IB Physics SL Topic B.3 — The Behaviour of Gases Paper 1 & 2 Boyle’s · Charles’s · Gay-Lussac’s Laws ~8 min read

Gas Laws

Hold one variable of a gas fixed and the other two lock into a simple relationship. Three separate experimental laws describe those relationships — and together, they combine into a single rule that governs how any ideal gas behaves.

📘 What you need to know

The Ideal Gas Relation

Each of the three empirical laws only holds one variable constant at a time. Combine all three together and you get a single relationship that links pressure, volume and temperature simultaneously:

Ideal gas relation PVTPV ÷ T = constant

This is the relationship an ideal gas is defined to obey. It’s the foundation the full ideal gas equation is built on — but before adding in the amount of substance, it’s worth getting comfortable with each of the three individual laws it’s built from.

Boyle’s Law — Constant Temperature

Squeeze a gas into a smaller space at constant temperature and its pressure rises — compress it into half the volume and the pressure roughly doubles. With less room to move, the same number of molecules collide with the walls more frequently, so pressure and volume move in opposite directions.

Boyle’s law P ∝ 1 ÷ VP1V1 = P2V2

Charles’s Law — Constant Pressure

Heat a gas at constant pressure and it expands. Faster-moving molecules would otherwise hit the walls harder and more often, raising the pressure — so to keep pressure unchanged, the gas has to occupy more volume, spacing the collisions back out.

Charles’s law VTV1 ÷ T1 = V2 ÷ T2

Gay-Lussac’s Law — Constant Volume

Seal a gas into a rigid, fixed-volume container and heat it, and the pressure climbs steadily. The molecules move faster and collide with the walls both more often and with greater force, and since the volume can’t change to compensate, that extra collision activity shows up directly as higher pressure.

Gay-Lussac’s law (pressure law) PTP1 ÷ T1 = P2 ÷ T2
P V Boyle’s law: P ∝ 1⁄V constant T V T Charles’s law: V ∝ T constant P P T Gay-Lussac’s law: P ∝ T constant V
Boyle’s law produces a curve (inverse proportion); Charles’s law and Gay-Lussac’s law both produce straight lines through the origin (direct proportion to thermodynamic temperature).

Reading a Pressure–Volume Diagram

Changes to a gas’s state — pressure, volume and temperature all shifting together — can be plotted on a pressure–volume (P–V) diagram. Curved lines called isotherms mark out every possible pressure–volume combination at one fixed temperature; isotherms further from the origin represent higher temperatures. The path a gas’s state traces across this diagram tells you exactly which variable was held constant during a process:

P V increasing T A B C D A — constant pressure (V, T change) B — constant temperature (P, V change, along an isotherm) C — constant volume (P, T change) D — general process (P, V and T all change)
Each path type on a P–V diagram corresponds to holding a different variable constant — horizontal for constant pressure, along an isotherm for constant temperature, vertical for constant volume.
Quick recap: Boyle’s (P∝1/V), Charles’s (V∝T) and Gay-Lussac’s (P∝T) laws each hold one variable fixed. Combined, they give PV∝T — the defining relationship of an ideal gas.
WE 1

A gas occupies 2.4 × 10⁻³ m³ at a pressure of 1.8 × 10⁵ Pa. It is compressed at constant temperature until its volume is 9.0 × 10⁻⁴ m³. Calculate the new pressure.

Step 1 — Identify the law Temperature is constant, so this is Boyle’s law: P₁V₁ = P₂V₂ Step 2 — Rearrange and substitute P₂ = (P₁V₁) ÷ V₂ = [(1.8 × 10⁵)(2.4 × 10⁻³)] ÷ (9.0 × 10⁻⁴) P₂ = 4.8 × 10⁵ Pa The volume shrank to less than half its original size, so the pressure more than doubled — exactly what Boyle’s law predicts.
WE 2

A sealed, rigid canister contains gas at 1.2 × 10⁵ Pa and 290 K. It is heated at constant volume to 350 K. Calculate the new pressure.

Step 1 — Identify the law Volume is fixed (rigid, sealed container), so this is Gay-Lussac’s law: P₁ ÷ T₁ = P₂ ÷ T₂ Step 2 — Rearrange and substitute P₂ = (P₁T₂) ÷ T₁ = [(1.2 × 10⁵)(350)] ÷ 290 P₂ ≈ 1.45 × 10⁵ Pa (3 s.f.) Both temperatures must be in kelvin — this calculation would go wrong immediately if left in °C.

💡 Top tips

⚠ Common mistakes

Up next: The Ideal Gas Equation — where we bring in the amount of substance and turn PV ∝ T into a full equation you can actually calculate with.

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