AQA A-Level Physics Paper 2, June 2025: Question 10

Award 1 mark for selecting option B (Assessment Objective: AO2).

💡 Key Knowledge

• Definition of Absolute Zero: The theoretical temperature at which particles have minimum internal energy and an ideal gas exerts zero pressure ( p = 0 ).
• Pressure Law: For a fixed mass of gas at constant volume, pressure varies linearly with temperature in degrees Celsius (°C):
  p = mθ + c
• Linear Extrapolation: Since the plot of p against θ is a straight line, the gradient m is strictly constant across the entire range, down to p = 0 .

📐 Step-by-Step Derivation

1. Find the gradient (m):
   From the two known points (θ₁, p₁) and (θ₂, p₂) :
   m = Δp / Δθ = (p₂ − p₁) / (θ₂ − θ₁)
2. Write the point-slope equation:
   Using point (θ₁, p₁) :
   p − p₁ = m (θ − θ₁)
3. Set pressure to zero (p = 0):
   At absolute zero ( θ = θ₀ ), p = 0 :
   0 − p₁ = m (θ₀ − θ₁)
   −p₁ = [(p₂ − p₁) / (θ₂ − θ₁)] (θ₀ − θ₁)
4. Rearrange for θ₀:
   Multiply by the reciprocal of the gradient:
   θ₀ − θ₁ = −p₁ [(θ₂ − θ₁) / (p₂ − p₁)]
   θ₀ = θ₁ − p₁ [(θ₂ − θ₁) / (p₂ − p₁)]

🧠 Exam Technique: Dimensional & Sign Checks

• Dimensional Analysis:
  • Absolute zero must have the unit °C.
  • In B, p₁ has units of kPa , and (θ₂ − θ₁) / (p₂ − p₁) has units of °C / kPa .
  • Multiplying them gives kPa × (°C / kPa) = °C . Thus, subtracting it from θ₁ (°C) is dimensionally valid!
  • Options C and D produce units of kPa² / °C or kPa , which can instantly be eliminated.
• Sanity Check with Numbers:
  Since θ₂ > θ₁ and p₂ > p₁ , the term in brackets is positive. Therefore, θ₁ − (positive value) must be less than θ₁ , which correctly places absolute zero far to the left on the horizontal axis.

❌ Common Errors & Pitfalls

• Inverting the Gradient: Forgetting that Δθ / Δp is 1 / gradient . Students who mistakenly write θ = m·p end up choosing inverted fractions like in C or D.
• Sign Confusion (Option A): Rearranging −p₁ = m(θ − θ₁) incorrectly to θ = p₂(1/m) − θ₁ leads to sign errors and mispairing coordinates ( p₂ with θ₁ ).
• Assuming the intercept is directly visible: The y-axis intercept shown on the given diagram is at temperature θ₁ , not 0 °C , so p₁ is not the c-value of the general equation y = mx + c unless θ₁ = 0 .