IB DP Physics: Ideal gases: pressure, the gas laws and kinetic theory (DP Physics) – Revision Notes

These notes condense topic B.3 (Gas laws) of the IB Physics guide, first assessment 2025, for Standard Level and Higher Level. The guide adds nothing extra for Higher Level in this topic. For full explanations and the derivations, use the study guide.

Then test yourself with the practice questions. Course page: IB DP Physics. Checklist: IB DP Physics checklist. For the molecular model and temperature scales, see the B.1 guide. If you are not sure where you stand, take the free 10-minute diagnostics.

Key definitions

• Pressure: force acting perpendicular to a surface per unit area, P = F/A. Unit: pascal, Pa = N m⁻².
• Amount of substance: n = N/NA, in mol, where NA is the Avogadro constant (6.02 × 10²³ mol⁻¹).
• Ideal gas: a modelled gas described by kinetic theory, used to approximate real gases.
• Empirical gas law: a relationship found by experiment for a fixed mass of gas.
• Monatomic gas: one whose particles are single atoms, with only translational kinetic energy.

Formulas

Quantity	Equation	Notes
Pressure	P = F/A	F perpendicular to the surface
Amount	n = N/NA	NA = 6.02 × 10²³ mol⁻¹
Combined gas law	PV/T = constant	Fixed mass; T in kelvin
Constant T	PV = constant	Pressure inversely proportional to V
Constant P	V/T = constant	V directly proportional to T
Constant V	P/T = constant	P directly proportional to T
Ideal gas, molecules	PV = NkBT	kB = 1.38 × 10⁻²³ J K⁻¹
Ideal gas, moles	PV = nRT	R = 8.31 J K⁻¹ mol⁻¹, R = NAkB
Kinetic theory	P = (1/3)ρv²	v² is the mean of squared speeds
Mean energy per molecule	(1/2)mv² = (3/2)kBT	Translational, per molecule
Internal energy	U = (3/2)NkBT = (3/2)nRT	Monatomic ideal gas only

Units to put in before you substitute: V in m³ (1 L = 10⁻³ m³; 1 cm³ = 10⁻⁶ m³), P in Pa, T in K.

Method in steps: a gas-law calculation

1. Identify what is fixed: the amount of gas, then T, P or V.
2. Convert every temperature to kelvin.
3. Choose the form: PV/T = constant for a change of state, PV = nRT for one state.
4. Rearrange before you substitute.
5. Check units, and give a sensible number of significant figures.
6. Ask whether the direction of change makes sense. Compression at constant temperature must raise pressure.

Worked reminder. How much gas is in a vessel of volume 3.1 × 10⁻³ m³ at 1.5 × 10⁵ Pa and 280 K? Rearrange: n = PV/(RT) = (1.5 × 10⁵ × 3.1 × 10⁻³) / (8.31 × 280) = 465 / 2327 = 0.20 mol.

How kinetic theory gives pressure

• A molecule rebounds from a wall; its momentum changes by 2mvₓ for that wall.
• The wall receives a force from many such collisions: force is the rate of change of momentum.
• Force on area gives pressure. More collisions per second, or harder collisions, give more pressure.
• Averaging over random directions gives ⟨v²⟩ = 3⟨vₓ²⟩, which leads to P = (1/3)ρv².
• Combine with PV = NkBT: mean kinetic energy per molecule = (3/2)kBT, so it depends only on the kelvin temperature.
• Root mean square speed goes as the square root of T. Quadruple T and the speed doubles. Check: a gas at 1200 K has twice the speed of the same gas at 300 K, since √(1200/300) = 2.

Ideal versus real gases

Feature	Ideal gas	Real gas
Molecular volume	Negligible	Small but not zero
Forces between molecules	None apart from during collisions	Attractive and repulsive forces exist
Collisions	Elastic	Close to elastic at moderate conditions
Obeys PV = nRT	Always	Approximately, under the conditions below

A real gas is closest to ideal at low pressure, low density and high temperature (well above its condensation point). Close to liquefying it departs from the model.

P-V diagrams

Process	Appearance	Reading the diagram
Constant pressure	Horizontal line	V changes in proportion to T
Constant volume	Vertical line	P changes in proportion to T
Constant temperature	Hyperbola, PV constant	Curves further from the axes are hotter

To get a temperature at any point, use T = PV/(nR) with the amount of gas fixed. Two points with equal PV are at the same temperature, even if they lie on different processes.

Linking the empirical laws to PV/T

Each law is one variable held fixed in PV/T = constant, for a fixed mass of gas.

• Constant T: the T cancels, leaving PV = constant. A graph of P against 1/V is a straight line through the origin.
• Constant P: V/T = constant. A graph of V against T in kelvin is a straight line through the origin. Extended back, it meets the temperature axis at 0 K.
• Constant V: P/T = constant. A graph of P against T in kelvin is a straight line through the origin.

If the graph uses degrees Celsius, the line does not pass through the origin; it cuts the temperature axis at about −273 °C.

Worked reminder. A fixed mass of gas at 2.0 × 10⁵ Pa and 300 K in a volume of 4.0 × 10⁻³ m³ is expanded to 6.0 × 10⁻³ m³ and cooled to 240 K. Then P₂ = P₁V₁T₂/(T₁V₂) = (2.0 × 10⁵ × 4.0 × 10⁻³ × 240) / (300 × 6.0 × 10⁻³) = 1.07 × 10⁵ Pa.

Must-know distinctions

• N (number of molecules) versus n (moles). PV = NkBT uses N with kB; PV = nRT uses n with R.
• Mean kinetic energy per molecule, (3/2)kBT, versus internal energy of the gas, (3/2)NkBT.
• Empirical gas laws (found by experiment) versus kinetic theory (a model that explains them).
• Ideal gas (a model) versus real gas (the thing being modelled).
• Pressure at a wall versus the pressure inside the gas: the force is always perpendicular to the surface.

Quick self-test

1. A force of 240 N acts perpendicular to a piston of area 0.020 m². Find the pressure.
2. How many molecules are in 0.75 mol?
3. Convert 27 °C to kelvin, using 273.
4. A fixed mass of gas has volume 5.0 L at 2.0 × 10⁵ Pa. At constant temperature, what is its volume at 5.0 × 10⁵ Pa?
5. A rigid container holds gas at 1.2 × 10⁵ Pa and 250 K. Find the pressure at 400 K.
6. A gas at constant pressure has volume 600 cm³ at 300 K. Find its volume at 375 K.
7. Find the amount of gas in 2.5 × 10⁻³ m³ at 1.0 × 10⁵ Pa and 300 K.
8. A gas has density 1.80 kg m⁻³ and a mean square speed of 2.5 × 10⁵ m² s⁻². Find the pressure.
9. Find the internal energy of 3.0 mol of monatomic ideal gas at 500 K.
10. Find the mean translational kinetic energy of a molecule at 700 K.
11. State the three conditions under which a real gas behaves most like an ideal gas.
12. By what factor does the root mean square speed change if the kelvin temperature is halved?

Answers

1. P = F/A = 240 / 0.020 = 1.2 × 10⁴ Pa.
2. N = nNA = 0.75 × 6.02 × 10²³ = 4.5 × 10²³ (4.515 × 10²³).
3. 27 + 273 = 300 K.
4. P₁V₁ = P₂V₂, so V₂ = 2.0 × 10⁵ × 5.0 / 5.0 × 10⁵ = 2.0 L.
5. P₂ = 1.2 × 10⁵ × 400/250 = 1.9 × 10⁵ Pa (1.92 × 10⁵).
6. V₂ = 600 × 375/300 = 750 cm³.
7. n = PV/(RT) = 250 / (8.31 × 300) = 0.100 mol.
8. P = (1/3) × 1.80 × 2.5 × 10⁵ = 1.5 × 10⁵ Pa.
9. U = 1.5 × 3.0 × 8.31 × 500 = 18 698 J, so 1.87 × 10⁴ J.
10. (3/2)kBT = 1.5 × 1.38 × 10⁻²³ × 700 = 1.45 × 10⁻²⁰ J.
11. Low pressure, low density, high temperature (well above condensation).
12. v is proportional to √T, so the factor is √(1/2) = 0.707, a fall of about 29%.

Where marks are usually lost

• Putting Celsius temperatures into PV/T = constant.
• Substituting volumes in litres or cm³ into PV = nRT with pressure in pascals.
• Mixing N with R, or n with kB, so the answer is out by a factor of 10²³.
• Quoting (3/2)nRT for a gas that the question does not describe as monatomic.
• Saying the pressure rises with temperature “because the molecules collide harder” and stopping there. Add that they also hit the walls more often, so the rate of change of momentum rises.
• Treating the derivation of P = (1/3)ρv² as a recall of the result: say where the 1/3 comes from (three equal directions).
• Describing real gases as ideal at high pressure.
• Reading a P-V diagram as if equal heights meant equal temperatures; temperature depends on PV.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page covers section B.3 (Gas laws).