Physics › Thermal physics › Molecular kinetic theory
Molecular kinetic theory
The gas laws were found by experiment; kinetic theory derives them from molecules bouncing elastically off the walls of a container. The derivation of pV as one third of Nm times the mean square speed is examinable in full, and comparing it with pV = NkT is what identifies temperature as a measure of mean molecular kinetic energy.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Ideal gases and the gas laws and Momentum and impulse.
IN THIS TOPIC
- Set the empirical gas laws against the kinetic theory model, with Brownian motion as the evidence for atoms.
- List the assumptions, then reproduce the derivation of pV = ⅓Nm(crms)2 from them.
- Use ½m(crms)2 = 3kT/2 = 3RT/2NA, and read off what temperature means.
COMMON MISCONCEPTION
Air pressure comes from air's weight pressing down.
A theory, and its evidence
The spec draws the philosophical line itself. Gas laws are empirical, generalisations from measurement, while kinetic theory arises from theory and derives the same relationships from a molecular model. When the derivation reproduces Boyle's hyperbola from first principles, the model has earned its keep.
The model's foundational evidence is Brownian motion. Pollen grains in water, or smoke in air, jitter along ceaseless random walks. A grain vastly heavier than any molecule visibly trembles because the molecular bombardment is random and, instant by instant, uneven.
Einstein's 1905 analysis turned that jitter into a quantitative prediction, tying how far a grain wanders to the size and number of the molecules doing the shoving. Jean Perrin spent the years that followed measuring exactly that, and got a value for the Avogadro constant out of it. Theory and experiment together moved the atom from a useful assumption to a measured object. There was no single decisive afternoon; the case was won by numbers that kept agreeing.
The demonstration is worth being able to describe, because it is asked for. A smoke cell is a small glass box holding smoke from a smouldering waxed taper, closed and set on a microscope stage. Light it from the side, so that each smoke particle scatters light up into the objective and shows as a bright speck against a dark field, and focus on the specks. What you see is not the smoke drifting as a body. Each individual speck jerks about on its own, changing direction constantly, at random, and never settling.
Then read the inference off it. The specks are smoke particles, far too large to be single molecules and visible only because they are that large. Nothing in the sealed cell is stirring them. So the only thing left to jolt them is the air molecules, arriving from every side and, at any instant, more of them on one side than the other. Their random jitter is a direct sight of the random motion the model assumes, and the smaller the particle you watch, the more violent the jitter, exactly as an uneven bombardment would predict.
The assumptions
The ideal-gas model rests on assumptions worth listing, since papers ask for them. The gas contains a very large number of identical molecules in random motion, and the molecules' own volume is negligible beside the container's. Collisions, with walls and each other, are elastic and of negligible duration, and no forces act between molecules except during collisions. Each assumption is an approximation to a real gas, and the approximations are closest, so the equation fits best, when a gas is hot and sparse.
The required derivation
AQA, CIE and Edexcel all examine this derivation in full, so here it is in five moves, with momentum and impulse from Year 12 as the tools. One. A molecule of mass m travels at velocity component vx towards a wall of a box of side L, and each elastic bounce reverses that component, a momentum change of 2mvx. Two. It returns after a round trip of 2L, so hits arrive every 2L/vx seconds, giving an average force on the wall of 2mvx/(2L/vx) = mvx2/L.
Three. Sum over all N molecules, replacing each vx2 by its average. Four. Random motion shares speed equally among the three directions, so the mean of vx2 is one third of the mean square speed. Five. Pressure is that total force over the wall's area L2, and with volume V = L3,
where crms, the root mean square speed, is the square root of the mean of the squared speeds, the appropriate average for quantities that matter through their squares.
What temperature is
Set the derived equation beside the empirical pV = NkT and the meaning of temperature follows.
The average translational kinetic energy per molecule is proportional to absolute temperature and to nothing else. Not the gas, not the pressure, only T. Absolute temperature is the molecular measure of that energy, at (3/2)kT per molecule, though the two are proportional rather than the same quantity, as their units alone make plain. An ideal gas has no intermolecular forces, so it keeps no potential-energy account at all, which makes its internal energy entirely kinetic.
The weight of the overlying air does set how large sea-level pressure comes out, but the pushing itself is molecular bombardment, striking floor and walls and ceiling alike. That is what makes the pressure act in every direction at once instead of only downwards.
WORKED EXAMPLE
The same temperature, two different speeds
Find the mean kinetic energy of a gas molecule at 300 K, and explain which moves faster at that temperature: hydrogen or oxygen molecules.
Mean KE = (3/2)kT = 1.5 × 1.38 × 10−23 × 300 = 6.2 × 10−21 J, and that number is the same for every gas. Absolute temperature sets it and nothing else does.
Equal ½m(crms)2 with a sixteenth of the mass means hydrogen's speed squared is sixteen times oxygen's.
So hydrogen moves four times faster at the same temperature. Lighter molecules sprint to hold the same energy, which is also why hydrogen leaks through everything and why Earth's atmosphere kept so little of it.
GUIDED PRACTICE
Doubling the speed
A gas is at 300 K. To what temperature must it be raised to double the root mean square speed of its molecules? Reason through the proportionality.
Show the working
Temperature tracks the mean kinetic energy, which goes as the speed squared, so doubling crms quadruples the energy.
T must therefore quadruple, to 1200 K. Speed doublings come expensive in kelvin, and an answer of 600 K is what forgetting the square looks like.
INDEPENDENT PRACTICE
Atmospheric pressure, from scratch
Air has density 1.2 kg m−3 and its molecules move with a root mean square speed of about 500 m s−1. Use the kinetic theory result p = ⅓ρ(crms)2 to estimate atmospheric pressure.
Show the working
p = ⅓ × 1.2 × 5002 = ⅓ × 3.0 × 105 = 1.0 × 105 Pa.
That is atmospheric pressure, recovered from molecular bombardment alone. The derivation you can be asked to reproduce is a description of the air in this room, molecule by molecule.
ASSESSMENT FOCUS
- Empirical against theoretical is quotable spec language. Gas laws come from experiment, kinetic theory derives them from a model. One sentence, one mark.
- Keep Brownian motion's inference chain in order. Visible random jitter of grains → uneven bombardment → by particles far too small to see → atoms exist.
- Describing the smoke cell earns its own marks, and side lighting is the detail candidates drop. Smoke in a sealed glass cell, lit from the side, viewed under a microscope, and each speck seen jerking randomly and independently of the rest.
- The derivation is examined stepwise, so practise until the five moves are automatic. Momentum change 2mvx, hit interval 2L/vx, force mvx2/L, sum and average, then the factor ⅓ from three dimensions.
- crms means root-mean-square. Square, average, then root. Averaging the speeds first is the classic error, and the two answers differ.
- Doubling the kelvin temperature doubles the mean KE but multiplies crms by only √2. A square root sits between energy and speed.
CHECK YOURSELF
Find the root mean square speed of nitrogen molecules (molar mass 0.028 kg mol−1) at 300 K. (R = 8.31 J K−1 mol−1.)
Show a hint
Write the average-KE equation with molar quantities, then solve for crms.
Show the answer
Work per mole. ½M(crms)2 = 3RT/2, so (crms)2 = 3RT/M.
That gives (crms)2 = 3 × 8.31 × 300 / 0.028 = 2.67 × 105 m2 s−2, so crms = 520 m s−1.
Set that beside the speed of sound in the same gas, which is a separate result with an equation of its own, c = √(γRT/M), quoted here for the comparison rather than for the exam. For nitrogen γ = 1.4, giving c = 350 m s−1. The two are different quantities and neither one caps the other; what they share is their scaling, both going as √(T/M), so both climb with temperature and fall with molar mass. Their ratio is the fixed number √(γ/3), about 0.68 for any diatomic gas.
Pressure is bombardment, derived from momentum.
Mean molecular kinetic energy is proportional to absolute temperature, and equals (3/2)kT.
crms is square, average, then root.
An ideal gas keeps no potential energy at all.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the molecular kinetic theory questions page.
WHERE TO GO NEXT
- Significant figures, units and checking is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Set the empirical gas laws against the kinetic theory model, with Brownian motion as the evidence for atoms.
- List the assumptions, then reproduce the derivation of pV = ⅓Nm(crms)2 from them.
- Use ½m(crms)2 = 3kT/2 = 3RT/2NA, and read off what temperature means.
Open the full revision checklist to track your progress across the whole unit.