The Kinetic Theory of Gases

By now you can use a gas: squeeze it and the pressure climbs, warm it and it swells or pushes harder, and the whole lot is tied together by the ideal gas equation, pV = NkT. But every one of those rules was discovered by measuring gases from the outside. None of them explains why a gas behaves that way. Where does the pressure actually come from? What is temperature, deep down?

The kinetic theory of gases answers those questions by taking one bold idea completely seriously: a gas is nothing but a vast swarm of tiny molecules in ceaseless, random motion, obeying ordinary Newtonian mechanics — bouncing off the walls, bouncing off each other, flying in straight lines in between. Astonishingly, if you apply Newton's laws to that swarm and average over its billions of members, the entire behaviour of the gas — pV, temperature, all of it — drops out of the mathematics. This page is that derivation: bulk gas behaviour, built from moving molecules.

The rules of the game: the assumptions

To turn "a box of bouncing molecules" into equations we need a clean, idealised model. The kinetic theory of an ideal gas rests on a short list of assumptions — each one a deliberate simplification that makes the maths tractable while staying close to how a real dilute gas actually behaves:

These are exactly the conditions under which a real gas is well described by pV = NkT — a hot, dilute gas well away from the point where it would condense into a liquid. Keep them in mind: every step below leans on one of them.

Where the pressure comes from

Picture the gas in a cubical box of side L, and follow a single molecule as it heads straight at the right-hand wall with an x-velocity u. Pressure is going to be the blur of countless such molecules drumming on that wall — so let's find the push from one, then add them up.

Step 1 — momentum change in one bounce. The collision is elastic, so the molecule rebounds with its x-velocity exactly reversed, from +u to -u. Its change of momentum is

\Delta p = m u - (-m u) = 2 m u.

Step 2 — how often it hits. After bouncing it must cross to the far wall and come back — a round trip of 2L — before it strikes this wall again. At speed u that takes a time

\Delta t = \dfrac{2L}{u}.

Step 3 — the average force from one molecule. By Newton's second law, force is the rate of change of momentum, so this molecule's average push on the wall is

F = \dfrac{\Delta p}{\Delta t} = \dfrac{2mu}{2L/u} = \dfrac{m u^2}{L}.

Step 4 — add up all the molecules. Every molecule contributes its own m u_i^2 / L, so the total force on the wall is F = \dfrac{m}{L}\left(u_1^2 + u_2^2 + \cdots + u_N^2\right). Writing the mean of the squared x-speeds as \overline{u^2}, this is F = \dfrac{Nm\,\overline{u^2}}{L}.

Step 5 — three directions, so bring in \overline{c^2}. A molecule's true speed c has components in all three directions, c^2 = u^2 + v^2 + w^2. Motion is random, so on average no direction is special: \overline{u^2} = \overline{v^2} = \overline{w^2}, and therefore \overline{u^2} = \tfrac13\,\overline{c^2}, where \overline{c^2} is the mean-square speed.

Step 6 — divide force by area to get pressure. The wall has area A = L^2 and the box has volume V = L^3. Putting \overline{u^2} = \tfrac13\overline{c^2} into the force and dividing by A:

p = \dfrac{F}{A} = \dfrac{Nm\,\overline{u^2}}{L^3} = \dfrac{1}{3}\,\dfrac{Nm\,\overline{c^2}}{V},

which rearranges into the central result of the whole theory:

pV = \tfrac13 N m \,\overline{c^2}.

Look at what just happened. We fed in nothing but Newton's laws and a box of bouncing points — and out came a relationship between the bulk pressure and volume of a gas and the microscopic mass, number and speed of its molecules. Everything else follows from here.

Temperature is molecular kinetic energy

We now have two expressions for the very same quantity pV: the one measured from the outside, and the one derived from the molecules inside.

pV = N k T \qquad\text{and}\qquad pV = \tfrac13 N m \,\overline{c^2}.

Set them equal — they describe the same gas — and the number of molecules N cancels straight off both sides:

\tfrac13 N m \,\overline{c^2} = N k T \;\;\Longrightarrow\;\; \tfrac13 m \,\overline{c^2} = k T.

Multiply through by \tfrac32 and the left-hand side becomes the average kinetic energy of a molecule, \tfrac12 m \overline{c^2}:

\tfrac12 m \,\overline{c^2} = \tfrac32 k T.

This is one of the most important sentences in all of thermal physics. The left-hand side is the mean translational kinetic energy of a single molecule; the right-hand side is just a constant times the absolute temperature. So temperature is not some separate substance or fluid — the absolute (kelvin) temperature of a gas is a direct measure of the average kinetic energy of its molecules. Double the kelvin temperature and you have doubled the mean molecular KE. Notice too that the mass has vanished: at a given temperature every gas, heavy or light, has the same mean translational KE per molecule.

And it explains absolute zero at last. As T \to 0\,\text{K}, the right-hand side goes to zero, so \tfrac12 m \overline{c^2} \to 0: molecular motion falls to its absolute minimum. You cannot cool further, because there is no kinetic energy left to remove.

For an ideal gas of N molecules, each of mass m, with mean-square speed \overline{c^2}:

Startlingly fast. The air around you sits at about 293\,\text{K}, and a nitrogen molecule (mass 4.7\times10^{-26}\,\text{kg}) at that temperature has a root-mean-square speed of very nearly 500\ \text{m s}^{-1} — roughly the speed of a rifle bullet, or one and a half times the speed of sound. Every molecule of the air you are breathing is hurtling about at that pace, changing direction billions of times a second as it smashes into its neighbours.

The same equation explains why the Earth keeps its atmosphere but the Moon has none, and why helium balloons are a slightly futile purchase. Since \tfrac12 m \overline{c^2} = \tfrac32 kT, a lighter molecule moves faster at the same temperature. Light, nimble helium and hydrogen molecules zip along so quickly that many exceed Earth's escape velocity and trickle away into space over millions of years — which is why free helium is rare in our atmosphere and why your helium balloon deflates as the tiny atoms squeeze out through the rubber. Heavier nitrogen and oxygen are too sluggish to escape, so they stay — and we get to breathe.

The root-mean-square speed

The mean-square speed \overline{c^2} is an average of squares, so its units are \text{m}^2\,\text{s}^{-2} — not a speed at all. To get back to something measured in \text{m s}^{-1} we take the square root, and the result is the root-mean-square speed:

c_{\text{rms}} = \sqrt{\overline{c^2}}.

It is a genuine three-step recipe, and the order matters: take every molecule's speed, square it, take the mean of all those squares, then take the square root — root of the mean of the squares. It is the natural "typical" speed to quote, because it is the one tied directly to the kinetic energy.

Rearranging the temperature result \tfrac12 m \overline{c^2} = \tfrac32 kT gives \overline{c^2} = \dfrac{3kT}{m}, and taking the root:

c_{\text{rms}} = \sqrt{\overline{c^2}} = \sqrt{\dfrac{3kT}{m}}.

Two features are worth burning in. First, c_{\text{rms}} \propto \sqrt{T}: to double the rms speed you must quadruple the absolute temperature. Second, c_{\text{rms}} \propto 1/\sqrt{m}: at a given temperature, lighter molecules are faster — exactly the fact behind the escaping helium.

Worked examples

Example 1 — rms speed from temperature. Find the root-mean-square speed of nitrogen molecules (mass m = 4.7\times10^{-26}\,\text{kg}) in air at 27\,^\circ\text{C}. Take k = 1.38\times10^{-23}\,\text{J K}^{-1}.

Step 1 — temperature in kelvin. T = 27 + 273 = 300\ \text{K}. Step 2 — use c_{\text{rms}} = \sqrt{3kT/m}.

c_{\text{rms}} = \sqrt{\dfrac{3 \times 1.38\times10^{-23} \times 300}{4.7\times10^{-26}}} = \sqrt{2.64\times10^{5}} \approx 514\ \text{m s}^{-1}.

Around 500\ \text{m s}^{-1} — faster than the speed of sound, just as promised.

Example 2 — mean kinetic energy from temperature. What is the mean translational kinetic energy of any ideal-gas molecule at 300\ \text{K}?

This one needs no mass at all — it depends only on temperature, through \tfrac12 m \overline{c^2} = \tfrac32 kT:

\overline{E_k} = \tfrac32 k T = \tfrac32 \times 1.38\times10^{-23} \times 300 = 6.21\times10^{-21}\ \text{J}.

A helium atom and a heavy carbon-dioxide molecule side by side at 300\ \text{K} carry the same mean KE — the helium simply reaches it by moving much faster.

Example 3 — pressure straight from the kinetic theory equation. A container of volume V = 0.025\ \text{m}^3 holds N = 1.0\times10^{24} molecules, each of mass m = 5.0\times10^{-26}\,\text{kg}, with a mean-square speed \overline{c^2} = 3.0\times10^{5}\ \text{m}^2\,\text{s}^{-2}. Find the pressure.

Step 1 — write pV = \tfrac13 N m \overline{c^2} and solve for p. p = \dfrac{N m \,\overline{c^2}}{3V}. Step 2 — substitute.

p = \dfrac{(1.0\times10^{24})(5.0\times10^{-26})(3.0\times10^{5})}{3 \times 0.025} = \dfrac{1.5\times10^{4}}{0.075} = 2.0\times10^{5}\ \text{Pa}.

That is about 200\ \text{kPa} — roughly twice atmospheric pressure — obtained without ever putting a gauge on the container, purely from the mechanics of the swarm inside.

See the swarm

Here is the box of molecules the whole theory is about. Each molecule carries a velocity arrow whose length is its speed. Drag the temperature slider and watch the arrows grow: raising the temperature raises the mean molecular kinetic energy, so the molecules move faster (the arrows lengthen as \sqrt{T}, since c_{\text{rms}} \propto \sqrt{T}). The Mean KE bar on the right rises in exact proportion to the absolute temperature — that is what temperature means. Drag the number slider and more molecules join the swarm; notice the mean KE and the rms speed don't change when you add molecules, because temperature is an average per molecule, not a total.