AQA A-Level Physics Paper 2, June 2023: Question 1

5 marks · Medium difficulty · Short Answer

Derive the pressure exerted by a single particle in a cubical box on a wall using the kinetic theory of gases and Newton's laws of motion.

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Question

Question 01 consists of four parts based on an ideal gas. Figure 1 shows a hollow cube with side length l and shaded face W. Inside, a single particle P of mass m travels with velocity c directly perpendicular toward face W. Part 01.1 asks to state what is meant by the internal energy of an ideal gas (1 mark). Part 01.2 asks to explain why P experiences a change in momentum of -2mc during a collision with W (1 mark). Part 01.3 asks to show that the collision frequency f is c/(2l) (1 mark). Part 01.4 asks to deduce an expression in terms of m, c, and V for the pressure exerted on W, referring to Newton's laws of motion (2 marks).
Question text

01.1 State what is meant by the internal energy of an ideal gas.

[1 mark]

Figure 1 shows a single gas particle P of an ideal gas inside a hollow cube.

Figure 1

The cube has side length l and volume V.

P has mass m and is travelling at a velocity c perpendicular to side W.

01.2 Explain why P has a change in momentum of −2mc during one collision with W.

[1 mark]

01.3 P collides repeatedly with W.

c

Show that the frequency f of collisions is .

2l

[1 mark]

01.4 Deduce an expression, in terms of m, c and V, for the contribution of P to the

pressure exerted on W.

Refer to appropriate Newton’s laws of motion.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for question 01: 01.1 awards 1 mark for stating the total kinetic energy of the particles. 01.2 awards 1 mark for showing change in momentum as delta p = -mc - mc = -2mc due to equal speed before and after an elastic collision. 01.3 awards 1 mark for calculating time between collisions T = 2l/c, giving frequency f = 1/T = c/(2l). 01.4 awards 1 mark for referencing a Newton law and P = F/A, and 1 mark for deriving P = mc^2 / V. Total marks: 5.

Question Answers Additional comments/Guidelines Mark AO

01.1 Condone “molecules” or “atoms” for 1 AO1

“particles”

Kinetic energy will be taken to mean total

total kinetic energy of the particles kinetic energy but do not accept use of mean

kinetic energy or reference to kinetic energy

of a single particle.

Do not allow any reference that implies there

is potential energy or any other energy added

to the kinetic energy.

01.2 (the speed before and after a collision is the same in the Use of subscripts i and f or before and after 1 AO1

elastic collision) do not need explanation.

Δp (= pf – pi) = − mc – mc = −2mc

Δ will be assumed to mean (final – initial).

Either the initial momentum or the final

momentum must be described clearly enough

to justify the negative final answer

01.3 𝑠𝑠 2l Must show evidence of a time calculation 1 AO1

Time between colliding with W (= ) = c

𝑐𝑐 using distance and speed

1 c

f = =

T 2l Do not allow any attempted use of v = f λ

01.4 F The reference to Newton law could be a 2 AO1

Reference to a Newton law AND P = 1 simple link between Newton’s name and an ×2

A

equation.

F mc2 1 mc2

P = = × 2 = 2

A l l V

Total 5

How to answer it

Ideal Gas Kinetic Theory: Deriving Pressure from First Principles

📋 Specification Focus

What this question tests

This question assesses your foundational understanding of the kinetic theory model and your ability to construct the theoretical derivation of gas pressure from first principles:

  • Thermal Physics Definitions: Defining internal energy specifically for an ideal gas (distinguishing between potential and kinetic energy).
  • Momentum & Collisions: Applying vector direction and sign conventions to elastic particle collisions ( Δp = -2mc ).
  • Kinematics in a Box: Relating particle speed, round-trip distance ( 2l ), time interval, and collision frequency ( f = 1/T ).
  • Newton’s Laws of Motion: Linking Newton's 2nd Law ( F = Δp/Δt ) and 3rd Law (action-reaction pairs) to deduce microscopic pressure ( P = F/A ).
Part 01.1 • 1 Mark

Internal Energy of an Ideal Gas

State what is meant by the internal energy of an ideal gas.

✅ Model Answer

The total kinetic energy of the gas particles (molecules/atoms).

Mark allocation: 1 mark for stating "total kinetic energy of the particles/molecules/atoms".

💡 Key Knowledge

By definition, internal energy is the sum of randomly distributed kinetic and potential energies of all particles in a system:

U = ΣEk + ΣEp

However, one core assumption of an ideal gas is that there are no intermolecular forces between particles (except during instantaneous collisions). Therefore, potential energy is zero ( Ep = 0 ), leaving internal energy composed entirely of kinetic energy.

❌ Common Errors & Lost Marks

  • Mentioning potential energy: Saying "the sum of kinetic and potential energy" scores 0 marks. While true for real gases, ideal gases have zero potential energy.
  • Mean vs Total: Writing "the average kinetic energy" scores 0. Average kinetic energy relates to temperature ( Ek ∝ T ), but internal energy represents the total sum over all particles.
  • Single particle: Stating the kinetic energy of "a single molecule" fails to earn the mark.

🧠 Exam Technique

Always spot the qualifier "ideal gas". If a question asks for the internal energy of a substance, quote both kinetic and potential energies. If it asks specifically for an ideal gas, only quote total kinetic energy.

Part 01.2 • 1 Mark

Change in Momentum on Wall Collision

Explain why P has a change in momentum of -2mc during one collision with W.

✅ Model Answer

Because the collision is elastic, speed is conserved. Taking the direction towards wall W as positive:

Initial momentum: pi = +mc

Final momentum after rebound: pf = -mc

Δp = pfinal - pinitial = (-mc) - (+mc) = -2mc

Mark allocation: 1 mark for clearly defining initial and final momentum to justify the negative sign: Δp = -mc - mc = -2mc .

💡 Key Knowledge

  • Momentum is a vector quantity ( p = mv ), meaning direction matters.
  • A change in any quantity is strictly defined as:
    Δ = (final value) - (initial value) .
  • In an elastic collision against a stationary wall of infinite mass, the particle rebounds with equal speed ( c ) in the opposite direction ( -c ).

❌ Common Errors

  • Failing to show directions: Just writing mc + mc = 2mc does not explain where the minus sign in -2mc comes from.
  • Omitting that the collision is elastic (speed remains unchanged).

🧠 Exam Technique

To justify a negative answer, you must state which direction is positive. State: "Taking direction towards wall W as positive, initial velocity = +c and final velocity = -c". Then write out (-mc) - (+mc) = -2mc .

Part 01.3 • 1 Mark

Collision Frequency

Show that the frequency f of collisions is c / (2l).

✅ Model Answer

1. Distance travelled between successive collisions with wall W is to the opposite wall and back: s = 2l .

2. Time taken between collisions (period T ):
T = distance / speed = 2l / c

3. Frequency is the reciprocal of the time period:
f = 1 / T = 1 / (2l / c) = c / (2l)

Mark allocation: 1 mark for showing clear evidence of calculating time from distance and speed ( t = 2l/c ), then inverting to find f .

📐 Step-by-Step Derivation

  1. Path: Particle starts at W, moves distance l to opposite face, rebounds, and travels distance l back to W. Total distance s = 2l .
  2. Time between impacts: Δt = s / c = 2l / c .
  3. Impact rate (frequency): f = 1 / Δt = c / (2l) .

❌ Examiner Pitfalls

  • Do NOT use wave equations: The mark scheme explicitly states: "Do not allow any attempted use of v = fλ". This is particle motion, not a wave phenomenon!
  • Missing the round trip: Using distance l instead of 2l gives c/l , which is the collision rate with any wall perpendicular to that axis, not with wall W specifically.
Part 01.4 • 2 Marks

Deducing Pressure and Applying Newton's Laws

Deduce an expression, in terms of m, c and V, for the contribution of P to the pressure exerted on W. Refer to appropriate Newton's laws of motion.

✅ Model Answer

Applying Newton's Laws:

  • By Newton's 2nd Law, force is the rate of change of momentum:
    F = Δp / Δt = (magnitude of Δp) × f = 2mc × (c / 2l) = mc² / l
  • By Newton's 3rd Law, the force exerted by particle P on wall W is equal and opposite to the force exerted by wall W on particle P.

Calculating Pressure:

Area of wall W is A = l² , and volume of cube is V = l³ .

Pressure P = F / A = (mc² / l) / l² = mc² / l³ = mc² / V

Mark allocation:
• Mark 1: Reference to a Newton's Law (e.g. Newton 2: F = Δp/Δt or Newton 3) AND P = F/A .
• Mark 2: Correct algebraic derivation leading to P = mc² / V .

📐 Step-by-Step Algebraic Deduction

  1. Rate of momentum change (Newton II):
    F = Δp × f = (2mc) × (c / 2l) = mc² / l
  2. Wall Surface Area:
    For a cube of side length l , A = l² .
  3. Pressure definition:
    P = F / A = (mc² / l) / l² = mc² / l³
  4. Cube Volume:
    Substitute V = l³ into the expression:
    P = mc² / V

🧠 Exam Technique: How Top Students Secured 2/2

  • Explicitly name the laws: Don't just write formulas without context. Write "From Newton's second law, F = rate of change of momentum" and "From Newton's third law, the force on the wall is equal in magnitude".
  • Sign clarity: Since pressure is a scalar magnitude, state that the force on the wall is +mc²/l (opposite to the force on the particle).
  • Final variables: Check that your final answer uses only the variables requested: m , c , and V . Leaving l³ unsimplified loses Mark 2.

❌ Common Errors

  • Forgetting to name or link Newton's Law, thus forfeiting the first mark even if the final equation is correct.
  • Omitting the area step: jumping directly from force to pressure without stating P = F/A .
  • Confusing the symbol for pressure ( P or p ) with momentum ( p ). Keep your notation distinct!

Topics

Physics · 3.4 Mechanics and materials · 3.6 Further mechanics and thermal physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.