Gases and the kinetic theory: collisions, pressure and the kelvin scale
By the end of this lesson you will be able to
- State the assumptions of the kinetic molecular theory of an ideal gas and explain what each one leaves out of the real world.
- Explain gas pressure as the total force of molecular collisions on each square metre of wall, and convert between Pa, kPa, bar and atm.
- Describe temperature as a measure of the average kinetic energy of the molecules, convert between °C and K, and explain why absolute zero is −273 °C.
- Predict, with the theory, what happens to the pressure of a gas when its temperature, volume or amount is changed.
- Process pressure–temperature data to a straight line and extrapolate it to absolute zero.
You met this before
- In a gas the particles are far apart, moving fast and at random in straight lines, and they bounce off each other and off the walls; most of a gas is empty space (Year 1 particle model).
- Heating a substance makes its particles move faster; temperature is a measure of how fast they move on average, while heat is energy transferred from hot to cold (Year 3 heat).
- Thermometers use three scales: Celsius (0 °C ice, 100 °C steam), Fahrenheit, and kelvin, whose zero, −273 °C, is the lowest possible temperature (Year 3 measuring temperature).
- Diffusion and Brownian motion are the evidence that gases are made of moving particles too small to see (Year 2 atoms).
- From Unit 2: one mole of any gas occupies about 24 dm³ at room temperature and pressure, because the spacing of the molecules, not their size, sets the volume.
A balloon is fitted over the neck of an empty glass bottle and the bottle is stood in hot water. Describe what happens and give the particle-level reason in two sentences.
Reveal
The balloon inflates. The air molecules in the bottle gain kinetic energy and move faster, so they hit the walls and the balloon skin harder and more often; the pressure inside rises above the pressure outside and the elastic skin is pushed out until the air inside has more room (the November 2021 Physics paper gave three marks for exactly this chain: faster molecules → more and harder collisions → greater pressure).
Revise it in full: Year 1 · Everyday materials · Lesson 1 — Solids, liquids and gases: the particle model · Year 3 · Heat · Lesson 2 — Measuring temperature: thermometers, three scales and absolute zero · Year 2 · Atoms and their structure · Lesson 1 — Everything is atoms: from a Greek argument to a jiggling pollen grain
The puzzle
Hold out your hand. The air is pressing on your palm with a force of about 1 500 newtons — the weight of a large motorbike — and you feel nothing.
The molecules doing the pressing are nitrogen and oxygen moving at about 500 metres per second, faster than a passenger jet, and each one is so light that a single hit is beyond any instrument to feel.
Yet in a bicycle tyre the same kind of molecules hold up a rider, in a car engine they drive a piston, and in an aerosol can left in the sun they can burst steel.
How does something you cannot see, feel or weigh push so hard — and what, exactly, does turning up the temperature change?
Predict which of these would double the pressure of the air in a sealed rigid can: (i) doubling the number of molecules; (ii) doubling the temperature from 20 °C to 40 °C; (iii) doubling the temperature from 293 K to 586 K. Write your reasoning before you read on.
Compare your answer
(i) and (iii) double it; (ii) does not. Twice as many molecules means twice as many collisions with every square centimetre of wall each second, so twice the pressure. Doubling the kelvin temperature doubles the average kinetic energy of the molecules, and the pressure rises in proportion.
But 20 °C to 40 °C is only 293 K to 313 K — a 7 % rise in kinetic energy, so about 7 % more pressure. The Celsius scale starts at the freezing point of water, which means nothing to a nitrogen molecule; the kelvin scale starts where the motion would stop.
The science — structured notes
1.1 The kinetic molecular theory: five assumptions
Everything in this unit follows from one picture of what a gas is. The kinetic molecular theory describes an ideal gas with five assumptions:
| Assumption | What it means | What it leaves out (real gases) |
|---|---|---|
| 1. The gas is a very large number of tiny particles (molecules, or atoms for the noble gases) in constant, rapid, random motion in straight lines. | Every direction is equally likely; a molecule changes direction only when it collides. Air at 20 °C: about 500 m s⁻¹ on average. | Nothing — this is true of every gas. |
| 2. The volume of the molecules themselves is negligible compared with the volume of the gas. | A gas is almost entirely empty space; at room pressure the molecules occupy about 0.1 % of the volume. | At very high pressure the molecules are crowded and their own size matters (Lesson 4). |
| 3. There are no forces between the molecules except during collisions. | Between collisions a molecule moves freely; nothing pulls it towards its neighbours. | Real molecules attract each other weakly; at low temperature or high pressure the attraction shows, and the gas can condense. |
| 4. Collisions — with each other and with the walls — are perfectly elastic. | Kinetic energy is not lost in a collision; the total kinetic energy of the gas stays the same unless heat enters or leaves. | Nothing measurable for a gas: molecules do not dent. |
| 5. The average kinetic energy of the molecules is proportional to the absolute (kelvin) temperature. | Double the kelvin temperature and the average kinetic energy doubles; at 0 K the motion would stop. | Quantum effects at extreme cold, far outside this course. |
“Ideal” does not mean “imaginary”: under ordinary conditions — around room temperature and pressure — nitrogen, oxygen, hydrogen, helium and even carbon dioxide behave to within about 1 % as the theory says. The theory earns its name by being simple enough to calculate with and accurate enough to design with.
1.2 Pressure: what the collisions add up to
A single nitrogen molecule hitting a wall delivers a push far too small to measure. But in one second, one square centimetre of wall in room air receives about 3 × 10²³ hits — and their total is a steady, measurable force. Pressure is force per unit area:
p = FA unit: newtons per square metre, N m⁻² = pascal, Pa
The pascal is tiny — the weight of a sheet of paper spread over a square metre — so gas pressures come in kilopascals: air at sea level pushes with about 101 kPa (101 000 N on every square metre). Two other units survive in everyday and industrial use, and the examination expects you to move between them:
| Unit | In pascals | Where you meet it |
|---|---|---|
| 1 kPa | 1 000 Pa | Laboratory sensors, weather (1 hPa = 100 Pa, so 1 013 hPa = 101.3 kPa) |
| 1 bar | 100 000 Pa = 100 kPa | Tyre gauges, gas cylinders (a full oxygen cylinder: 150–200 bar), hydrogen tanks (700 bar) |
| 1 atm (standard atmosphere) | 101 325 Pa = 101.325 kPa | Older data tables; “at atmospheric pressure”; diving (each 10 m of sea water adds about 1 atm) |
| 1 mmHg (torr) | 133.3 Pa | Blood pressure (“120 over 80”), vacuum pumps |
Atmospheric pressure is not a constant: it is the weight of the air above you. In Gaborone, about 1 000 m above sea level, the air presses with about 90 kPa; on the summit of Everest, 8 849 m, with about 34 kPa; at the coast in Durban or Walvis Bay, about 101 kPa.
Every gas-law calculation in this unit needs the absolute pressure of the gas, which is why a tyre gauge — which reads how much more than the atmosphere is inside — is a trap that Lesson 2 disarms.
A weather report gives 1 013 hPa; an oxygen cylinder gauge reads 150 bar; a textbook states 2.5 × 10⁵ Pa; a barometer in Gaborone reads 0.89 atm. Express all four in kPa.
hPa → kPa. 1 hPa = 100 Pa
1 013 hPa = 101 300 Pa
1 013 hPa = 101.3 kPa
bar → kPa. 1 bar = 100 kPa
150 bar = 15 000 kPa, which is 1.5 × 10⁷ Pa (150 times atmospheric).
Pa → kPa. 2.5 × 10⁵ Pa ÷ 1 000 = 250 kPa.
atm → kPa. 0.89 × 101.325 = 90.2 kPa. (Rounded to the two significant figures of the data: 90 kPa.)
The valve of a 150 bar oxygen cylinder has a circular opening 4.0 mm in diameter. Calculate the force the gas exerts on the closed valve, and comment on why cylinders are stored with the valve protected.
Convert. p = 150 bar, which is 1.5 × 10⁷ Pa
r = 2.0 mm, which is 2.0 × 10⁻³ m
Area. A = πr²
A = π × (2.0 × 10⁻³ m)²
A = 1.26 × 10⁻⁵ m²
Force. F = pA
F = 1.5 × 10⁷ Pa × 1.26 × 10⁻⁵ m²
F = 189 N ≈ 190 N — about the weight of a 19 kg child, pressing on an opening the size of a grain of rice.
The outside air pushes back with only 101 000 × 1.26 × 10⁻⁵ = 1.3 N. If the valve is sheared off in a fall, that force accelerates the cylinder like a rocket; the cap is there to stop it.
Convert: (i) 90 kPa to Pa and to atm; (ii) 3.0 atm to kPa; (iii) 2.4 × 10⁶ Pa to bar. Then calculate the force of the atmosphere (101 kPa) on a window 1.2 m by 0.80 m, and explain why the window does not break.
Reveal answer
(i) 90 kPa = 9.0 × 10⁴ Pa
90 ÷ 101.325 = 0.89 atm
(ii) 3.0 × 101.325 = 304 kPa (300 kPa if 1 atm is taken as 100 kPa — say which you used).
(iii) 2.4 × 10⁶ ÷ 10⁵ = 24 bar.
Window: A = 1.2 m × 0.80 m = 0.96 m²
F = pA
F = 1.01 × 10⁵ Pa × 0.96 m²
F = 9.7 × 10⁴ N — nearly ten tonnes’ weight.
But the air inside the room pushes back on the other face with the same pressure, so the resultant force is zero.
It is differences in pressure that break things: an aircraft cabin at 75 kPa inside with 26 kPa outside carries 49 kPa across every window.
1.3 Temperature: the average kinetic energy of the molecules
What does a thermometer actually measure? In the kinetic theory the answer is exact: the absolute temperature is proportional to the average kinetic energy of the molecules. Warm a gas and its molecules do not become “hotter” in any private way — they simply move faster.
Cool it and they slow. Keep cooling and the theory predicts a floor: a temperature at which the molecules have no kinetic energy left to lose. That floor is absolute zero, and experiments of every kind — the pressure and volume of gases (Lessons 2 and 3), the speed of sound, the glow of hot objects — all point to the same place: −273.15 °C.
A temperature scale that starts there is the kelvin scale. Its steps are the same size as Celsius degrees, so converting is a shift, not a multiplication:
T / K = θ / °C + 273 θ / °C = T / K − 273 (273.15 if the data justify it)
| Event | °C | K | Note |
|---|---|---|---|
| Absolute zero | −273.15 | 0 | Molecular motion at its minimum; never reached, approached to within billionths of a kelvin |
| Liquid nitrogen boils | −196 | 77 | Used to store cells and vaccines |
| Dry ice (solid CO₂) sublimes | −78 | 195 | Cold enough to freeze skin |
| Ice melts | 0 | 273 | The Celsius zero — a property of water, not of matter |
| Room temperature (RTP) | 20 | 293 | The 24 dm³ mol⁻¹ of Unit 2 belongs here |
| Human body | 37 | 310 | |
| Water boils at sea level | 100 | 373 | About 97 °C in Gaborone, where the pressure is lower |
| Bunsen flame (blue) | ≈ 1 500 | ≈ 1 770 |
“At 0 °C the molecules of a gas have no kinetic energy, which is why water freezes.”
Reveal answer
False. 0 °C is 273 K: the molecules of air at 0 °C have 273 ÷ 293 = 93 % of the kinetic energy they have at room temperature and are still moving at nearly 500 m s⁻¹.
Water freezes at 0 °C because at that temperature the attractions between water molecules can hold them in a lattice against their motion — a property of water, not a stopping of motion.
Molecular motion would stop only at 0 K, and every gas has condensed and frozen long before that.
1.4 How fast is a molecule? Mass, speed and the same kinetic energy
Assumption 5 says the average kinetic energy is fixed by the temperature — not the speed. Kinetic energy is ½mv², so at the same temperature a light molecule must move faster than a heavy one to carry the same energy: the speed goes as 1 ÷ √M. At 20 °C the typical (root-mean-square) speeds are:
| Gas | M / g mol⁻¹ | Typical speed at 20 °C / m s⁻¹ | Compare |
|---|---|---|---|
| Hydrogen, H₂ | 2.02 | 1 900 | Fast enough to escape from the top of the atmosphere over geological time — why Earth has almost none |
| Helium, He | 4.00 | 1 350 | |
| Nitrogen, N₂ | 28.02 | 510 | Air: about 1.5 times the speed of sound (343 m s⁻¹), which is no coincidence — sound is carried by these molecules |
| Oxygen, O₂ | 32.00 | 480 | |
| Carbon dioxide, CO₂ | 44.01 | 410 | Slowest of the common gases; diffuses slowest |
Not every molecule moves at the typical speed. In any gas there is a spread — a few almost still, a few moving several times faster than average — and the shape of the spread, worked out by James Clerk Maxwell in 1860, is the same for every gas; it just stretches to higher speeds for lighter molecules and for higher temperatures.
At the same temperature, the average kinetic energy ½mv² is the same for every gas. Calculate how many times faster a hydrogen molecule moves than an oxygen molecule, and use the oxygen speed in the table to check the hydrogen speed.
Equal energies. ½m(H₂)v(H₂)² = ½m(O₂)v(O₂)²
v(H₂)² ÷ v(O₂)² = m(O₂) ÷ m(H₂)
v(H₂)² ÷ v(O₂)² = 32.00 ÷ 2.02 = 15.8
Square root. v(H₂) ÷ v(O₂) = √15.8 = 3.98 ≈ 4 times faster.
Check. 480 × 3.98 = 1 910 m s⁻¹, matching the 1 900 m s⁻¹ in the table. This is why hydrogen leaks through the smallest gaps and why, in Unit 2, a hydrogen balloon goes down within a day.
Helium (M = 4.00) and methane (M = 16.05) are at the same temperature. Which moves faster, and by what factor? Then explain why a helium balloon deflates overnight while a nitrogen-filled tyre holds its pressure for months.
Reveal answer
Helium: v ∝ 1 ÷ √M
v(He) ÷ v(CH₄) = √(16.05 ÷ 4.00)
v(He) ÷ v(CH₄) = √4.01 = 2.0 — helium atoms move twice as fast.
Helium atoms are also tiny and single, so they slip through the microscopic pores of a rubber balloon far more often than nitrogen molecules (M = 28, 2.6 times slower and larger); the balloon loses gas within hours.
Tyre rubber is thicker and butyl-lined, and nitrogen is slow and large, so the loss is slow.
1.5 Using the theory to predict: three ways to raise a pressure
The theory turns every gas-law question into a question about collisions. Pressure depends on how often molecules hit a square metre of wall and how hard each hits. Three conditions change those:
| Change (others held constant) | Collisions per second per area | Force per collision | Pressure | Named law (lesson) |
|---|---|---|---|---|
| Raise the temperature, fixed volume and amount | Up — molecules cross the container faster | Up — they hit harder | Rises in proportion to T in kelvin | Pressure law (Lesson 3) |
| Halve the volume, fixed temperature and amount | Doubles — the same molecules have half the space, so each wall is reached twice as often | Unchanged — same speed | Doubles: p ∝ 1 ÷ V | Boyle law (Lesson 2) |
| Double the amount (moles), fixed volume and temperature | Doubles — twice as many molecules | Unchanged | Doubles: p ∝ n | Ideal gas equation (Lesson 4) |
| Change the gas (same n, V, T) | Lighter molecules hit more often… | …but less hard, exactly compensating | Unchanged — the pressure does not depend on which gas | Avogadro (Unit 2) |
A sealed, rigid 500 cm³ steel flask of air at 20 °C and 101 kPa is placed in a freezer at −18 °C. Predict the new pressure with a calculation, and explain at the molecular level why the flask does not “suck in” anything when there is no hole.
Reveal answer
T falls from 293 K to 255 K; pressure is proportional to kelvin temperature at fixed V and n, so p = 101 × 255 ÷ 293 = 88 kPa.
The molecules have lost 13 % of their average kinetic energy, so they hit the walls less often and less hard. Nothing “sucks”: the outside air, still at 101 kPa, now pushes harder on the flask than the inside pushes out, a net 13 kPa inward — 650 N on a 0.05 m² surface — which a steel flask resists but a plastic bottle does not (it crumples).
1.6 Collisions between molecules: how a gas mixes and why it takes time
The theory says a molecule moves at 500 m s⁻¹ — so why does a smell take a minute to cross a room? Because assumption 1 includes collisions with each other.
In air at room pressure a nitrogen molecule travels on average only about 70 nanometres — a few hundred molecular diameters — before it hits another molecule, and it suffers several billion collisions every second.
Its path is a zigzag of astonishing complexity, and its net progress across the room is slow: this is diffusion, and its rate rises with temperature (faster molecules) and falls with molecular mass.
It is also Brownian motion seen from the other side: a smoke particle jitters because the invisible air molecules hitting it from all sides do not exactly balance from moment to moment — the evidence Einstein analysed in 1905 and Perrin measured in 1908 to convince the last doubters that molecules are real.
“Deodorant sprayed in one corner of a warm room is smelt in the far corner within a minute because the scent molecules travel straight across at about 400 m s⁻¹. In a cold room at 4 °C the molecules stop moving, so the scent does not spread.” Find three errors and rewrite the passage correctly.
Reveal answer
(1) The molecules do not travel straight across: they collide billions of times a second with air molecules and follow a random zigzag, which is why diffusion across a room takes about a minute, not a hundredth of a second.
(2) At 4 °C (277 K) the molecules do not stop — they have 277 ÷ 293 = 95 % of the kinetic energy they have at 20 °C and move only about 3 % more slowly.
(3) The scent still spreads at 4 °C, just more slowly, because slower molecules diffuse more slowly (and air currents, not diffusion alone, do most of the mixing in a real room).
The November 2021 Chemistry paper gave three marks for: scent molecules move randomly; they diffuse until evenly mixed, faster at higher temperature; because higher temperature means faster molecules with more kinetic energy.
A sealed syringe of air is compressed to half its volume so quickly that no heat can escape. A sensor shows the temperature of the air rises from 20 °C to about 110 °C. Using the kinetic theory, explain where the extra kinetic energy came from and why the pressure ends up more than double.
Reveal answer
The plunger moving inward is a moving wall: a molecule that bounces off a wall coming towards it rebounds faster than it arrived (like a ball hit by a moving bat), so the compression does work on the gas and raises the average kinetic energy — the temperature rises.
Halving the volume alone would double the collision rate and double the pressure at constant temperature (Boyle); the higher temperature adds harder and more frequent collisions on top, so the pressure more than doubles (about 2 × 383 ÷ 293 = 2.6 times).
This is why a bicycle pump gets hot and why the Boyle lab in Lesson 2 makes you wait before reading.
Air at 20 °C and 101 kPa contains about 2.5 × 10²⁵ molecules per cubic metre. Estimate (i) the average volume of space “belonging” to one molecule and the average spacing between molecules; (ii) the number of moles in 1 m³ and hence the volume of one mole, and compare it with the 24 dm³ of Unit 2. Then use the spacing and the molecular diameter (about 0.35 nm for N₂) to justify assumption 2.
Reveal answer
(i) Volume per molecule = 1 ÷ (2.5 × 10²⁵)
Volume per molecule = 4.0 × 10⁻²⁶ m³
Side of a cube of that volume = ³√(4.0 × 10⁻²⁶ m³)
Side = 3.4 × 10⁻⁹ m, which is 3.4 nm — so molecules are about 3.4 nm apart on average.
(ii) n = 2.5 × 10²⁵ ÷ 6.02 × 10²³
n = 41.5 mol per m³
Volume of one mole = 1 ÷ 41.5 = 0.0241 m³, which is 24.1 dm³ — the molar volume at RTP.
(iii) The spacing, 3.4 nm, is about ten molecular diameters, so the volume actually filled by molecules is roughly (0.35 ÷ 3.4)³ ≈ 0.1 % of the gas — negligible, as assumption 2 requires; at 700 bar the spacing shrinks to about 0.4 nm and the assumption fails (Lesson 5).
Try it — does pressure really follow the kelvin temperature?
The theory makes a numerical prediction: in a rigid, sealed flask the pressure should rise in a straight line with temperature and reach zero at −273 °C. The lab seals a 250 cm³ round-bottomed flask of air with a bung carrying a pressure sensor and stands it in a water bath you control from ice to boiling.
Two hidden choices are yours: where the flask was sealed (sea level or Gaborone — a different amount of air) and how long the connecting tube is (a systematic error you will meet again in Lesson 3).
Read the pressure once the flask has had time to reach the bath temperature.
Suppose your mean pressures (sealed at 101 kPa, short tube) were: 0 °C → 94.2; 20 °C → 101.0; 40 °C → 107.8; 60 °C → 114.9; 80 °C → 121.7; 100 °C → 128.5 kPa.
(a) Calculate the gradient with its unit and the value of absolute zero the line predicts.
(b) Test with numbers whether pressure is proportional to the kelvin temperature.
(c) With the long tube the readings were 95.0 and 124.1 kPa at 0 and 100 °C. Explain, at the molecular level, why the gradient is smaller and in which direction the extrapolated absolute zero moves.
Reveal answer
(a) Gradient = (128.5 − 94.2) ÷ (100 − 0)
Gradient = 0.343 kPa °C⁻¹
Absolute zero: the line reaches p = 0 where 94.2 + 0.343 × θ = 0
θ = −94.2 ÷ 0.343
θ = −275 °C (accept −270 to −280 from a drawn line), within 1 % of −273 °C.
(b) Test p ÷ T in kelvin:
94.2 ÷ 273 = 0.345 kPa K⁻¹
101.0 ÷ 293 = 0.345 kPa K⁻¹
128.5 ÷ 373 = 0.345 kPa K⁻¹ — constant, so p ∝ T(K).
Against Celsius the ratio is not constant (101.0 ÷ 20 = 5.1 but 128.5 ÷ 100 = 1.3): pressure is linear in °C but proportional only in kelvin.
(c) The air in the tube stays at room temperature: its molecules do not speed up when the bath is heated, so only 85 % of the molecules contribute the extra collisions, and the pressure rise per degree is smaller (0.291 instead of 0.343 kPa °C⁻¹).
The line is shallower but still starts near 95 kPa at 0 °C, so it reaches zero further away: −95.0 ÷ 0.291 = −326 °C — a false absolute zero, too cold.
A systematic error shifts the intercept in one direction; averaging more trials cannot remove it.
Common mistakes to avoid
- Using °C in a proportion — “double the temperature from 20 °C to 40 °C, double the pressure”. Convert to kelvin first; 293 K to 313 K is a 7 % rise.
- “The particles expand” or “get bigger” when heated — molecules move faster; they do not grow. The gas expands because the molecules push harder.
- “The particles want to escape / move up” — molecules have no intentions and no preferred direction; the motion is random and pressure is the same on every wall.
- Mixing up temperature and heat — temperature measures the average kinetic energy of the molecules; heat is energy transferred because of a temperature difference.
- Pressure as a force — pressure is force per unit area, in Pa or kPa; a force is in N. “The gas exerts a pressure of 500 N” loses the mark.
- Unit slips — 1 bar = 100 kPa, not 1 kPa; 1 atm = 101.325 kPa; 150 bar is 1.5 × 10⁷ Pa. Write the conversion line.
- Ignoring the atmosphere — the pressure “in” a gauge reading or on a piston with a mass on it must have atmospheric pressure added (Lesson 2).
- Vague “particles” — say molecules of N₂ and O₂ when you know the gas; “particle” alone is acceptable only when the gas is unnamed.
Study smart: how this is examined
Study routine. Today: write the five assumptions and the three-link chain on a card, then convert 90 kPa, 150 bar, 0.89 atm and 1 013 hPa to Pa from memory.
In two days: redo the valve EXAMPLE and the freezer What-if without looking, and draw the thermometer diagram with both scales. In a week: run the lab as a full investigation, including the long-tube error and the Gaborone extension, then do the Think task and the quiz.
How we found out — a theory nobody could see
The idea that a gas is a swarm of flying particles is older than the evidence for it. In 1738 Daniel Bernoulli, a Swiss mathematician, wrote in his Hydrodynamica that air pressure was the drumming of tiny corpuscles on the walls of their container, and showed that squeezing them into half the space should double the drumming — Boyle’s law, derived from collisions seventy-six years after Boyle measured it.
Almost nobody believed him, because the rival theory said heat was a fluid, “caloric”, that seeped between particles and pushed them apart.
A French instrument-maker, Guillaume Amontons, had already found in 1702 that the pressure of air in a fixed volume rose steadily with temperature, and made the unsettling observation that if the trend continued downwards the pressure would vanish at some very low temperature — about −240 °C on our scale, the first estimate of absolute zero.
The theory came back in the 1850s, when heat had been shown to be a form of energy (Year 3). Rudolf Clausius published “The kind of motion we call heat” in 1857 and, a year later, explained why a 500 m s⁻¹ molecule takes minutes to cross a room: it collides, and he introduced the mean free path.
In 1860 James Clerk Maxwell worked out how the speeds must be distributed — the curves in Section 1.4 — and Ludwig Boltzmann spent the 1870s and 1880s showing that the theory explained not only pressure but the whole of thermodynamics.
Meanwhile William Thomson, later Lord Kelvin, had proposed in 1848 a temperature scale with its zero at the point where a perfect engine could extract no more work: absolute zero, which the gas experiments placed at −273 °C. Even so, many chemists refused to believe that atoms were more than a bookkeeping device.
What ended the argument was Brownian motion: Albert Einstein showed in 1905 exactly how far a pollen grain should wander if it was being bombarded by molecules, and Jean Perrin measured it in 1908 and got the Avogadro constant out of the wandering.
Perrin’s Nobel lecture in 1926 was titled, with some satisfaction, “Discontinuous structure of matter”. In 2019 the kelvin itself was redefined in terms of the Boltzmann constant — the theory’s own number — so that the unit of temperature is now, officially, a unit of molecular energy.
Amontons measured only over a narrow range of ordinary temperatures, yet he predicted a lowest possible temperature. Explain how a straight-line graph allowed him to do it, and why his value (−240 °C) differs from the modern −273 °C.
Reveal answer
His pressure readings against temperature lay on a straight line; extending the line beyond his data (extrapolation) to the point where the pressure would be zero gives a temperature the experiment never reached.
The value depends on the gradient. His −240 °C is too warm, so his line was too steep for its height: his temperature scale and fixed points were crude, and moist air adds evaporating water vapour at the hot end, raising the hot readings and steepening the line.
(A dead space at room temperature, like the lab’s long tube, has the opposite effect: a shallower line and an intercept that is too cold.)
The extrapolation is only as good as the gradient — which is why the lab asks you to compare your intercept with −273 °C and name the error that moved it.
Where it is used now
- Cooking at altitude. In Gaborone, at about 90 kPa, water boils near 97 °C, so beans and pap take longer; a pressure cooker holds steam at about 200 kPa, raising the boiling point to about 120 °C and cutting the time by more than half. Both are the kinetic theory at the stove.
- Tyres, footballs and bicycle pumps. A pump gets hot because the moving plunger speeds up the molecules; a tyre’s pressure rises on a hot road; a football inflated indoors goes soft on a cold pitch — every one is a change in molecular speed or spacing (Lesson 5).
- Vacuum and pressure in industry. Freeze-drying vaccines and instant coffee at a few hundred pascals; vacuum-packing food so that the outside atmosphere (101 kPa) crushes the bag; compressed air at about 700 kPa driving pneumatic drills and tools in mines and workshops.
- Weather. A barometer falling from 1 020 to 990 hPa is the mass of air overhead decreasing; winds are air molecules moving from high pressure to low — the same idea the November 2023 Physics paper used for finding leaks in an airtight house.
- Anaesthetics and diving. Gases are dosed in partial pressures in kPa; a diver at 30 m breathes air at about 400 kPa and its molecules dissolve in the blood four times as readily — the chemistry behind “the bends” in Lesson 5.
Science now — and the scientist you need to be
What is happening in 2026
The kinetic theory is now something you can watch. Molecular-dynamics software follows millions of molecules colliding, one femtosecond at a time, and machine-learned models of the forces between molecules — trained on quantum-chemistry calculations — have in the last few years made those simulations accurate enough to design gas separation membranes, carbon-capture solvents and hydrogen storage materials before anything is synthesised.
At the cold end, laboratories cool clouds of atoms to a few billionths of a kelvin, where the motion is so slow that the atoms’ wave nature takes over; the same technology gives the atomic clocks that time satellite navigation and the dilution refrigerators, at about ten thousandths of a kelvin, in which quantum computers run.
At the hot end, the “temperature” of the plasma in fusion experiments is quoted in millions of kelvin — average kinetic energy again, with the same definition as a bottle of air.
Since 20 May 2019 the kelvin has been defined by fixing the Boltzmann constant at exactly 1.380 649 × 10⁻²³ J K⁻¹, so temperature is officially a measure of molecular energy, and a laboratory anywhere can realise it by measuring the speed of sound in argon or the pressure of a known gas.
The scientist the next decade needs can move without a pause between a thermometer reading, a pressure in kilopascals and a picture of what the molecules are doing — and can tell, in an unfamiliar problem, which of the three conditions has been changed.
The mindshift
What you need to be able to do
Write the particle-level chain for any change in a gas without prompting: what changed, what the molecules do differently, how the collisions change, what the pressure or volume does. Convert pressures and temperatures before you calculate and write the unit after.
Read a p–T graph for its gradient and its intercept and say what each means physically. These are the marks that the examination awards for reasoning rather than recall — and, according to the reports, the ones most often left on the table.
See an example
A car tyre sidewall: 220 kPa (gauge) is 2.2 × 10⁵ Pa, or 2.2 bar, or 2.17 atm — and the absolute pressure inside is about 320 kPa at the coast, 310 kPa in Gaborone.
A weather app: 1 016 hPa is 101.6 kPa, or 1.003 atm.
A full answer for the bottle: “In the cold room the air molecules inside lose kinetic energy and move more slowly; they collide with the bottle walls less often and less hard, so the pressure inside falls below the outside atmospheric pressure, and the net inward force of the outside air crumples the plastic.”
If the assistant said only “the air inside contracts”, it described the result and skipped the mechanism — the examiner’s oldest complaint.
Key terms
| Kinetic molecular theory | The model of a gas as a huge number of tiny molecules in constant random straight-line motion, colliding elastically, with negligible volume and no forces between them, whose average kinetic energy is proportional to the kelvin temperature. |
| Ideal gas | A gas that obeys the kinetic theory exactly; real gases come close at ordinary temperatures and pressures. |
| Pressure, p | Force per unit area exerted by molecular collisions on a wall: p = F ÷ A, in pascals (N m⁻²). |
| Pascal, kilopascal, bar, atmosphere | 1 kPa = 1 000 Pa; 1 bar = 100 kPa; 1 atm = 101.325 kPa. |
| Atmospheric pressure | The pressure of the air, about 101 kPa at sea level and about 90 kPa at 1 000 m. |
| Absolute temperature, T | Temperature measured from absolute zero, in kelvin: T = θ + 273. |
| Absolute zero | −273.15 °C = 0 K, the temperature at which molecular motion would be at its minimum and an ideal gas would exert no pressure. |
| Elastic collision | A collision in which no kinetic energy is lost. |
| Mean free path | The average distance a molecule travels between collisions, about 70 nm in air at room conditions. |
| Diffusion | The net spreading of molecules from where they are concentrated to where they are not, caused by random motion; faster for lighter molecules and at higher temperature. |
| Brownian motion | The random jitter of a visible particle caused by uneven bombardment by invisible molecules. |
Think like a scientist
Criterion C A student investigates how the pressure of a fixed volume of air depends on temperature. She seals a 500 cm³ flask of air with a bung and pressure sensor at 20 °C and 101.3 kPa, then stands it in water baths at six temperatures, waiting three minutes before each reading. Her results:
| Bath temperature / °C | Pressure, trial 1 / kPa | Trial 2 / kPa | Trial 3 / kPa |
|---|---|---|---|
| 10 | 97.9 | 97.6 | 97.8 |
| 25 | 103.1 | 102.9 | 103.0 |
| 40 | 108.3 | 108.0 | 108.4 |
| 55 | 113.5 | 113.2 | 113.6 |
| 70 | 118.4 | 118.8 | 121.9 |
| 85 | 123.9 | 123.6 | 123.7 |
(a) Identify the anomalous reading and suggest a specific cause. [2]
(b) Calculate the mean pressure at each temperature to the precision of the data. [2]
(c) Describe the graph you would draw so that the line can be extrapolated to zero pressure, and state the gradient with its unit. [3]
(d) Use the gradient and one point to determine the temperature at which the pressure would be zero, and compare it with the accepted value. [3]
(e) Show numerically that pressure is proportional to the kelvin temperature but not to the Celsius temperature. [2]
(f) Explain one random and one systematic error, each with its effect on the intercept, and one improvement. [4]
Model answer
(a) Trial 3 at 70 °C, 121.9 kPa, is 3.3 kPa above the other two trials and above the steady rise of about 5.2 kPa per 15 °C [1].
Cause: the bath was hotter than the thermometer showed at that moment — for example the reading was taken just after hot water was added and before stirring, so the flask was at a higher temperature than recorded (a leak or a loose bung would lower the pressure, not raise it) [1].
(b) Means, excluding the anomaly and stating that two trials were used at 70 °C: 10 °C → 97.8; 25 → 103.0; 40 → 108.2; 55 → 113.4; 70 → 118.6; 85 → 123.7 kPa [1], all to one decimal place, matching the 0.1 kPa sensor [1].
(c) Temperature (independent) on the horizontal axis, mean pressure on the vertical, both labelled with unit (“Bath temperature / °C”, “Mean pressure / kPa”) [1].
The temperature axis scaled from −300 to +100 °C in even steps (for example 50 °C per two large squares) and the pressure axis from 0 to 130 kPa, so that the line can be extended to p = 0 — not an axis starting at 10 °C or 95 kPa [1].
Six points as crosses, one straight line of best fit extended as a dashed line to the axis.
Gradient = (123.7 − 97.8) ÷ (85 − 10)
Gradient = 25.9 ÷ 75
Gradient = 0.345 kPa °C⁻¹ (accept 0.34–0.35) [1].
(d) Using the 10 °C point: p = 0 when 97.8 − 0.345 × (10 − θ) = 0
10 − θ = 97.8 ÷ 0.345 = 283
θ = 10 − 283 = −273 °C [2, one for the method, one for the value in the range −265 to −280].
This agrees with the accepted absolute zero of −273.15 °C to within the uncertainty of the gradient — the extrapolation spans nearly 300 °C from 75 °C of data, so a 1 % error in the gradient moves the intercept by about 3 °C [1].
(e) Kelvin: 97.8 ÷ 283 = 0.3456 kPa K⁻¹
123.7 ÷ 358 = 0.3455 kPa K⁻¹ — the same ratio.
Check the other way: 358 ÷ 283 = 1.265 and 123.7 ÷ 97.8 = 1.265, so pressure is proportional to kelvin temperature [1].
Celsius: 97.8 ÷ 10 = 9.8 but 123.7 ÷ 85 = 1.46 — not constant; the pressure is a straight-line (linear) function of °C but not proportional to it, because the line does not pass through 0 °C [1].
(f) Random: the bath temperature drifting by ±0.5 °C during the three-minute wait and the sensor’s ±0.1 kPa, scattering the trials by about 0.3 kPa about the mean (97.6 to 97.9 at 10 °C) — a random scatter does not shift the intercept systematically but widens its uncertainty; improvement: a stirred, thermostatically controlled bath and the mean of three trials [2].
Systematic: the short length of tube between bung and sensor, and the air in the sensor itself, stay at room temperature and do not warm with the bath, so the pressure rises less per degree than it should — the gradient is slightly low and the intercept extrapolates too cold (below −273 °C); improvement: use the shortest possible connection and immerse it, or correct for the dead volume [2].
(Also acceptable: the flask expanding slightly when heated, so V is not quite constant — every hot reading slightly low, same direction. Not acceptable: “the student made a mistake”, “more repeats”.)
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth. Take 1 atm = 101.325 kPa and absolute zero = −273 °C.
Convert 37 °C and −196 °C to kelvin, and 77 K and 1 770 K to degrees Celsius.
Reveal answer
37 + 273 = 310 K
−196 + 273 = 77 K [1]
77 − 273 = −196 °C
1 770 − 273 = 1 497 °C (about 1 500 °C) [1]
A gas cylinder gauge reads 200 bar. Express this in kPa and in Pa (standard form), and calculate the force on the 3.0 cm² end of the valve inside the cylinder.
Reveal answer
p = 200 bar, which is 20 000 kPa or 2.0 × 10⁷ Pa [1]
A = 3.0 cm², which is 3.0 × 10⁻⁴ m² [1]
F = pA
F = 2.0 × 10⁷ Pa × 3.0 × 10⁻⁴ m²
F = 6 000 N (6.0 × 10³ N) [1]
State two assumptions of the kinetic molecular theory and explain, for one of them, a condition under which a real gas would not obey it.
Reveal answer
Any two of: molecules in constant random straight-line motion; negligible molecular volume; no intermolecular forces except in collisions; elastic collisions; average kinetic energy proportional to kelvin temperature [2].
Where it fails: at very high pressure (hundreds of bar) the molecules are so close that their own volume is a significant fraction of the gas volume, so the gas is harder to compress than the theory predicts; or at low temperature the weak attractions between molecules slow them near each other and the gas can condense [1].
A sealed steel can of gas is left in the sun and its temperature rises from 20 °C to 50 °C. Use the kinetic theory to explain why the pressure inside rises, and calculate the percentage rise.
Reveal answer
The molecules gain kinetic energy and move faster [1]; they collide with the walls more often and each collision exerts more force, and force per area is pressure, so the pressure rises [1]; 293 K → 323 K, a factor of 323 ÷ 293 = 1.10, so the pressure rises by 10 % — not by 150 % as a Celsius ratio would suggest [1].
Explain why the pressure of a gas is the same on every wall of its container, and why a gas fills any container completely while a liquid does not.
Reveal answer
Molecular motion is random, with no preferred direction, so on average as many molecules hit each square metre of the top, sides and bottom with the same average force [1]; there are no forces holding gas molecules together (assumption 3), so they move apart until stopped by the walls, spreading through any volume [1].
In a liquid the molecules attract each other strongly enough to stay in contact, so the liquid keeps its volume and only takes the shape of the part of the container it occupies [1].
Oxygen molecules in a room move at about 480 m s⁻¹. Calculate the typical speed of chlorine molecules, Cl₂ (M = 70.9), in the same room.
Reveal answer
Speed ∝ 1 ÷ √M
v(Cl₂) = 480 × √(32.0 ÷ 70.9) [1]
v(Cl₂) = 480 × 0.672
v(Cl₂) = 320 m s⁻¹ [1]
7. A weather balloon’s sensor records, at the same moment, a temperature of −56 °C and a pressure of 26 kPa at 10 km. Compare the average kinetic energy and the number of molecules per cubic metre of the air there with the air at ground level (20 °C, 101 kPa), giving each as a ratio.
Then explain why a passenger aircraft cabin at that height must be pressurised even though the oxygen fraction of the air (21 %) is unchanged.
Reveal answer
Average KE ∝ T
217 ÷ 293 = 0.74 — the molecules have 74 % of the ground-level kinetic energy [1].
Molecules per m³ ∝ p ÷ T (pressure per collision energy):
(26 ÷ 217) ÷ (101 ÷ 293)
0.120 ÷ 0.345 = 0.35 — only 35 % as many molecules per cubic metre [1].
Each breath at 10 km therefore contains 35 % as many oxygen molecules as at the ground, since 21 % of a third as many molecules is a third as much oxygen [1].
The lungs need a sufficient number of oxygen molecules per breath (a partial pressure of about 21 kPa at ground level, 5.5 kPa at 10 km), so the cabin is pressurised to about 75 kPa, the equivalent of 2 400 m [1].
8. A student claims: “If I heat a sealed flask of hydrogen and a sealed flask of carbon dioxide, each at the same starting temperature and pressure, by the same number of degrees, the hydrogen pressure will rise more because its molecules speed up more.” Evaluate the claim with the kinetic theory, using numbers for a rise from 300 K to 330 K.
Reveal answer
The average kinetic energy of both gases rises by the same factor, 330 ÷ 300 = 1.10, because it depends only on temperature [1]; hydrogen molecules do speed up more in m s⁻¹ (from about 1 930 to 2 020 m s⁻¹, against 410 to 430 for CO₂), but they are 22 times lighter, so each collision delivers correspondingly less force.
The product — collision rate × force per collision — depends on the kinetic energy, not the speed, and rises by 10 % for both [1].
At the same starting p, V and T the two flasks also hold the same number of molecules (Avogadro), so both pressures rise from, say, 100 to 110 kPa [1]; the claim is false: the pressure law is the same for every ideal gas, which is exactly why it is a law and not a table of gases [1].
Summary
- A gas is a huge number of tiny molecules in constant, random, straight-line motion, colliding elastically, with negligible volume and no forces between them: the kinetic molecular theory of an ideal gas.
- Pressure is the total force of molecular collisions per unit area of wall: p = F ÷ A in pascals. The atmosphere presses with about 101 kPa at sea level and 90 kPa in Gaborone.
- Units: 1 kPa is 1 000 Pa; 1 bar is 100 kPa; 1 atm is 101.325 kPa.
- Temperature measures the average kinetic energy of the molecules; the kelvin scale starts at absolute zero, −273 °C, where that motion would stop: T = θ + 273. Every gas-law proportion uses kelvin.
- At a given temperature every gas has the same average kinetic energy, so lighter molecules move faster: speed ∝ 1 ÷ √M — hydrogen four times faster than oxygen.
- Raising the temperature, shrinking the volume or adding molecules all raise the pressure by making collisions more frequent or harder; a p–T graph of a fixed volume is a straight line that extrapolates to −273 °C.
Next time → Boyle law: squeezing a gas at constant temperature — why pressure × volume is a constant, how to straighten the curve with 1 ÷ V, the trap of gauge pressure, and a syringe lab whose product pV tells you whether you waited long enough.