4PH1

Ideal Gases

Solids, Liquids & Gases · 1 question type

Practise

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 7% of your exam marks.

stable
Low
Stable7%

Gas law calculations (Boyle's Law, pressure-temperature) and kinetic theory explanations appear regularly.

Statement

  • At constant and for a fixed amount of an ideal gas:

P × V = constant

  • Equivalently, between two states:

p₁ × V₁ = p₂ × V₂

  • where:
    • p₁, p₂ = pressures in Pa
    • V₁, V₂ = volumes in m³ (or any consistent unit, as long as both match)
  • p and V are inversely proportional: doubling the pressure halves the volume; halving the pressure doubles the volume
Boyle's law: compressing a fixed mass of gas at constant temperature from state P₁,V₁ into a smaller volume P₂,V₂ raises its pressure
Source: Boyle's law by Save My Exams

Why it works microscopically

  • At constant temperature the molecules have the same average kinetic energy, so each individual collision with the wall carries the same average force
  • Shrink the container to half its volume and the molecules have to travel only half as far between wall collisions
  • They therefore hit each wall twice as often per second, and the pressure doubles
  • The same number of molecules, moving at the same average speed, in half the space, gives twice the pressure
Boyle's law at constant temperature: increasing the volume spreads the molecules further apart so they collide with the walls less often, lowering the pressure (P ∝ 1/V)
Source: Boyle's law by Save My Exams

Checking the answer makes sense

  • A useful sanity check after any gas-law calculation:
    • if the volume has shrunk (compression), expect the new pressure to be larger than the original
    • if the gas has been heated at fixed volume, the new pressure should be larger than before
    • if you get the opposite trend, you've probably substituted the temperatures the wrong way round, or forgotten to convert °C to K
Worked example

Boyle's law calculation

A gas is compressed from a volume of 0.60 m³ at a pressure of 100 kPa to a new volume of 0.40 m³. The temperature does not change. Calculate the new pressure.

Solution:

  • Identify the known values: p₁ = 100 kPa, V₁ = 0.60 m³, V₂ = 0.40 m³
  • Substitute into p₁V₁ = p₂V₂: 100 × 0.60 = p₂ × 0.40
  • Rearrange: p₂ = (100 × 0.60) / 0.40
  • p₂ = 150 kPa
  • Sanity check: the volume shrank, so the pressure should be larger — 150 kPa > 100 kPa. ✓