Free access until 31 December 2026 · Physics and Maths questions and notes, updated continuously · Build a test · Get an email before anything changes

ExaminerPrep › Question bank › IB Physics HL › B.4 Thermodynamics (HL)

IB Physics Higher Level · Topic B.4

B.4 Thermodynamics (HL)

63 free, high-quality past-paper-style questions on Thermodynamics (HL) for IB Physics HL, each with a full mark scheme. Covers: Work done when the pressure is not constant; The second law of thermodynamics; Isothermal and adiabatic processes; and more. For when you have finished the past papers, or want to practise one topic at a time.Read the B.4 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.338 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1B-1A-27Work done when the pressure is not constant Paper 1AEasy1 mark

A fixed mass of ideal gas is compressed from state X to state Y along the straight line shown on the p–V diagram.

The first law of thermodynamics is written Q = ΔU + W. What is the value of W for this change?

012345V / 10⁻³ m³01234p / 10⁵ PaXY↖
A fixed mass of ideal gas is compressed from X to Y along the straight line (drawn to scale).
Show mark scheme
Marking pointMarkNotes
Step 1The magnitude of the work is the area under the p–V line between X and Y. The line is straight, so the area is a trapezium: ½(1.0 + 3.0) × 105 × (4.0 − 1.0) × 10−3 = 600 J.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2In Q = ΔU + W, W is the work done BY the gas. The volume decreases, so work is done on the gas and W is negative.—
Step 3W = −600 J.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis uses the final, highest pressure for the whole change: 3.0 × 105 × 3.0 × 10−3 = 900 J. The pressure was lower for most of the compression.
  • BCorrect: the area under the line is 600 J, and W is negative because the gas is compressed.
  • CThis uses the initial pressure for the whole change (W = pΔV at 1.0 × 105 Pa): it counts only the rectangle below the line and ignores the rise in pressure.
  • DThe area is right, but the sign is not: the gas is compressed, so the work done by the gas is negative.
2B-1A-50The second law of thermodynamics Paper 1AEasy1 mark

Which process is impossible according to the second law of thermodynamics?

Show mark scheme
Marking pointMarkNotes
Step 1Kelvin form of the second law: no cyclic process can take thermal energy from a single reservoir and convert it completely into work.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Clausius form: thermal energy cannot pass spontaneously from a colder to a hotter body; it can if work is done (a refrigerator).—
Step 3A local decrease of entropy is allowed if the surroundings gain at least as much entropy.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AAllowed: the Clausius form forbids only a spontaneous transfer from cold to hot; here work is done on the refrigerator.
  • BCorrect: this breaks the Kelvin form. An engine must reject some thermal energy to a colder reservoir.
  • CAllowed: free expansion is irreversible and increases the entropy of the isolated gas; the second law requires this, it does not forbid it.
  • DAllowed: the entropy of the water decreases, but the thermal energy released to the kitchen increases the entropy of the surroundings by at least as much.
3B-1A-68Isothermal and adiabatic processes Paper 1AEasy1 mark

A fixed mass of a monatomic ideal gas is in state X. The diagram shows four changes from X, labelled P, Q, R and S. One change is isovolumetric, one is isobaric, one is isothermal and one is adiabatic.

For which change is the thermal energy supplied to the gas equal to the work done by the gas?

0.00.51.01.52.02.53.0V / 10⁻³ m³050100150200250300350400p / kPaXPQRS
Four changes of state from X on a pressure–volume diagram (drawn to scale)
Show mark scheme
Marking pointMarkNotes
Step 1First law: Q = ΔU + W, so Q = W requires ΔU = 0, i.e. no change of temperature for an ideal gas: the isothermal change.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2P is vertical (isovolumetric) and Q horizontal (isobaric). Of the two expansion curves, R keeps pV constant (240 × 1.0 = 96 × 2.5) while S is steeper, with pV falling.—
Step 3So R is the isothermal change: ΔU = 0 and Q = W.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AP is isovolumetric: W = 0, so the thermal energy supplied equals ΔU, not the work.
  • BQ is isobaric: the temperature rises (pV increases), so Q = ΔU + W is greater than W.
  • CCorrect: along R the product pV, and so the temperature and internal energy, stay constant.
  • DS is the adiabatic change: Q = 0 while the gas does work, so the work comes from a fall in internal energy. The steeper curve has been mistaken for the isotherm.
4IBP2H-015-01-01Entropy and microstates Paper 1AEasy1 mark

In a simple model of a polymer strand inside a drug-delivery capsule, warming the capsule increases the number of microstates available to the strand by a factor of 8. What is the change in entropy of the strand? (kB is the Boltzmann constant.)

Show mark scheme
Marking pointMarkNotes
Entropy is S = kB ln Ω, so ΔS = kB ln(Ω2/Ω1) = kB ln 8 = 3kB ln 2. Option A treats entropy as proportional to Ω. Option B takes the difference in Ω rather than the logarithm of the ratio. Option D wrongly uses the power 3 inside the logarithm instead of ln 2.✓ 1Answer C

Answer: C

Common error: Entropy depends on the logarithm of the number of microstates, so multiplying Ω by a factor adds a fixed amount to S.

5IBP2H-015-01-02The second law of thermodynamics Paper 1AEasy1 mark

Which of the following processes is forbidden by the second law of thermodynamics?

Show mark scheme
Marking pointMarkNotes
The Kelvin form of the second law states that no cyclic process can convert thermal energy entirely into work. Option B describes exactly that. Option A is allowed because external work drives the transfer. Option C is a single process rather than a cycle, so complete conversion is allowed. Option D is allowed because the surroundings gain more entropy than the water loses.✓ 1Answer B

Answer: B

Exam tip: The ban on 100% conversion of heat to work applies to complete cycles. A single isothermal expansion can do it.

6IBP2H-015-00-01First law of thermodynamics Paper 1AEasy1 mark

An ideal gas in a thermally insulated bicycle-pump barrel is compressed rapidly by the piston. The first law is written as Q = ΔU + W, where W is the work done by the gas. Which row describes the compression?

Show mark scheme
Marking pointMarkNotes
The barrel is insulated and the compression is rapid, so no energy is transferred by heating: Q = 0. The gas volume decreases, so the work done by the gas is negative: W < 0. Then ΔU = Q − W = −W > 0. The internal energy, and therefore the temperature, increases. Option A describes an adiabatic expansion. Option C describes an isothermal compression. Option D wrongly assumes that a temperature rise needs heating.✓ 1Answer B

Answer: B

Common error: A rise in temperature does not mean energy was supplied by heating. In an adiabatic compression, the temperature rises only because work is done on the gas.

7IBP2H-015-00-04Cyclic processes Paper 1AEasy1 mark

The working gas in a heat engine is taken once round a closed clockwise loop on a pressure–volume diagram. Which statement is correct for one complete cycle?

Show mark scheme
Marking pointMarkNotes
Internal energy is a function of state. The gas returns to its initial state, so ΔU = 0 over the cycle. From Q = ΔU + W, the net Q equals the net W, which is the area enclosed by the loop. Only ΔU must be zero. Q and W depend on the path, so options B and D are wrong. Option C contradicts the first law.✓ 1Answer A

Answer: A

Exam tip: Over a complete cycle, only state quantities (U, T, p, V, S of the gas) return to their starting values. Q and W do not.

8B-1A-04First law of thermodynamics Paper 1AMedium1 mark

A fixed mass of a monatomic ideal gas at pressure p and volume V expands at constant pressure to a volume 2V.

What is the thermal energy supplied to the gas?

Show mark scheme
Marking pointMarkNotes
Step 1Work done by the gas at constant pressure: W = pΔV = p(2V − V) = pV.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a monatomic ideal gas U = ³⁄₂nRT = ³⁄₂pV, so ΔU = ³⁄₂Δ(pV) = ³⁄₂(2pV − pV) = ³⁄₂pV.—
Step 3First law: Q = ΔU + W = ³⁄₂pV + pV = 5pV/2.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is only the work done by the gas; the internal energy also rises because the temperature doubles.
  • BThis is only the change in internal energy; the gas also does work pushing back the surroundings.
  • CCorrect: Q = ΔU + W = ³⁄₂pV + pV.
  • DThis uses the final internal energy ³⁄₂p(2V) instead of the change in internal energy.
9B-1A-05Adiabatic processes Paper 1AMedium1 mark

A monatomic ideal gas expands adiabatically until its volume is 8 times the initial volume.

By what factor does the pressure decrease?

Show mark scheme
Marking pointMarkNotes
Step 1For an adiabatic change of a monatomic gas pV5/3 = constant.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2p1/p2 = (V2/V1)5/3 = 85/3.—
Step 381/3 = 2, so 85/3 = 2⁵ = 32.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is Boyle's law, which applies to an isothermal change. Adiabatic expansion also cools the gas, so the pressure falls further.
  • BThis is 84/3, which would need an exponent of 4/3.
  • CCorrect: 85/3 = 2⁵ = 32.
  • DThis is 8², an exponent of 2.
10B-1A-06Entropy and the second law Paper 1AMedium1 mark

A quantity Q of thermal energy passes by conduction from a large reservoir at absolute temperature 2T to a large reservoir at absolute temperature T. The temperatures of the reservoirs do not change.

What is the total change in entropy of the two reservoirs?

Show mark scheme
Marking pointMarkNotes
Step 1Each reservoir stays at constant temperature, so ΔS = ΔQ/T applies to each separately.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Hot reservoir loses Q: ΔShot = −Q/(2T). Cold reservoir gains Q: ΔScold = +Q/T.—
Step 3Total: −Q/(2T) + Q/T = +Q/(2T) — positive, as the second law requires for a spontaneous transfer.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the cold reservoir gains more entropy than the hot one loses because it is at the lower temperature.
  • BThis swaps the signs of the two terms (+Q/(2T) − Q/T): an irreversible transfer cannot decrease the total entropy.
  • CThis adds the magnitudes Q/(2T) + Q/T, forgetting that the hot reservoir loses entropy.
  • DThis is the entropy gained by the cold reservoir alone; the hot reservoir's loss of Q/(2T) has been ignored.
11B-1A-11Entropy and microstates Paper 1AMedium1 mark

A container is divided into two equal halves by an imaginary plane. It holds N molecules of an ideal gas that move freely through the whole container. Every microstate — every possible specification of which half each molecule is in — is equally probable.

What is the probability that, at a given instant, all N molecules are in the left half of the container?

Show mark scheme
Marking pointMarkNotes
Step 1Each molecule can be in either of two halves, independently of the others, so the total number of microstates is 2 × 2 × … = 2N.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Only one of these microstates has every molecule in the left half.—
Step 3All microstates are equally probable, so the probability is 1/2N = (½)N. Even for N = 20 this is about 1 × 10−6; for a real gas it is so small that the gas never returns spontaneously to one half — the macrostate with the most microstates (the highest entropy) is the one observed.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: one favourable microstate out of 2N equally probable microstates.
  • BThis adds the possibilities (2 for each of N molecules gives 2N) instead of multiplying them (2N).
  • CThis treats the N + 1 macrostates (0, 1, 2, …, N molecules in the left half) as equally probable. It is the microstates that are equally probable, and the macrostate "all in the left half" contains only one of them.
  • DThis is the probability for a single molecule; it has not been applied independently to each of the N molecules.
12B-1A-14Isothermal and adiabatic processes Paper 1AMedium1 mark

Two identical samples of a monatomic ideal gas start in the same state. One sample expands isothermally and the other expands adiabatically. Both changes are reversible and both samples end with the same final volume.

Which statements are correct?

I. The final pressure is greater after the adiabatic expansion.
II. More work is done by the gas in the isothermal expansion.
III. The entropy of the gas does not change in the adiabatic expansion.

Show mark scheme
Marking pointMarkNotes
Step 1I: in the adiabatic expansion the gas does work with Q = 0, so its internal energy and temperature fall; p = nRT/V then gives a lower final pressure than the isothermal one (pV5/3 constant against pV constant). I is false.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: the adiabatic curve lies below the isotherm throughout the expansion, so the area under it is smaller: more work is done in the isothermal expansion. II is true.—
Step 3III: in a reversible adiabatic change no thermal energy is transferred, so ΔS = ΔQ/T = 0. III is true.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AIncludes I, which is false: the adiabatic expansion cools the gas, so its final pressure is lower.
  • BIncludes I and omits II: a lower pressure at every volume means less area under the adiabatic curve.
  • CCorrect: II and III are true; I has the comparison the wrong way round.
  • DIncludes I, which is false: for a doubling of volume, for example, the pressure falls by a factor of 25/3 ≈ 3.2 adiabatically but only 2 isothermally.
13B-1A-20Cyclic processes Paper 1AMedium1 mark

A fixed mass of ideal gas is taken once round the cycle X → Y → Z → X shown on the p–V diagram.

Which statements about one complete cycle are correct?

I. The change in internal energy of the gas is zero.
II. The net work done by the gas is −200 J.
III. The net thermal energy transferred to the gas is positive.

01234V / 10⁻³ m³01234p / 10⁵ PaXYZ→↖↓
The cycle X → Y → Z → X on a p–V diagram (drawn to scale).
Show mark scheme
Marking pointMarkNotes
Step 1I: the gas returns to its initial state, so U (a function of state) returns to its initial value: ΔU = 0. True.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: the enclosed area is ½ × (2 × 10−3) × (2 × 105) = 200 J. The cycle is anticlockwise (expansion X → Y at low pressure, compression Y → Z at higher pressure), so the net work done by the gas is −200 J. True.—
Step 3III: Q = ΔU + W = 0 + (−200 J) = −200 J, so there is a net transfer of thermal energy out of the gas. False.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: ΔU = 0 and W = −200 J, so Q = −200 J (net energy leaves the gas as heat).
  • BIncludes III and omits II: over the cycle Q = W, and W = −200 J for this anticlockwise cycle.
  • COmits I: internal energy depends only on the state, which is the same at the start and end of a cycle.
  • DIncludes III, which assumes the gas behaves as an engine; for an anticlockwise cycle work is done on the gas and the same energy leaves as heat.
14B-1A-47First law of thermodynamics Paper 1AMedium1 mark

During one compression stroke of a bicycle pump, 35 J of work is done on the air in the pump and 12 J of thermal energy is transferred from the air to the barrel of the pump.

The first law of thermodynamics is written Q = ΔU + W. Which row gives the values of W and ΔU for the air?

WΔU
Show mark scheme
Marking pointMarkNotes
Step 1In Q = ΔU + W, W is the work done by the gas and Q is the thermal energy transferred to the gas.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Work is done on the air, so W = −35 J; thermal energy leaves the air, so Q = −12 J.—
Step 3ΔU = Q − W = −12 − (−35) = +23 J.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: W = −35 J and ΔU = −12 + 35 = +23 J.
  • BThe sign of W is right but Q is taken as +12 J although energy leaves the air: 12 + 35 = 47 J.
  • CThis treats W as the work done on the gas: ΔU = −12 − 35 = −47 J.
  • DΔU is right (from ΔU = Q + work done on the gas), but W in this form of the law is the work done by the gas, so it is −35 J.
15B-1A-48Adiabatic processes Paper 1AMedium1 mark

0.250 mol of neon, a monatomic ideal gas, is in a thermally insulated cylinder. The gas expands and does 180 J of work on a piston. R = 8.31 J mol−1 K−1.

What is the change in temperature of the gas?

Show mark scheme
Marking pointMarkNotes
Step 1The cylinder is insulated, so Q = 0 (adiabatic) and ΔU = −W = −180 J.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2ΔU = (3/2)nRΔT, so ΔT = −180/(1.5 × 0.250 × 8.31) = −58 K.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis uses ΔU = nRΔT, omitting the factor 3/2: 180/(0.250 × 8.31) = 87 K.
  • BCorrect: ΔT = −2W/(3nR) = −58 K.
  • CThis uses a factor 5/2 instead of 3/2 (as if the gas were heated at constant pressure): 180/(2.5 × 0.250 × 8.31) = 35 K.
  • DThis has the wrong sign: the gas does work with no thermal energy supplied, so its internal energy and temperature fall.
16B-1A-49Entropy and microstates Paper 1AMedium1 mark

A box contains ten gas molecules. Each molecule is equally likely to be found in the left half or the right half of the box, and every arrangement (microstate) specifying which half each molecule is in is equally probable. At first 8 molecules are in the left half and 2 in the right half; later 5 are in each half.

Treating the positions only, what is the change in the entropy S = kB ln Ω of the system of molecules? (kB = 1.38 × 10−23 J K−1)

Show mark scheme
Marking pointMarkNotes
Step 1Number of ways of choosing which 2 of the 10 molecules are in the right half: Ω1 = 10!/(2! 8!) = 45.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Number of ways of having 5 molecules in each half: Ω2 = 10!/(5! 5!) = 252.—
Step 3ΔS = kB ln Ω2 − kB ln Ω1 = kB ln(252/45) = 1.38 × 10−23 × ln 5.6 = +2.4 × 10−23 J K−1.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the ratio, ln(45/252). As the molecules spread out the system moves to the macrostate with more microstates, so its entropy increases.
  • BCorrect: kB ln(252/45) = +2.4 × 10−23 J K−1.
  • CThis takes the logarithm of the difference in microstates: kB ln(252 − 45) = kB ln 207 = 7.4 × 10−23 J K−1.
  • DThis multiplies kB by the difference in microstates without the logarithm: 207kB = 2.9 × 10−21 J K−1.
17B-1A-69Heat engines and the Carnot cycle Paper 1AMedium1 mark

An inventor proposes three engines, I, II and III, each working between a hot reservoir at 227 °C and a cold reservoir at 27 °C. In each cycle each engine takes 500 J of thermal energy from the hot reservoir. The table gives the work done by each engine and the thermal energy it transfers to the cold reservoir in each cycle.

Which engines are impossible?

EngineWork done / JThermal energy transferred to cold reservoir / J
I250250
II150300
III180320
Show mark scheme
Marking pointMarkNotes
Step 1First law: work + thermal energy rejected must equal 500 J. II gives 150 + 300 = 450 J, so II is impossible.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Second law: ηCarnot = 1 − 300/500 = 0.40 (kelvin temperatures). I has η = 250/500 = 0.50 > 0.40, so I is impossible.—
Step 3III: 180 + 320 = 500 J and η = 180/500 = 0.36 ≤ 0.40, so III is possible (its entropy change, −500/500 + 320/300 = +0.07 J K−1, is positive).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis applies only the second law and misses that engine II does not conserve energy.
  • BThis applies only the first law, or uses Celsius temperatures (1 − 27/227 = 0.88), so engine I seems to be within the limit.
  • CCorrect: II breaks the first law and I breaks the second law; III obeys both.
  • DThis calculates the efficiency as work/(energy rejected): 250/250, 150/300 and 180/320 all exceed 0.40. Efficiency is work/(energy input) = work/500 J.
18IBP2H-015-01-03Heat engines and the Carnot cycle Paper 1AMedium1 mark

A Carnot engine in an orbiting power unit works between a hot reservoir at 600 K and a radiator at 300 K. A designer can either raise the hot reservoir temperature by 50 K or lower the radiator temperature by 50 K. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
η = 1 − TC/TH, and the initial efficiency is 0.50. Raising TH to 650 K gives η = 1 − 300/650 ≈ 0.54. Lowering TC to 250 K gives η = 1 − 250/600 ≈ 0.58. Lowering TC reduces the ratio TC/TH more, so B is correct. The temperature difference is 350 K in both cases, yet the efficiencies differ, so C is wrong. The ratios also differ, so D is wrong.✓ 1Answer B

Answer: B

Exam tip: Carnot efficiency depends on the ratio TC/TH, not on the difference. Test each option numerically.

19IBP2H-015-01-04Entropy and the second law Paper 1AMedium1 mark

In a car engine, 1200 J of thermal energy flows from the engine block at 400 K to the coolant at 300 K. Both temperatures stay effectively constant. What is the total entropy change of the block and the coolant?

Show mark scheme
Marking pointMarkNotes
The block loses entropy: ΔS = −1200/400 = −3.0 J K−1. The coolant gains entropy: ΔS = +1200/300 = +4.0 J K−1. The total is +1.0 J K−1, a positive value, as expected for heat flowing irreversibly from hot to cold. Option B has the signs reversed. Option C adds the magnitudes. Option D divides by the temperature difference.✓ 1Answer A

Answer: A

Common error: Find the entropy change of each body separately, with its own sign and temperature. Never divide by the temperature difference.

20IBP2H-015-01-05Isothermal and adiabatic processes Paper 1AMedium1 mark

In an idealised engine, a monatomic ideal gas is compressed adiabatically to half its original volume. What is the ratio of final temperature to initial temperature?

Show mark scheme
Marking pointMarkNotes
For an adiabatic change, pV5/3 is constant. Combining this with pV = nRT gives TV2/3 = constant. So T2/T1 = (V1/V2)2/3 = 22/3. Option B is the pressure ratio, not the temperature ratio. Option C assumes T ∝ 1/V, as if the pressure were constant. Option D assumes that no heat transfer means no temperature change.✓ 1Answer A

Answer: A

Common error: An adiabatic process involves no heat transfer, but the temperature still changes because work is done on or by the gas.

21IBP2H-015-01-06Cyclic processes Paper 1AMedium1 mark

A Stirling engine at the focus of a solar dish takes a fixed mass of ideal gas around a closed clockwise cycle on a pressure–volume diagram. Which statement is correct for one complete cycle?

Show mark scheme
Marking pointMarkNotes
Internal energy is a state function, so ΔU = 0 over a complete cycle. The first law then gives net Q = net W. The net work equals the area enclosed by the loop, which is not zero, so B is wrong. Net heat transfer is not zero, so C is wrong. Entropy is also a state function, so the entropy change of the gas over a cycle is zero. Any entropy increase happens in the surroundings, so D is wrong.✓ 1Answer A

Answer: A

Exam tip: State functions (U, S, T, p, V) return to their starting values after a cycle. Q and W depend on the path and generally do not.

22IBP2H-015-00-02First law of thermodynamics Paper 1AMedium1 mark

A monatomic ideal gas in a balloon is heated slowly at constant pressure. What fraction of the energy supplied by heating goes into increasing the internal energy of the gas?

Show mark scheme
Marking pointMarkNotes
At constant pressure, ΔU = (3/2)nRΔT and W = pΔV = nRΔT. So Q = (5/2)nRΔT, and ΔU/Q = (3/2)/(5/2) = 0.60. Option A, 0.40, is the fraction that becomes work. Option C, 0.67, is the ratio W/ΔU. Option D, 1.0, applies only at constant volume.✓ 1Answer B

Answer: B

Exam tip: For a monatomic ideal gas at constant pressure, split Q in the ratio ΔU : W = 3 : 2.

23IBP2H-015-00-03Isothermal and adiabatic processes Paper 1AMedium1 mark

A sealed sample of argon, a monatomic ideal gas, has pressure p. It expands reversibly and adiabatically until its volume has doubled. Take 22/3 ≈ 1.6. What is the final pressure?

Show mark scheme
Marking pointMarkNotes
For an adiabatic change, pV5/3 is constant. So p2 = p(1/2)5/3 = p/(2 × 22/3) ≈ p/3.2 ≈ 0.31p. Option A inverts the ratio, giving a pressure rise. Option B uses the exponent 2/3, which belongs to the TVγ−1 relation for temperature. Option C is the isothermal result.✓ 1Answer D

Answer: D

Common error: Using Boyle's law for an adiabatic change. Adiabatic pressure falls faster than isothermal pressure because the temperature also falls.

24IBP2H-015-00-05Isothermal and adiabatic processes Paper 1AMedium1 mark

A fixed mass of helium in a cryocooler undergoes several separate processes. In which process is the entropy change of the helium zero?

Show mark scheme
Marking pointMarkNotes
ΔS = ΔQ/T for a reversible change. In a reversible adiabatic process, ΔQ = 0 at every stage, so ΔS = 0. Isothermal compression rejects energy by heating, so ΔS < 0, even though T is constant. Heating at constant pressure gives ΔS > 0. Cooling at constant volume gives ΔS < 0.✓ 1Answer C

Answer: C

Common error: Thinking that constant temperature means constant entropy. It is the absence of heat transfer, in a reversible process, that keeps entropy constant.

25IBP2H-015-00-07First law of thermodynamics Paper 1AMedium1 mark

A sealed, rigid steel container holds 0.50 mol of helium, a monatomic ideal gas. While standing in a laboratory, it warms by 40 K. Take R ≈ 8.3 J K−1 mol−1. What is the energy transferred to the helium by heating?

Show mark scheme
Marking pointMarkNotes
The container is rigid, so W = 0 and Q = ΔU. Then ΔU = (3/2)nRΔT = 1.5 × 0.50 × 8.3 × 40 ≈ 250 J. Option A confuses zero work with zero heating. Option B, nRΔT, is the work that would be done at constant pressure. Option D uses (5/2)nRΔT, which applies at constant pressure.✓ 1Answer C

Answer: C

Exam tip: In a rigid container, all of the energy supplied by heating becomes internal energy: Q = (3/2)nRΔT for a monatomic gas.

26IBP2H-015-00-08Work done when the pressure is not constant Paper 1AMedium1 mark

In a car airbag test rig, gas expands so that its pressure falls linearly from 4.0 × 105 Pa to 1.0 × 105 Pa. Over the same change, its volume increases from 2.0 × 10−3 m3 to 5.0 × 10−3 m3. What is the work done by the gas?

0123456012345V / 10⁻³ m³p / 10⁵ Pa
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The work done is the area under the straight-line p–V graph, which is a trapezium. The mean pressure is 2.5 × 105 Pa, and ΔV = 3.0 × 10−3 m3. So W = 2.5 × 105 × 3.0 × 10−3 = 750 J. Option A uses only the final pressure. Option D uses only the initial pressure. Option B counts only the triangular part of the area and omits the rectangle below it.✓ 1Answer C

Answer: C

Common error: Using W = pΔV with a single pressure when the pressure changes. For a linear change, use the mean pressure.

27B-1A-07Heat engines and the Carnot cycle Paper 1AHard1 mark

A Carnot engine operating between a hot reservoir at absolute temperature Th and a cold reservoir has efficiency η.

The absolute temperature of the hot reservoir is doubled; the cold reservoir is unchanged. What is the new efficiency?

Show mark scheme
Marking pointMarkNotes
Step 1η = 1 − Tc/Th, so the original ratio is Tc/Th = 1 − η.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Doubling Th halves the ratio: Tc/(2Th) = (1 − η)/2.—
Step 3New efficiency = 1 − (1 − η)/2 = (1 + η)/2. (Check: η = 0.30 gives 0.65.)✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis takes the efficiency to be proportional to Th. It is 1 − Tc/Th, which can never exceed 1; 2η would exceed 1 for η > 0.5.
  • BThis is the new value of Tc/Th, the fraction of the input rejected to the cold reservoir, not the efficiency.
  • CThis doubles the cold temperature instead of the hot one: 1 − 2(1 − η) = 2η − 1.
  • DCorrect: 1 − (1 − η)/2 = (1 + η)/2.
28B-1A-25Efficiency of a gas cycle Paper 1AHard1 mark

A monatomic ideal gas is the working substance of an engine that follows the cycle A → B → C → D → A shown. The pressure and volume at A are p0 and V0.

What is the efficiency of the engine?

0.00.51.01.52.02.5V / V₀0.00.51.01.52.02.5p / p₀ABCD↑→↓←
The engine cycle on a p–V diagram (drawn to scale).
Show mark scheme
Marking pointMarkNotes
Step 1Net work done by the gas per cycle = enclosed area = (2p0 − p0)(2V0 − V0) = p0V0.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Thermal energy is supplied in A → B and B → C. A → B (constant volume): Q = ΔU = ³⁄₂V0 Δp = ³⁄₂p0V0. B → C (constant pressure): Q = ΔU + W = ³⁄₂(2p0)(V0) + (2p0)(V0) = 5p0V0.—
Step 3Input = 6.5p0V0, so efficiency = p0V0/6.5p0V0 = 2/13 (≈ 0.15). In C → D and D → A energy leaves the gas.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: useful work p0V0 divided by the thermal energy input 6.5p0V0.
  • BThis counts only the input in B → C (5p0V0) and leaves out the 1.5p0V0 supplied in A → B.
  • CThis uses only the changes in internal energy (1.5p0V0 + 3p0V0) as the input, forgetting the work done by the gas in B → C.
  • DThis is the Carnot efficiency between the highest (4TA, at C) and lowest (TA, at A) temperatures. It is only an upper limit; this cycle is not a Carnot cycle.
29IBP2H-015-01-07Efficiency of a gas cycle Paper 1AHard1 mark

A combined-cycle power station is modelled as two Carnot engines in series. The first engine operates between 800 K and 400 K. All the energy it rejects is absorbed by the second engine, which operates between 400 K and 200 K. What is the overall efficiency of the combination?

Show mark scheme
Marking pointMarkNotes
Suppose the first engine absorbs Q. It does 0.5Q of work and rejects 0.5Q at 400 K. The second engine converts half of this, 0.25Q, into work. The total work is 0.75Q, so η = 0.75. This equals 1 − 200/800, the efficiency of a single Carnot engine between the extreme temperatures. Option A multiplies the two efficiencies. Option B averages them. Option D adds them.✓ 1Answer C

Answer: C

Exam tip: Follow a definite amount of energy through each stage instead of combining efficiencies by guesswork.

30IBP2H-015-01-08Entropy and microstates Paper 1AHard1 mark

A rigid, thermally insulated container is divided into two equal halves. One half holds an ideal gas and the other half is a vacuum. The partition is punctured, and the gas expands to fill the whole container. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Q = 0 because the container is insulated. W = 0 because the gas expands into a vacuum. So ΔU = 0, and the temperature of the ideal gas is unchanged. Each molecule now has twice the volume available, so Ω increases and S = kB ln Ω increases. The process is irreversible. ΔS = Q/T applies only to reversible transfers, so A is wrong. No work is done, so B is wrong. ΔU = 0, so D is wrong.✓ 1Answer C

Answer: C

Common error: Q = 0 does not mean ΔS = 0. For irreversible processes, work out the entropy change from microstates or along a reversible path.

31IBP2H-015-00-06Isothermal and adiabatic processes Paper 1AHard1 mark

Two identical samples, X and Y, of a monatomic ideal gas start at the same pressure, volume and temperature. X expands isothermally to twice its volume. Y expands reversibly and adiabatically to twice its volume. Which compares the work done by the gases, W, and their final pressures, p?

Show mark scheme
Marking pointMarkNotes
In the adiabatic expansion, the gas does work using its own internal energy, so its temperature falls. At every volume beyond the start, the pressure of Y is therefore lower than that of X: p ∝ T/V. The adiabatic curve lies below the isotherm. So the area under it, the work done, is smaller, and the final pressure of Y is lower. Option D assumes that equal ΔV means equal work, which is true only at constant pressure. Option C wrongly assumes that drawing on internal energy makes the gas do more work.✓ 1Answer A

Answer: A

Exam tip: Picture both curves on one p–V diagram from the same start point. The adiabatic curve is steeper, so it gives less area for the same expansion.

32B-1B-18Heating at constant volume and constant pressure Paper 1BMedium7 marks

A sample of argon, a monatomic ideal gas, is enclosed in a vertical cylinder by a piston of mass 2.00 kg and cross-sectional area 5.00 × 10−3 m². Atmospheric pressure is 1.01 × 105 Pa. A heater inside the cylinder supplies thermal energy Q, measured with a joulemeter, and the temperature rise ΔT is recorded. In experiment 1 the piston is clamped (constant volume). In experiment 2 it moves freely without friction (constant pressure); for ΔT = 20 K it rises by 12.6 cm. The graph shows Q against ΔT with the lines of best fit.

ΔT / K510152025
Q (constant volume) / J285278103128
Q (constant pressure) / J4487127170211
051015202530ΔT / K050100150200250Q / Jconstant pressureconstant volume
Q against ΔT with the lines of best fit (drawn to scale)
(a)
(i)

Determine the amount of argon using the constant-volume line.

(2)
(b)
(i)

Determine the gradient of the constant-pressure line and show that the ratio of the two gradients is consistent with the theory for a monatomic ideal gas.

(2)
(c)
(i)

Calculate the work done by the argon in experiment 2 when ΔT = 20 K, using the movement of the piston, and compare it with the value of nRΔT.

(2)
(d)
(i)

Thermal energy is also lost through the walls of the cylinder at a rate that increases with the temperature of the argon. State and explain the effect of this on your answer to (a).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient = (153 − 2.5)/30 = 5.0 J K−1✓ 1Accept 4.9–5.1 J K−1.
Gradient = (3/2)nR, so n = 5.0/(1.5 × 8.31) = 0.40 mol✓ 1Accept 0.39–0.41 mol.
Part (b)(i)
Gradient = 8.3 J K−1✓ 1Accept 8.2–8.5 J K−1.
Ratio = 8.3/5.0 = 1.66, consistent with (5/2)nR ÷ (3/2)nR = 5/3 = 1.67✓ 1Allow ECF from the gradient in (a). Theory must be quoted: Q = ΔU + pΔV.
Part (c)(i)
p = 1.01 × 105 + 2.00 × 9.81/5.00 × 10−3 = 1.05 × 105 Pa; W = pΔV = 1.05 × 105 × 5.00 × 10−3 × 0.126 = 66.1 J✓ 1The weight of the piston must be included (using 1.01 × 105 Pa alone gives 63.6 J: [0] for this mark).
nRΔT = 0.40 × 8.31 × 20 = 66.5 J; the two agree to within about 1 %✓ 1Allow ECF from (a).
Part (d)(i)
The measured Q includes the loss, which is larger for larger ΔT, so the gradient is too large and n is overestimated✓ 1Allow ECF from (a). "Too large" must be linked to the gradient, not to the intercept.

Answers: (a)(i) 0.40 mol  ·  (b)(i) 8.3 J K−1; ratio 1.66  ·  (c)(i) 66.1 J (66.5 J predicted) (the remaining parts are explanations — see the table above)

33B-1B-02Adiabatic processes Paper 1BHard7 marks

A light piston of mass m = 16.5 g fits closely in a vertical glass tube of internal diameter 16.0 mm on top of a flask of volume V filled with argon, a monatomic gas. When displaced, the piston performs simple harmonic oscillations. The compressions are rapid, so the gas is treated as adiabatic, and the period T is then

T = 2π √(mV / (γpA²))

where p = 1.02 × 105 Pa is the gas pressure and A the cross-sectional area of the tube. The student times 20 oscillations with a hand-held stopwatch for flasks of different volume. The graph shows T² against V with the line of best fit.

V / dm³2.004.006.008.0010.00
time for 20 oscillations / s8.812.315.017.519.4
T² / s²0.1940.5620.7660.941
0246810V / dm³0.00.20.40.60.81.0T² / s²
T² against V with the line of best fit (drawn to scale)
(a)
(i)

Calculate the missing value of T² in the table.

(1)
(ii)

The uncertainty in each time for 20 oscillations is ±0.2 s. Determine the absolute uncertainty in your answer to (a)(i).

(1)
(b)
(i)

Determine the gradient of the line. Give a unit for your answer.

(2)
(c)
(i)

Determine γ.

(2)
(ii)

The uncertainty in the gradient is ±4 %. Deduce whether the result is consistent with the value of γ for a monatomic ideal gas.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
T = 12.3/20 = 0.615 s, T² = 0.378 s²✓ 1Accept 0.38 s².
Part (a)(ii)
ΔT = 0.2/20 = 0.01 s; fractional uncertainty in T² = 2 × 0.01/0.615 = 3.3 %, so Δ(T²) = ±0.01 s²✓ 1Absolute uncertainty to 1 s.f. Allow ECF from (a)(i). Award [0] for ±0.2 s² or for not doubling (±0.006 s²).
Part (b)(i)
Gradient from two well-separated points on the line, e.g. 0.946/10.0 = 0.095✓ 1Accept 0.092–0.096.
Unit s² dm−3 (= 95 s² m−3)✓ 1Unit mark is independent of the value.
Part (c)(i)
Gradient = 4π²m/(γpA²), so γ = 4π² × 0.0165/(95 × 1.02 × 105 × (2.01 × 10−4)²)✓ 1Requires A = π(8.0 × 10−3)² = 2.01 × 10−4 m² and the gradient in s² m−3.
γ = 1.67✓ 1Accept 1.65–1.72. Allow ECF from (b). A power-of-ten error in the gradient unit loses this mark.
Part (c)(ii)
γ = 1.67 ± 0.07, a range that includes 5/3 = 1.67 (and excludes 1.40 for a diatomic gas), so it is consistent✓ 1Allow ECF from (c)(i). The conclusion must follow from the candidate's own range.

Answers: (a)(i) 0.378 s²  ·  (a)(ii) ±0.01 s²  ·  (b)(i) 0.095 s² dm−3  ·  (c)(i) 1.67 (the remaining parts are explanations — see the table above)

34B-1B-04Heat engines and the Carnot cycle Paper 1BHard7 marks

A small model heat engine runs between a hot reservoir kept at Th = 368 K and a cold reservoir whose temperature Tc is set with water baths. The engine winds a string onto a drum and raises a load at constant speed. For each value of Tc the student measures the thermal energy taken from the hot reservoir with an electrical heater and joulemeter, and the work done on the load. The graph shows the efficiency η of the engine against Tc with the line of best fit, extended to η = 0.

270280290300310320330340350360370Tc / K0.000.010.020.030.040.050.060.070.08efficiency η
Efficiency η against Tc; the line of best fit is extended as a dashed line (drawn to scale)
(a)
(i)

At Tc = 288 K the engine raises a load of mass 2.00 kg through a height of 1.69 m while 600 J of thermal energy enters it from the hot reservoir. Calculate η.

(2)
(b)
(i)

Compare your answer to (a) with the maximum possible efficiency at these reservoir temperatures.

(2)
(c)
(i)

Use the graph to determine the cold-reservoir temperature at which the engine would just stop doing useful work.

(1)
(ii)

The Carnot model predicts zero efficiency only when Tc = Th. Explain why the value in (c)(i) is lower than 368 K.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Useful work = mgh = 2.00 × 9.81 × 1.69 = 33.2 J✓ 1
η = 33.2/600 = 0.055✓ 1Accept 0.055 or 5.5 %.
Part (b)(i)
ηCarnot = 1 − 288/368 = 0.22✓ 1
The engine achieves only about 25 % of the Carnot limit (0.055/0.22 = 0.25)✓ 1Allow ECF from (a). A comparison (ratio, or a statement of how many times smaller) is required.
Part (c)(i)
Extrapolated line meets η = 0 at Tc ≈ 356 K✓ 1Accept 350–362 K.
Part (c)(ii)
Part of the work done by the gas each cycle is used against friction in the engine (and in the drum), so the useful work falls to zero while the ideal work is still positive✓ 1Allow ECF from (c)(i).
The ideal work per cycle falls as Tc approaches Th, so it equals the frictional losses at a Tc below Th (356 K < 368 K)✓ 1Answers that say only "energy is lost" or "the engine is not ideal" score [0].

Answers: (a)(i) 0.055  ·  (b)(i) ηCarnot = 0.22  ·  (c)(i) ≈ 356 K (the remaining parts are explanations — see the table above)

35B-1B-19Entropy change from experimental data Paper 1BHard7 marks

Two identical funnels are filled with crushed ice at 0 °C and stand in a laboratory whose air is at 20 °C. Funnel H contains an immersed electrical heater of power 60.0 W; funnel C has no heater and acts as a control. The mass m of melt water collected from each funnel is recorded at times t after the heater is switched on. While ice remains, the ice–water mixture in each funnel stays at 273 K, and the laboratory air may be treated as a reservoir at a constant 293 K.

t / s0120240360480600
m (funnel H) / g0.024.247.672.195.7120.0
m (funnel C) / g0.02.54.77.29.711.9
0100200300400500600020406080100120t / sm / gfunnel H (heater)funnel C (control)
Data from the table
(a)
(i)

Test the hypothesis that the ice in funnel H melts at a constant rate. Use at least three values from the table.

(2)
(b)
(i)

Determine the specific latent heat of fusion of ice from the data.

(2)
(c)
(i)

For funnel C during the 600 s, determine the entropy change of the ice–water mixture and the total entropy change of the mixture and the laboratory air.

(2)
(d)
(i)

The entropy of the laboratory air decreases. Explain, using your answer to (c), why this is consistent with the second law of thermodynamics.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
m/t = 0.202, 0.200, 0.200 g s−1 (e.g. at 120 s, 360 s and 600 s)✓ 1A calculation from two values only scores this mark only.
The values agree to within about 1 %, so the rate is constant (and the line passes through the origin)✓ 1A conclusion is required.
Part (b)(i)
Rates from the whole 600 s: H = 0.200 g s−1, C = 0.0198 g s−1; heater alone = 0.180 g s−1✓ 1Allow ECF from (a). Using the first interval only is capped at [1 max]; the control must be subtracted.
L = 60.0/1.80 × 10−4 = 3.33 × 105 J kg−1✓ 1Accept 3.30–3.36 × 105 J kg−1.
Part (c)(i)
Q = 11.9 × 10−3 × 3.33 × 105 = 3.96 × 103 J; ΔSice = +3.96 × 103/273 = +14.5 J K−1✓ 1Allow ECF from (b). ΔS = ΔQ/T applies because each body stays at constant temperature.
ΔSair = −3.96 × 103/293 = −13.5 J K−1; total = +0.99 J K−1✓ 1Signs required. Accept +0.9 to +1.1 J K−1.
Part (d)(i)
The air is not an isolated system; its decrease (−13.5 J K−1) is more than compensated by the increase of the ice–water mixture, so the total entropy increases (+0.99 J K−1)✓ 1Allow ECF from (c).

Answers: (b)(i) 3.33 × 105 J kg−1  ·  (c)(i) +14.5 J K−1; total +0.99 J K−1 (the remaining parts are explanations — see the table above)

36IBP2H-015-01-09Heat engines and the Carnot cycle Paper 2Easy5 marks

A geothermal binary-cycle power plant takes thermal energy from underground brine at 180 °C. It rejects energy to air-cooled condensers at 25 °C. The plant delivers a useful output power of 2.0 MW with an actual efficiency of 0.21.

(a)

Calculate the maximum theoretical efficiency of an engine operating between these temperatures.

(2)
(b)

State one reason why the actual efficiency is lower than the maximum theoretical efficiency.

(1)
(c)

Calculate the rate at which energy is rejected to the condensers.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Temperatures converted to kelvin: TH = 453 K and TC = 298 K .✓ 1
ηmax = 1 − 298/453 = 0.34✓ 1
Part (b)
Any one of: real processes are irreversible, whereas the Carnot limit assumes a reversible engine; friction or turbulence in moving parts or the working fluid; heat transfer across finite temperature differences, which is irreversible; energy losses from pipes to the surroundings✓ 1
Part (c)
Input power = 2.0/0.21 = 9.5 MW .✓ 1
Rejected power = 9.5 − 2.0 = 7.5 MW✓ 1

Common error: Using Celsius temperatures in 1 − TC/TH gives a badly wrong efficiency. Always convert to kelvin.

37IBP2H-015-00-09First law of thermodynamics Paper 2Easy5 marks

An office chair contains a sealed gas spring. When a person sits down, the gas is compressed. During this compression, 45 J of work is done on the gas, and 12 J is transferred from the gas to the surroundings by heating. Assume the gas behaves as an ideal gas.

(a)

State the first law of thermodynamics, defining each symbol you use.

(2)
(b)

Calculate the change in internal energy of the gas.

(2)
(c)

Deduce whether the temperature of the gas increases or decreases.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Q = ΔU + W✓ 1
where Q is the energy transferred to the gas by heating, ΔU is the change in internal energy of the gas and W is the work done by the gas✓ 1
(Accept ΔU = Q + W if W is defined as the work done on the gas.)—
Part (b)
Q = −12 J and W = −45 J, with signs consistent with the stated convention✓ 1
ΔU = Q − W = −12 − (−45) = +33 J✓ 1
Part (c)
For an ideal gas, the internal energy is the total random kinetic energy of the molecules. ΔU is positive, so the mean random kinetic energy of the molecules increases and the temperature increases (ECF from (b))✓ 1

Common error: Getting the signs wrong. Work done ON the gas makes W (work done by the gas) negative, and energy lost by heating makes Q negative.

38B-2-10Work from a cyclic p–V diagram Paper 2Medium11 marks

The graph is the indicator diagram (pressure p against volume V) of the gas in one cylinder of a car engine during one cycle. The cycle is traversed clockwise.

0.00.10.20.30.40.5V / 10⁻³ m³010203040506070p / 10⁵ Pa
Indicator diagram of one cylinder (drawn to scale)
(a)
(i)

Estimate the net work done by the gas in one cycle.

(2)
(b)
(i)

Explain why the net work done by the gas in one cycle is equal to the area enclosed by the loop.

(2)
(c)
(i)

The engine has four identical cylinders and its crankshaft turns at 3000 revolutions per minute. Each cylinder completes one cycle every two revolutions. Determine the mean power output of the engine.

(2)
(d)
(i)

The car travels at a constant speed of 27 m s−1 on a level road. The engine delivers the mean power found in (c), and 80 % of this power is transferred to the driving wheels. Determine the total resistive force on the car.

(2)
(e)
(i)

The fuel burned in one cylinder in each cycle releases 820 J. Determine the efficiency of the engine.

(1)
(f)
(i)

Calculate the efficiency of a Carnot engine operating between reservoirs at 2300 K, the peak temperature of the gas, and 300 K. Suggest one reason why the efficiency in (e) is much smaller.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Area enclosed found by counting squares: about 50 small squares, each worth 0.025 × 10−3 m³ × 2.5 × 105 Pa = 6.25 J✓ 1Any valid method (e.g. triangles/trapezia); the value of one square must be correct.
Net work ≈ 3.2 × 102 J✓ 1Accept 280–350 J.
Part (b)(i)
During the expansion the gas does work equal to the area under the upper (expansion) curve; during the compression work equal to the area under the lower curve is done on the gas✓ 1
The net work done by the gas is the difference between these areas, i.e. the area enclosed (positive because the cycle is clockwise)✓ 1
Part (c)(i)
Cycles per second for each cylinder = 3000/60/2 = 25 s−1✓ 1Award [1 max] for 6.3 × 104 W (one cycle per revolution).
P = 4 × 25 × 317 = 3.2 × 104 W✓ 1Allow ECF from (a).
Part (d)(i)
Power at the wheels = 0.80 × 3.17 × 104 = 2.54 × 104 W; at constant speed the driving force equals the total resistive force✓ 1Allow ECF from (c).
F = P/v = 2.54 × 104/27 = 9.4 × 102 N✓ 1Accept 8.3–10.4 × 102 N. Award [1 max] for 1.2 × 103 N (the 80 % ignored).
Part (e)(i)
η = 317/820 = 0.39✓ 1Allow ECF from (a).
Part (f)(i)
ηCarnot = 1 − 300/2300 = 0.87✓ 1
Any one: the thermal energy is not all supplied at the highest temperature (the gas is heated through a range of temperatures); thermal energy is lost through the cylinder walls to the cooling system; the exhaust gas leaves while still hot; the real processes are rapid and irreversible (friction, turbulence)✓ 1Accept any valid reason. "Energy is wasted" alone is not enough.

Answers: (a)(i) 3.2 × 102 J  ·  (c)(i) 3.2 × 104 W  ·  (d)(i) 9.4 × 102 N  ·  (e)(i) 0.39  ·  (f)(i) 0.87 (the remaining parts are explanations — see the table above)

39B-2-27Internal energy and irreversible processes Paper 2Medium10 marks

A rigid, thermally insulated container holds 0.500 mol of argon, a monatomic ideal gas, at 290 K. The volume of the container is 1.20 × 10−2 m³. A paddle wheel inside the container is turned by a cord attached to a 2.0 kg mass, which falls at a small constant speed through a height of 0.90 m. The mass is wound back up and the fall is repeated until it has fallen 12 times. Neglect friction in the pulley and the heat capacity of the container and paddle.

argon, 0.500 molinsulated, 1.20×10⁻² m³, 290 K2.0 kg0.90 mpaddle wheel
Diagram not to scale
(a)
(i)

Show that the energy transferred to the argon by the paddle wheel is about 210 J.

(1)
(b)
(i)

State the values of Q and W for the argon in the first law of thermodynamics, Q = ΔU + W, and hence deduce ΔU.

(2)
(c)
(i)

Determine the rise in the temperature of the argon.

(1)
(d)
(i)

Show that the rise in the pressure of the argon is given by Δp = 2ΔU/3V.

(2)
(ii)

Calculate the rise in pressure.

(1)
(e)
(i)

Explain, with reference to the second law of thermodynamics, why the argon cannot cool back to 290 K by lifting the mass to its original height.

(1)
(ii)

The insulation is removed and the argon cools to 290 K in contact with surroundings that stay at 290 K. Determine the entropy change of the surroundings, and explain why the total entropy increases although the entropy of the argon decreases.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The mass falls at constant speed, so all the gravitational potential energy it loses is transferred by the paddle: W = 12mgh = 12 × 2.0 × 9.81 × 0.90 = 212 J✓ 1An answer to at least 3 s.f. is required.
Part (b)(i)
Q = 0 because the container is insulated✓ 1
W = −212 J, since the work is done on the gas, not by it; so ΔU = Q − W = +212 J✓ 1Allow ECF from (a). The negative sign of W is required.
Part (c)(i)
ΔU = (3/2)nRΔT, so ΔT = 212/(1.5 × 0.500 × 8.31); ΔT = 34.0 K✓ 1Allow ECF from (b). Accept 33–35 K. 20.4 K (factor 5/2 used) scores 0.
Part (d)(i)
U = (3/2)nRT = (3/2)pV, using pV = nRT✓ 1
V is constant, so ΔU = (3/2)VΔp, giving Δp = 2ΔU/3V✓ 1The constancy of V must be stated or used.
Part (d)(ii)
Δp = 2 × 212/(3 × 1.20 × 10−2) = 1.18 × 104 Pa✓ 1Allow ECF from (a). Or (nR/V)ΔT with ΔT from (c).
Part (e)(i)
That would convert thermal energy taken from the gas entirely into work (gravitational potential energy) with no other change, which the Kelvin form of the second law forbids; equivalently the entropy of the isolated system would decrease✓ 1"Energy cannot be destroyed" scores 0: the first law allows the reverse process.
Part (e)(ii)
ΔSsurroundings = +212/290 = +0.73 J K−1✓ 1Allow ECF from (a). The positive sign is required.
The argon gives out the same energy while at temperatures above 290 K (between 324 K and 290 K), so its entropy falls by less than 212/290; the total change is positive✓ 1The comparison of temperatures is required; "entropy always increases" alone scores 0.

Answers: (a)(i) 212 J  ·  (b)(i) ΔU = +212 J  ·  (c)(i) 34.0 K  ·  (d)(ii) 1.18 × 104 Pa  ·  (e)(ii) +0.73 J K−1 (the remaining parts are explanations — see the table above)

40IBP2H-015-01-10Entropy change from experimental data Paper 2Medium7 marks

A 350 W immersion heater warms 0.50 kg of water in a laboratory sterilising bath. Assume all the energy from the heater goes into the water. The specific heat capacity of water is 4200 J kg−1 K−1. The table shows the temperature of the water at intervals.

time / s: 0, 60, 120, 180, 240
temperature / °C: 20.0, 30.0, 40.0, 50.0, 60.0

(a)

Outline why the entropy change of the water over the full 240 s cannot be found by dividing the total energy supplied by a single temperature.

(1)
(b)

Show that the energy supplied in each 60 s interval is consistent with the temperature rise recorded in the table.

(2)
(c)

Calculate the entropy change of the water during the first 60 s, using the mean temperature of the interval.

(2)
(d)

Determine the total entropy change of the water during the 240 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The relation ΔS = ΔQ/T applies at constant temperature. The temperature of the water changes continuously, so the heating must be split into small intervals (or integrated)✓ 1
Part (b)
Energy supplied = 350 × 60 = 2.1×104 J .✓ 1
Energy needed = mcΔT = 0.50 × 4200 × 10 = 2.1×104 J, so the two agree✓ 1
Part (c)
Mean temperature = 25.0 °C = 298 K .✓ 1
ΔS = 2.1×104/298 = 70 J K−1✓ 1
Part (d)
Sum ΔQ/T over the four intervals using mean temperatures: 2.1×104 × (1/298 + 1/308 + 1/318 + 1/328). Equally valid: mc ln(T2/T1) = 2100 ln(333/293) .✓ 1
ΔS ≈ 2.7×102 J K−1 (accept 268–270 J K−1)✓ 1

Exam tip: When the temperature changes, split the heating into intervals and use the mean kelvin temperature of each. Then add the separate ΔQ/T values.

41IBP2H-015-01-11Entropy and microstates Paper 2Medium6 marks

A rubber band is modelled as a chain of 10 identical links. Each link points either left or right along the band. The length of the band is proportional to (number of links pointing right − number of links pointing left). The band has its maximum length when all 10 links point right. Its length is zero when 5 links point right and 5 point left. It can be shown that the zero-length macrostate has 252 microstates.
(kB = 1.38×10−23 J K−1)

(a)

State the entropy of the band in the maximum-length macrostate.

(1)
(b)

Calculate the entropy of the band in the zero-length macrostate.

(2)
(c)

A stretched band is released and allowed to contract. Explain, with reference to microstates and the second law of thermodynamics, why it contracts.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Only one arrangement is possible (Ω = 1), so S = kB ln 1 = 0✓ 1
Part (b)
Uses S = kB ln Ω with Ω = 252✓ 1
S = 1.38×10−23 × ln 252 = 1.38×10−23 × 5.53 = 7.6×10−23 J K−1 (accept 7.63×10−23 J K−1)✓ 1
Part (c)
Shorter macrostates correspond to far more microstates than the fully stretched macrostate✓ 1
Random thermal motion changes link directions, so the band is overwhelmingly likely to end up in a macrostate with larger Ω, so its entropy increases✓ 1
The second law states that the total entropy of the band and its surroundings tends to increase in a spontaneous process, so contraction is the spontaneous direction✓ 1

Exam tip: Connect three ideas in a chain: more microstates, higher probability, greater entropy. Name the second law explicitly and apply it to the total entropy of system plus surroundings.

42IBP2H-015-01-14Entropy and the Clausius form of the second law Paper 2Medium8 marks

During resurfacing, an ice rink's refrigeration plant freezes 1500 kg of water at 0 °C. The plant rejects energy to the outside air at 288 K. The specific latent heat of fusion of water is 3.34×105 J kg−1.

(a)

Calculate the energy that must be removed from the water to freeze it.

(1)
(b)

Calculate the entropy change of the water as it freezes.

(2)
(c)

Explain why the entropy of the outside air must increase by at least the magnitude of the entropy change of the water.

(2)
(d)

The work needed is least when the plant operates reversibly. Determine the minimum work that the plant must do to freeze the water.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Q = mL = 1500 × 3.34×105 = 5.01×108 J✓ 1
Part (b)
ΔS = −Q/T, with T = 273 K and a negative sign because energy is removed .✓ 1
ΔS = −5.01×108/273 = −1.84×106 J K−1✓ 1
Part (c)
By the second law, the total entropy of the water, the plant and the air cannot decrease .✓ 1
The plant works in a cycle, so its own entropy change is zero. The air's entropy gain must therefore be at least the water's entropy loss, with equality only for a reversible process✓ 1
Part (d)
Minimum energy rejected to the air: Qout = 288 × 1.835×106 = 5.29×108 J (allow ECF from the entropy change of the water) .✓ 1
W = Qout − Qin = 5.29×108 − 5.01×108. Accept the equivalent W = Qin(288/273 − 1) .✓ 1
Wmin = 2.8×107 J✓ 1

Common error: Carry unrounded values into the final part. The work is a small difference between two large energies. Rounding the entropy change to 1.8×10⁶ J K⁻¹ first gives about 1.7×10⁷ J instead of 2.8×10⁷ J.

43IBP2H-015-01-15Heat engines and the Carnot cycle Paper 2Medium8 marks

A coastal power station delivers 500 MW of electrical power with an overall efficiency of 0.38. Steam enters its turbines at 820 K, and waste energy is rejected to sea water at 290 K. Treat both as reservoirs at constant temperature.

(a)

Calculate the Carnot efficiency for these reservoir temperatures.

(1)
(b)

Calculate the rate at which energy is rejected to the sea water.

(2)
(c)

Determine the net rate of entropy change of the two reservoirs.

(3)
(d)

Explain how your answer for the net rate of entropy change is consistent with the second law of thermodynamics, and state the net rate for an ideal Carnot engine working between the same reservoirs.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
ηC = 1 − 290/820 = 0.65✓ 1
Part (b)
Input power = 500/0.38 = 1316 MW .✓ 1
Rejected power = 1316 − 500 = 816 MW✓ 1
Part (c)
Hot reservoir: −1316×106/820 = −1.60×106 W K−1 (allow error carried forward from the previous part) .✓ 1
Sea water: +816×106/290 = +2.81×106 W K−1 .✓ 1
Net rate = +1.21×106 W K−1 (accept 1.2×106)✓ 1
Part (d)
The net entropy change is positive, so the entropy of the universe increases, as the second law requires for a real, irreversible engine (efficiency 0.38 < 0.65) .✓ 1
For a reversible Carnot engine, the net rate would be zero✓ 1

Exam tip: Entropy production is a useful measure of how far a real engine falls short of the Carnot ideal. It is zero only for reversible operation.

44IBP2H-015-00-10First law of thermodynamics Paper 2Medium8 marks

The bellows of a hospital ventilator compress a fixed mass of air slowly. The air can be treated as an ideal gas. The table gives measurements taken during the compression.

V / 10−4 m3: 5.0, 4.0, 3.0, 2.0
p / kPa: 100, 125, 167, 250

0123456050100150200250300V / 10⁻⁴ m³p / kPa
Data from the table
(a)

Show that the data are consistent with an isothermal compression.

(2)
(b)

Estimate the work done on the air as its volume changes from 5.0 × 10−4 m3 to 2.0 × 10−4 m3, using the trapezium method on the tabulated values.

(3)
(c)

Explain whether your answer to (b) is an overestimate or an underestimate.

(1)
(d)

Determine the energy transferred by heating during the compression, stating its direction.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
pV is calculated for each row: 50 J, 50 J, 50.1 J and 50 J✓ 1
pV is constant within rounding, and for a fixed amount of ideal gas pV ∝ T, so T is constant✓ 1
Part (b)
Each strip's work = mean pressure × 1.0 × 10−4 m3✓ 1
11.25 J + 14.6 J + 20.85 J✓ 1
Total ≈ 47 J of work done on the air✓ 1
Part (c)
The p–V curve is concave upward and lies below the straight chords, so the trapezia include extra area: it is an overestimate✓ 1
Part (d)
For an isothermal change of an ideal gas, ΔU = 0✓ 1
Using Q = ΔU + W, where W is the work done by the gas (= −47 J), gives Q = −47 J. Equivalently, ΔU = Q + Won with Won = +47 J gives the same result. About 47 J is transferred from the air to the surroundings (ECF from (b))✓ 1

Exam tip: When the pressure changes, work is the area under the p–V curve. Strips or trapezia give a good estimate from tabulated data. State whether W means work done on or by the gas before you apply the first law.

45IBP2H-015-00-11Adiabatic processes Paper 2Medium8 marks

In a hypersonic wind-tunnel facility, a free piston rapidly compresses helium, a monatomic ideal gas. The compression can be treated as adiabatic. Initially, the helium has pressure 1.0 × 105 Pa, volume 2.4 × 10−3 m3 and temperature 300 K. It is compressed to a volume of 3.0 × 10−4 m3.

(a)

Outline why the compression can be treated as adiabatic.

(1)
(b)

Show that the final temperature of the helium is about 1200 K.

(2)
(c)

Calculate the final pressure of the helium.

(2)
(d)

Determine the work done on the helium during the compression.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The compression is so rapid that there is negligible time for energy to be transferred by heating between the gas and its surroundings✓ 1
Part (b)
TV2/3 is constant (or pV5/3 is constant combined with pV = nRT)✓ 1
T = 300 × 82/3 = 300 × 4 = 1200 K✓ 1
Part (c)
p = 1.0 × 105 × 85/3, or p = p1(V1/V2)(T2/T1) = 1.0 × 105 × 8 × 4✓ 1
p = 3.2 × 106 Pa✓ 1
Part (d)
Q = 0, so the work done on the gas = ΔU✓ 1
ΔU = (3/2)nRΔT = (3/2)(p2V2 − p1V1) = 1.5 × (960 − 240)✓ 1
= 1.1 × 103 J (1080 J) (ECF from (c))✓ 1

Exam tip: For a monatomic gas, (3/2)nRΔT can be written as (3/2)Δ(pV), which avoids calculating n.

46IBP2H-015-00-12Heating at constant volume and constant pressure Paper 2Medium8 marks

Engineers testing the thermal control system of a satellite use a cylinder containing 0.80 mol of argon, a monatomic ideal gas. A frictionless piston keeps the gas at constant pressure when required. Assume that all the energy from any heater used is transferred to the gas.

(a)

Show that the energy needed to raise the temperature of the argon by 1 K at constant volume is about 10 J.

(2)
(b)

Explain why more energy is needed to raise the temperature by 1 K at constant pressure.

(2)
(c)

A 12 W heater operates for 30 s with the argon at constant pressure. Determine the rise in temperature of the argon.

(3)
(d)

Calculate the work done by the argon during this heating.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
At constant volume no work is done, so Q = ΔU = (3/2)nRΔT✓ 1
= 1.5 × 0.80 × 8.31 × 1 = 9.97 J ≈ 10 J✓ 1
Part (b)
At constant pressure the gas expands, so it does work on the surroundings✓ 1
The energy supplied must provide this work as well as the same increase in internal energy, since ΔU depends only on ΔT✓ 1
Part (c)
Q = 12 × 30 = 360 J✓ 1
Q = (5/2)nRΔT✓ 1
ΔT = 360/(2.5 × 0.80 × 8.31) = 21.7 K ≈ 22 K✓ 1
Part (d)
W = pΔV = nRΔT = 0.80 × 8.31 × 21.7 ≈ 144 J, or W = (2/5) × 360 J = 144 J. Accept 144–146 J, including 146 J from using 22 K. Allow error carried forward from (c).✓ 1

Common error: Using (3/2)nRΔT for the energy supplied at constant pressure. That expression gives only ΔU, so you must add the work pΔV = nRΔT.

47B-2-02Thermodynamic cycles Paper 2Hard13 marks

In a closed-cycle power unit for a spacecraft, a fixed amount of helium, a monatomic ideal gas, is taken round the cycle ABCDA shown on the graph.

A → B: adiabatic compression; B → C: heating at a constant pressure of 8.0 × 105 Pa; C → D: adiabatic expansion; D → A: cooling at a constant pressure of 2.0 × 105 Pa. The temperature at A is 300 K and the temperature at C is 1100 K.

051015202530V / 10⁻³ m³0123456789p / 10⁵ PaABCD
The cycle ABCDA on a pressure–volume diagram (drawn to scale)
(a)
(i)

Show, using the graph, that the amount of helium is about 1.0 mol.

(2)
(b)
(i)

Show that, for an adiabatic change of a monatomic ideal gas, T ∝ p2/5.

(2)
(c)
(i)

Determine the temperature at B and the temperature at D.

(2)
(d)
(i)

Show that the thermal energy supplied to the gas from B to C is given by Q = (5/2)nRΔT, and determine its value.

(2)
(e)
(i)

Determine the efficiency of the cycle.

(2)
(f)
(i)

The spacecraft needs 150 kW of mechanical power from the unit. The thermal energy is supplied by the fission of uranium-235 in a small reactor; each fission releases 200 MeV. Determine the mass of uranium-235 that undergoes fission in one year.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
VA read from the graph = 12.5 × 10−3 m³✓ 1Accept 12.0–13.0 × 10−3 m³.
n = pV/RT = 2.0 × 105 × 12.5 × 10−3/(8.31 × 300) = 1.00 mol✓ 1Full substitution or an answer to at least 2 s.f. is required (0.96–1.04 mol).
Part (b)(i)
Substituting V = nRT/p into pV5/3 = constant gives p(nRT/p)5/3 = constant, i.e. p−2/3T5/3 = constant✓ 1
so T5/3 ∝ p2/3 and T ∝ p2/5✓ 1Any valid algebraic route.
Part (c)(i)
TB = 300 × (8.0/2.0)2/5 = 522 K✓ 1Allow ECF from (b). Accept 520–525 K.
TD = 1100 × (2.0/8.0)2/5 = 632 K✓ 1Accept 630–635 K. ALT: either temperature from pV = nRT with volumes read from the graph.
Part (d)(i)
Q = ΔU + W = (3/2)nRΔT + pΔV, and pΔV = nRΔT at constant pressure, so Q = (5/2)nRΔT✓ 1
Q = 2.5 × 1.0 × 8.31 × (1100 − 522) = 1.20 × 104 J✓ 1Allow ECF from (a) and (c).
Part (e)(i)
Thermal energy removed from D to A = 2.5 × 1.0 × 8.31 × (632 − 300) = 6.9 × 103 J✓ 1No thermal energy is transferred in the adiabatic stages. Allow ECF from (c).
η = 1 − 6.9 × 103/1.20 × 104 = 0.43✓ 1Accept 0.42–0.44. Allow ECF from (d). Using net work/input energy is equivalent.
Part (f)(i)
Thermal power = 150 × 103/0.43 = 3.5 × 105 W✓ 1Allow ECF from (e).
Fissions per second = 3.5 × 105/(200 × 106 × 1.60 × 10−19) = 1.1 × 1016 s−1✓ 1
Mass per year = 1.1 × 1016 × 3.16 × 107 × 235 × 1.66 × 10−27 = 0.13 kg✓ 1Accept 0.13–0.15 kg. Award [2 max] for 0.025 kg (efficiency multiplied instead of divided).

Answers: (a)(i) 1.00 mol  ·  (c)(i) 522 K, 632 K  ·  (d)(i) 1.20 × 104 J  ·  (e)(i) 0.43  ·  (f)(i) 0.13 kg (the remaining parts are explanations — see the table above)

48B-2-05Heat engines and the Carnot cycle Paper 2Hard9 marks

An ocean thermal energy conversion (OTEC) plant runs a heat engine between warm surface sea water and cold sea water pumped up from a depth of 1000 m. The graph shows how the temperature θ of the sea varies with depth z at the site. Specific heat capacity of sea water = 4.2 × 103 J kg−1 K−1.

01002003004005006007008009001000z / m051015202530θ / °C
Temperature θ of the sea against depth z (drawn to scale)
(a)
(i)

Use the graph to determine the maximum possible efficiency of a heat engine operating between the surface water and the water at a depth of 1000 m.

(2)
(b)
(i)

The plant generates 5.0 MW of electrical power with an overall efficiency of 2.5 %. Determine the rate at which thermal energy is transferred to the cold water.

(2)
(c)
(i)

The cold water warms by 3.0 K as it passes through the plant. Determine the mass of cold water that must be pumped up each second.

(1)
(d)
(i)

Treat the warm and the cold water as thermal reservoirs at the temperatures read in (a). Determine the rate of change of the total entropy of the two reservoirs.

(2)
(e)
(i)

Show that, for any engine working between reservoirs at Th and Tc, (Carnot power − actual power) = Tc × (rate of increase of total entropy), and verify this for the plant.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Temperatures read: 26 °C = 299 K and 4 °C = 277 K✓ 1Accept 25–27 °C and 3–5 °C.
ηCarnot = 1 − 277/299 = 0.074✓ 1Accept 0.067–0.080. Award [0] for 1 − 4/26 (Celsius temperatures).
Part (b)(i)
Rate of energy input from the warm water = 5.0/0.025 = 200 MW✓ 1
Rate of transfer to the cold water = 200 − 5.0 = 195 MW✓ 1Award [1 max] for 200 MW.
Part (c)(i)
m/t = 1.95 × 108/(4.2 × 103 × 3.0) = 1.5 × 104 kg s−1✓ 1Allow ECF from (b).
Part (d)(i)
ΔS/Δt = −2.00 × 108/299 + 1.95 × 108/277✓ 1Allow ECF from (a) and (b). The negative term for the warm water is required.
= +3.5 × 104 W K−1✓ 1Accept 3.4–3.6 × 104 W K−1; a positive sign is required.
Part (e)(i)
TcΔS = Tc(Qc/Tc − Qh/Th) = Qc − QhTc/Th = (Qh − W) − Qh(1 − ηCarnot) = ηCarnotQh − W✓ 1Per unit time throughout.
277 × 3.5 × 104 = 9.7 MW and 0.0736 × 200 − 5.0 = 9.7 MW✓ 1Allow ECF from (a) and (d).

Answers: (a)(i) 0.074  ·  (b)(i) 195 MW  ·  (c)(i) 1.5 × 104 kg s−1  ·  (d)(i) +3.5 × 104 W K−1 (the remaining parts are explanations — see the table above)

49B-2-07Adiabatic processes Paper 2Hard11 marks

In a light-gas gun used to test materials, 2.0 × 10−4 m³ of helium, a monatomic ideal gas, at a pressure of 5.0 × 106 Pa and a temperature of 300 K is released behind a projectile of mass 10 g. The projectile starts from rest in a barrel of cross-sectional area 1.0 × 10−4 m² and length 1.20 m. The expansion of the helium is so rapid that it can be treated as adiabatic. Friction is negligible and atmospheric pressure is 1.01 × 105 Pa.

helium2.0×10⁻⁴ m³5.0×10⁶ Pa300 K10 gbarrel length 1.20 mbarrel cross-section 1.0×10⁻⁴ m²air at1.01×10⁵ Pa
Diagram not to scale
(a)
(i)

Calculate the amount of helium, in mol.

(1)
(b)
(i)

Show that the pressure of the helium when the projectile leaves the barrel is about 2.3 × 106 Pa.

(2)
(c)
(i)

Determine the temperature of the helium at that moment.

(1)
(d)
(i)

Show that the work done by a monatomic ideal gas in an adiabatic expansion from (p0, V0) to (p1, V1) is W = (3/2)(p0V0 − p1V1).

(2)
(e)
(i)

Determine the speed of the projectile as it leaves the barrel.

(3)
(f)
(i)

The gun has a mass of 3.0 kg and is free to recoil. Estimate the recoil speed of the gun, stating one assumption you make.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
n = 5.0 × 106 × 2.0 × 10−4/(8.31 × 300) = 0.401 mol✓ 1
Part (b)(i)
Final volume = 2.0 × 10−4 + 1.0 × 10−4 × 1.20 = 3.2 × 10−4 m³✓ 1The hidden step: the gas fills the reservoir and the barrel.
p = 5.0 × 106 × (2.0/3.2)5/3 = 2.28 × 106 Pa✓ 1At least 3 s.f. or full substitution is required.
Part (c)(i)
T = 300 × (2.28 × 106 × 3.2 × 10−4)/(5.0 × 106 × 2.0 × 10−4); T = 219 K✓ 1Allow ECF from (b). ALT: T = pV/nR with n from (a). Accept 218–221 K.
Part (d)(i)
Q = 0, so the first law gives W = −ΔU✓ 1
U = (3/2)nRT = (3/2)pV, so W = (3/2)(p0V0 − p1V1)✓ 1
Part (e)(i)
Work done by the helium = 1.5 × (1000 − 731) = 403 J✓ 1Allow ECF from (b) and (d).
The projectile also does work pushing the air in the barrel out against atmospheric pressure: 1.01 × 105 × 1.2 × 10−4 = 12 J, so its kinetic energy is 391 J✓ 1
v = √(2 × 391/0.010) = 280 m s−1✓ 1Accept 278–282 m s−1. Award [2 max] for 284 m s−1 (atmosphere ignored).
Part (f)(i)
Total momentum stays zero: v = 0.010 × 280/3.0 = 0.93 m s−1✓ 1Allow ECF from (e).
Assumption, e.g.: the momentum of the helium (and of the expelled air) is negligible / no external horizontal force acts on the gun✓ 1

Answers: (a)(i) 0.401 mol  ·  (b)(i) 2.28 × 106 Pa  ·  (c)(i) 219 K  ·  (e)(i) 280 m s−1  ·  (f)(i) 0.93 m s−1 (the remaining parts are explanations — see the table above)

50B-2-08Entropy and microstates Paper 2Hard11 marks

A space-station module of volume 12.0 m³ contains air at 1.00 × 105 Pa and 290 K. It is joined through a closed valve to an empty (evacuated) module of volume 6.0 m³. Both modules have rigid, thermally insulating walls. The valve is opened and the air spreads into both modules. Treat the air as an ideal gas.

In a simple model of the microstates, the space available to the gas is divided into a very large number of identical small cells. Each molecule can occupy any cell, independently of the other molecules, and all arrangements are equally probable.

air12.0 m³1.00×10⁵ Pa290 Kevacuated6.0 m³valve(closed)rigid, thermally insulated walls
Diagram not to scale
(a)
(i)

Explain why the temperature of the air is the same after the expansion as before it.

(2)
(b)
(i)

Calculate the pressure of the air after the expansion.

(1)
(c)
(i)

Use the model to show that the entropy change of the air is ΔS = NkB ln(Vf/Vi), where N is the number of molecules.

(2)
(d)
(i)

Determine the entropy change of the air.

(2)
(e)
(i)

The same change of state could be produced reversibly by a slow isothermal expansion against a piston, in which the work done by an ideal gas is W = nRT ln(Vf/Vi). Show that ΔS = Q/T for this process gives the same result as (c).

(2)
(f)
(i)

By considering the entropy change of the surroundings in each process, explain why the free expansion is irreversible but the slow isothermal expansion is reversible.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Q = 0 (insulated walls) and W = 0 (the air expands into a vacuum and the walls are rigid, so no work is done on the surroundings), so ΔU = 0✓ 1Both reasons are needed.
The internal energy of an ideal gas depends only on its temperature, so the temperature is unchanged✓ 1
Part (b)(i)
p = 1.00 × 105 × 12.0/18.0 = 6.67 × 104 Pa✓ 1Allow ECF from (a) (constant temperature).
Part (c)(i)
The number of cells is proportional to the volume, so each molecule has Vf/Vi times as many positions; for N independent molecules Ωf/Ωi = (Vf/Vi)N✓ 1
ΔS = kB ln Ωf − kB ln Ωi = kB ln(Ωf/Ωi) = NkB ln(Vf/Vi)✓ 1
Part (d)(i)
N = pV/kBT = 1.00 × 105 × 12.0/(1.38 × 10−23 × 290) = 3.00 × 1026✓ 1
ΔS = 3.00 × 1026 × 1.38 × 10−23 × ln 1.5 = 1.68 × 103 J K−1✓ 1Allow ECF from (c). Award [1 max] for ln 0.5 or ln 3.
Part (e)(i)
ΔU = 0, so Q = W = nRT ln(Vf/Vi) and ΔS = Q/T = nR ln(Vf/Vi)✓ 1
nR = (N/NA)(NAkB) = NkB, so the two expressions are identical✓ 1Allow ECF from (c).
Part (f)(i)
Free expansion: no energy is exchanged, so the surroundings' entropy is unchanged and the total entropy increases by 1.7 × 103 J K−1: the reverse change would decrease the entropy of an isolated system, so it never happens spontaneously✓ 1Allow ECF from (d).
Isothermal expansion: the surroundings supply Q at T and lose entropy Q/T equal to the gain of the air, so the total change is zero and the process can be reversed✓ 1

Answers: (b)(i) 6.67 × 104 Pa  ·  (d)(i) 1.68 × 103 J K−1 (the remaining parts are explanations — see the table above)

51B-2-16First law of thermodynamics Paper 2Hard10 marks

A monatomic ideal gas is enclosed in a horizontal cylinder by a frictionless piston of area A = 2.0 × 10−3 m². The piston is attached to a spring whose other end is fixed; at A the spring has its natural length. The atmosphere outside the piston is at p0 = 1.0 × 105 Pa.

At A the gas has volume V0 = 1.0 × 10−3 m³, pressure p0 and temperature 300 K. The gas is heated slowly until it reaches state B. The graph shows the pressure p against the volume V during the heating.

0.80.91.01.11.21.31.41.51.6V / 10⁻³ m³0.00.51.01.52.02.5p / 10⁵ PaAB
Pressure p against volume V during the heating (drawn to scale)
(a)
(i)

Show that, during the heating, p = p0 + k(V − V0)/A², where k is the spring constant.

(2)
(b)
(i)

Use the graph to determine k.

(2)
(c)
(i)

Determine the temperature of the gas at B.

(1)
(d)
(i)

Determine the work done by the gas from A to B.

(1)
(e)
(i)

Determine the elastic potential energy stored in the spring at B, and identify what the rest of the work in (d) has been done on.

(2)
(f)
(i)

Determine the thermal energy supplied to the gas from A to B.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The piston is in equilibrium: pA = p0A + kx, where x is the compression of the spring✓ 1
x = (V − V0)/A, so p = p0 + k(V − V0)/A²✓ 1
Part (b)(i)
Gradient = (2.0 − 1.0) × 105/0.40 × 10−3 = 2.5 × 108 Pa m−3✓ 1
k = gradient × A² = 2.5 × 108 × (2.0 × 10−3)² = 1.0 × 103 N m−1✓ 1Allow ECF from (a).
Part (c)(i)
TB = 300 × (2.0 × 1.4)/(1.0 × 1.0) = 840 K✓ 1
Part (d)(i)
Area under the line = ½(1.0 + 2.0) × 105 × 0.40 × 10−3 = 60 J✓ 1Area of the trapezium; pΔV with a single pressure scores [0].
Part (e)(i)
x = 0.40 × 10−3/2.0 × 10−3 = 0.20 m; EH = ½ × 1.0 × 103 × 0.20² = 20 J✓ 1Allow ECF from (b).
The remaining 40 J is work done pushing back the atmosphere (p0ΔV = 1.0 × 105 × 0.40 × 10−3 = 40 J)✓ 1Allow ECF from (d).
Part (f)(i)
ΔU = (3/2)(pBVB − pAVA) = 1.5 × (280 − 100) = 270 J✓ 1Allow ECF from (c) (or (3/2)nRΔT).
Q = ΔU + W = 270 + 60 = 330 J✓ 1Allow ECF from (d).

Answers: (b)(i) 1.0 × 103 N m−1  ·  (c)(i) 840 K  ·  (d)(i) 60 J  ·  (e)(i) 20 J  ·  (f)(i) 330 J (the remaining parts are explanations — see the table above)

52B-2-17Heat engines and the Carnot cycle Paper 2Hard13 marks

A radioisotope power source on a space probe contains plutonium-238, which decays by alpha emission. Each decay releases 5.59 MeV, all of which becomes thermal energy in the source. The graph shows how the thermal power P of the source is predicted to vary over the first 40 years. The mass of a plutonium-238 atom is 238 u.

0510152025303540t / years1500175020002250250027503000P / W
Thermal power P of the source against time t (drawn to scale)
(a)
(i)

Use the graph to show that the decay constant of plutonium-238 is about 7.9 × 10−3 yr−1.

(2)
(ii)

Calculate the half-life of plutonium-238.

(1)
(b)
(i)

Determine the activity of the source at t = 0.

(2)
(c)
(i)

Determine the mass of plutonium-238 in the source at t = 0.

(3)
(d)
(i)

A thermoelectric converter turns part of the thermal energy into electrical energy. It operates between the source at 1300 K and a radiator at 500 K. Calculate the maximum possible efficiency of the converter.

(1)
(e)
(i)

The actual efficiency of the converter is 6.5 %. The radiator has an emissivity of 0.85 and must radiate all the thermal energy that is not converted into electrical energy. Determine the area of the radiator needed at t = 0.

(2)
(f)
(i)

Determine the electrical power available after 20 years, assuming that the efficiency of the converter does not change.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
P at t = 0 and at t = 40 years read as 2720 W and 1980 W✓ 1Accept ±20 W on each reading.
P ∝ e−λt, so λ = ln(2720/1980)/40 = 7.94 × 10−3 yr−1✓ 1An answer to at least 3 s.f. or full substitution is required. Any valid pair of points.
Part (a)(ii)
T1/2 = ln 2/λ = ln 2/7.9 × 10−3 = 88 years✓ 1Allow ECF from (a)(i). Accept 84–92 years.
Part (b)(i)
Energy per decay = 5.59 × 106 × 1.60 × 10−19 = 8.94 × 10−13 J✓ 1
A0 = 2720/8.94 × 10−13 = 3.04 × 1015 Bq✓ 1Allow ECF from (a)(i) for the initial power.
Part (c)(i)
λ = 7.9 × 10−3/3.16 × 107 = 2.5 × 10−10 s−1✓ 1Allow ECF from (a)(i). The decay constant must be in s−1.
N0 = A0/λ = 1.2 × 1025✓ 1Allow ECF from (b).
Mass = 1.2 × 1025 × 238 × 1.66 × 10−27 = 4.8 kg✓ 1Accept 4.7–4.9 kg.
Part (d)(i)
ηCarnot = 1 − 500/1300 = 0.62✓ 1
Part (e)(i)
Power to be radiated = 0.935 × 2720 = 2.54 × 103 W✓ 1Allow ECF from (a)(i). Award [1 max] for 0.90 m² (all 2720 W radiated).
A = 2.54 × 103/(0.85 × 5.67 × 10−8 × 500⁴) = 0.84 m²✓ 1
Part (f)(i)
Initial electrical power = 0.065 × 2720 = 177 W✓ 1
P = 177 × e−7.9 × 10−3 × 20 = 151 W✓ 1Allow ECF from (a)(i). Accept 150–152 W; reading P at 20 years from the graph and multiplying by 0.065 is equivalent.

Answers: (a)(i) 7.94 × 10−3 yr−1  ·  (a)(ii) 88 years  ·  (b)(i) 3.04 × 1015 Bq  ·  (c)(i) 4.8 kg  ·  (d)(i) 0.62  ·  (e)(i) 0.84 m²  ·  (f)(i) 151 W (the remaining parts are explanations — see the table above)

53B-2-26Entropy and the Clausius form of the second law Paper 2Hard11 marks

A copper bar of length 0.250 m and cross-sectional area 4.0 × 10−4 m² is lagged along its length. One end is in boiling water at 373 K and the other end is in a mixture of ice and water at 273 K. After a short time the temperature gradient along the bar is uniform and steady. The boiling water and the ice–water mixture stay at constant temperatures and may be treated as thermal reservoirs. Thermal conductivity of copper = 390 W m−1 K−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1.

boiling water, 373 Kice and water, 273 Klagging (dashed)copper bar, area 4.0×10⁻⁴ m²0.250 m
Diagram not to scale
(a)

Conduction.

(i)

Calculate the rate at which thermal energy is conducted along the bar.

(2)
(ii)

Calculate the mass of ice that melts in 10.0 minutes.

(1)
(b)

Entropy.

(i)

For the 10.0 minutes, determine the entropy change of the boiling water, the entropy change of the ice–water mixture and the total entropy change.

(3)
(ii)

Explain why the entropy of the copper bar itself does not change once the temperature gradient is steady.

(1)
(iii)

State the Clausius form of the second law of thermodynamics and explain how your answer to (b)(i) is consistent with it.

(2)
(c)

Reversing the transfer.

(i)

A heat pump is used to take the same quantity of thermal energy from the ice–water mixture and deliver energy to the boiling water. Determine, using the second law, the minimum work that must be done on the heat pump.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ΔQ/Δt = kAΔT/Δx = 390 × 4.0 × 10−4 × 100/0.250✓ 1
= 62.4 W✓ 1Accept 62 W.
Part (a)(ii)
Q = 62.4 × 600 = 3.74 × 104 J; mass = 3.74 × 104/3.34 × 105 = 0.112 kg✓ 1Allow ECF.
Part (b)(i)
ΔShot = −3.74 × 104/373 = −100 J K−1✓ 1Negative sign required. ΔS = ΔQ/T applied only at constant temperature (phase change / reservoir).
ΔScold = +3.74 × 104/273 = +137 J K−1✓ 1ΔS = ΔQ/T applied only at constant temperature (phase change / reservoir).
Total = +37 J K−1✓ 1Accept 36–37 J K⁻¹. Allow ECF.
Part (b)(ii)
In the steady state the temperature at each point of the bar is constant, so its state (and internal energy) does not change: it receives and passes on energy at the same rate, and entropy is a function of state✓ 1
Part (b)(iii)
Thermal energy cannot spontaneously transfer from a colder body to a hotter body (without work being done)✓ 1
The spontaneous flow from hot to cold gives a positive total entropy change; a flow from cold to hot on its own would give a negative total, which is not possible✓ 1
Part (c)(i)
For the minimum work the total entropy change is zero: Qh/373 = Qc/273, so Qh = 3.74 × 104 × 373/273 = 5.12 × 104 J✓ 1Accept an argument using the Carnot relation for a reversible machine.
Wmin = Qh − Qc = 5.12 × 104 − 3.74 × 104 = 1.37 × 104 J✓ 1Accept 1.4 × 10⁴ J.

Answers: (a)(i) 62.4 W  ·  (a)(ii) 0.112 kg  ·  (b)(i) −100 J K−1, +137 J K−1, +37 J K−1  ·  (c)(i) 1.37 × 104 J (the remaining parts are explanations — see the table above)

54B-2-28A Stirling engine cycle Paper 2Hard14 marks

An ideal Stirling engine uses 0.150 mol of helium, a monatomic ideal gas. Its cycle ABCDA has four stages:

A → B: heating at a constant volume of 1.00 × 10−3 m³ from 320 K to 720 K;
B → C: isothermal expansion at 720 K to 2.50 × 10−3 m³;
C → D: cooling at constant volume to 320 K;
D → A: isothermal compression at 320 K back to 1.00 × 10−3 m³.

The work done by an ideal gas in an isothermal change is W = nRT ln(Vf/Vi). R = 8.31 J mol−1 K−1.

(a)

The cycle.

(i)

Show that the pressure at A is about 4.0 × 105 Pa.

(1)
(ii)

Sketch the cycle on a p–V diagram. Label A, B, C and D and show the direction of the cycle.

(2)
(b)

Energy transfers.

(i)

Determine the thermal energy transferred to the gas during A → B.

(2)
(ii)

Determine the thermal energy transferred to the gas during B → C.

(2)
(iii)

Show that the net work done by the gas in one cycle is about 460 J.

(2)
(c)

Efficiency.

(i)

Calculate the efficiency of the engine, assuming that all the thermal energy supplied in A → B and B → C comes from the hot reservoir.

(1)
(ii)

Calculate the efficiency of a Carnot engine working between reservoirs at 720 K and 320 K.

(1)
(iii)

In a real Stirling engine a "regenerator" stores the thermal energy released by the gas during C → D and returns it to the gas during A → B. Show that an ideal regenerator raises the efficiency to the Carnot value.

(2)
(iv)

Explain, with reference to the second law of thermodynamics, why even this engine must transfer thermal energy to the cold reservoir.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
pA = nRT/V = 0.150 × 8.31 × 320/1.00 × 10−3 = 3.99 × 105 Pa✓ 1
Part (a)(ii)
Vertical line A → B (upwards) at the smaller volume and vertical line C → D (downwards) at the larger volume✓ 1Pressures: A 4.0, B 9.0, C 3.6, D 1.6 × 10⁵ Pa.
Two curves (isotherms) B → C and D → A, the 720 K curve above the 320 K curve; cycle clockwise✓ 1
Part (b)(i)
W = 0 (constant volume), so Q = ΔU = (3/2)nRΔT = 1.5 × 0.150 × 8.31 × 400✓ 1
Q = +748 J✓ 1Accept 750 J.
Part (b)(ii)
ΔU = 0 (isothermal), so Q = W = 0.150 × 8.31 × 720 × ln 2.5✓ 1
Q = +822 J✓ 1Accept 820–823 J.
Part (b)(iii)
WDA = 0.150 × 8.31 × 320 × ln(1/2.5) = −365 J; no work in the constant-volume stages✓ 1
Wnet = 822 − 365 = 457 J ≈ 460 J✓ 1
Part (c)(i)
η = 457/(748 + 822) = 0.29 (29 %)✓ 1Accept 0.29–0.30. Allow ECF.
Part (c)(ii)
ηCarnot = 1 − 320/720 = 0.56 (56 %)✓ 1
Part (c)(iii)
The energy released in C → D (U falls by 748 J) equals the energy needed in A → B, so with the regenerator the only energy from the hot reservoir is the 822 J of B → C✓ 1
η = 457/822 = 0.556 = nR(Th − Tc) ln 2.5/(nRTh ln 2.5) = 1 − Tc/Th, the Carnot value✓ 1The algebraic cancellation or the numerical agreement earns the mark.
Part (c)(iv)
Kelvin form: no cyclic engine can take thermal energy from a single reservoir and convert it all into work; energy must be rejected to a colder reservoir (equivalently, the entropy the gas gains at 720 K must be passed to the cold reservoir so that the total entropy does not decrease)✓ 1

Answers: (a)(i) 3.99 × 105 Pa  ·  (b)(i) +748 J  ·  (b)(ii) +822 J  ·  (b)(iii) 457 J  ·  (c)(i) 29 %  ·  (c)(ii) 56 %  ·  (c)(iii) 56 % (the remaining parts are explanations — see the table above)

Model sketch

00.511.522.53012345678910V / 10⁻³ m³p / 10⁵ PaABCD720 K320 K
55B-2-29Adiabatic processes Paper 2Hard14 marks

In an expansion cloud chamber, argon mixed with a small amount of alcohol vapour is suddenly expanded by a piston. The argon cools, the vapour becomes supersaturated, and charged particles leave visible tracks. The argon is initially at 1.00 × 105 Pa and 293 K and occupies 8.00 × 10−4 m³. It is expanded rapidly to 1.00 × 10−3 m³. Treat the argon as a monatomic ideal gas and ignore the vapour.

(a)

The expansion.

(i)

State what is meant by an adiabatic process, and suggest why the rapid expansion can be modelled as adiabatic.

(2)
(ii)

Show that the pressure after the expansion is about 6.9 × 104 Pa.

(2)
(iii)

Determine the temperature of the argon after the expansion.

(2)
(iv)

Determine the work done by the argon during the expansion.

(2)
(b)

After the expansion.

(i)

The piston is then held fixed and the argon slowly warms back to 293 K by thermal energy transferred from the walls of the chamber. Calculate the final pressure of the argon.

(1)
(ii)

Calculate the thermal energy transferred to the argon while it warms.

(1)
(iii)

Sketch, on one p–V diagram, the adiabatic expansion and the warming. Add the isotherm for 293 K through the initial state, and explain why the adiabatic curve is steeper than the isotherm.

(2)
(c)

Entropy.

(i)

Explain why the warming of the argon by the walls increases the total entropy of the argon and the walls.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
No thermal energy is transferred between the gas and its surroundings (Q = 0)✓ 1
The expansion is so fast that there is no time for significant thermal energy transfer through the walls✓ 1
Part (a)(ii)
p2 = p1(V1/V2)5/3 = 1.00 × 105 × (0.800)5/3✓ 1From pV5/3 = constant.
= 6.89 × 104 Pa✓ 1At least 3 s.f. required for a "show that".
Part (a)(iii)
T2 = T1p2V2/(p1V1) = 293 × (6.89 × 104 × 1.00 × 10−3)/(1.00 × 105 × 8.00 × 10−4)✓ 1
T2 = 252 K✓ 1Accept 252–253 K.
Part (a)(iv)
Q = 0, so W = −ΔU = (3/2)(p1V1 − p2V2) = 1.5 × (80.0 − 68.9)✓ 1Or (3/2)nRΔT with n = 0.0329 mol.
W = +16.6 J (done by the gas)✓ 1Accept 16–17 J.
Part (b)(i)
Constant volume: p3 = 6.89 × 104 × 293/252 = 8.0 × 104 Pa✓ 1Also from p₁V₁ = p₃V₂, since the final temperature equals the initial one.
Part (b)(ii)
W = 0, so Q = ΔU = (3/2)V(p3 − p2) = 1.5 × 1.00 × 10−3 × (8.00 − 6.89) × 104 = 16.6 J✓ 1Equal to the work in (a)(iv) because the argon returns to 293 K. Accept 16–17 J.
Part (b)(iii)
Adiabatic curve from (0.80, 1.00) down to (1.00, 0.69) [× 10⁻³ m³, × 10⁵ Pa], below the isotherm ending at (1.00, 0.80); vertical line up from (1.00, 0.69) to (1.00, 0.80), ending on the isotherm✓ 1
During an adiabatic expansion the gas does work with no thermal energy supplied, so its temperature falls as well as its volume increasing; the pressure therefore falls more than on the isotherm, where the temperature stays constant✓ 1
Part (c)(i)
Thermal energy flows from the walls at 293 K to the colder argon (252–293 K); the argon gains entropy ΔQ/T at a lower temperature than the walls lose it✓ 1
So the argon gains more entropy than the walls lose (16.6/293 = 0.057 J K−1 lost by the walls; about 16.6/273 ≈ 0.061 J K−1 gained by the argon); the transfer across a temperature difference is irreversible and the total entropy increases✓ 1Numerical estimate not required for the mark if the argument is clear.

Answers: (a)(ii) 6.89 × 104 Pa  ·  (a)(iii) 252 K  ·  (a)(iv) 16.6 J  ·  (b)(i) 8.0 × 104 Pa  ·  (b)(ii) 16.6 J (the remaining parts are explanations — see the table above)

Model sketch

0.60.70.80.911.11.20.50.60.70.80.911.1V / 10⁻³ m³p / 10⁵ Painitial stateafter expansion293 K againisotherm 293 Kadiabat
56B-2-41Gas springs and oscillations Paper 2Hard20 marks

An optical table of mass 48 kg rests on four identical gas springs that isolate it from vibrations of the floor. Each spring is a vertical cylinder, closed at the bottom, containing argon (a monatomic ideal gas) beneath a frictionless piston of area 3.0 × 10−3 m² and negligible mass. The table rests on the four pistons, and the atmosphere, at 1.01 × 105 Pa, acts on the top of each piston.

In equilibrium the column of argon in each cylinder is 0.080 m high and the argon is at 293 K.

0.00.10.20.30.40.50.60.70.80.91.0t / s-2.5-2.0-1.5-1.0-0.50.00.51.01.52.02.5x / mm
Displacement x of the table against time t after release (drawn to scale)
(a)
(i)

Show that the pressure of the argon in each spring is about 1.4 × 105 Pa.

(2)
(ii)

Calculate the amount of argon, in mol, in one spring.

(1)
(b)

The table is pushed down slowly through 2.0 mm, so that the argon stays at 293 K.

(i)

Determine the increase in the total upward force exerted on the table by the four springs.

(2)
(ii)

Show that, for a small slow displacement Δh, the four springs together behave like a single spring of spring constant k = 4pA/h, where h is the equilibrium height of the gas column, and evaluate k.

(2)
(c)
(i)

The table is instead pushed down quickly through 2.0 mm, so that the compression is adiabatic. Determine the increase in the total upward force.

(2)
(ii)

Explain, using the first law of thermodynamics, why the increase in force is greater for the quick compression.

(2)
(d)

The table is pushed down by 2.0 mm and released. The graph shows its displacement x from the equilibrium position against time t.

(i)

Determine the period of the oscillation.

(2)
(ii)

Deduce whether the oscillations of the table are better modelled as isothermal or as adiabatic.

(3)
(e)
(i)

The floor vibrates at a frequency of 25 Hz. Explain why only a small fraction of this vibration reaches the table, and suggest why gas springs with taller gas columns are used when even better isolation is needed.

(2)
(ii)

Calculate the initial energy of the oscillation shown in the graph. This energy is eventually transferred to the surroundings at 293 K. Determine the resulting entropy change of the surroundings.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Each piston supports a quarter of the weight: 48 × 9.81/4 = 118 N, which needs an extra pressure of 118/3.0 × 10−3 = 3.92 × 104 Pa✓ 1The hidden step (four supports) must be seen.
p = 1.01 × 105 + 3.92 × 104 = 1.40 × 105 Pa✓ 1Atmospheric pressure must be added. Answer to at least 3 s.f.
Part (a)(ii)
n = pV/RT = 1.40 × 105 × 3.0 × 10−3 × 0.080/(8.31 × 293) = 1.38 × 10−2 mol✓ 1Allow ECF from (a)(i).
Part (b)(i)
Boyle's law with V ∝ height: p = 1.40 × 105 × 0.080/0.078 = 1.438 × 105 Pa✓ 1Allow ECF from (a)(i).
ΔF = 4 × (1.438 × 105 − 1.402 × 105) × 3.0 × 10−3 = 43.2 N✓ 1Accept 42–44 N. Award [1 max] for one spring only (10.8 N).
Part (b)(ii)
p′(h − Δh) = ph gives Δp = pΔh/(h − Δh) ≈ pΔh/h when Δh ≪ h✓ 1The small-displacement approximation must be stated.
k = 4AΔp/Δh = 4pA/h = 4 × 1.40 × 105 × 3.0 × 10−3/0.080 = 2.1 × 104 N m−1✓ 1Allow ECF from (a)(i). Consistent with (b)(i): 43.2/0.0020 = 2.2 × 104 N m−1.
Part (c)(i)
pV5/3 = constant: p = 1.40 × 105 × (0.080/0.078)5/3 = 1.463 × 105 Pa✓ 1Allow ECF from (a)(i).
ΔF = 4 × (1.463 × 105 − 1.402 × 105) × 3.0 × 10−3 = 72.5 N✓ 1Accept 71–74 N.
Part (c)(ii)
Work is done on the argon and Q = 0, so ΔU = −W > 0: the internal energy and the temperature of the argon increase✓ 1
p = nRT/V rises both because V decreases and because T increases, whereas in the slow compression only V changes (energy leaves as thermal energy at constant T)✓ 1
Part (d)(i)
Time for four oscillations read from the graph ≈ 0.93 s (fourth trough)✓ 1Measuring over several oscillations is required for this mark.
T = 0.93/4 = 0.23 s✓ 1Accept 0.22–0.245 s.
Part (d)(ii)
Isothermal: k = 43.2/0.0020 = 2.16 × 104 N m−1, T = 2π√(48/2.16 × 104) = 0.30 s✓ 1Allow ECF from (b)(i). Or with k from (b)(ii): 0.30 s.
Adiabatic: k = 72.5/0.0020 = 3.63 × 104 N m−1, T = 2π√(48/3.63 × 104) = 0.23 s✓ 1Allow ECF from (c)(i).
The measured period (0.23 s) agrees with the adiabatic model: the compressions are too fast for significant thermal energy transfer✓ 1Allow ECF from (d)(i). A conclusion consistent with the candidate's values.
Part (e)(i)
The natural frequency of the table is about 1/0.23 ≈ 4.3 Hz, far below the 25 Hz driving frequency, so the table is driven far from resonance and its forced amplitude is small✓ 1Allow ECF from (d)(i).
k ∝ 1/h from (b)(ii), so f ∝ √k ∝ 1/√h: a taller column lowers the natural frequency further below the driving frequency✓ 1Dependent on the use of the relation from (b)(ii).
Part (e)(ii)
E = ½kx0² = ½ × 3.63 × 104 × (2.0 × 10−3)² = 7.3 × 10−2 J✓ 1Allow ECF from (c)(i). Accept 0.07 J; 4.3 × 10−2 J with the isothermal constant scores this mark.
ΔS = 7.3 × 10−2/293 = 2.5 × 10−4 J K−1✓ 1ΔS = ΔQ/T is valid because the surroundings stay at 293 K.

Answers: (a)(i) 1.40 × 105 Pa  ·  (a)(ii) 1.38 × 10−2 mol  ·  (b)(i) 43.2 N  ·  (b)(ii) 2.1 × 104 N m−1  ·  (c)(i) 72.5 N  ·  (d)(i) 0.23 s  ·  (d)(ii) adiabatic  ·  (e)(ii) 7.3 × 10−2 J; 2.5 × 10−4 J K−1 (the remaining parts are explanations — see the table above)

57IBP2H-015-01-12Efficiency of a gas cycle Paper 2Hard11 marks

A prototype engine for a Mars rover takes a fixed mass of monatomic ideal gas through a cycle of four stages:
A → B: expansion at a constant pressure of 3.0×105 Pa, from 2.0×10−3 m3 to 5.0×10−3 m3
B → C: cooling at constant volume until the pressure is 1.0×105 Pa
C → D: compression at a constant pressure of 1.0×105 Pa, back to 2.0×10−3 m3
D → A: heating at constant volume, back to state A

012345601234V / 10⁻³ m³p / 10⁵ PaABCD
Diagram drawn to scale
(a)

Determine the net work done by the gas in one cycle.

(2)
(b)

Show that the energy absorbed as heat during D → A is about 600 J.

(2)
(c)

Calculate the energy absorbed as heat during A → B.

(3)
(d)

Determine the efficiency of the cycle.

(2)
(e)

The highest and lowest temperatures in the cycle occur at B and D. Determine the efficiency of a Carnot engine operating between these two temperatures, and compare it with your answer to (d).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The cycle is a rectangle on a p–V diagram, so W = Δp × ΔV = 2.0×105 × 3.0×10−3 .✓ 1
W = 600 J✓ 1
Part (b)
No work is done at constant volume, so Q = ΔU = (3/2)VΔp .✓ 1
Q = 1.5 × 2.0×10−3 × 2.0×105 = 600 J✓ 1
Part (c)
ΔU = (3/2)pΔV = 1.5 × 3.0×105 × 3.0×10−3 = 1350 J .✓ 1
W = pΔV = 900 J .✓ 1
Q = ΔU + W = 2250 J✓ 1
Part (d)
Heat is absorbed only in D → A and A → B, so Qin = 600 + 2250 = 2850 J .✓ 1
η = 600/2850 = 0.21; ECF from (a)–(c)✓ 1
Part (e)
TB/TD = pBVB/(pDVD) = 1500/200 = 7.5 .✓ 1
ηCarnot = 1 − 1/7.5 = 0.87, which is much greater than 0.21; ECF from (d)✓ 1

Common error: Use only the stages where heat enters the gas to find Qin. Including stages where heat is rejected, or using the net heat, gives the wrong efficiency.

58IBP2H-015-01-13Efficiency of a gas cycle Paper 2Hard9 marks

A deep-space probe uses a Carnot engine with a monatomic ideal gas as its working substance. A radioisotope source holds the hot reservoir at 900 K, and a radiator holds the cold reservoir at 300 K. In each cycle the gas absorbs 1800 J during its isothermal expansion.

(a)

State the change in internal energy of the gas during the isothermal expansion, and hence determine the work done by the gas in this stage.

(2)
(b)

Calculate the entropy change of the gas during the isothermal expansion.

(1)
(c)

Explain why the entropy of the gas must decrease by the same amount during the isothermal compression, and hence determine the energy rejected per cycle.

(3)
(d)

Show that the efficiency is consistent with the Carnot efficiency.

(1)
(e)

Determine the ratio of final volume to initial volume for the adiabatic expansion from 900 K to 300 K.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔU = 0, because the internal energy of an ideal gas depends only on temperature .✓ 1
First law: Q = ΔU + W, so W = Q = 1800 J✓ 1
Part (b)
ΔS = Q/T = 1800/900 = 2.0 J K−1✓ 1
Part (c)
The adiabatic stages are reversible with Q = 0, so the entropy of the gas does not change during them .✓ 1
Entropy is a state function, so ΔS = 0 over the complete cycle and the compression must give ΔS = −2.0 J K−1 .✓ 1
QC = TC|ΔS| = 300 × 2.0 = 600 J; allow error carried forward from the entropy change calculated earlier✓ 1
Part (d)
η = (1800 − 600)/1800 = 0.667 and 1 − 300/900 = 0.667, so the two agree✓ 1
Part (e)
For a monatomic gas γ = 5/3, so TV2/3 = constant, giving V2/V1 = (T1/T2)3/2 .✓ 1
V2/V1 = 31.5 = 5.2✓ 1

Exam tip: In a Carnot cycle, the entropy gained at TH equals the entropy lost at TC. This gives QC/QH = TC/TH directly.

59IBP2H-015-01-16Entropy and microstates Paper 2Hard9 marks

In a Carnot engine, 0.050 mol of monatomic ideal gas expands isothermally at 420 K until its volume has tripled. Assume the number of positions available to each molecule is proportional to the volume of the gas. (R = 8.31 J mol−1 K−1)

(a)

Sketch a Carnot cycle on axes of pressure p against volume V. (Describe your sketch in words.)

(3)
(b)

Explain why the number of microstates of the gas increases by a factor of 3N, where N is the number of molecules.

(2)
(c)

Show that the entropy change of the gas is ΔS = nR ln 3.

(2)
(d)

Calculate the energy absorbed by the gas as heat during the expansion.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Two isotherms are drawn as curves of decreasing pressure, with the higher-temperature (TH) curve further from the origin .✓ 1
Two adiabats are drawn steeper than the isotherms. One is an expansion and one is a compression, and they join the ends of the isotherms to form a closed loop .✓ 1
The cycle runs clockwise, with the isothermal expansion at TH along the upper isotherm and the isothermal compression at TC along the lower isotherm✓ 1
Part (b)
Each molecule has three times as many available positions, because the number of positions is proportional to volume .✓ 1
The molecules are independent, so the factors multiply: 3 × 3 × … × 3 = 3N✓ 1
Part (c)
ΔS = kB ln(Ωf/Ωi) = kB ln 3N = NkB ln 3 .✓ 1
NkB = nNAkB = nR, so ΔS = nR ln 3✓ 1
Part (d)
Either: the process is reversible and isothermal, so Q = TΔS. Or: ΔU = 0, so Q = W = nRT ln(Vf/Vi) .✓ 1
Q = 420 × 0.050 × 8.31 × ln 3 = 1.9×102 J. Allow ECF from the entropy change found in the previous part✓ 1

Model sketch

VpABCDisothermal (hot)isothermal (cold)adiabaticadiabatic

Exam tip: The microscopic definition S = kB ln Ω and the macroscopic relation ΔS = Q/T give the same result here. Showing that link is a strong way to check your answer.

60IBP2H-015-00-13A Stirling engine cycle Paper 2Hard10 marks

A small engine for a remote weather station uses helium, a monatomic ideal gas, in the following cycle.
A → B: expansion at a constant pressure of 3.2 × 105 Pa, from 1.0 × 10−3 m3 to 8.0 × 10−3 m3.
B → C: cooling at constant volume.
C → A: reversible adiabatic compression back to state A.

1.0×10⁻³ m³8.0×10⁻³ m³3.2VpABC3.2×10⁵ Pa
Diagram drawn to scale
(a)

Show that the pressure at C is 1.0 × 104 Pa.

(2)
(b)

Calculate the energy transferred to the helium by heating during A → B.

(3)
(c)

Determine the work done on the helium during C → A.

(2)
(d)

Determine the efficiency of the engine.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
pCVB5/3 = pAVA5/3✓ 1
pC = 3.2 × 105 × (1/8)5/3 = 3.2 × 105/32 = 1.0 × 104 Pa✓ 1
Part (b)
W = pΔV = 3.2 × 105 × 7.0 × 10−3 = 2240 J✓ 1
ΔU = (3/2)pΔV = 3360 J✓ 1
Q = 5600 J✓ 1
Part (c)
Q = 0, so work done on the gas = ΔU = (3/2)(pAVA − pCVB)✓ 1
= 1.5 × (320 − 80) = 360 J✓ 1
Part (d)
Net work done by the gas = 2240 − 360 = 1880 J, since no work is done in B → C✓ 1
Energy is absorbed by heating only in A → B, because B → C rejects 3720 J✓ 1
Efficiency = 1880/5600 = 0.34 (ECF from (b) and (c))✓ 1

Exam tip: In efficiency calculations, the input is only the energy ABSORBED by heating. Do not subtract the energy rejected in other stages from it.

61IBP2H-015-00-14Thermodynamic cycles Paper 2Hard10 marks

Helium, a monatomic ideal gas, is used to lift a scientific instrument package. In a test, 2.0 mol of helium is taken from an initial state at 290 K and 0.040 m3 to a final state at 290 K and 0.10 m3. Two routes are compared.
Route 1: reversible isothermal expansion.
Route 2: reversible adiabatic expansion to 0.10 m3, followed by reversible heating at constant volume back to 290 K.
The work done by an ideal gas in a reversible isothermal expansion is nRT ln(V2/V1).

(a)

Calculate the work done by the helium in route 1.

(2)
(b)

Determine the entropy change of the helium in route 1.

(2)
(c)

Show that the temperature at the end of the adiabatic stage of route 2 is about 160 K.

(2)
(d)

Determine the energy supplied by heating in the constant-volume stage of route 2.

(2)
(e)

Route 2 absorbs less energy by heating than route 1. Explain why the entropy change of the helium is nevertheless the same for both routes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = 2.0 × 8.31 × 290 × ln(0.10/0.040) = 2.0 × 8.31 × 290 × ln(2.5)✓ 1
= 4.4 × 103 J✓ 1
Part (b)
ΔU = 0, so Q = W = 4.4 × 103 J✓ 1
ΔS = Q/T = 4416/290 = 15 J K−1 (allow ECF from (a))✓ 1
Part (c)
Uses pV5/3 = constant with pV = nRT, or uses TV2/3 = constant directly✓ 1
T = 290 × (0.040/0.10)2/3 = 157 K ≈ 160 K✓ 1
Part (d)
W = 0, so Q = ΔU = (3/2)nRΔT✓ 1
= 1.5 × 2.0 × 8.31 × (290 − 157) = 3.3 × 103 J (allow ECF from (c))✓ 1
Part (e)
Entropy is a function of state, and both routes have the same initial and final states, so ΔS is the same✓ 1
The adiabatic stage has ΔS = 0. In route 2 the energy is absorbed at temperatures below 290 K, so each joule contributes a larger ΔQ/T, which makes up for the smaller Q✓ 1

Exam tip: Q and W depend on the route, but ΔU and ΔS depend only on the end states. Comparing two routes between the same states is a quick check on your working.

62IBP2H-015-00-15Work done when the pressure is not constant Paper 2Hard10 marks

A pneumatic artificial muscle in a robotic prosthetic hand contains a fixed amount of ideal gas. During one movement, the pressure of the gas falls linearly with volume from state A to state B.
State A: p = 6.0 × 105 Pa, V = 1.0 × 10−4 m3, T = 300 K.
State B: p = 2.0 × 105 Pa, V = 3.0 × 10−4 m3.

00.511.522.533.501234567V / 10⁻⁴ m³p / 10⁵ PaAB
Diagram not to scale
(a)

Show that the temperature at B is equal to the temperature at A.

(2)
(b)

Calculate the work done by the gas from A to B.

(2)
(c)

Determine the net energy transferred to the gas by heating from A to B.

(2)
(d)

Outline why this process is not isothermal, even though the initial and final temperatures are equal.

(1)
(e)

Determine the maximum temperature of the gas during the process.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
pAVA = 60 J and pBVB = 60 J✓ 1
For a fixed amount of ideal gas T ∝ pV, so TB = 300 K✓ 1
Part (b)
Area under the line = ½(6.0 + 2.0) × 105 × 2.0 × 10−4✓ 1
= 80 J✓ 1
Part (c)
TA = TB, so ΔU = 0✓ 1
Q = W = 80 J, transferred into the gas (ECF from (b))✓ 1
Part (d)
pV is not constant at intermediate states: for example, at V = 2.0 × 10−4 m3, p = 4.0 × 105 Pa, so pV = 80 J, which is greater than 60 J✓ 1
Part (e)
The line is p = 8.0 × 105 − 2.0 × 109V✓ 1
pV = 8.0 × 105V − 2.0 × 109V2, which is a maximum at V = 2.0 × 10−4 m3, where pV = 80 J✓ 1
Tmax = 300 × 80/60 = 400 K✓ 1

Common error: Assuming that equal start and end temperatures mean an isothermal process. Isothermal means the temperature is constant at every stage.

63IBP2H-015-00-16Thermodynamic cycles Paper 2Hard10 marks

A proposed engine for a lunar rover takes 0.10 mol of helium, a monatomic ideal gas, round the cycle A → B → C → D → A. A → B and C → D are isothermal; B → C and D → A are adiabatic. The table gives the state of the gas at each point.

State: A, B, C, D
p / kPa: 499, 249, 44.0, 88.1
V / 10−3 m3: 1.00, 2.00, 5.66, 2.83
T / K: 600, 600, 300, 300

The work done by an ideal gas in a reversible isothermal expansion is nRT ln(V2/V1).

01234560100200300400500V / 10⁻³ m³p / kPaABCD600 K300 K
Diagram drawn to scale
(a)

Use the data to show that process B → C is adiabatic.

(2)
(b)

Calculate the work done on the gas during D → A.

(2)
(c)

Calculate the energy absorbed by heating during A → B.

(2)
(d)

Determine the efficiency of the cycle.

(3)
(e)

Comment on your answer to (d) with reference to the second law of thermodynamics.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
TV2/3 at B = 600 × (2.00 × 10−3)2/3 = 9.52, and at C = 300 × (5.66 × 10−3)2/3 = 9.52 (or pV5/3 ≈ 7.9 at both)✓ 1
The values are equal, as required for an adiabatic change of a monatomic ideal gas✓ 1
Part (b)
Q = 0, so work done on the gas = ΔU = (3/2)nRΔT✓ 1
= 1.5 × 0.10 × 8.31 × 300 = 374 J✓ 1
Part (c)
ΔU = 0, so Q = W✓ 1
= 0.10 × 8.31 × 600 × ln 2 = 346 J✓ 1
Part (d)
Energy rejected in C → D = 0.10 × 8.31 × 300 × ln(5.66/2.83) = 173 J✓ 1
The adiabatic works cancel, so net work = 346 − 173 = 173 J✓ 1
Efficiency = 173/346 = 0.50 (ECF from (c))✓ 1
Part (e)
The Carnot efficiency is 1 − 300/600 = 0.50. The cycle is reversible and reaches the maximum efficiency allowed, so it is consistent with the second law✓ 1

Exam tip: In a Carnot cycle, the work in the two adiabatic stages is equal and opposite. The net work is just Qin − Qout from the isothermal stages.