AQA A-Level Physics Paper 2, June 2023: Question 2

📐 Step-by-Step Calculation

Step 1: Calculate the Volume of Gas ( V )
The gas occupies a hollow cylinder (annulus of width 370 mm):
• Outer diameter D = 660 mm = 0.660 m → radius R = 0.330 m
• Inner rim diameter d = 330 mm = 0.330 m → radius r = 0.165 m
• Depth/width h = 370 mm = 0.370 m
V_outer = π × (0.330)² × 0.370 ≈ 0.1265 m³
V_inner = π × (0.165)² × 0.370 ≈ 0.0316 m³
V = 0.1265 − 0.0316 = 0.0949 m³

Step 2: Convert Temperature to Kelvin
T = 100 °C = 100 + 273 = 373 K

Step 3: Calculate Added Amount of Substance ( Δn )
Using Δp = 2.11 × 10⁵ − 1.01 × 10⁵ = 1.10 × 10⁵ Pa :
Δn = (Δp × V) / (R × T)
Δn = (1.10 × 10⁵ × 0.0949) / (8.31 × 373) ≈ 3.37 mol
(Alternatively: n₁ = 3.09 mol, n₂ = 6.46 mol → Δn = 6.46 − 3.09 = 3.37 mol)

Step 4: Calculate Added Mass ( Δm )
Δm = 14.991 − 14.897 = 0.094 kg

Step 5: Determine Molar Mass ( M )
M = Δm / Δn = 0.094 kg / 3.37 mol = 0.028 kg mol⁻¹ (2 s.f.)

✅ Mark Scheme Breakdown

• Mark 1: Attempt to calculate either volume using the correct equation πh(d/2)² or (πhd²)/4 . (Condones power of ten error here).
• Mark 2: Correct subtraction of the two cylinder volumes: V = 0.1265 − 0.0316 = 0.0949 m³ .
• Mark 3: Correct attempt to calculate moles using n = pV / RT with absolute temperature in kelvin ( 373 K ). Yields n₁ = 3.09 mol or n₂ = 6.46 mol .
• Mark 4: Correct calculation of Δn = 3.37 mol .
• Mark 5: Correct final value of molar mass: 0.028 kg mol⁻¹ (or 28 g mol⁻¹ converted to kg mol⁻¹, representing nitrogen gas, N₂).

❌ Common Errors & Pitfalls

• Treating the dimensions as radii instead of diameters: 660 mm and 330 mm are given as diameters; forgetting to divide by 2 gives a volume 4 times too large.
• Forgetting to convert Celsius to Kelvin: Using T = 100 instead of 373 K in pV = nRT loses multiple marks.
• Unit conversion for millimetres: 1 mm = 10⁻³ m , so mm³ = 10⁻⁹ m³ . Working directly in metres avoids power-of-ten mistakes.
• Molar mass units: The answer is required in kg mol⁻¹, not g mol⁻¹. Writing 28 without converting to 0.028 loses the final mark.

🧠 Exam Technique & Tips

• Combine into Δp early: Because V and T remain constant, use Δp × V = Δn × R × T in one calculation step to save time and minimise rounding errors.
• Sanity-check the result: Formula 1 tyres are inflated with dry air or pure nitrogen (N₂). Nitrogen has a molar mass of 28 g mol⁻¹ = 0.028 kg mol⁻¹ . Recognizing this confirms your answer is physically reasonable!

💡 Key Knowledge

From the ideal gas relation p = (nRT) / V (with constant volume V ):

• Pressure is directly proportional to absolute temperature ( p ∝ T ) for a fixed amount of gas.
• For a given pressure, increasing temperature reduces the number of moles needed: n = pV / (RT) . Higher T means a smaller n produces the exact same measured pressure.

✅ Model Answer (2 Marks)

Point 1 (1 Mark): Carrying out the check at a higher temperature increases the gas pressure inside the tyre (as p ∝ T ).

Point 2 (1 Mark): Therefore, a tyre at a higher temperature could reach the required minimum pressure regulation with a smaller amount (fewer moles) of gas than intended.

❌ Common Errors

• Focusing on safety instead of rules: Saying "the tyre might explode if it gets too hot" receives zero marks. The question asks about checking minimum gas content.
• Failing to link pressure back to moles/amount of gas: Merely stating "pressure increases when temperature increases" only scores 1 of the 2 marks. You must explain how this circumvents the minimum amount requirement.

🧠 Examiner Commentary

Top students directly referenced the purpose of the rule stated in the prompt: "Motorsport regulations specify a minimum amount of gas". They realised that measuring pressure is just an indirect proxy for measuring n . When heated, less gas is needed to produce the same threshold pressure, allowing teams to cheat the minimum mass requirement.