Specification: AQA A-level Physics (7408) 3.6.2.2 | Paper: 2 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Boyle’s law: for a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume: pV = constant.
Charles’s law: for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute temperature: V/T = constant. Pressure law: for a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature: p/T = constant.
Absolute zero: the temperature at which the particles of a substance have minimum internal energy (for an ideal gas, the extrapolated temperature at which pressure and volume would be zero). It is 0 K = −273 °C, and T/K = θ/°C + 273.
The equation
combined gas law (fixed mass of ideal gas) p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂ which gives pV = constant (T fixed), V ÷ T = constant (p fixed), p ÷ T = constant (V fixed)
| Quantity | Symbol | Unit |
|---|---|---|
| pressure | p | pascals (Pa), equivalent to N m⁻² |
| volume | V | cubic metres (m³) |
| absolute temperature | T | kelvin (K) |
Rearranged: p₂ = p₁V₁T₂ ÷ (T₁V₂) | V₂ = p₁V₁T₂ ÷ (T₁p₂) | T₂ = p₂V₂T₁ ÷ (p₁V₁) | Always convert °C to K (add 273) before you substitute. p and V only need the same units on both sides, so kPa and cm³ can be used for ratios, but T must be in kelvin. Pressure at depth h in a liquid: p = patm + ρgh.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A fixed mass of gas has a pressure of 2.0 × 105 Pa and a volume of 3.0 × 10−3 m³. It is compressed slowly at constant temperature to a volume of 1.2 × 10−3 m³. Calculate its new pressure.
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F: p₁V₁ = p₂V₂ (Boyle’s law)
I: 2.0 × 105 × 3.0 × 10−3 = p₂ × 1.2 × 10−3
F: p₂ = (2.0 × 105 × 3.0 × 10−3) ÷ (1.2 × 10−3) = 600 ÷ (1.2 × 10−3)
A: 5.0 × 105 Pa (2 s.f.)
Q2 (Core) A gas occupies 0.50 m³ at 27 °C. It is heated at constant pressure to 177 °C. Calculate its new volume.
Show solution
Convert first: 27 °C = 300 K; 177 °C = 450 K
F: V₁ ÷ T₁ = V₂ ÷ T₂ (Charles’s law)
I: 0.50 ÷ 300 = V₂ ÷ 450
F: V₂ = 0.50 × 450 ÷ 300
A: 0.75 m³ (2 s.f.)
Q3 (Core) A sealed rigid container holds air at 1.0 × 105 Pa and 20 °C. It is heated to 100 °C. Calculate the new pressure, to 3 significant figures.
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Convert first: 20 °C = 293 K; 100 °C = 373 K
F: p₁ ÷ T₁ = p₂ ÷ T₂ (pressure law)
I: 1.0 × 105 ÷ 293 = p₂ ÷ 373
F: p₂ = 1.0 × 105 × 373 ÷ 293
A: 127 303… Pa = 1.27 × 105 Pa (3 s.f.)
Watch out: examiners regularly see temperatures left in °C. Using 100 ÷ 20 gives 5.0 × 105 Pa – five times the pressure, which is clearly wrong for an 80 °C rise.
Q4 (Core) A sealed gas syringe contains 60 cm³ of air at 101 kPa. The plunger is pushed in slowly, at constant temperature, until the volume is 25 cm³. Calculate the new pressure in Pa, to 3 significant figures.
Show solution
Convert first: 101 kPa = 1.01 × 105 Pa (the volumes can stay in cm³ because only their ratio matters)
F: p₁V₁ = p₂V₂
I: 1.01 × 105 × 60 = p₂ × 25
F: p₂ = 1.01 × 105 × 60 ÷ 25
A: 242 400 Pa = 2.42 × 105 Pa (3 s.f.)
Watch out: pushing the plunger in slowly lets energy transfer to the surroundings, so the temperature stays constant. A rapid compression warms the gas and Boyle’s law does not apply.
Q5 (Exam) A car tyre contains air at an absolute pressure of 2.4 × 105 Pa at 12 °C. After a long drive the air temperature in the tyre is 35 °C. Assuming the volume of the tyre does not change, calculate the new pressure, to 3 significant figures.
Show solution
Convert first: 12 °C = 285 K; 35 °C = 308 K
F: p₁ ÷ T₁ = p₂ ÷ T₂
I: 2.4 × 105 ÷ 285 = p₂ ÷ 308
F: p₂ = 2.4 × 105 × 308 ÷ 285
A: 259 368… Pa = 2.59 × 105 Pa (3 s.f.)
Watch out: a tyre gauge reads pressure above atmospheric. The gas laws need the absolute pressure (gauge reading + 1.01 × 105 Pa).
Q6 (Exam) A weather balloon contains 4.0 m³ of helium at 1.01 × 105 Pa and 20 °C at ground level. It rises to a height where the pressure is 2.6 × 104 Pa and the temperature is −40 °C. Calculate the volume of the balloon at this height, to 2 significant figures. Assume no gas escapes.
Show solution
Convert first: 20 °C = 293 K; −40 °C = 233 K
F: p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂
I: (1.01 × 105 × 4.0) ÷ 293 = (2.6 × 104 × V₂) ÷ 233
F: V₂ = 1.01 × 105 × 4.0 × 233 ÷ (293 × 2.6 × 104)
A: 12.35… m³ = 12 m³ (2 s.f.)
Q7 (Exam) In a constant-volume gas thermometer, the pressure of the gas is 1.000 × 105 Pa at 0 °C and 1.366 × 105 Pa at 100 °C. Assuming pressure varies linearly with temperature, extrapolate to find the temperature in °C at which the pressure would be zero, to 3 significant figures. State what this temperature represents.
Show solution
Step 1 – gradient of the p–θ graph
gradient = (1.366 × 105 − 1.000 × 105) ÷ 100 = 366 Pa °C⁻¹
Step 2 – extrapolate to p = 0
F: p = p₀ + (gradient × θ)
I: 0 = 1.000 × 105 + 366θ
F: θ = −1.000 × 105 ÷ 366 = −273.2… °C
A: −273 °C (3 s.f.). This is absolute zero (0 K), the temperature at which the gas particles would have minimum internal energy.
Q8 (Exam) An aerosol can contains gas at 3.0 × 105 Pa at 20 °C. The can will burst if the pressure reaches 8.0 × 105 Pa. Calculate the temperature, in K and in °C, at which the can bursts, to 3 significant figures. Assume the volume of the can stays constant.
Show solution
Convert first: 20 °C = 293 K
F: p₁ ÷ T₁ = p₂ ÷ T₂
I: 3.0 × 105 ÷ 293 = 8.0 × 105 ÷ T₂
F: T₂ = 293 × 8.0 × 105 ÷ (3.0 × 105)
A: 781.3… K = 781 K (3 s.f.), and 781.3… − 273 = 508 °C (3 s.f.)
Watch out: using 20 °C instead of 293 K gives 53 °C – a dangerously wrong answer for a safety warning.
Q9 (Stretch) An air bubble of volume 2.0 cm³ is released by a diver at a depth of 15 m in a freshwater lake, where the temperature is 8.0 °C. The bubble rises to the surface, where the temperature is 18 °C. Calculate the volume of the bubble at the surface, to 2 significant figures. (density of water = 1000 kg m⁻³; atmospheric pressure = 1.01 × 105 Pa; g = 9.81 m s⁻²)
Show solution
Convert first: 8.0 °C = 281 K; 18 °C = 291 K
Step 1 – pressure at 15 m depth
p₁ = patm + ρgh = 1.01 × 105 + 1000 × 9.81 × 15 = 248 150 Pa
Step 2 – combined gas law
F: p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂
I: (248 150 × 2.0) ÷ 281 = (1.01 × 105 × V₂) ÷ 291
F: V₂ = 248 150 × 2.0 × 291 ÷ (281 × 1.01 × 105)
A: 5.08… cm³ = 5.1 cm³ (2 s.f.)
Watch out: the pressure at depth is ρgh plus atmospheric pressure. Using ρgh alone (147 150 Pa) gives 3.0 cm³.
Q10 (Stretch) A gas syringe is sealed with 50 cm³ of air inside at 1.01 × 105 Pa and 20 °C. The syringe is placed in a water bath at 30 °C and the plunger is pushed in until the volume is 20 cm³. (a) Calculate the new pressure of the air, to 3 significant figures. (b) The plunger has a cross-sectional area of 2.0 cm² and atmospheric pressure acts on its outside. Calculate the force that must be applied to hold the plunger in place, to 2 significant figures. Ignore friction.
Show solution
Convert first: 20 °C = 293 K; 30 °C = 303 K; 2.0 cm² = 2.0 × 10−4 m²
Step 1 – (a) new pressure
F: p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂
I: (1.01 × 105 × 50) ÷ 293 = (p₂ × 20) ÷ 303
F: p₂ = 1.01 × 105 × 50 × 303 ÷ (293 × 20)
A: 261 117… Pa = 2.61 × 105 Pa (3 s.f.)
Step 2 – (b) force on the plunger
pressure difference = 261 117… − 1.01 × 105 = 160 117… Pa
F = Δp × A = 160 117… × 2.0 × 10−4 = 32.02… N
A: 32 N (2 s.f.)
Watch out: in (b) the atmosphere pushes on the outside of the plunger, so only the pressure difference has to be supplied. Using the full 2.61 × 105 Pa gives 52 N.
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