10 AQA GCSE Physics questions on gas pressure and volume (pV = constant, p1V1 = p2V2) that get harder as you go, from Foundation to Higher, each with a full worked solution.
Course: Separate Physics only | Tier: Foundation + Higher | Equation sheet: given in the exam
📌 KEY DEFINITIONS – learn them word for word
Gas pressure: gas particles move randomly and collide with the walls of their container. Each collision exerts a force, and the total force per unit area is the pressure. The force acts at right angles to the walls.
Squashing a gas into a smaller volume (at constant temperature) makes the particles hit the walls more often, so the pressure increases.
The equation
pressure × volume = constant (for a fixed mass of gas at constant temperature) p V = constant, so p₁V₁ = p₂V₂
| Quantity | Symbol | Unit |
|---|---|---|
| pressure | p | pascals (Pa) |
| volume | V | cubic metres (m³) |
Rearranged: p₂ = p₁V₁ ÷ V₂ | V₂ = p₁V₁ ÷ p₂ | Tip: as long as both pressures use the same unit and both volumes use the same unit, you can keep kPa, cm³ or litres.
Use FIFA for every answer
- F – Formula: write p₁V₁ = p₂V₂.
- I – Insert: put in the numbers (matching units on both sides).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Foundation: aim for Q1–8. Higher: aim for Q4–10. Every question needs rearranging.
Questions
Q1 (F) A gas at 100 kPa has a volume of 2.0 m³. It is slowly squashed to 1.0 m³ at constant temperature. Calculate the new pressure.
Show solution
F: p₁V₁ = p₂V₂
I: 100 × 2.0 = p₂ × 1.0
F: p₂ = 200 ÷ 1.0
A: 200 kPa
Q2 (F) A bicycle pump holds 120 cm³ of air at 100 000 Pa. The air is squashed until its pressure is 300 000 Pa. Calculate its new volume.
Show solution
F: p₁V₁ = p₂V₂
I: 100 000 × 120 = 300 000 × V₂
F: V₂ = 12 000 000 ÷ 300 000
A: 40 cm³
Q3 (F) A sealed syringe holds 50 cm³ of air at 100 kPa. The plunger is pulled out until the volume is 125 cm³. Calculate the new pressure.
Show solution
F: p₁V₁ = p₂V₂
I: 100 × 50 = p₂ × 125
F: p₂ = 5000 ÷ 125
A: 40 kPa
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Q4 (F) A scuba tank holds 0.012 m³ of air at 20 000 kPa. What volume would the air take up at 100 kPa? Give your answer in litres.
Show solution
F: p₁V₁ = p₂V₂
I: 20 000 × 0.012 = 100 × V₂
F: V₂ = 240 ÷ 100 = 2.4 m³
A: 2.4 × 1000 = 2400 litres
Q5 (F) A sealed bag of crisps holds 500 ml of air at 100 kPa. It is taken up a mountain where the pressure is 80 kPa. Calculate the new volume of air.
Show solution
F: p₁V₁ = p₂V₂
I: 100 × 500 = 80 × V₂
F: V₂ = 50 000 ÷ 80
A: 625 ml
Q6 (F/H) A weather balloon contains 5.0 m³ of helium at 100 kPa. High in the atmosphere it expands to 25 m³. Calculate the pressure there (assume constant temperature).
Show solution
F: p₁V₁ = p₂V₂
I: 100 × 5.0 = p₂ × 25
F: p₂ = 500 ÷ 25
A: 20 kPa
Q7 (F/H) A 2.0 cm³ air bubble at the bottom of a lake is at 200 kPa. It rises to the surface where the pressure is 100 kPa. Calculate its volume at the surface.
Show solution
F: p₁V₁ = p₂V₂
I: 200 × 2.0 = 100 × V₂
F: V₂ = 400 ÷ 100
A: 4.0 cm³
Q8 (F/H) A gas cylinder holds 5.0 × 10−2 m³ of gas at 1.2 × 107 Pa. What volume would the gas take up at 1.0 × 105 Pa?
Show solution
F: p₁V₁ = p₂V₂
I: 1.2 × 107 × 5.0 × 10−2 = 1.0 × 105 × V₂
F: V₂ = 6.0 × 105 ÷ 1.0 × 105
A: 6.0 m³
Q9 (H) A diver 10 m deep in fresh water breathes out a 3.0 cm³ bubble. Atmospheric pressure is 100 000 Pa. Calculate the bubble’s volume at the surface, to 2 significant figures. (ρ water = 1000 kg/m³, g = 9.8 N/kg)
Show solution
Step 1 – pressure at 10 m: water adds p = h ρ g = 10 × 1000 × 9.8 = 98 000 Pa
total p₁ = 100 000 + 98 000 = 198 000 Pa
Step 2 – volume at surface
F: p₁V₁ = p₂V₂
I: 198 000 × 3.0 = 100 000 × V₂
F: V₂ = 594 000 ÷ 100 000 = 5.94 cm³
A: 5.9 cm³ (2 s.f.)
Q10 (H) 0.75 litres of air at 101 kPa is slowly compressed to 0.32 litres at constant temperature. Calculate the new pressure, to 2 significant figures.
Show solution
F: p₁V₁ = p₂V₂
I: 101 × 0.75 = p₂ × 0.32
F: p₂ = 75.75 ÷ 0.32 = 236.7 kPa
A: 240 kPa (2 s.f.)
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