1.2 The kinetic-theory picture

Air, at rest, is a gas of mostly nitrogen and oxygen molecules at room temperature, moving randomly at thermal speeds around 500 m/s, colliding constantly with each other and with whatever boundaries enclose them. The macroscopic state we care about — the three numbers from the previous lesson — is the time-averaged consequence of those microscopic motions.

Pressure, kinetically

A molecule has mass and velocity, so it carries momentum $mv$ and kinetic energy $\tfrac12 m v^2$ — the work needed to bring it from rest to speed $v$ (refresher: kinetic energy →) — where $m$ is the molecular mass and $v$ its speed. When a molecule strikes a wall and rebounds, it hands the wall some of that momentum as an impulse; the countless impulses arriving each second add up to a steady force, and that force spread over the wall’s area is the pressure. Pressure is therefore bookkeeping on momentum:

p \;=\; \frac{F}{A}, \qquad F \;=\; \frac{\Delta(\text{momentum})}{\Delta t}, \qquad\Longrightarrow\qquad p \;=\; \frac{\text{momentum delivered to the wall}}{\text{area}\times\text{time}}.

The middle step — force is the rate of momentum delivery — is Newton’s second law in its general form, $F = \Delta(\text{momentum})/\Delta t$ . Everything that follows is a count of that momentum.

Counting the momentum

Put a flat patch of wall of area $A$ in the $y$ – $z$ plane, with gas on the $+x$ side; only the $x$ -velocity $v_x$ carries a molecule toward or away from it. A molecule moving toward the wall at $x$ -speed $v_x$ travels $v_x\,\Delta t$ in a time $\Delta t$ , so it can reach the wall within $\Delta t$ only if it starts within that distance. The molecules close enough to hit therefore fill a slab of thickness $v_x\,\Delta t$ against the wall, of volume

V_\text{slab} \;=\; A\, v_x\, \Delta t.

Take, for a first pass, a single $x$ -speed $v_x$ and treat every molecule in the slab as heading straight at the wall. With number density $n$ (molecules per unit volume) the slab holds $n\,A\,v_x\,\Delta t$ of them, and each rebounds elastically from $+v_x$ to $-v_x$ , handing the wall momentum of exactly $2 m v_x$ . The momentum delivered in time $\Delta t$ is

\underbrace{(2 m v_x)}_{\text{per molecule}}\;\underbrace{(n\, A\, v_x\, \Delta t)}_{\text{molecules}} \;=\; 2\, n\, m\, A\, v_x^2\, \Delta t.

Divide by the time for the force, then by the area for the pressure, and both $A$ and $\Delta t$ cancel — exactly as the bookkeeping above said they would:

p \;=\; \frac{1}{A}\,\frac{2\, n\, m\, A\, v_x^2\, \Delta t}{\Delta t} \;=\; 2\, n\, m\, v_x^2.

This first pass overcounts in two correctable ways. First, the molecules neither share one speed nor all head for the wall: averaging $v_x^2$ over the real velocity distribution while keeping only the half moving toward the wall replaces $v_x^2$ by exactly $\tfrac12\langle v_x^2\rangle$ , where $\langle v_x^2\rangle$ is the mean-square $x$ -velocity — turning the leading $2$ into $1$ and leaving $p = n m\langle v_x^2\rangle$ . Second, no direction is special, so $\langle v_x^2\rangle = \tfrac13\langle v^2\rangle$ . Together they give the result:

\boxed{\;p \;=\; n\, m\, \langle v_x^2\rangle \;=\; \frac{1}{3}\, n\, m\, \langle v^2 \rangle.\;}

Both corrections are exact for a gas in equilibrium.

▶ The factor of ½: only molecules moving toward the wall count Derivation

The single-speed sketch used $n\,A\,v_x\,\Delta t$ as the molecule count and a bare $v_x^2$ . Both are repaired at once by sorting the molecules by their $x$ -velocity and integrating.

Let $n(v_x)\,dv_x$ be the number per unit volume with $x$ -velocity between $v_x$ and $v_x + dv_x$ . Only molecules with $v_x>0$ move toward the wall, and of those, the ones within $v_x\,\Delta t$ strike it in time $\Delta t$ — a slab of volume $A\,v_x\,\Delta t$ . The momentum from this group is the per-impact $2 m v_x$ times the count:

d(\Delta\,\text{mom}) \;=\; (2 m v_x)\,\big(n(v_x)\,A\,v_x\,\Delta t\big)\,dv_x \;=\; 2 m A\,\Delta t\; v_x^2\, n(v_x)\,dv_x.

Sum over the molecules that actually approach, $v_x>0$ :

\Delta\,\text{mom} \;=\; 2 m A\,\Delta t \int_{0}^{\infty} v_x^2\, n(v_x)\,dv_x.

In equilibrium the velocity distribution is symmetric — as many molecules move in $+x$ as in $-x$ — so $v_x^2\, n(v_x)$ is an even function and the integral over $v_x>0$ is exactly half the integral over all $v_x$ . That full integral is, by definition of the average, $n\,\langle v_x^2\rangle$ :

\int_{0}^{\infty} v_x^2\, n(v_x)\,dv_x \;=\; \tfrac12\, n\,\langle v_x^2\rangle.

So the $\tfrac12$ that turns the leading $2$ into $1$ is exact, and it comes from restricting the sum to approaching molecules. Dividing the momentum by $\Delta t$ (force) and by $A$ (pressure):

p \;=\; \frac{1}{A}\,\frac{\Delta\,\text{mom}}{\Delta t} \;=\; n\, m\, \langle v_x^2\rangle. \;✓

One subtlety in the heuristic: “half the molecules move toward the wall” is the unweighted fraction with $v_x>0$ , which is also $\tfrac12$ . It matches the $v_x^2$ -weighted fraction in the integral only because the distribution is symmetric — which, in equilibrium, it is.

▶ Why one direction carries a third of the motion Derivation

Nothing distinguishes the $x$ , $y$ , and $z$ axes for a gas at rest, so the mean-square velocity components are equal:

\langle v_x^2\rangle \;=\; \langle v_y^2\rangle \;=\; \langle v_z^2\rangle.

The speed obeys $v^2 = v_x^2 + v_y^2 + v_z^2$ . Averaging both sides and using the equality of the three components,

\langle v^2\rangle \;=\; \langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle \;=\; 3\,\langle v_x^2\rangle \qquad\Longrightarrow\qquad \langle v_x^2\rangle \;=\; \tfrac13\,\langle v^2\rangle.

This is the $\tfrac13$ in the boxed result: of a molecule’s motion, the share pressing on any single wall is one direction out of three. ✓

Pressure as an energy density

The boxed result has a cleaner reading in terms of energy. Multiplying any per-molecule quantity by the number density $n$ rescales it from “per molecule” to “per unit volume” — the same step that turns the mass of one molecule into the mass density of the gas, $\rho = n m$ (molecules per volume times mass per molecule). Applied to kinetic energy, with each molecule carrying $\big\langle\tfrac12 m v^2\big\rangle$ on average, it gives the kinetic-energy density:

u \;=\; \underbrace{n}_{\text{molecules per volume}}\;\underbrace{\Big\langle \tfrac12 m v^2 \Big\rangle}_{\text{energy per molecule}} \;=\; \tfrac12\, n\, m\, \langle v^2\rangle.

Comparing this with the boxed pressure gives a clean ratio:

\frac{p}{u} \;=\; \frac{\tfrac13\, n\, m\, \langle v^2\rangle}{\tfrac12\, n\, m\, \langle v^2\rangle} \;=\; \frac{1/3}{1/2} \;=\; \frac{2}{3}, \qquad\text{so}\qquad p \;=\; \tfrac23\, u.

Pressure is two-thirds of the kinetic-energy density. The $\tfrac23$ is the $\tfrac13$ from the three directions divided by the $\tfrac12$ in the definition of kinetic energy — the factor of two between $m v^2$ and $\tfrac12 m v^2$ .

Temperature and the ideal-gas law

The last ingredient is the temperature $T$ . Boltzmann’s constant

k_B \;=\; 1.380649\times 10^{-23}\ \text{J/K}

is the fixed conversion factor between temperature and energy: it sets how much kinetic energy corresponds to a given absolute temperature $T$ (in kelvin, K). The equipartition theorem of statistical mechanics fixes that energy at $\tfrac32 k_B T$ of translational kinetic energy per molecule,

\Big\langle \tfrac12 m v^2 \Big\rangle \;=\; \tfrac32 k_B T, \qquad\text{so}\qquad m\,\langle v^2\rangle \;=\; 3 k_B T.

▶ Equipartition: where the factor 3/2 comes from Derivation

Equipartition (developed in Physics → Kinetic theory) assigns, on average, an energy $\tfrac12 k_B T$ to each independent quadratic term in a system’s energy. A molecule’s translational kinetic energy is a sum of three such terms, one per direction,

\tfrac12 m v^2 \;=\; \tfrac12 m v_x^2 + \tfrac12 m v_y^2 + \tfrac12 m v_z^2,

so each carries $\tfrac12 k_B T$ and the total is

\Big\langle \tfrac12 m v^2 \Big\rangle \;=\; 3\cdot\tfrac12 k_B T \;=\; \tfrac32 k_B T.

Per direction this reads $\big\langle\tfrac12 m v_x^2\big\rangle = \tfrac12 k_B T$ , i.e. $\langle v_x^2\rangle = k_B T/m$ , consistent with $\langle v_x^2\rangle = \tfrac13\langle v^2\rangle$ above. ✓

Substituting $m\langle v^2\rangle = 3 k_B T$ into the boxed result $p = \tfrac13 n m\langle v^2\rangle$ cancels the $\tfrac13$ against the $3$ and leaves the ideal-gas law,

p \;=\; n\, k_B\, T.

The units confirm it: $n$ is a count per volume ( $\text{m}^{-3}$ ), $k_B$ is energy per temperature ( $\text{J/K}$ ), and $T$ is a temperature ( $\text{K}$ ), so $n\,k_B\,T$ carries units of $\text{J}/\text{m}^3 = \text{N}/\text{m}^2 = \text{Pa}$ — an energy per unit volume, which is exactly a pressure.

The simulation

The interactive below runs a 2-D ideal gas: $N$ identical particles, mass $1$ , bouncing elastically off the walls of a box of area $V$ and off each other. Each particle’s velocity components $v_x, v_y$ are drawn from independent Gaussians of width $\sigma\propto\sqrt{T}$ , so the speed $|v|$ is distributed according to the Maxwell–Boltzmann law

f(v) \;=\; \frac{m}{k_B T}\, v\, \exp\!\Big(-\frac{m v^2}{2 k_B T}\Big),

shown as the red curve on the right. The gray histogram is the measured speed distribution from the simulation, time-averaged as the dynamics run. The two converge to each other — the macroscopic distribution is what the microscopic dynamics produce.

N particles

T temperature

300 K

V volume

100%

⟨v⟩ (sim units/s): 0.0
P (sim units): 0.00
PV / NkT: 0.00

Every collision is time-reversible, and energy is conserved exactly. The macroscopic gas law emerges as a statistical statement about a system whose microscopic dynamics are perfectly deterministic: macroscopic thermodynamic equilibrium is consistent with, and indeed implies, vigorous microscopic motion.

What one molecule looks like

Pick any single molecule in the simulation. Its trajectory is a sequence of free flights at constant velocity, punctuated by elastic collisions that scatter it into a new direction with about the same speed. Over many collisions the directions become uncorrelated, and the molecule’s position random-walks through the box.

A caution on language: the next lesson treats a different random-walk-like phenomenon — Brownian motion — that is often conflated with this one. The wandering trajectory of one bath molecule is a microscopic random walk emerging from deterministic, time-reversible physics. What Robert Brown saw under his microscope in 1827 was the motion of a particle very much larger than the bath molecules, jostled by them. That phenomenon needs its own treatment, which the next lesson gives.