AQA A-Level Physics Paper 3 (3BC), June 2025: Question 3

15 marks · Hard difficulty · Extended Answer

Analyze a three-process thermodynamic cycle on a p–V diagram by determining the amount of gas, work done, values of Q, W, and ΔU, and evaluating engine efficiency and practicality.

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Question

Question text

03.1 State what is meant by an isothermal change.

[1 mark]

Figure 6 shows a p–V diagram in which a fixed mass of an ideal gas is taken through

a cycle of three processes.

A → B: adiabatic change

B → C: isothermal change

C → A: reduction in volume at constant pressure

Figure 6

The maximum temperature of the gas in the cycle is 536 K and the minimum

temperature is 291 K.

03.2 Calculate, in mol, the amount of gas.

[2 marks]

amount of gas9 = mol

03.3 Show that the work done in process C → A is about 100 J.

[1 mark]

03.4 Complete Table 2 for each process and for the whole cycle.

[3 marks]

Table 2

Q / J W / J ΔU / J

Process A → B −253

Process B → C 473

Process C → A −253

Whole cycle 119

03.5 Two criteria for a heat engine to achieve the maximum theoretical efficiency are:

1. the gas must always be at the same temperature as the hot source when energy is

being transferred to the gas by heat transfer

2. the gas must always be at the same temperature as the cold sink when energy is

being transferred from the gas by heat transfer.

The temperatures of the source and sink do not change.

Deduce whether the cycle in Figure 6 satisfies either or both of these criteria.

[2 marks]

03.6 The overall efficiency of a heat engine is defined as

net work output in one cycle

overall efficiency =

energy supplied by heat transfer in one cycle

An engineer suggests designing a real engine based on the cycle shown in Figure 6.

Discuss the engineer’s suggestion.

In your answer you should:

• use Table 2 to determine the overall efficiency of a theoretical engine that uses

the cycle

• compare this overall efficiency with the efficiency of an ideal engine working

between the same temperatures as the cycle

• discuss with reasons the problems that would need to be overcome in making a

real engine work on this cycle

• explain whether the engineer’s suggestion would lead to a useful engine.

[6 marks]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidance Mark AO

03.1 A change (of state that occurs) at constant temperature ✓ Acceptable alternatives: 1 AO1

process in which the temperature remains

constant

compression or expansion of a gas at constant

temperature

Treat reference to pressure and volume as

neutral.

Do not allow ‘no change in internal energy’

OR ΔU = 0

03.2 pV ‘Use of’ is by substitution; condone value of R not 2 AO1

Use of n = ✓ substituted.

291 K Correct answer gains both marks.

at A using

OR Allow either T in MP1 for any correct pair of pV

B or C using 536 K values.

giving n = 4.97 × 10−2 (mol) ✓

Accept 4.96, 0.050 but not 0.05

– A-LEVEL PHYSICS – –

03.3 W = pΔV For mark to be awarded, correct equation or 1 AO1

✓ correct substitution must be seen.

= (−)101 J

E.g. accept

[1.00 ×] 105 × (2.21 − 1.20) × 10−3 = 100 J

W = pΔV = (−)101 J

Ignore sign

– A-LEVEL PHYSICS – –

03.4 Any row correct ✓ up to Max 3 3 1× AO1

Q / J W / J ΔU / J

1× AO2

Process A → B 0 −253 253 1× AO3

Process B → C 473 473 0

Process C → A −354 −101 −253

Whole cycle 119 119 0

Give ECF in C → A if their answer to 03.3 OR 100

(J) used for W.

If no sign take as +ve.

If no marks for rows look for any column in which

whole cycle figure is consistent with process

figures and award 1 Max.

– A-LEVEL PHYSICS – –

Penalise incorrect physics. E.g. getting adiabatic

03.5 Links criterion 1 to process B → C 2 1× AO1

and isothermal processes the wrong way round.

AND 1× AO3

Links criterion 2 to process C → A✓

Criterion 1 met

AND

Criterion 2 not met ✓

– A-LEVEL PHYSICS – –

The mark scheme gives some guidance as to what In each area, a partial answer is given for 1 statement and

03.6 6 1× AO

statements are expected to be seen in a 1- or 2-mark a full answer for more than half the statements.

(L1), 3- or 4-mark (L2) and 5- or 6-mark (L3) answer. Area 1 (1st two bullets in question) 2× AO

Guidance provided in section 3.10 of the ‘Mark 1. max theoretical efficiency = (536 − 291)/536 = 0.46 or 46% 3× AO

Scheme Instructions’ document should be used to 2. overall efficiency = 119/473 = 0.25 or 25%

assist in marking this question. 3. identifies that their theoretical efficiency is greater than

their overall efficiency. Do not accept idea that max

Mark Criteria theoretical efficiency is 100%

All three areas (as grouped alongside)

covered in some detail. Area 2 (3rd bullet)

66 marks can be awarded even if there is an 4. AB / adiabatic stroke would have to be very fast to ensure

error and/or parts of one aspect missing, no time for heat transfer

provided points 1 and 3 are included. 5. BC / isothermal stroke would have to be very slow to

ensure constant temperature

All three areas covered, at least two in detail. 6. very difficult to arrange a slow ‘stroke’ and a fast ‘stroke’ in 13

5 Whilst there will be gaps, there should only one engine OR output speed would vary over a cycle

be an occasional error. 7. difficulty in arranging for end of expansion and start of

compression to occur at one point OR idea of need for

Two areas successfully discussed, or one induction/exhaust/pumping strokes

discussed and two others covered 8. engine valves take time/energy to open and close OR

partially. Whilst there will be several gaps, incomplete fuel combustion

there should only be an occasional error.

One area discussed and one discussed Area 3 (4th bullet)

partially, or all three covered partially. There NO because:

3 9. idea that very low net work per cycle OR area of the loop is

are likely to be several errors and omissions

in the discussion. small

10. real efficiency will be much lower than their overall

Only one area discussed, or makes a partial efficiency (and much lower than engines currently

attempt at two areas. available) (due to friction etc.)

11. low speed or cycle s−1 hence low power

None of the three areas covered without −1

1 12. because power = cycle s × work done per cycle

significant error.

– A-LEVEL PHYSICS – –

0 No relevant comments Accept ideas shown by sketches.

Total 15

How to answer it

Thermodynamic Cycles & Heat Engine Analysis

📋 WHAT THIS QUESTION TESTS

This multi-step question assesses core principles of thermodynamics, the ideal gas law, and practical engine design:

Definition of isothermal processes and application of the ideal gas state equation ( pV = nRT ).
Calculation of work done during isobaric compression ( W = pΔV ).
Applying the First Law of Thermodynamics ( Q = ΔU + W ) across closed cycles.
Carnot engine efficiency criteria (isothermal heat absorption and rejection).
Evaluating the theoretical vs practical feasibility of a real engine cycle (friction, finite cycle speeds, valve operation, and net power).

Question 03.1 • 1 Mark

Definition of an Isothermal Change

A-Level Core Vocabulary & Thermodynamic Conditions

✅ Correct Answer

A change (of state) that occurs at constant temperature.

💡 Key Knowledge

Isothermal: Temperature remains strictly constant ( ΔT = 0 ).
For an ideal gas, internal energy depends only on temperature: ΔU = 0 .
Heat transfer must be slow enough to allow equilibrium with the surroundings.

❌ Common Errors

Writing "no change in internal energy" or " ΔU = 0 " alone — the mark scheme explicitly states: Do not allow 'no change in internal energy' OR ΔU = 0.
Confusing isothermal ( ΔT = 0 ) with adiabatic ( Q = 0 ).

Mark Breakdown: [1 mark] for explicitly stating "constant temperature" or "temperature remains constant".

Question 03.2 • 2 Marks

Calculate the Amount of Gas ( n )

Ideal Gas Equation of State: pV = nRT

📐 Step-by-Step Calculation

Identify state values at a known corner:
At point A: p = 1.00 × 10⁵ Pa , V = 12.0 × 10⁻⁴ m³ , minimum T = 291 K .
Rearrange ideal gas law for moles n :
n = pV / RT
Substitute values ( R = 8.31 J mol⁻¹ K⁻¹ ):
n = (1.00 × 10⁵ × 12.0 × 10⁻⁴) / (8.31 × 291)
n = 120 / 2418.21 = 0.04962... mol
State answer to 3 significant figures:
n = 4.97 × 10⁻² mol (or 0.050 mol ).

🧠 Exam Technique: Choosing Coordinates

You can also evaluate at points B or C using the maximum temperature T = 536 K :

At B: p = 8.51 × 10⁵ Pa , V = 2.60 × 10⁻⁴ m³
pV = 221.26 J
n = 221.26 / (8.31 × 536) = 0.04968 mol ≈ 4.97 × 10⁻² mol
At C: p = 1.00 × 10⁵ Pa , V = 22.1 × 10⁻⁴ m³
n = (1.00 × 10⁵ × 22.1 × 10⁻⁴) / (8.31 × 536) = 0.04962 mol

❌ Examiner Trap

Do not write 0.05 with only 1 significant figure without working; the mark scheme accepts 4.97 × 10⁻² , 4.96 × 10⁻² , or 0.050 , but explicitly rejects unrounded 0.05 .

Question 03.3 • 1 Mark

Work Done During Isobaric Compression (C → A)

Constant Pressure Work: W = pΔV

📐 "Show That" Proof

State the formula: W = pΔV
Substitute values from Figure 6:
p = 1.00 × 10⁵ Pa
ΔV = (22.1 - 12.0) × 10⁻⁴ m³ = 10.1 × 10⁻⁴ m³
Calculate magnitude of work:
W = 1.00 × 10⁵ × 10.1 × 10⁻⁴ = 101 J ≈ 100 J

🧠 Exam Technique: "Show That" Rules

For a "show that" question, the examiner needs to see the exact formula and complete substitution. Simply writing 100 J gets zero marks. Both the formula and explicit values must be clearly written. The sign ( -101 J vs +101 J ) is ignored for this specific mark.

Question 03.4 • 3 Marks

Completing the First Law Cycle Table

Applying Q = ΔU + W and Cyclic Constraints

Using the sign convention: Q is heat absorbed by the gas, W is work done by the gas ( Q = ΔU + W ).

Process	Q / J	W / J	ΔU / J
Process A → B (Adiabatic)	0	-253	+253
Process B → C (Isothermal)	473	+473	0
Process C → A (Isobaric)	-354	-101	-253
Whole Cycle	119	119	0

💡 Step-by-Step Logic per Row

A → B (Adiabatic): By definition, no heat transfer occurs, so Q = 0 . Since Q = ΔU + W , ΔU = -W = -(-253) = +253 J .
B → C (Isothermal): Temperature is constant, so ΔU = 0 . Therefore, W = Q = +473 J .
C → A (Isobaric): Compression means work is done on gas: W = -101 J . Using Q = ΔU + W = -253 + (-101) = -354 J .
Whole Cycle: The system returns to state A, so total ΔU = 0 . Net heat equals net work: Q_net = W_net = 119 J .

🧠 Consistency Checks

Sum of columns:
ΣΔU = +253 + 0 + (-253) = 0 (Mandatory for any closed cycle!)
ΣW = -253 + 473 - 101 = 119 J (Matches table!)
ΣQ = 0 + 473 - 354 = 119 J (Matches table!)
1 mark awarded for each fully correct row (up to max 3 marks).

Question 03.5 • 2 Marks

Carnot Efficiency Criteria Deduction

Conditions for Maximum Theoretical (Carnot) Efficiency

✅ Mark Scheme Model Answer

Linking processes:
Criterion 1 applies to process B → C (where heat is transferred to the gas).
Criterion 2 applies to process C → A (where heat is rejected from the gas).
Deduction:
Criterion 1 IS met (process B → C is isothermal at hot source temperature).
Criterion 2 IS NOT met (process C → A is isobaric, so temperature changes as volume drops from 536 K to 291 K, meaning gas is not kept at cold sink temperature).

❌ Common Misconceptions

Confusing adiabatic process A → B with isothermal heat transfer. During A → B, heat transfer is zero.
Failing to recognise that process C → A is non-isothermal: because V decreases at constant p , T drops from 536 K to 291 K, violating criterion 2.

Question 03.6 • 6 Marks

Extended Response: Feasibility of the Proposed Engine

6-Mark Level-of-Response Masterclass

🧠 How to Score 5–6 Marks (Level 3)

A Level 3 response must address all three areas in detail, with points 1 and 3 included in the quantitative analysis:

Area 1: Quantitative Efficiencies (Carnot theoretical vs proposed cycle).
Area 2: Practical Mechanical Problems (competing speed demands, valving, cooling/heating).
Area 3: Conclusion on Usefulness (low net power output, friction losses).

Area 1: Efficiency Calculations

Ideal (Carnot) Efficiency:
η_max = (T_H - T_C) / T_H
η_max = (536 - 291) / 536 = 245 / 536 = 0.457 (46%)
Proposed Cycle Efficiency:
η_cycle = W_net / Q_in = 119 / 473 = 0.252 (25%)
Comparison:
The cycle's theoretical efficiency (25%) is substantially lower than the maximum theoretical Carnot limit (46%).

Area 2: Practical Engineering Barriers

Conflicting stroke speeds:
Adiabatic stroke (A → B) must be extremely rapid to prevent heat transfer ( Q ≈ 0 ).
Isothermal stroke (B → C) must be extremely slow to maintain constant temperature with the heat source.
Alternating between fast and slow strokes in a single reciprocation is mechanically impractical.
Gas exchange & timing:
Requires precisely timed valves and intake/exhaust strokes to reset between expansion and compression.
Heat transfer through cylinder walls is too slow to achieve true isothermal expansion.

Area 3: Overall Conclusion & Usefulness Verdict

The engineer's suggestion would NOT lead to a useful engine because:

Tiny Net Work: The enclosed loop area on the p-V diagram is small, producing only 119 J of net work per cycle.
Extremely Low Power Output: Because stroke B → C must be very slow, the cycle frequency (cycles per second) will be minimal, resulting in negligible power ( Power = work × frequency ).
Friction Losses: Mechanical friction in pistons, bearings, and fluid drag would consume most or all of the modest 119 J output, dropping real-world efficiency close to zero.

Topics

Physics · Optional topics · 3.6 Further mechanics and thermal physics (A-level only) · 3.11 Engineering physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3BC), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.