Kinetic theory of pressure

A tyre gauge reads a number, and it feels like a fixed property of the gas inside — but that number is really a statistical accident of violence. Every second, an astronomical number of molecules slam into the tyre's inner wall, bounce off, and hand over a tiny scrap of momentum each time. Add up those countless tiny shoves over one second and one square metre, and you get something steady enough to read off a dial: pressure. This article rebuilds that connection from scratch, using nothing beyond Newton's laws of motion. Starting from a short list of assumptions about how gas molecules behave, we derive the single most important result of kinetic theory, $p = \tfrac13\rho\overline{c^2}$ , and use it to move between pressure, density and molecular speed in numerical problems.

The core insight

Gas pressure is the macroscopic footprint of countless elastic molecular collisions with a container's walls, and applying Newton's laws to those collisions gives $p = \tfrac13\rho\overline{c^2}$ .

The core idea

To see why this matters, put a number on it: a litre of air at room conditions holds roughly $10^{22}$ molecules, each moving at hundreds of metres per second and colliding with its neighbours billions of times a second. No single collision is remarkable, but the rate at which momentum arrives at a wall, averaged over that enormous number of impacts, is what a pressure gauge actually measures. Kinetic theory replaces the real, messy gas with an idealised model built on a handful of assumptions, precisely so that this averaging can be done with algebra instead of by tracking every molecule individually:

A gas is a very large number of identical molecules in random, ceaseless motion, each effectively a point (negligible size compared with the distance between molecules).
Collisions — molecule with molecule, and molecule with wall — are perfectly elastic, so kinetic energy is conserved.
Between collisions the molecules move in straight lines; intermolecular attraction is negligible, and collision time is negligible next to the time between collisions.
Molecules obey Newton's laws of motion.

Picture one molecule of mass $m$ bouncing back and forth inside a cubical box of side $l$ , striking the wall perpendicular to the $x$ -axis over and over. Each elastic bounce reverses its $x$ -velocity and delivers a fixed scrap of momentum to that wall; the time between successive bounces on the same wall is fixed by how far the molecule must travel there and back. Multiply one molecule's contribution by the number of molecules, divide by the wall's area, and a force per unit area — a pressure — falls out.

Figure

A molecule bouncing between two opposite walls of a cube along the x axis showing its velocity reversing at each elastic collision

Carrying this through (worked in full below) gives the key kinetic-theory result:

p = \frac{1}{3}\rho\,\overline{c^2},

where: $p$ is the pressure exerted on the walls, $\rho = Nm/V$ is the gas density (total mass over volume), and $\overline{c^2}$ is the mean of the squared speed of all $N$ molecules — not the square of the mean speed. Equivalently, since $M = Nm$ is the total mass of gas, $pV = \tfrac13 M\overline{c^2}$ .

Derivation APressure from molecular collisions

The goal is to show, starting only from Newton's second and third laws applied to one molecule bouncing between two walls of a cube, that the pressure exerted by $N$ identical molecules of mass $m$ in a container of volume $V$ takes the specific form $p = \tfrac13\rho\overline{c^2}$ , with the factor of one-third arising purely from the random, direction-independent nature of molecular motion.

Step 1. Take one molecule with velocity components $(c_x, c_y, c_z)$ inside a cube of side $l$ (so $V = l^3$ ), where its speed satisfies $c^2 = c_x^2 + c_y^2 + c_z^2$ . Focus on the wall perpendicular to the $x$ -axis. An elastic collision reverses only the $x$ -component, $c_x \to -c_x$ , so the molecule's momentum changes by $\Delta p_x = mc_x - (-mc_x) = 2mc_x$ . By Newton's third law the wall receives momentum $2mc_x$ from this one collision.

Step 2. Between successive collisions with this same wall, the molecule must cross the cube and return, a round trip of distance $2l$ , taking time $\Delta t = 2l/c_x$ . The rate at which this one molecule delivers momentum to the wall is therefore $\Delta p_x/\Delta t = 2mc_x/(2l/c_x) = mc_x^2/l$ , which by Newton's second law is the force this molecule exerts on the wall.

Step 3. Summing this force over all $N$ molecules and dividing by the wall's area $l^2$ gives the pressure: $p = \dfrac{1}{l^2}\sum \dfrac{mc_x^2}{l} = \dfrac{m}{l^3}\sum c_x^2 = \dfrac{Nm}{V}\overline{c_x^2}$ , where $\overline{c_x^2}$ is the average of $c_x^2$ over all $N$ molecules.

Step 4. Because molecular motion is completely random, no direction is preferred, so $\overline{c_x^2} = \overline{c_y^2} = \overline{c_z^2} = \tfrac13\overline{c^2}$ . Substituting gives $p = \dfrac13\dfrac{Nm}{V}\overline{c^2} = \dfrac13\rho\overline{c^2}$ , since $\rho = Nm/V$ . This is the boxed result: pressure is not an independent given of nature but the direct consequence of Newton's laws applied to elastic molecular collisions, with the one-third coming purely from isotropy.

Worked example

A cubical vessel of side $0.5\ \text{m}$ contains $2.0\times10^{25}$ molecules of a gas, each of mass $5.3\times10^{-26}\ \text{kg}$ , with mean square speed $\overline{c^2} = 2.5\times10^5\ \text{m}^2/\text{s}^2$ . Find the pressure on the walls.

Step 1. Volume of the cube: $V = (0.5)^3 = 0.125\ \text{m}^3$ .

Step 2. Total mass and density: $M = Nm = (2.0\times10^{25})(5.3\times10^{-26}\ \text{kg}) = 1.06\ \text{kg}$ , so $\rho = M/V = 1.06/0.125 = 8.48\ \text{kg/m}^3$ .

Step 3. Apply the formula: $p = \tfrac13\rho\overline{c^2} = \tfrac13(8.48)(2.5\times10^5) = 7.07\times10^5\ \text{Pa}$ .

The gas exerts a pressure of about $7.1\times10^5\ \text{Pa}$ (roughly $7$ atmospheres) on the container walls.

Competitive stretch

Nitrogen gas at STP has $\rho = 1.25\ \text{kg/m}^3$ and exerts $p = 1.01\times10^5\ \text{Pa}$ . Rearranging, $\overline{c^2} = 3p/\rho = (3\times1.01\times10^5)/1.25 = 2.424\times10^5\ \text{m}^2/\text{s}^2$ , so the rms speed is $c_{rms} = \sqrt{\overline{c^2}} \approx 492\ \text{m/s}$ — matching the known order of magnitude (~500 m/s) for nitrogen at room temperature, a good sanity check.

Common traps

Writing $p = \rho\overline{c^2}$ or dropping the factor of one-third — the $\tfrac13$ comes specifically from isotropy ( $\overline{c_x^2}=\overline{c_y^2}=\overline{c_z^2}=\tfrac13\overline{c^2}$ ) and is not optional.
Using $\Delta p_x = mc_x$ instead of $2mc_x$ — an elastic bounce reverses momentum, it does not merely stop it.
Treating $\overline{c^2}$ (mean of the squared speed) as the same thing as $(\bar c)^2$ (square of the mean speed) — they are numerically different, even though $c_{rms}=\sqrt{\overline{c^2}}$ is the quantity actually used in calculation.
Believing this derivation proves the ideal gas law $pV = nRT$ — it does not. The ideal gas law is a separate experimental result; kinetic theory only explains why pressure takes the form it does.

Ask the tutor: if you would like to see how this same derivation is redone for a spherical container, or how $\overline{c^2}$ later gets connected to temperature, ask the AI tutor to walk through it.

Why does the derivation use a cube instead of some other container shape?

A cube is chosen purely because it makes the geometry easy: a molecule bouncing perpendicular to one wall travels a clean distance $2l$ between successive hits on that same wall, so the time between collisions is simple to write down. The final result, $p=\tfrac13\rho\overline{c^2}$ , does not depend on this choice — the isotropy assumption already guarantees that pressure is the same on every wall of any shaped container, since no direction is preferred. The cube is a teaching convenience, not a physical requirement.

Does this derivation assume the molecules collide with each other, or only with the walls?

It only tracks collisions with the walls; molecule-molecule collisions are not used directly in the derivation. However, they are essential to the underlying picture — they are what keeps the motion random and constantly redistributes velocity among all directions, which is exactly what justifies the isotropy step $\overline{c_x^2}=\overline{c_y^2}=\overline{c_z^2}=\tfrac13\overline{c^2}$ . Without frequent, randomising molecule-molecule collisions, a gas could have an ordered, non-isotropic flow, and the one-third factor would not apply.

If pressure comes from kinetic theory, why do we still need the ideal gas equation pV equals nRT separately?

Because the two results answer different questions and neither can be derived from the other. $p=\tfrac13\rho\overline{c^2}$ is a mechanical result: given a molecule's mass, density and mean square speed, it tells you the pressure. $pV=nRT$ is an experimental law connecting pressure, volume, amount of gas and temperature, discovered independently by measuring real gases. Combining the two later (by identifying $\overline{c^2}$ with temperature) is what lets kinetic theory explain what temperature physically is — but that combination is a separate step, not something this derivation alone provides.

Kinetic theory of pressure

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads