Pressure in gases: Boyle’s law
Key idea: Learn Boyle’s law for a fixed mass of gas: absolute pressure, isothermal compression, p–V graphs, particle reasoning, and worked A Level examples.
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The core idea
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Learning objectives
- Use ideal-gas equations with particles, moles and SI units.
- Apply the kinetic model to gas pressure and mean translational kinetic energy.
- Derive pV = ⅓Nm⟨c²⟩ from the definition of pressure and a one-dimensional model of molecular collisions extended to three dimensions.
Boyle’s law is not required for O Level Physics (6091). If you only need the particle explanation of gas pressure, see Gas Pressure (Particle Model).
1. Definitions (Must Know)
A. Pressure, p
- Pressure, p: force per unit area normal to a surface, p = F/A (unit: Pa).
B. Gas pressure (particle model)
Gas pressure is due to molecules colliding with the container walls and changing momentum.
At fixed temperature, gas pressure increases if:
- there are more molecules per unit volume (higher number density), or
- molecules move faster (higher average speed).
C. Boyle’s law
Boyle’s law states that the pressure of a fixed amount of gas is inversely proportional to the volume of the gas when the temperature is held constant.
pV = constant
p₁V₁ = p₂V₂
where p₁,p₂ are pressures and V₁,V₂ are volumes.
The pressures in a gas law must be absolute pressures, measured from a vacuum. If a question gives gauge pressure, first use: p_absolute = p_gauge + p_atmospheric
The diagram shows compression of gas in a cylinder while its temperature is kept constant. As the volume decreases, the absolute pressure increases so that pV remains constant.
If you plot p against 1/V, you get a straight line through the origin (gradient k).
An intuitive explanation comes from the particle model: pressure is due to molecules colliding with the container walls and changing momentum.
At constant temperature, the mean molecular kinetic energy is unchanged. If the volume is halved, the number density doubles, so the rate of momentum transfer to each unit area of wall doubles. Hence V → (1/2)V ⇒ p → 2p.
2. Key Ideas (What Earns Marks)
- State the conditions: fixed mass and constant temperature.
- Use Boyle’s law:
- p₁V₁ = p₂V₂
- Fast factor check:
- V decreases by a factor f ⇒ p increases by a factor f.
- Graphs:
- p vs V: inverse curve,
- p vs 1/V: straight line through the origin.
- Use absolute pressure, not gauge pressure.
3. Detailed Explanations
A. What “fixed mass” and “constant temperature” mean
- Fixed mass: no gas enters or leaves, so the amount of gas (n or N) is constant.
- Constant temperature (isothermal): the gas stays at the same thermodynamic temperature T (in K).
B. Deriving Boyle’s law from the ideal gas equation
For an ideal gas: pV = nRT
If the mass is fixed (n constant) and temperature is constant (T constant), then nRT is constant, so: pV = constant ⇒ p ∝ 1/V
C. Particle explanation (qualitative)
At constant temperature, the average kinetic energy of the molecules stays constant, so typical molecular speeds are similar.
When the gas is compressed into a smaller volume:
- number density increases (more molecules per m³),
- collision rate with the walls increases,
- so the total rate of momentum transfer to the walls increases,
- therefore the pressure increases.
D. Units you must be careful with
- SI units are Pa and m³.
- In p₁V₁ = p₂V₂, pressure units may cancel and volume units may cancel, but each type of unit must be consistent on both sides.
- Useful conversions:
- 1 L = 10⁻³ m³
- 1 cm³ = 10⁻⁶ m³
4. Common Mistakes
- Using Boyle’s law even though temperature changes (e.g. rapid compression without time to cool).
- Treating “fixed mass” as optional (Boyle’s law fails if gas leaks).
- Mixing pressure or volume units between the two states.
- Substituting gauge pressure directly into Boyle’s law instead of converting to absolute pressure.
5. Exam Tips
- Write “fixed mass, constant temperature” before you use p₁V₁ = p₂V₂.
- In factor questions, use ratios to save time:
- p₂ = p₁V₁/V₂
- If asked to verify Boyle’s law from data, show that pV is (approximately) constant.
- Label graph axes precisely: p against V is a curve, while p against 1/V is linear.
6. Worked Examples
Modelled example 1
Compressing air with a piston (factor change)
Problem
Study the worked solution
Check the conditions
Method
Use Boyle’s law because the gas mass and temperature are fixed.Reason
Only under these conditions is pV constant.Working
p₁V₁ = p₂V₂Express the volume factor
Method
Write V₂ = V₁/5.Reason
The final volume is one-fifth of the initial volume.Working
V₁/V₂ = 5Apply the inverse change
Method
The pressure increases by a factor of five to 5.0 × 10⁵ Pa.Reason
At constant temperature, reducing volume by five increases collision frequency per unit wall area enough to make pressure five times larger.Working
p₂ = p₁V₁/V₂ = 5(1.0 × 10⁵) = 5.0 × 10⁵ Pa
Guided practice 2
Find the new volume after a pressure change
Problem
Try this before viewing the solution
Hints
Hint 1: state the invariant
View solution step by step
Rearrange
Method
Make the final volume the subject.Reason
Boyle’s law relates the two equilibrium states.Working
V₂ = p₁V₁/p₂Calculate
Method
The final volume is 20 cm³.Reason
The pressure doubles, so the volume halves at constant temperature.Working
V₂ = ((1.0 × 10⁵)(40))/(2.0 × 10⁵) = 20 cm³
Common misconception 3
Check whether data obeys Boyle’s law
Learner claim
Try this before viewing the solution
View solution step by step
Choose the correct test
Method
Compare pV rather than pressure alone.Reason
Boyle’s law predicts pressure changes inversely with volume; it does not predict equal pressures at different volumes.Working
pV = constantTest the first state
Method
The first product is 4.8 × 10⁶ Pa cm³.Reason
Multiplying the paired pressure and volume obtains the Boyle constant in consistent units.Working
(1.6 × 10⁵)(30) = 4.8 × 10⁶ Pa cm³Test the second state
Method
The second product is also 4.8 × 10⁶ Pa cm³.Reason
Equal products show the expected inverse relationship.Working
(1.2 × 10⁵)(40) = 4.8 × 10⁶ Pa cm³
Challenge 4
Convert gauge pressure before using Boyle’s law
Independent transfer
Try this before viewing the solution
Hints
Hint 1: identify the pressure used by the gas law
View solution step by step
Set the initial absolute pressure
Method
The initial absolute pressure is 100 kPa.Reason
A gas exposed initially to atmospheric pressure has absolute pressure equal to the stated atmospheric pressure.Working
p₁ = 100 kPa absoluteApply Boyle's law
Method
The final absolute pressure is 150 kPa.Reason
Absolute pressures must be used in the inverse pressure–volume relation.Working
p₂ = p₁V₁/V₂ = (100)(60)/40 = 150 kPaConvert to gauge pressure
Method
The final gauge pressure is 50 kPa.Reason
Gauge pressure measures the excess above atmospheric pressure.Working
p_gauge = 150-100 = 50 kPa
7. Mind Stretchers
Mind stretcher 1: Same temperature, different volumeExtension
Containers A and B contain the same type of gas, with the same number of molecules, at the same temperature. Container B has a greater volume than container A. Compare the gas pressures.
Show Answer
Using the ideal gas equation pV = NkT:
For both containers, N and T are the same, so pV is the same.
Therefore p ∝ 1/V. Since V_B > V_A, we have p_B < p_A.
Mind stretcher 2: Why must Boyle’s law be isothermal?Extension
Explain why Boyle’s law requires constant temperature.
Show Answer
From the ideal gas equation pV = nRT, if the mass is fixed then pV is proportional to T.
So for pV to stay constant when V changes, T must stay constant. In a rapid compression, temperature can rise, so pressure increases more than Boyle’s law predicts.
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Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027