speed. Both exceed the speed of sound in their gas, as they must — sound is carried by these very molecules and cannot outrun them. The formula also marks where the model stops. It assumes a molecule arrives at full speed and leaves at full speed; if its neighbours attract it, it arrives slower and delivers less, so a dense or cold real gas exerts less than ⅓ ρ ⟨v²⟩. It also takes n as N per volume of box; when the molecules themselves fill a noticeable share of that volume the free space is smaller than V and the wall is struck more often than the point model counts. Those two corrections are the van der Waals terms met at the end of this section.