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.

  • GCE A-Level H2 Physics 2027
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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.
O Level note

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

Isothermal compression of a fixed amount of gasTwo piston cylinders contain the same eight gas particles. The second cylinder has half the gas volume. At the same temperature, its pressure is twice as large.Initial stateCompressed isothermallyvolume V, pressure ppiston moves downvolume V/2, pressure 2pfixed amount of gas at constant temperature T
Scroll diagram horizontally to read all labels.
For the same amount of gas at constant temperature, halving the volume doubles the absolute pressure. The number of particles is unchanged.

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).

p ∝ 1/V; p = k/V; pV = k
Boyle's law graph formsThe left graph shows absolute pressure against volume as a decreasing inverse curve. The right graph shows absolute pressure against reciprocal volume as a straight line through the origin.p against VVpinverse curvep against 1/V1/Vpstraight line through origin
Scroll diagram horizontally to read all labels.
At fixed mass and constant temperature, p against V is an inverse curve. Plotting absolute pressure p against 1/V linearises the relationship.

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)

Core

Problem

Air at 1.0 × 10⁵ Pa is compressed to one-fifth of its original volume, with the mass and temperature unchanged. Calculate its new pressure.
Study the worked solution
  1. 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 = p₂V₂
  2. Express the volume factor

    Method

    Write V₂ = V₁/5.

    Reason

    The final volume is one-fifth of the initial volume.

    Working

    V₁/V₂ = 5
  3. Apply 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

About 4 min

Problem

A fixed mass of gas occupies 40 cm³ at 1.0 × 10⁵ Pa. Determine its volume at 2.0 × 10⁵ Pa, assuming constant temperature.

Try this before viewing the solution

Unit: cm^3

Hints

Hint 1: state the invariant
For fixed mass and temperature, p₁V₁ = p₂V₂.
View solution step by step
  1. Rearrange

    Method

    Make the final volume the subject.

    Reason

    Boyle’s law relates the two equilibrium states.

    Working

    V₂ = p₁V₁/p₂
  2. Calculate

    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

Find and correct the mistake

Learner claim

For a fixed mass at constant temperature, one reading is V = 30 cm³, p = 1.6 × 10⁵ Pa and another is V = 40 cm³, p = 1.2 × 10⁵ Pa. A learner says the data cannot obey Boyle’s law because the pressures differ. Diagnose the claim.

Try this before viewing the solution

View solution step by step
  1. 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 = constant
  2. Test 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 cm³
  3. Test 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

Minimal support

Independent transfer

Air occupies 60 cm³ at atmospheric pressure 100 kPa. It is compressed isothermally to 40 cm³. Find the final absolute and gauge pressures, taking atmospheric pressure as 100 kPa.

Try this before viewing the solution

Hints

Hint 1: identify the pressure used by the gas law
Apply Boyle’s law to absolute pressure, then convert the final result to gauge pressure.
View solution step by step
  1. 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 absolute
  2. Apply 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 kPa
  3. Convert 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.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027