AQA A-Level Physics Paper 2, June 2025: Question 2
11 marks · Medium difficulty · Extended Answer
Explain the kinetic theory derivation assumptions, thermal expansion at constant pressure, and deduce if water vapour behaves as an ideal gas.
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Kinetic Theory of Gases & Thermal Physics
📋 Revision Overview
What this question tests
- Microscopic Assumptions: Precise definitions of kinetic theory postulates (random motion, molecular volumes, intermolecular forces, collision durations).
- Phenomenological Derivation & Gas Laws: Applying Newton's laws and momentum principles ( F = Δp/Δt , p = F/A ) to justify isobaric expansion with temperature.
- Microscopic Energy: Calculating molecular mean kinetic energy using Ek = (3/2)kT .
- Thermodynamic Reasoning: Quantitative deduction comparing the theoretical specific heat capacity of a monoatomic ideal gas with real gas data.
Part 02.1 · 2 Marks
Meaning of Random Particle Motion
Explaining the kinetic theory postulate
✅ Correct Answer (Mark Scheme)
- Mark 1: Particles possess a range of speeds / kinetic energies.
- Mark 2: There is no preferred direction of motion (all directions of movement are equally probable).
❌ Common Errors & Pitfalls
- "Range of directions": Too vague. You must express that all directions are equally likely, or that there is no preferred direction.
- Unpredictability / Chaos: Saying particles "do not follow Newton's laws" or "move unpredictably" caps your total mark to max 1! Gas particles strictly obey deterministic Newtonian mechanics.
- Average speed: Stating "particles have an average speed" does not earn Mark 1; you must specify a distribution / range of speeds.
Examiner Insight: Mathematical equivalent for Mark 2: average velocity components along Cartesian axes are equal, i.e., v̅x = v̅y = v̅z . Do not confuse random particle motion with Brownian motion.
Part 02.2 · 2 Marks
Assumptions of the Kinetic Theory Model
Additional postulates required to derive pV = ⅓Nm(crms)²
✅ Any TWO of the following (1 mark each)
- Total particle volume is negligible compared to the total volume occupied by the gas ( volume of particles << gas volume / particles are point masses).
- Time of collision is negligible compared to the time spent between collisions ( tcollision << tbetween / collisions are instantaneous).
- No intermolecular / non-contact forces act between particles except during collisions.
- All collisions (particle-particle and particle-wall) are perfectly elastic.
- Particles obey Newton's laws of motion.
🧠 Exam Technique: The "List Principle"
This question uses the strict list principle. If you provide three statements and one is incorrect or contradicts another, you will lose marks. Stick to the two clearest, most textbook-accurate assumptions.
- Saying merely "particles are very small" is insufficient for full credit.
- Saying merely "collision time is short" is insufficient without comparing it to time between collisions.
Part 02.3 · 3 Marks
Kinetic Theory Explanation of Gas Expansion at Constant Pressure
Explaining Charles's Law at the molecular scale
✅ Marking Points Structure (3 Marks)
Step 1: Effect of higher temperature (Any TWO points):
- Higher temperature increases the mean speed / root mean square speed (crms) / mean kinetic energy of particles.
- Greater speed leads to a larger rate of change of momentum ( Δ(mv) per collision increases).
- Collision frequency with container walls increases (if volume remained constant).
- Resulting force increases because F = Δ(mv)/Δt .
Step 2: Reason volume must expand (ONE point):
- Since pressure = force / area , a larger surface area is required to keep pressure constant when force increases.
- OR: Larger volume increases time between collisions, lowering the collision frequency to keep the total rate of momentum change (and pressure) constant.
❌ Critical Examiner Traps
- Velocity vs Speed: Never write "mean velocity increases" — average velocity is zero in a gas! Always specify mean speed or mean kinetic energy.
- Gas Law equations: Merely quoting V/T = constant or pV = nRT scores zero. You are asked to explain using the kinetic theory model (mechanics of particles).
- Particle collisions: Collisions between particles do not directly create pressure on the container; focus specifically on wall collisions.
💡 Alternative Mathematical Route Accepted by Mark Scheme
1. State that an increase in temperature increases crms .
2. State the kinetic theory equation: pV = ⅓Nm(crms)² .
3. Explain that because p , N , and m are held constant, V must increase proportionally with (crms)² .
Part 02.4 · 1 Mark
Change in Average Kinetic Energy of an Atom
Calculation using temperature change
📐 Step-by-Step Calculation
Step 1: Determine temperature change in Kelvin
Notice that a change in temperature in °C is identical to a change in K:
ΔT = 77 °C - 27 °C = 50 K
Notice that a change in temperature in °C is identical to a change in K:
ΔT = 77 °C - 27 °C = 50 K
Step 2: Apply the average translational kinetic energy formula
ΔEk = (3/2) k ΔT
Where Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹
ΔEk = (3/2) k ΔT
Where Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹
Step 3: Calculate value
ΔEk = 1.5 × (1.38 × 10⁻²³) × 50
ΔEk = 1.035 × 10⁻²¹ J ≈ 1.0 × 10⁻²¹ J
ΔEk = 1.5 × (1.38 × 10⁻²³) × 50
ΔEk = 1.035 × 10⁻²¹ J ≈ 1.0 × 10⁻²¹ J
Award 1 mark: Answer of 1.0 × 10⁻²¹ J or 1.04 × 10⁻²¹ J .
Part 02.5 · 3 Marks
Deducing Whether Water Vapour Behaves as an Ideal Gas
Quantitative deduction from specific heat capacity
📐 Method 1: Compare Theoretical vs Actual Specific Heat Capacity
Step 1: Link molecular kinetic energy to specific heat capacity
For an ideal gas, all energy supplied increases kinetic energy:
Energy per molecule per kelvin = (3/2) k
Mass of one molecule: m = 3.0 × 10⁻²⁶ kg
For an ideal gas, all energy supplied increases kinetic energy:
Energy per molecule per kelvin = (3/2) k
Mass of one molecule: m = 3.0 × 10⁻²⁶ kg
Step 2: Calculate theoretical c for an ideal gas
cideal = (3/2 × k) / m = (1.5 × 1.38 × 10⁻²³) / (3.0 × 10⁻²⁶)
cideal = 690 J kg⁻¹ K⁻¹
cideal = (3/2 × k) / m = (1.5 × 1.38 × 10⁻²³) / (3.0 × 10⁻²⁶)
cideal = 690 J kg⁻¹ K⁻¹
Step 3: Compare and conclude
Actual c = 1400 J kg⁻¹ K⁻¹ .
Since 690 J kg⁻¹ K⁻¹ ≠ 1400 J kg⁻¹ K⁻¹ (actual is over twice the ideal value), water vapour does not behave as an ideal gas.
Actual c = 1400 J kg⁻¹ K⁻¹ .
Since 690 J kg⁻¹ K⁻¹ ≠ 1400 J kg⁻¹ K⁻¹ (actual is over twice the ideal value), water vapour does not behave as an ideal gas.
📐 Method 2: Compare Energy Values (Using Part 02.4)
Step 1: Energy supplied to one molecule for ΔT = 50 K
Q = m × c × ΔT = (3.0 × 10⁻²⁶ kg) × 1400 × 50
Q = 2.1 × 10⁻²¹ J
Q = m × c × ΔT = (3.0 × 10⁻²⁶ kg) × 1400 × 50
Q = 2.1 × 10⁻²¹ J
Step 2: Compare with ideal kinetic energy change from 02.4
Ideal ΔEk = 1.035 × 10⁻²¹ J
Ideal ΔEk = 1.035 × 10⁻²¹ J
Step 3: Comparison and deduction
2.1 × 10⁻²¹ J ≠ 1.035 × 10⁻²¹ J (supplied heat is greater than kinetic energy increase).
Therefore, water vapour does not behave as an ideal gas.
2.1 × 10⁻²¹ J ≠ 1.035 × 10⁻²¹ J (supplied heat is greater than kinetic energy increase).
Therefore, water vapour does not behave as an ideal gas.
🧠 Mark Breakdown
- Mark 1: Finding number of molecules in 1 kg ( N = 1 / (3.0 × 10⁻²⁶) = 3.33 × 10²⁵ ) OR using (3/2)kT / (3/2)RT .
- Mark 2: Correct use of mcΔθ for the identical temperature change.
- Mark 3: Explicit comparison of the calculated quantities leading to a clear "No" conclusion.
💡 Physical Insight: Why does it differ?
Water ( H₂O ) is a non-linear triatomic molecule. The supplied thermal energy is absorbed not only into translational kinetic energy, but also into rotational and vibrational modes, as well as overcoming significant intermolecular forces (hydrogen bonding/dipole interactions).
Topics
Physics · 3.6 Further mechanics and thermal physics (A-level only)
Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.