Lesson 2 · 25 min

Kinetic theory and the speed distribution

Temperature measures the average kinetic energy of molecules. Molecules do not all share one speed; their speeds follow the Maxwell–Boltzmann distribution.

In this lesson

State the assumptions of the kinetic molecular theory
Relate temperature to average translational kinetic energy: $\bar{E}_\mathrm{k} = \tfrac{3}{2}kT$
Calculate root-mean-square speeds and explain their dependence on molar mass and temperature
Read the Maxwell–Boltzmann speed distribution and how it changes with temperature

The nitrogen molecules in the air you breathe fly at an average of about 500 metres per second — faster than an airliner. Yet open a bottle of perfume and it takes seconds or minutes before someone across the room smells it. No contradiction: the molecules are fast, but each collides billions of times a second, and its path zigzags like a drunkard’s walk.

Predict first

At a given temperature, do all the molecules in a gas move at the same speed?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Yes — temperature fixes a single speed No — speeds are spread over a distribution Only at high temperature do they differ

Five assumptions of kinetic theory

A gas consists of vast numbers of molecules in constant random motion.
The molecules’ own volume is negligible compared with the container.
Apart from collisions, molecules exert no forces on one another.
Collisions between molecules and with the walls are perfectly elastic.
The average kinetic energy of the molecules depends only on temperature.

Starting from these and adding up the momentum delivered to the walls gives $PV = \tfrac{1}{3}Nm\overline{v^2}$ . Comparing with $PV = nRT$ reveals what temperature means at the molecular level.

\bar{E}_\mathrm{k} = \tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}kT,\qquad v_\mathrm{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{3RT}{M}}

Temperature measures the average translational kinetic energy, whatever the molecule.

$k$: Boltzmann constant $1.380649\times10^{-23}$ J/K
$M$: molar mass in kg/mol
$v_\mathrm{rms}$: root-mean-square speed

Worked example

How fast are air molecules?

Find the rms speeds of $\ce{N2}$ (28.0 g/mol) and $\ce{H2}$ (2.02 g/mol) at 25 °C.

The shape of the speed distribution depends on both temperature and molecular mass: heat the gas or make the molecules lighter and the distribution broadens and shifts to higher speeds, and its high-speed tail grows rapidly. That tail matters enormously for chemistry: only molecules with enough energy can cross a reaction barrier — which is why a 10 °C rise can double the rate of many reactions (see Molecules over the mountain).

Concept check

At the same temperature, which quantity is the same for helium and xenon atoms?

Root-mean-square speed Average translational kinetic energy Momentum Collision frequency with the walls

Try the calculation

At what temperature (K) does oxygen (32.0 g/mol) have an rms speed of 500 m/s?

Key takeaways

Kinetic theory: many randomly moving molecules, negligible size, no forces, elastic collisions, average energy set by temperature.
Temperature measures average translational kinetic energy: $\bar{E}_\mathrm{k} = \tfrac{3}{2}kT$ .
$v_\mathrm{rms} = \sqrt{3RT/M}$ : lighter molecules and higher temperatures mean faster speeds.
Speeds follow the Maxwell–Boltzmann distribution; heating grows the high-energy tail rapidly.

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Data and references

P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 1B (the kinetic model).
J. C. Maxwell, “Illustrations of the dynamical theory of gases”, Phil. Mag. 19, 19–32 (1860).