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

This study guide follows the IB Physics guide, first assessment 2025, and covers topic B.3 (Gas laws) in the theme on the particulate nature of matter. B.3 applies to Standard Level and Higher Level alike, and the guide lists no additional higher level content in it. Everything here is for both.

Pair it with the revision notes and the practice set. The course page is IB DP Physics and the printable list is the IB DP Physics checklist. The molecular picture of solids, liquids and gases, temperature and internal energy is in the B.1 guide. To find weak spots first, try the free 10-minute diagnostics.

What this unit covers

B.3 understanding	What you must be able to do	Level
Pressure, P = F/A	Use F as the force perpendicular to the surface	SL and HL
Amount of substance, n = N/NA	Convert between molecules and moles	SL and HL
Ideal gas as a model	Describe it through kinetic theory and as an approximation to real gases	SL and HL
Empirical gas laws	Link constant P, constant V and constant T results to PV/T = constant	SL and HL
Equations of state	Use PV = NkBT and PV = nRT	SL and HL
Pressure from collisions	Relate momentum change at a surface to P = (1/3)ρv²	SL and HL
Internal energy	Use U = (3/2)NkBT = (3/2)nRT for a monatomic ideal gas	SL and HL
Real versus ideal	Give the temperature, pressure and density conditions for a good approximation	SL and HL
P-V diagrams	Represent changes of state of an ideal gas	SL and HL

The guide restricts the laws you need to three empirical results (one quantity held fixed each time) together with the full ideal gas equation. Temperature in every gas equation is in kelvin.

Pressure and amount of substance

Pressure is the force acting perpendicular to a surface divided by the area of that surface:

P = F/A, in pascals (Pa), where 1 Pa = 1 N m⁻².

Only the component of the force at right angles to the surface counts. A gas pushes on a wall at right angles to it, so for a gas the whole force is usually the one you want.

Worked example 1. A crate of mass 18 kg rests on its smallest face, 0.30 m by 0.20 m. Find the pressure it exerts on the floor. Use g = 9.81 N kg⁻¹.

1. Force perpendicular to the floor = weight = mg = 18 × 9.81 = 176.6 N.
2. Area = 0.30 × 0.20 = 0.060 m².
3. P = F/A = 176.6 / 0.060 = 2943 Pa, so 2.9 × 10³ Pa.

The amount of substance n, in moles, counts particles in bundles of the Avogadro constant NA = 6.02 × 10²³ mol⁻¹:

n = N/NA, so N = nNA.

Worked example 2. A sample contains 2.408 × 10²⁴ molecules. n = 2.408 × 10²⁴ / 6.02 × 10²³ = 4.00 mol. In reverse, 0.250 mol has 0.250 × 6.02 × 10²³ = 1.51 × 10²³ molecules.

The ideal gas model

An ideal gas is a modelled system. It is described by kinetic theory and it approximates the behaviour of real gases. In the model:

• the gas is a very large number of identical molecules in constant random motion
• the molecules are treated as points, so their own volume is negligible compared with the container
• molecules exert no forces on one another apart from when they collide
• collisions with each other and with the walls are elastic, and each collision is very brief
• between collisions, molecules move in straight lines at constant velocity

Real gases break these assumptions. Molecules have size and attract one another. The last section of this guide says when the approximation is still good.

The empirical gas laws and PV/T = constant

“Empirical” means found by experiment. For a fixed mass of gas, three results hold. Always use kelvin.

Held constant	Result	Plot that is a straight line through the origin
Temperature T	PV = constant (pressure inversely proportional to volume)	P against 1/V
Pressure P	V/T = constant (volume directly proportional to T)	V against T
Volume V	P/T = constant (pressure directly proportional to T)	P against T

Combining them. Take a fixed mass of gas from state 1 (P₁, V₁, T₁) to state 2 (P₂, V₂, T₂) in two steps.

• Step A: heat it at constant pressure P₁ until the volume is V₂. Then V₁/T₁ = V₂/T_x, so the temperature reached is T_x = T₁V₂/V₁.
• Step B: change the pressure at constant volume V₂ until the temperature is T₂ and the pressure is P₂. Then P₁/T_x = P₂/T₂.

Substituting T_x gives P₁V₁/(T₁V₂) = P₂/T₂, which rearranges to P₁V₁/T₁ = P₂V₂/T₂.

So for a fixed mass of gas, PV/T = constant. The constant equals nR, or NkB, as the next section shows.

Worked example 3. A fixed mass of gas has volume 2.4 × 10⁻³ m³ at 1.0 × 10⁵ Pa and 290 K. It is compressed to 1.5 × 10⁻³ m³ while heated to 360 K. Find the final pressure.

1. P₁V₁/T₁ = P₂V₂/T₂, so P₂ = P₁V₁T₂ / (T₁V₂).
2. P₂ = (1.0 × 10⁵ × 2.4 × 10⁻³ × 360) / (290 × 1.5 × 10⁻³).
3. Numerator = 86 400. Denominator = 0.435. P₂ = 1.99 × 10⁵ Pa.

The equations of state

The constant in PV/T depends on how much gas there is. Written with the number of molecules N:

PV = NkBT, where kB is the Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹.

Written with the amount of substance n, using N = nNA:

PV = nRT, where R = NAkB is the molar gas constant, 8.31 J K⁻¹ mol⁻¹.

Check: 6.02 × 10²³ × 1.38 × 10⁻²³ = 8.31, so the two equations agree.

Worked example 4. Find the pressure of 0.50 mol of an ideal gas in a rigid vessel of volume 8.0 × 10⁻³ m³ at 310 K, and the number of molecules.

1. P = nRT/V = (0.50 × 8.31 × 310) / (8.0 × 10⁻³) = 1288 / 0.0080 = 1.61 × 10⁵ Pa.
2. N = nNA = 0.50 × 6.02 × 10²³ = 3.01 × 10²³.
3. Check with PV = NkBT: N = PV/(kBT) = (1.61 × 10⁵ × 8.0 × 10⁻³) / (1.38 × 10⁻²³ × 310) = 3.01 × 10²³. The two routes agree.

Units matter. Volume must be in m³ (1 litre = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³), pressure in Pa, temperature in K.

Pressure from molecular collisions

Pressure arises because molecules bounce off the walls. Each collision changes a molecule’s momentum. By Newton’s second law that needs a force from the wall, and by the third law the molecule pushes back on the wall equally. Very many collisions each second give a steady force, and force per unit area is pressure.

Here is the argument for a cube of side L containing N molecules of mass m.

• A molecule with velocity component vₓ towards a wall rebounds with −vₓ, so its momentum changes by 2mvₓ.
• It crosses the cube and returns every 2L/vₓ, so it hits that wall vₓ/2L times per second.
• The average force from one molecule is (2mvₓ)(vₓ/2L) = mvₓ²/L.
• For N molecules, the total force is Nm⟨vₓ²⟩/L, where ⟨vₓ²⟩ is the mean of vₓ².
• Random motion gives no preferred direction, so ⟨v²⟩ = ⟨vₓ²⟩ + ⟨vᵧ²⟩ + ⟨v_z²⟩ = 3⟨vₓ²⟩.
• Pressure = force / L² = Nm⟨v²⟩/(3L³). The volume is V = L³, and Nm/V is the density ρ.

P = (1/3)ρv²

Here v² stands for the mean of the squared speeds, so v is the root mean square speed, the quantity the guide calls the average translational speed. The result holds for any container shape.

Worked example 5. Helium at 1.00 × 10⁵ Pa has density 0.164 kg m⁻³. Find the root mean square speed.

1. v² = 3P/ρ = 3 × 1.00 × 10⁵ / 0.164 = 1.829 × 10⁶ m² s⁻².
2. v = 1.35 × 10³ m s⁻¹.

Temperature and internal energy of a monatomic gas

Combine P = (1/3)ρv² with PV = NkBT. Since ρ = Nm/V, PV = (1/3)Nm v². Setting this equal to NkBT gives (1/3)mv² = kBT, so the mean translational kinetic energy of a molecule is

(1/2)mv² = (3/2)kBT.

The mean kinetic energy of a molecule is proportional to the absolute temperature. For a monatomic ideal gas the molecules have only translational kinetic energy and there are no intermolecular forces, so the internal energy U is the total of the N kinetic energies:

U = (3/2)NkBT = (3/2)nRT

U depends only on temperature and the amount of gas. Doubling the kelvin temperature doubles U.

Worked example 6. A monatomic ideal gas of 2.0 mol is at 400 K. Then 0.80 mol is warmed from 300 K to 450 K.

1. U = (3/2)nRT = 1.5 × 2.0 × 8.31 × 400 = 9972 J, so 1.0 × 10⁴ J.
2. Change in U for the 0.80 mol = (3/2) × 0.80 × 8.31 × 150 = 1496 J, so 1.5 × 10³ J.
3. Mean kinetic energy per atom at 320 K = 1.5 × 1.38 × 10⁻²³ × 320 = 6.62 × 10⁻²¹ J.

When a real gas behaves like an ideal gas

A real gas is a good approximation to an ideal gas at:

• low pressure, so the molecules are far apart and the forces between them matter little
• low density, for the same reason, so their own volume is small compared with the container
• high temperature, well above the temperature at which the gas would condense, so the molecules move fast and attractions have little effect on their motion

Close to condensing, forces between molecules and molecular size both matter, and PV = nRT fails.

Pressure-volume diagrams

A P-V diagram plots pressure on the vertical axis and volume on the horizontal axis. Each point is a state; a line is a change of state.

Change	Shape on P-V diagram	What stays fixed
Constant pressure	Horizontal line	P
Constant volume	Vertical line	V
Constant temperature	Curve of PV = constant (a hyperbola)	T

On an isothermal curve PV is the same everywhere. A curve further from the axes has a higher PV and so a higher temperature (for a fixed amount of gas).

Worked example 7. An ideal gas of 0.12 mol goes round a cycle. A: 1.8 × 10⁵ Pa, 2.0 × 10⁻³ m³. A to B: expands at constant pressure to 4.0 × 10⁻³ m³. B to C: pressure falls at constant volume to 0.90 × 10⁵ Pa. C to A: isothermal compression. Find the temperature at each state.

1. T = PV/(nR). At A, PV = 360 J, so T = 360 / (0.12 × 8.31) = 361 K.
2. At B, PV = 720 J, so T = 722 K. The volume doubled at constant pressure, so T doubled.
3. At C, PV = 360 J again, so T = 361 K, equal to A. This confirms C to A is isothermal.

Common errors

• Putting °C into a gas equation. Convert with T/K = θ/°C + 273 (or 273.15).
• Leaving volume in litres or cm³ when pressure is in pascals.
• Writing U = (3/2)nRT for a gas that is not monatomic. The guide gives it for monatomic ideal gases only.
• Saying that a high-pressure gas is “more ideal”. The reverse is true.
• Forgetting that v² in P = (1/3)ρv² is the mean of squared speeds.
• Using (3/2)kBT as the energy of the whole gas rather than of one molecule.

Official syllabus

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