Physics/Lessons/Thermodynamics/The Kinetic Theory of Gases
Physics · Thermodynamics · A2 Level

The Kinetic Theory of Gases

The gas laws work brilliantly, but they don't explain anything — they're pattern-matching on experimental data. Kinetic theory is the payoff: start from molecules bouncing off the walls of a container, apply Newton's laws to a single collision, and the pressure of a gas, the ideal gas equation, and the true meaning of temperature all fall out of the same short argument.

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The assumptions of the kinetic theory model

Kinetic theory models a gas as an enormous number of identical, tiny particles in constant random motion, and derives macroscopic properties like pressure directly from their mechanics. The model rests on a specific set of simplifying assumptions:

  • A gas consists of a very large number of identical molecules, moving with random directions and speeds.
  • The volume of the molecules themselves is negligible compared with the volume of the container they occupy.
  • Collisions — both between molecules, and between molecules and the container walls — are perfectly elastic: kinetic energy is conserved overall in every collision, none is lost.
  • Intermolecular forces are negligible, except during the brief moment of a collision itself — between collisions, molecules move in straight lines at constant velocity.
  • The time taken for a collision is negligible compared with the time between collisions.

Every one of these assumptions is what makes a gas "ideal" in the sense of the previous lesson — a real gas violates them slightly (molecules do have some volume, and attract each other weakly), which is why real gases deviate from $pV=nRT$ under extreme conditions.

The kinetic theory model: a large number of molecules moving freely in random directions inside a container, colliding elastically with the walls. Each individual collision transfers momentum to the wall; summed over all molecules, this produces the steady macroscopic pressure of the gas.

Deriving pressure from molecular collisions

Pressure is a macroscopic idea — force spread over an area — but underneath it, at the molecular scale, pressure is the accumulated effect of vast numbers of molecules colliding with the container walls and bouncing back, each delivering a tiny impulse. The full derivation (beyond A-level rigour, but worth following in outline) considers one molecule bouncing between two opposite walls of a cubical box.

DerivationpV = ⅓Nm⟨c²⟩, from a single molecule's collisions
1
Consider one molecule of mass $m$ in a cubical box of side $L$, moving with velocity component $u$ perpendicular to one pair of walls. It collides elastically with a wall, so its velocity component reverses: $u \to -u$.
Elastic collision with a much more massive wall reverses the molecule's velocity component without loss of speed.
2
The molecule's momentum change in that collision is $\Delta p = mu - (-mu) = 2mu$.
3
Between hitting this wall and hitting it again (after bouncing off the opposite wall and returning), the molecule travels a distance $2L$, taking time $t = \dfrac{2L}{u}$.
This is the time between successive collisions with the same wall.
4
The average force this one molecule exerts on the wall is momentum change over time: $F = \dfrac{2mu}{2L/u} = \dfrac{mu^2}{L}$.
Newton's second law in the impulse form, $F = \Delta p / \Delta t$.
5
Summing this over all $N$ molecules in the box, and replacing each molecule's individual $u^2$ with the mean square value $\langle u^2 \rangle$ averaged over all of them, the total force on one wall is $F_{\text{total}} = \dfrac{Nm\langle u^2\rangle}{L}$.
6
Pressure is force over area, and the wall has area $L^2$: $p = \dfrac{F_{\text{total}}}{L^2} = \dfrac{Nm\langle u^2\rangle}{L^3} = \dfrac{Nm\langle u^2\rangle}{V}$.
$L^3$ is just the volume $V$ of the cubical box.
7
Molecules move in three dimensions, not just along one axis, so their true speed $c$ satisfies $c^2 = u^2+v^2+w^2$ for the three perpendicular velocity components. By symmetry, $\langle u^2\rangle = \langle v^2\rangle = \langle w^2\rangle$, so $\langle u^2\rangle = \tfrac13\langle c^2\rangle$.
Random motion means no direction is preferred over any other, on average.
8
Substituting gives $p = \dfrac{Nm}{3V}\langle c^2\rangle$, or rearranged: $pV = \tfrac13Nm\langle c^2\rangle$.
$\langle c^2 \rangle$ is the mean square speed, the average of the square of each molecule's speed — not the square of the average speed (see the next lesson for why this distinction matters).
▪ derived

This result, $pV = \frac13Nm\langle c^2\rangle$, is one of the most important in the whole topic — it connects a macroscopic, measurable quantity (pressure) directly to the microscopic motion of individual molecules, with nothing else needed.

Linking kinetic theory to the ideal gas equation

The kinetic theory result and the ideal gas equation both describe the same gas, so they must be consistent. Setting them equal gives a result of huge conceptual importance.

Derivation½m⟨c²⟩ = (3/2)kT — average molecular KE depends only on T
1
From kinetic theory: $pV = \tfrac13Nm\langle c^2\rangle$. From the ideal gas equation (molecules form): $pV = NkT$.
2
Both expressions equal $pV$ for the same gas, so set them equal: $\tfrac13Nm\langle c^2\rangle = NkT$.
3
The number of molecules $N$ cancels from both sides: $\tfrac13m\langle c^2\rangle = kT$.
4
Multiply both sides by $\tfrac32$: $\tfrac12m\langle c^2\rangle = \tfrac32kT$.
The left-hand side, $\tfrac12m\langle c^2\rangle$, is exactly the average translational kinetic energy of one molecule.
▪ derived
The single most exam-favourite result in this topic

$\tfrac12m\langle c^2\rangle = \tfrac32kT$ says the average translational kinetic energy of a gas molecule depends on absolute temperature alone — not pressure, volume, what gas it is, or the molecule's mass individually (though mass does affect the corresponding speed, since a heavier molecule needs a lower speed to carry the same kinetic energy). This is the precise, quantitative version of the earlier statement that "temperature measures average random kinetic energy" — kinetic theory shows exactly what the constant of proportionality is, and confirms that doubling the absolute temperature of a gas exactly doubles the average kinetic energy of its molecules.

Worked exampleFinding a molecule's r.m.s. speed at a given temperature
1
Estimate the root-mean-square speed of a nitrogen molecule ($N_2$, molar mass $28\,\text{g}\,\text{mol}^{-1}$) at room temperature, $293\,\text{K}$.
2
Find the mass of one molecule: $m = \dfrac{28\times10^{-3}}{6.02\times10^{23}} = 4.65\times10^{-26}\,\text{kg}$.
3
Rearrange $\tfrac12m\langle c^2\rangle = \tfrac32kT$ for $\langle c^2 \rangle$: $\langle c^2\rangle = \dfrac{3kT}{m}$.
4
$\langle c^2\rangle = \dfrac{3\times1.38\times10^{-23}\times293}{4.65\times10^{-26}} \approx 2.61\times10^{5}\,\text{m}^2\text{s}^{-2}$.
5
$c_{rms} = \sqrt{\langle c^2\rangle} \approx \sqrt{2.61\times10^{5}} \approx 511\,\text{m}\,\text{s}^{-1}$.
Around 500 m/s — comfortably faster than the speed of sound in air, which is a genuinely useful order-of-magnitude fact about how fast air molecules move.
▪ solved

Key takeaways

  • Kinetic theory models a gas as many identical molecules in random motion, negligible molecular volume, elastic collisions, negligible intermolecular forces except during collisions, and negligible collision duration.
  • Pressure arises from the accumulated momentum change of molecules colliding elastically with the container walls: $pV = \tfrac13Nm\langle c^2\rangle$, where $\langle c^2\rangle$ is the mean square speed.
  • Combining this with the ideal gas equation $pV=NkT$ gives $\tfrac12m\langle c^2\rangle = \tfrac32kT$: the average translational kinetic energy of a gas molecule depends only on absolute temperature.
  • This result is the microscopic explanation of what temperature actually measures — doubling absolute temperature exactly doubles the average kinetic energy per molecule, regardless of the gas.
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