Last page we said temperature measures the average kinetic energy of a substance’s particles. This page turns “measures” into an actual equation — one clean line linking absolute temperature to the average energy of a molecule. And once you have it, you can even work out roughly how fast those molecules are flying around.
📚 What you need to know
Molecules in a gas don’t all move at one speed — they have a range of speeds.
The average kinetic energy of a molecule is E̅k = 3/2 kBT, with kB = 1.38 × 10−23 J K−1 and T in kelvin.
So average KE is directly proportional to absolute temperature — a graph of E̅k against T (in K) is a straight line through the origin.
Tmust be in kelvin (proportional ⇒ through zero; °C would break it).
Equate with ½mv2 to get a typical speed: v = √(3kBT / m). Lighter molecules move faster at the same temperature.
Doubling absolute temperature doubles the average KE but multiplies the speed by only √2.
A whole range of speeds
Picture the molecules in a gas: billions of them, colliding constantly and endlessly swapping energy. After each collision one speeds up and another slows down, so at any instant there’s a huge range of speeds — some crawling, some racing, most somewhere in between. That’s exactly why we talk about the average.
Molecules share a wide spread of speeds. Heating the gas shifts the whole distribution to higher speeds — the average rises, and so does the temperature.
Because temperature tracks the average, a handful of unusually fast or slow molecules doesn’t shift it. Heat the gas and the whole spread slides rightward to higher speeds — the average climbs, and the temperature climbs with it.
The equation that links them
For an ideal gas, the average kinetic energy of a molecule depends on just one thing — the absolute temperature:
Average kinetic energy of a moleculeE̅k = 3/2 kBT
Here kB = 1.38 × 10−23 J K−1 is the Boltzmann constant — the fixed bridge between temperature and energy per particle — and T is in kelvin. Since everything else is constant, this says the average KE is directly proportional to the absolute temperature: plot one against the other and you get a straight line through the origin.
Average KE against absolute temperature is a straight line through the origin — proof that E̅k ∝ T, with gradient 3/2 kB. This only holds in kelvin.
Where does the “3/2” come from? Particles move in three independent directions — x, y and z — and each contributes ½kBT of energy. Three of them add up to 3/2 kBT. You won’t be asked to derive it, but that’s where the number lives.
From temperature to a speed
That same average kinetic energy is also just ½mv2 written the ordinary way. Set the two expressions equal and you can solve for a typical molecular speed:
From temperature to a typical speed3/2 kBT = ½mv2v = √(3kBT / m)
Two things fall out of this. Raise the temperature and the speed rises — but only as √T. And at a given temperature a lighter molecule moves faster, because v ∝ 1/√m.
Same temperature means the same average KE — but since KE = ½mv2, a lighter molecule must move faster to match. v ∝ 1/√m.
This is why the lightest gases — hydrogen and helium — leak out of a planet’s atmosphere first. At the same temperature they carry the same average KE as heavier molecules, but their tiny mass means a far higher speed — fast enough for some to reach escape velocity and drift away into space.
Worked examples
WE 1
Find the average kinetic energy of a single gas molecule at room temperature, 27 °C. (kB = 1.38 × 10−23 J K−1)
Temperature in kelvin: T = 27 + 273 = 300 KAverage KE: E̅k = 3/2 kBT= 1.5 × (1.38 × 10−23) × 300E̅k = 6.2 × 10−21 JTiny for one molecule — but a mole has about 6 × 1023 of them, so it soon adds up.
WE 2
A helium atom has mass 6.6 × 10−27 kg. Estimate the typical (rms) speed of helium atoms at 300 K.
Equate the two energy expressions: 3/2 kBT = ½mv²Rearrange: v = √(3kBT / m)v = √(3 × 1.38 × 10−23 × 300 ÷ 6.6 × 10−27)= √(1.88 × 106)v ≈ 1370 m s−1Over a kilometre every second — and remember that’s an average; plenty of atoms move much faster.
WE 3
A fixed sample of gas is heated so its absolute temperature doubles. By what factor does (a) the average kinetic energy of a molecule change, and (b) the typical molecular speed change?
(a) E̅k = 3/2 kBT, so E̅k ∝ T. Double T → double E̅k.(a) average KE × 2(b) v = √(3kBT/m), so v ∝ √T. Double T → × √2.√2 ≈ 1.41(b) speed × 1.41KE keeps step with temperature, but speed only with its square root — you’d need to quadruple T to double the speed.
🔧 Using E̅k = 3/2 kBT
Temperature in kelvin — always; the relation is with absolute temperature.
Average KE per molecule: multiply, E̅k = 3/2 kBT.
Need a speed? Set 3/2 kBT = ½mv2 and rearrange to v = √(3kBT/m).
Use one molecule’s mass (in kg) — not the mass of the whole gas.
Ratios: KE ∝ T, speed ∝ √T, and at fixed T, speed ∝ 1/√m.
T / K absolute temp
× 3/2 kB
average KE E̅k
= ½mv2
speed v √(3kBT/m)
Quick recap: A molecule’s average kinetic energy is E̅k = 3/2 kBT, so average KE is directly proportional to absolute temperature (a straight line through the origin — kelvin only). Molecules share a spread of speeds, and heating shifts it faster. Equate with ½mv2 to get v = √(3kBT/m), so lighter molecules are quicker at the same temperature.
💡 Top tips
Kelvin, always. The proportionality only works from absolute zero — never plug in °C.
It’s an average. Real molecules range from nearly still to very fast; the equation gives the mean.
Speed lags temperature. KE ∝ T but v ∝ √T, so quadrupling T is needed to double the speed.
Lighter = faster. At the same T, v ∝ 1/√m — why H2 and He escape atmospheres first.
Single-molecule mass. Use the mass of one atom or molecule in kg — not the molar mass.
⚠ Common mistakes
Using °C instead of kelvin in E̅k = 3/2 kBT — the answer comes out nonsense (even negative)
Assuming doubling temperature doubles the speed — it multiplies speed by √2, not 2
Dropping the ½ or the 3/2 when equating 3/2 kBT = ½mv2
Using the molar mass or the whole gas’s mass instead of one molecule’s mass
Treating the calculated speed as everyone’s speed — it’s an average of a wide spread
So temperature pins down the average kinetic energy of the particles — but that’s only half of a substance’s energy. The particles also carry potential energy from the forces between them. Add both, for every particle, and you get the substance’s total internal energy. Coming up: internal energy — and why heating something doesn’t always raise its temperature.
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