Thermodynamics is an HL-only topic built on the first law, Q = ΔU + W. You need to find the work done from the area under a p–V graph, recognise isovolumetric, isobaric, isothermal and adiabatic changes, and use pV^(5/3) = constant for adiabatic changes of a monatomic ideal gas.
The second law appears through entropy, both as ΔS = ΔQ/T and in the statistical form S = k ln Ω, and through heat engines and heat pumps. Cycle questions ask for the net work, the heat exchanged at each stage, the efficiency and how it compares with the Carnot limit.
52 questions
283 marks
Paper 1A: 28
Paper 1B: 7
Paper 2: 17
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29 practice questions on B.4 Thermodynamics
1B-1A-04
First law of thermodynamics·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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?
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Notes
Step 1Work done by the gas at constant pressure: W = pΔV = p(2V − V) = pV.
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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.
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Step 3First law: Q = ΔU + W = ³⁄₂pV + pV = 5pV/2.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system; that the work done by or on a closed system as given by W = PΔV; that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔTCommand term: Determine
2B-1A-05
Adiabatic processes·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A monatomic ideal gas expands adiabatically until its volume is 8 times the initial volume.
By what factor does the pressure decrease?
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Notes
Step 1For an adiabatic change of a monatomic gas pV5/3 = constant.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2p1/p2 = (V2/V1)5/3 = 85/3.
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Step 381/3 = 2, so 85/3 = 2⁵ = 32.
✓ 1
Answer 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.
Syllabus understandingB.4 — adiabatic processes in monatomic ideal gases as given by pV5/3 = constant Command term: Determine
3B-1A-06
Entropy and the second law·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
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Notes
Step 1Each reservoir stays at constant temperature, so ΔS = ΔQ/T applies to each separately.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3Total: −Q/(2T) + Q/T = +Q/(2T) — positive, as the second law requires for a spontaneous transfer.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the entropy change ΔS of a system at constant temperature is given by ΔS = ΔQ/T; that the second law of thermodynamics refers to the change in entropy of an isolated system Command term: Deduce
4B-1A-07
Heat engines and the Carnot cycle·B.4 Thermodynamics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
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Notes
Step 1η = 1 − Tc/Th, so the original ratio is Tc/Th = 1 − η.
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All 3 steps must be completed — there is no mark for a part-answer.
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.
Syllabus understandingB.4 — that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/ThCommand term: Deduce
5B-1A-11
Entropy and microstates·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
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Notes
Step 1Each molecule can be in either of two halves, independently of the others, so the total number of microstates is 2 × 2 × … = 2N.
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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.
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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.
✓ 1
Answer 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.
Syllabus understandingB.4 — that entropy can be determined in terms of the properties of individual particles of the system as given by S = kB ln Ω, where Ω is the number of possible microstates of the system; that the microstates of a system are equally probable; that processes in real isolated systems are almost always irreversible and consequently the entropy of a real isolated system always increases Command term: Deduce
6B-1A-14
Isothermal and adiabatic processes·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
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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.
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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.
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Step 3III: in a reversible adiabatic change no thermal energy is transferred, so ΔS = ΔQ/T = 0. III is true.
✓ 1
Answer 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.
Syllabus understandingB.4 — isovolumetric, isobaric, isothermal and adiabatic processes; adiabatic processes in monatomic ideal gases as given by pV5/3 = constant; ΔS = ΔQ/TCommand term: Deduce
7B-1A-20
Cyclic processes·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
The cycle X → Y → Z → X on a p–V diagram (drawn to scale).Show mark scheme
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Notes
Step 1I: the gas returns to its initial state, so U (a function of state) returns to its initial value: ΔU = 0. True.
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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.
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Step 3III: Q = ΔU + W = 0 + (−200 J) = −200 J, so there is a net transfer of thermal energy out of the gas. False.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system; that the work done by or on a closed system as given by W = PΔV can be described in terms of pressure and changes of volume of the system; that cyclic gas processes are used to run heat engines (Guidance: work done on a system is taken to be negative) Command term: Deduce
8B-1A-25
Efficiency of a gas cycle·B.4 Thermodynamics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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?
The engine cycle on a p–V diagram (drawn to scale).Show mark scheme
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Notes
Step 1Net work done by the gas per cycle = enclosed area = (2p0 − p0)(2V0 − V0) = p0V0.
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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.
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Step 3Input = 6.5p0V0, so efficiency = p0V0/6.5p0V0 = 2/13 (≈ 0.15). In C → D and D → A energy leaves the gas.
✓ 1
Answer 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.
Syllabus understandingB.4 — the first law of thermodynamics as given by Q = ΔU + W; the change in internal energy as given by ΔU = (3/2)nRΔT; that a heat engine can respond to different cycles and is characterized by its efficiency as given by η = useful work/input energy; that the Carnot cycle sets a limit for the efficiency of a heat engine as given by ηCarnot = 1 − Tc/ThCommand term: Determine
9B-1A-27
Work done when the pressure is not constant·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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?
A fixed mass of ideal gas is compressed from X to Y along the straight line (drawn to scale).Show mark scheme
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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.
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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.
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Step 3W = −600 J.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the work done by or on a closed system as given by W = PΔV when its boundaries are changed can be described in terms of pressure and changes of volume of the system (Guidance: work done on a system is taken to be negative; quantitative problems include situations where pressure is not constant) Command term: Determine
10B-1A-47
First law of thermodynamics·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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
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Notes
Step 1In Q = ΔU + W, W is the work done by the gas and Q is the thermal energy transferred to the gas.
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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.
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Step 3ΔU = Q − W = −12 − (−35) = +23 J.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work Command term: Determine
11B-1A-48
Adiabatic processes·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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?
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Step 1The cylinder is insulated, so Q = 0 (adiabatic) and ΔU = −W = −180 J.
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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.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔT of a system is related to the change of its temperature; adiabatic processes Command term: Determine
12B-1A-49
Entropy and microstates·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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)
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Step 1Number of ways of choosing which 2 of the 10 molecules are in the right half: Ω1 = 10!/(2! 8!) = 45.
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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.
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.
Syllabus understandingB.4 — that entropy can be determined in terms of the properties of individual particles of the system as given by S = kB ln Ω where Ω is the number of possible microstates of the system; entropy relates to the degree of disorder of the particles in a system; the microstates of a system are equally probable (simple combinatorial model) Command term: Determine
13B-1A-50
The second law of thermodynamics·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Which process is impossible according to the second law of thermodynamics?
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Step 1Kelvin form of the second law: no cyclic process can take thermal energy from a single reservoir and convert it completely into work.
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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).
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Step 3A local decrease of entropy is allowed if the surroundings gain at least as much entropy.
✓ 1
Answer 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.
Syllabus understandingB.4 — that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes (Kelvin and Clausius forms); that the entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase of the entropy of the surroundings Command term: Deduce
14B-1A-68
Isothermal and adiabatic processes·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
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?
Four changes of state from X on a pressure–volume diagram (drawn to scale)Show mark scheme
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Notes
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.
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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.
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Step 3So R is the isothermal change: ΔU = 0 and Q = W.
✓ 1
Answer 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.
Syllabus understandingB.4 — that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; the first law of thermodynamics Q = ΔU + W; ΔU = (3/2)nRΔTCommand term: Identify
15B-1A-69
Heat engines and the Carnot cycle·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
Engine
Work done / J
Thermal energy transferred to cold reservoir / J
I
250
250
II
150
300
III
180
320
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Step 1First law: work + thermal energy rejected must equal 500 J. II gives 150 + 300 = 450 J, so II is impossible.
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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.
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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).
✓ 1
Answer 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.
Syllabus understandingB.4 — the first law of thermodynamics Q = ΔU + W; that a heat engine is characterized by its efficiency η = useful work/input energy; that the Carnot cycle sets a limit for the efficiency of a heat engine as given by ηCarnot = 1 − Tc/Th; the second law sets constraints on possible physical processes Command term: Deduce
16B-1A-112
Microstates of a coin model·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Four identical coins are tossed. A microstate states which face (head or tail) each individual coin shows. A macrostate states only the total number of heads. Every microstate is equally probable.
Which statements are correct?
I. The macrostate "two heads and two tails" contains six microstates. II. Every macrostate is equally probable because every microstate is equally probable. III. The entropy S = kB ln Ω of the macrostate "four heads" is zero.
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Notes
Step 1I: the two heads can be any 2 of the 4 coins: HHTT, HTHT, HTTH, THHT, THTH, TTHH — six microstates. True.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: the five macrostates contain 1, 4, 6, 4 and 1 of the 16 equally probable microstates, so their probabilities are 1/16, 4/16, 6/16, 4/16 and 1/16 — not equal. False.
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Step 3III: "four heads" is reached in only one way, Ω = 1, so S = kB ln 1 = 0. True.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis rejects III: with only one microstate Ω = 1 and ln 1 = 0, so the entropy of that macrostate is zero.
BThis confuses microstates with macrostates: a macrostate with more microstates is more probable, so II is false; it also rejects III, which is true.
CCorrect: I and III are true; II is false because the macrostates contain different numbers of microstates.
DThis accepts II. Equal probability of microstates makes the macrostate "two heads" six times as probable as "four heads".
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; that entropy S is a thermodynamic quantity that relates to the degree of disorder of the particles in a system; Guidance: the microstates of a system are equally probable; simple combinatorial models (coins) Command term: Deduce
17B-1A-113
Entropy and disorder·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
In which process does the entropy of the substance named in bold decrease?
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Notes
Step 1Entropy relates to the disorder of the particles: it increases when the particles have more possible arrangements (more microstates).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Free expansion, melting and the spreading of ink all give the particles more possible positions, so the entropy of the named substance increases.
—
Step 3Condensation turns a disordered gas into a much more ordered liquid, and thermal energy leaves the water (ΔS = ΔQ/T < 0): its entropy decreases. The window and the air gain at least as much entropy.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe gas has a larger volume to move in, so the number of microstates and the entropy increase, even though no energy enters it.
BMelting needs thermal energy to enter the ice at constant temperature, so ΔS = ΔQ/T is positive; the liquid is less ordered than the crystal.
CThe ink particles spread out over more possible positions, so their disorder and entropy increase; this is an irreversible mixing process.
DCorrect: condensation orders the water molecules and removes thermal energy from them, so the entropy of the water decreases (a local decrease, compensated by the surroundings).
Syllabus understandingB.4 — that entropy S is a thermodynamic quantity that relates to the degree of disorder of the particles in a system; that the entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase of the entropy of the surroundings; that processes in real isolated systems are almost always irreversible and consequently the entropy of a real isolated system always increases Command term: Identify
18B-1A-114
Entropy change in a phase change·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
A mass of 0.150 kg of liquid nitrogen at its boiling point, −196 °C, is completely vaporized at constant pressure. The specific latent heat of vaporization of nitrogen is 2.0 × 105 J kg−1.
What is the change in entropy of the nitrogen?
Show mark scheme
Marking point
Mark
Notes
Step 1Thermal energy supplied at constant temperature: Q = mL = 0.150 × 2.0 × 105 = 3.0 × 104 J.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The temperature must be absolute: −196 °C = 77 K.
—
Step 3ΔS = ΔQ/T = 3.0 × 104/77 = +390 J K−1 (positive: energy enters and the gas is more disordered).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis divides by the Celsius temperature, −196, which gives a negative value although energy enters the nitrogen.
BThis converts the temperature with the wrong sign, 273 + 196 = 469 K.
CCorrect: ΔS = mL/T = 3.0 × 104 J/77 K.
DThis multiplies the energy by the temperature instead of dividing by it.
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; B.1 — specific latent heat; phase changes occur at constant temperature; the Kelvin and Celsius scales Command term: Calculate
19B-1A-115
Processes on a volume–temperature graph·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
A fixed mass of an ideal gas is taken round the cycle J → K → L → M → J shown on the graph of volume V against absolute temperature T. The dashed lines are extensions of KL and MJ to the origin.
During which stage is work done on the gas while its internal energy stays constant?
The cycle J → K → L → M → J on a graph of volume against absolute temperature (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1The internal energy of an ideal gas depends only on T, so it is constant only along the vertical lines J → K and L → M (isothermal).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Work is done on the gas when its volume decreases: J → K is an expansion, L → M a compression (6.0 → 2.0 × 10−3 m³).
—
Step 3So L → M, an isothermal compression, is the answer (the energy leaves as heat: Q = W < 0).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AJ → K is isothermal, but the volume increases, so the gas does work on its surroundings; the sign of the work has been reversed.
BK → L lies on a line through the origin, so V ∝ T: an isobaric expansion in which the temperature, and so the internal energy, increases.
CCorrect: constant temperature (ΔU = 0) and decreasing volume (work done on the gas).
DM → J is an isobaric compression, so work is done on the gas, but the temperature falls from 600 K to 300 K, so the internal energy decreases.
Syllabus understandingB.4 — that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔT of a system is related to the change of its temperature; B.3 — changes of state of an ideal gas represented on graphs Command term: Identify
20B-1A-116
Isothermal compression and entropy·B.4 Thermodynamics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A fixed mass of an ideal gas is compressed slowly at constant temperature.
Which row gives the sign of the thermal energy Q transferred to the gas and the sign of the change in entropy of the gas?
Qchange in entropy of the gas
Show mark scheme
Marking point
Mark
Notes
Step 1Constant temperature: ΔU = 0, so the first law gives Q = W.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The gas is compressed, so the work done by the gas is negative; hence Q < 0: thermal energy leaves the gas.
—
Step 3At constant temperature ΔS = ΔQ/T, so the entropy of the gas also decreases (its molecules have less space, fewer microstates).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: Q = W < 0 and ΔS = Q/T < 0.
BThis describes an adiabatic compression (no thermal energy transferred); here the temperature is held constant, so energy must leave as heat.
CThis takes "constant temperature" to mean "constant entropy". ΔS = ΔQ/T is not zero because Q is not zero.
DThis treats the work done on the gas as thermal energy entering it; in fact the energy given as work leaves again as heat.
Syllabus understandingB.4 — that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln ΩCommand term: Deduce
21B-1A-117
Internal energy during an isobaric expansion·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A monatomic ideal gas expands at constant pressure from state X to state Y. The graph shows how the internal energy U of the gas varies with its volume V during the expansion.
What is the thermal energy supplied to the gas?
Internal energy against volume for the isobaric expansion X → Y (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1From the graph: ΔU = 630 − 180 = 450 J.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a monatomic ideal gas U = (3/2)nRT = (3/2)pV, so at constant pressure the gradient is (3/2)p: p = (2/3) × 720/(4.0 × 10−3) = 1.2 × 105 Pa.
—
Step 3W = pΔV = 1.2 × 105 × 2.5 × 10−3 = 300 J, so Q = ΔU + W = 450 + 300 = 750 J.
✓ 1
Answer 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 increases.
BThis is only the change in internal energy; the gas also does work as it expands.
CCorrect: Q = ΔU + W = 450 J + 300 J.
DThis takes the gradient of the graph (1.8 × 105 J m−3) as the pressure, forgetting U = (3/2)pV, so W = 450 J and Q = 900 J.
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that the work done by or on a closed system as given by W = PΔV when its boundaries are changed can be described in terms of pressure and changes of volume of the system; that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔT of a system is related to the change of its temperature; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed Command term: Determine
22B-1A-118
Isothermal and adiabatic changes in sequence·B.4 Thermodynamics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A monatomic ideal gas at absolute temperature T0 and volume V0 expands isothermally to a volume 8V0. It is then compressed adiabatically back to volume V0.
What is the final temperature of the gas?
Show mark scheme
Marking point
Mark
Notes
Step 1After the isothermal expansion the temperature is still T0, at volume 8V0.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Adiabatic change: substituting p = nRT/V into pV5/3 = constant gives TV2/3 = constant.
—
Step 3T0(8V0)2/3 = TfV02/3, so Tf = 82/3T0 = 4T0.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes that returning to the starting volume restores the starting state; the adiabatic compression heats the gas, so the final state is not the initial one.
BCorrect: TV2/3 = constant, and 82/3 = 4.
CThis uses TV = constant (an exponent of 1 instead of 2/3), i.e. takes the temperature to be inversely proportional to the volume.
DThis uses the pressure ratio 85/3 = 32 of an adiabatic change starting at the same pressure, not the temperature ratio.
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by the equation as given by pV5/3 = constant; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; B.3 — the ideal gas equation pV = nRTCommand term: Deduce
23B-1A-119
Identifying a process from a logarithmic graph·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A fixed mass of a monatomic ideal gas changes from state X to state Y. The graph shows lg p against lg V for the change, where p is the pressure in Pa and V the volume in m³.
What change does the graph represent?
lg p against lg V for the change from X to Y (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1A straight line on a lg p–lg V graph means p ∝ Vk, where k is the gradient.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3lg V increases from X to Y, so the volume increases: an adiabatic expansion (in which the gas cools).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: gradient −5/3 and increasing volume.
BThe gradient is right, but the change goes from X to Y, towards larger volume: it is an expansion.
CAn isothermal change (pV constant) would have a gradient of −1; this line has a gradient of −5/3.
DThis inverts the gradient (Δlg V/Δlg p = −0.6), so that pV appears to increase; with the correct gradient −5/3, pV decreases and the gas cools.
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by the equation as given by pV5/3 = constant; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; Tools — linearising a power law with logarithms; gradient of a graph Command term: Deduce
24B-1A-120
Irreversible processes·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In which processes does the total entropy of the system and its surroundings increase?
I. A metal block slides along a rough horizontal floor and comes to rest. II. A gas in a thermally insulated cylinder is compressed very slowly by a frictionless piston. III. Thermal energy is conducted through a window from a warm room to the colder air outside.
Show mark scheme
Marking point
Mark
Notes
Step 1I: friction converts the kinetic energy of the block into internal energy of the block and floor; the reverse never happens spontaneously, so the process is irreversible and the total entropy increases.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: Q = 0 at every stage of a slow, frictionless compression, so ΔS = ΔQ/T = 0 for the gas and nothing else is affected: reversible, no increase.
—
Step 3III: the room loses Q/Troom and the outside air gains Q/Tout, which is larger because Tout < Troom: the total increases.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis includes II, treating the work done on the gas as dissipated energy; in a slow, frictionless adiabatic compression no entropy is produced. It also omits III.
BCorrect: friction and conduction across a temperature difference are irreversible; the slow frictionless adiabatic compression is reversible.
CThis omits I: the ordered kinetic energy of the block becomes disordered internal energy, which increases the total entropy.
DThis assumes every process increases the total entropy; the idealised process II is reversible and leaves it unchanged.
Syllabus understandingB.4 — that processes in real isolated systems are almost always irreversible and consequently the entropy of a real isolated system always increases; that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes and on the overall evolution of the system; that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; that adiabatic processes in monatomic ideal gases can be modelled by the equation as given by pV5/3 = constant Command term: Deduce
25B-1A-121
Carnot engines in series·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two Carnot engines X and Y operate in series. X takes thermal energy from a reservoir at 600 K and rejects thermal energy at 400 K. Y takes in all the thermal energy rejected by X, at 400 K, and rejects thermal energy to a reservoir at 200 K.
What is the ratio (work done by X)/(work done by Y) in each cycle?
All 3 steps must be completed — there is no mark for a part-answer.
Step 2If X takes in Q, it does work Q/3 and rejects 2Q/3, which is the input of Y.
—
Step 3Y does work (1/2)(2Q/3) = Q/3, so the ratio is 1.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the ratio of the efficiencies; it ignores that Y receives less energy than X.
BCorrect: X does Q/3 and Y does (1/2)(2Q/3) = Q/3.
CThis inverts the ratio of the efficiencies and also ignores the smaller input of Y.
DThis takes the efficiency as Tc/Th instead of 1 − Tc/Th: X does 2Q/3, Y does (1/2)(Q/3) = Q/6.
Syllabus understandingB.4 — that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/Th; that a heat engine can respond to different cycles and is characterized by its efficiency as given by η = useful work/input energy; that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work Command term: Deduce
26B-1A-122
Entropy change on melting·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Gallium melts at 30 °C. A sample of solid gallium at its melting point is supplied slowly with thermal energy. The graph shows the increase in entropy ΔS of the gallium against the mass m of gallium that has melted.
What is the specific latent heat of fusion of gallium?
Entropy gained by the gallium against the mass that has melted (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1The melting takes place at constant temperature, T = 30 + 273 = 303 K, so ΔS = mL/T.
—
All 3 steps must be completed — there is no mark for a part-answer.
BThis is the gradient of the graph, L/T, in J K−1 kg−1; it has not been multiplied by the temperature.
CThis multiplies the gradient by the Celsius temperature, 30, instead of 303 K.
DCorrect: L = T × gradient = 303 K × 265 J K−1 kg−1.
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; B.1 — specific latent heat; the Kelvin and Celsius scales Command term: Determine
27B-1A-123
Path dependence of heat and work·B.4 Thermodynamics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A fixed mass of a monatomic ideal gas is taken from state X to state Y by two different paths, 1 and 2, shown on the p–V diagram.
Which row compares the change in internal energy ΔU and the thermal energy Q supplied to the gas along path 1 with those along path 2?
Two paths from X to Y on a pressure–volume diagram (drawn to scale)
ΔU along path 1 compared with path 2Q along path 1 compared with path 2
Show mark scheme
Marking point
Mark
Notes
Step 1U is a function of state, so ΔU is the same for both paths (here pV = 300 J at X and at Y, so ΔU = 0 for both).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Work done by the gas = area under the path: path 1, 3.0 × 105 × 2.0 × 10−3 = 600 J; path 2, 1.0 × 105 × 2.0 × 10−3 = 200 J.
—
Step 3Q = ΔU + W, so Q is greater along path 1 (600 J against 200 J).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: same ΔU, but more work and so more thermal energy along path 1.
BThis treats Q like U, as a function of state; the thermal energy depends on the path because the work does.
CThis assumes that the larger area also changes the internal energy; U depends only on the end states.
DThis reverses the effect of the work: with ΔU fixed, more work done by the gas requires more thermal energy, not less.
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that the work done by or on a closed system as given by W = PΔV when its boundaries are changed can be described in terms of pressure and changes of volume of the system; that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔT of a system is related to the change of its temperature Command term: Deduce
28B-1A-124
Entropy production by a real engine·B.4 Thermodynamics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In each cycle a heat engine takes thermal energy Q from a hot reservoir at absolute temperature Th and rejects thermal energy to a cold reservoir at absolute temperature Tc. Its efficiency is half of the Carnot efficiency for these temperatures.
What is the total change in entropy of the two reservoirs in each cycle?
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes and on the overall evolution of the system; that processes in real isolated systems are almost always irreversible and consequently the entropy of a real isolated system always increases; that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/Th; that a heat engine can respond to different cycles and is characterized by its efficiency as given by η = useful work/input energy Command term: Deduce
29B-1B-02
Adiabatic processes·B.4 Thermodynamics (HL)
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
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 / dm³
2.00
4.00
6.00
8.00
10.00
time for 20 oscillations / s
8.8
12.3
15.0
17.5
19.4
T² / s²
0.194
0.562
0.766
0.941
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.
Requires A = π(8.0 × 10−3)² = 2.01 × 10−4 m² and the gradient in s² m−3.
γ = 1.67
✓ 1
Accept 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
✓ 1
Allow 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)
Syllabus understandingB.4 — adiabatic processes in monatomic ideal gases as given by pV5/3 = constant; C.1 — the conditions that lead to simple harmonic motion and its period; Tools — linearising, gradient with unit, propagation of uncertainties Command term: Determine
30B-1B-04
Heat engines and the Carnot cycle·B.4 Thermodynamics (HL)
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
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.
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 point
Mark
Notes
Part (a)(i)
Useful work = mgh = 2.00 × 9.81 × 1.69 = 33.2 J
✓ 1
η = 33.2/600 = 0.055
✓ 1
Accept 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)
✓ 1
Allow 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
✓ 1
Accept 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
✓ 1
Allow 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)
✓ 1
Answers 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)
Syllabus understandingB.4 — that a heat engine … is characterized by its efficiency as given by η = useful work/input energy; that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/Th; A.3 — work done and gravitational potential energy; Tools — extrapolation Command term: Determine
31B-1B-18
Heating at constant volume and constant pressure·B.4 Thermodynamics (HL)
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
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 / K
5
10
15
20
25
Q (constant volume) / J
28
52
78
103
128
Q (constant pressure) / J
44
87
127
170
211
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 point
Mark
Notes
Part (a)(i)
Gradient = (153 − 2.5)/30 = 5.0 J K−1
✓ 1
Accept 4.9–5.1 J K−1.
Gradient = (3/2)nR, so n = 5.0/(1.5 × 8.31) = 0.40 mol
✓ 1
Accept 0.39–0.41 mol.
Part (b)(i)
Gradient = 8.3 J K−1
✓ 1
Accept 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
✓ 1
Allow ECF from the gradient in (a). Theory must be quoted: Q = ΔU + pΔV.
The 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 %
✓ 1
Allow 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
✓ 1
Allow 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)
Syllabus understandingB.4 — the first law of thermodynamics as given by Q = ΔU + W; the work done by a closed system as given by W = pΔV; ΔU = (3/2)nRΔT; isovolumetric and isobaric processes; A.2 — weight and force balance on the piston; Tools — gradients and systematic errors Command term: Determine
32B-1B-19
Entropy change from experimental data·B.4 Thermodynamics (HL)
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
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 / s
0
120
240
360
480
600
m (funnel H) / g
0.0
24.2
47.6
72.1
95.7
120.0
m (funnel C) / g
0.0
2.5
4.7
7.2
9.7
11.9
(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 point
Mark
Notes
Part (a)(i)
m/t = 0.202, 0.200, 0.200 g s−1 (e.g. at 120 s, 360 s and 600 s)
✓ 1
A 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)
✓ 1
A 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
✓ 1
Allow ECF from (a). Using the first interval only is capped at [1 max]; the control must be subtracted.
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)
✓ 1
Allow 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)
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T; that the entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase of the entropy of the surroundings; B.1 — specific latent heat; Tools — control experiment, testing a hypothesis, using the full data range Command term: Determine
33B-1B-35
Efficiency of a model Stirling engine·B.4 Thermodynamics (HL)
Paper 1BEasy7 marks
Data-based question7 steps to full marksDetermine
A small Stirling engine stands on a hot plate. Its hot end is at temperature θh, measured with a thermocouple, and its cooling fins keep its cold end at the room temperature of 20 °C. The engine turns a small generator connected to a resistor of resistance R = 12.0 Ω. For several values of θh a student measures the rate Pin at which thermal energy enters the engine and the potential difference V across the resistor. The student treats the electrical power in the resistor as the useful power output of the engine, so its efficiency is η = V2/(RPin).
The uncertainty in each value of V is ±0.01 V and the uncertainty in each value of Pin is ±3 %.
θh / °C
Pin / W
V / V
η
60
8.00
0.96
0.0096
80
12.0
1.40
0.0136
100
16.0
1.81
120
20.0
2.21
0.0204
140
24.0
2.57
0.0229
(a)
(i)
Calculate the missing value of η for θh = 100 °C.
(1)
(ii)
Determine the absolute uncertainty in your answer to (a)(i).
(2)
(b)
(i)
The student suggests that the efficiency of the engine is proportional to the Carnot efficiency for the temperatures of its hot and cold ends. Test this hypothesis, using three rows of the table.
(2)
(c)
(i)
Suggest one reason why the efficiency of the engine is much smaller than the Carnot efficiency.
(1)
(d)
(i)
Predict the useful power output of the engine when θh = 160 °C and Pin = 28 W.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
η = 1.81²/(12.0 × 16.0) = 0.0171
✓ 1
Accept 0.0171 or 1.71 %. Answer to at least 2 s.f.
Allow ECF from (a)(i). Absolute uncertainty to 1 s.f. (accept ±0.0007 to ±0.0008).
Part (b)(i)
Carnot efficiency in kelvin for three rows, e.g. 60 °C: 1 − 293/333 = 0.120; 100 °C: 1 − 293/373 = 0.214; 140 °C: 1 − 293/413 = 0.291, and the ratio η/ηCarnot for each: 0.0799, 0.0796, 0.0789
✓ 1
Allow ECF from (a)(i). Celsius temperatures (1 − 20/60 etc.) score [0] for this mark: the ratios then vary from 0.014 to 0.027.
The ratios are constant to within about 1.5 %, so the hypothesis is supported (η ≈ 0.080ηCarnot)
✓ 1
A conclusion consistent with the candidate's ratios is required. A test using two rows only scores [1 max].
Part (c)(i)
Any one: friction in the moving parts (engine and generator); thermal energy conducted directly through the body of the engine from the hot end to the cold end without doing work; electrical losses in the generator coil; the gas is not at the temperature of the hot end while thermal energy enters it (the processes are not reversible)
✓ 1
"Energy is lost" or "the engine is not ideal" alone scores [0].
Part (d)(i)
η = 0.080 × (1 − 293/433) = 0.080 × 0.323 = 0.0259, so Pout = 0.0259 × 28 = 0.72 W
✓ 1
Allow ECF from (b). Accept 0.70–0.75 W.
Answers: (a)(i) 0.0171 · (a)(ii) ±0.0007 · (b)(i) η/ηCarnot ≈ 0.080 · (d)(i) 0.72 W (the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that a heat engine can respond to different cycles and is characterized by its efficiency as given by η = useful work/input energy; that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/Th; that cyclic gas processes are used to run heat engines; B.5 — electrical power P = V2/R; Tools — propagation of uncertainties, testing a hypothesis with at least three data points Command term: Determine
34B-1B-36
Entropy produced by conduction·B.4 Thermodynamics (HL)
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
An aquarium full of water stands in a room whose air is kept at 20.0 °C, measured with a calibrated thermometer. A thermostat switches an electrical heater in the water so that the water stays at a steady temperature. When the temperature is steady, the mean electrical power P supplied to the heater is equal to the rate at which thermal energy passes from the water through the glass walls to the room. A student sets the thermostat to several temperatures and records the reading θ of a digital thermometer in the water and the mean power P. The graph shows P against θ with the line of best fit.
θ / °C
24.0
26.0
28.0
30.0
32.0
P / W
7.7
12.6
17.4
22.4
27.0
Heater power against thermometer reading with the line of best fit (drawn to scale)
(a)
(i)
Determine the gradient of the line. Give a unit for your answer.
(2)
(b)
(i)
The line meets the θ axis at a temperature that is not 20.0 °C. Explain what this suggests about the digital thermometer.
(1)
(c)
(i)
For the thermometer reading of 30.0 °C, determine the rate at which entropy is produced by the transfer of thermal energy from the water to the room. Use your answer to (b).
(2)
(d)
(i)
The answer to (c) is not the total rate at which entropy is produced in the aquarium and the room. Explain why the total rate is greater.
(1)
(e)
(i)
Suggest one improvement to the procedure that would remove the error identified in (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (27.1 − 0)/(32.0 − 20.8) = 2.42
✓ 1
Accept 2.35–2.50.
Unit: W K−1 (or W °C−1)
✓ 1
The unit mark is independent of the value.
Part (b)(i)
When P = 0 the water must be at the temperature of the room, 20.0 °C, but the line meets the axis at about 20.8 °C: the digital thermometer reads about 0.8 K too high (a systematic/zero error)
✓ 1
Accept 0.6–1.0 K. The direction (reads too high) is required.
Part (c)(i)
Water temperature = 30.0 − 0.8 = 29.2 °C = 302.2 K; rate = P(1/Troom − 1/Twater) = 22.4 × (1/293.0 − 1/302.2)
✓ 1
Allow ECF from (b). Both terms with the room term positive are required. Using 303 K (uncorrected) gives 2.5 × 10−3 W K−1: [1 max].
= 2.3 × 10−3 W K−1
✓ 1
Accept 2.2–2.4 × 10−3 W K−1. Allow ECF from (b).
Part (d)(i)
The conversion of electrical energy into thermal energy in the heater (dissipation in its resistance) is itself irreversible and produces entropy, at a rate P/Twater ≈ 22.4/302 ≈ 7.4 × 10−2 W K−1, in addition to that produced by the conduction (total P/Troom ≈ 7.6 × 10−2 W K−1)
✓ 1
The irreversible heating in the heater must be identified; the numerical values are not required. "Energy is lost as heat" alone scores [0].
Part (e)(i)
Measure the room temperature with the same digital thermometer (so that only temperature differences are used) / calibrate the digital thermometer against the calibrated thermometer, e.g. in the room air or in melting ice, and correct the readings
✓ 1
"Use a better thermometer" alone scores [0].
Answers: (a)(i) 2.42 W K−1 · (b)(i) reads about 0.8 K too high · (c)(i) 2.3 × 10−3 W K−1(the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes and on the overall evolution of the system; that processes in real isolated systems are almost always irreversible and consequently the entropy of a real isolated system always increases; B.1 — thermal energy transfer by conduction; rate proportional to the temperature difference; Tools — gradient with unit, intercept, systematic (zero) error Command term: Determine
35B-1B-37
Pressure changes after a rapid compression·B.4 Thermodynamics (HL)
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
A cylinder fitted with a piston contains 100.0 cm³ of argon, a monatomic ideal gas, at 101 kPa and 293 K. The piston is pushed in very rapidly, reducing the volume to 50.0 cm³, and is then clamped. A pressure sensor records the pressure p of the argon at intervals of 0.2 s, starting 0.2 s after the end of the compression (t = 0). The walls of the cylinder stay at 293 K. The graph shows the data, with error bars, and a best-fit curve drawn from 0.2 s to 3.0 s.
Pressure of the argon against time after the compression, with error bars and the best-fit curve (drawn to scale)
(a)
(i)
Use the graph to estimate the pressure of the argon immediately after the compression, at t = 0.
(1)
(b)
(i)
Deduce whether your answer to (a) is consistent with an adiabatic compression of the argon.
(2)
(c)
(i)
Determine the temperature of the argon immediately after the compression.
(2)
(d)
(i)
Explain, using the first law of thermodynamics, why the pressure then falls and approaches a constant value of about 202 kPa.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Curve extended back to t = 0: p ≈ 316 kPa
✓ 1
Accept 308–324 kPa. Reading the first data point (294 kPa) scores [0].
316 kPa is within about 2 % of 321 kPa (within the uncertainty of the extrapolation), so the compression is consistent with (almost) adiabatic
✓ 1
Allow ECF from (a). The conclusion must agree with the candidate's own value.
Part (c)(i)
T = 293 × (316 × 50.0)/(101 × 100.0)
✓ 1
Allow ECF from (a). ALT: T = 293 × 22/3 = 465 K from TV2/3 = constant (accept).
= 458 K
✓ 1
Accept 450–466 K.
Part (d)(i)
The volume is constant, so W = 0; thermal energy flows from the hot argon to the cooler walls (Q < 0), so ΔU < 0 and the temperature falls; at constant volume p ∝ T, so p falls
✓ 1
The link ΔU = Q (with W = 0) to a falling temperature is required.
The transfer stops when the argon reaches the wall temperature, 293 K; then pV = 101 × 100.0 at 50.0 cm³ gives 202 kPa, twice the initial pressure (the end state is the same as for a slow isothermal compression)
✓ 1
Allow ECF from (c). "Heat is lost" without the link to the final temperature scores [0] for this mark.
Answers: (a)(i) 316 kPa · (b)(i) 321 kPa predicted; consistent · (c)(i) 458 K (the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by the equation as given by pV5/3 = constant; that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; B.3 — the ideal gas equation; Tools — extrapolation of a best-fit curve, comparison with a model within uncertainty Command term: Determine
36B-2-02
Thermodynamic cycles·B.4 Thermodynamics (HL)
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
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.
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.
Thermal energy removed from D to A = 2.5 × 1.0 × 8.31 × (632 − 300) = 6.9 × 103 J
✓ 1
No thermal energy is transferred in the adiabatic stages. Allow ECF from (c).
η = 1 − 6.9 × 103/1.20 × 104 = 0.43
✓ 1
Accept 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
✓ 1
Allow 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
✓ 1
Accept 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)
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by pV5/3 = constant; the first law Q = ΔU + W; W = pΔV; that cyclic gas processes are used to run heat engines characterized by their efficiency; B.3 — pV = nRT and changes of state on pressure–volume diagrams; E.4 — energy released in fission Command term: Determine
37B-2-05
Heat engines and the Carnot cycle·B.4 Thermodynamics (HL)
Paper 2Hard9 marks
Short answer & extended response9 steps to full marksDetermine
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.
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 point
Mark
Notes
Part (a)(i)
Temperatures read: 26 °C = 299 K and 4 °C = 277 K
✓ 1
Accept 25–27 °C and 3–5 °C.
ηCarnot = 1 − 277/299 = 0.074
✓ 1
Accept 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
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)
Syllabus understandingB.4 — that the Carnot cycle sets a limit for the efficiency of a heat engine as given by ηCarnot = 1 − Tc/Th; entropy change ΔS = ΔQ/T; that the second law refers to the change in entropy of an isolated system; B.1 — Kelvin and Celsius scales; Q = mcΔTCommand term: Determine
38B-2-07
Adiabatic processes·B.4 Thermodynamics (HL)
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksDetermine
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.
(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.
The hidden step: the gas fills the reservoir and the barrel.
p = 5.0 × 106 × (2.0/3.2)5/3 = 2.28 × 106 Pa
✓ 1
At 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
Accept 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
✓ 1
Allow 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)
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by pV5/3 = constant; the first law Q = ΔU + W; ΔU = (3/2)nRΔT; B.3 — pV = nRT; A.3 — that work done by the resultant force equals the change in energy, Ek = ½mv²; A.2 — conservation of momentum in explosions Command term: Determine
39B-2-08
Entropy and microstates·B.4 Thermodynamics (HL)
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksExplain
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.
(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 point
Mark
Notes
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
✓ 1
Both 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
✓ 1
Allow 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
Allow 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
✓ 1
Allow 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
✓ 1
Allow 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)
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities as ΔS = ΔQ/T and in terms of microstates as S = kB ln Ω; that the microstates of a system are equally probable; that processes in real isolated systems are almost always irreversible and the entropy of a real isolated system always increases; the first law; B.3 — pV = NkBTCommand term: Explain
40B-2-10
Work from a cyclic p–V diagram·B.4 Thermodynamics (HL)
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksDetermine
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.
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
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Notes
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
✓ 1
Any valid method (e.g. triangles/trapezia); the value of one square must be correct.
Net work ≈ 3.2 × 102 J
✓ 1
Accept 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
✓ 1
Award [1 max] for 6.3 × 104 W (one cycle per revolution).
P = 4 × 25 × 317 = 3.2 × 104 W
✓ 1
Allow 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
✓ 1
Allow ECF from (c).
F = P/v = 2.54 × 104/27 = 9.4 × 102 N
✓ 1
Accept 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
✓ 1
Allow 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)
✓ 1
Accept 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)
Syllabus understandingB.4 — that the work done by or on a closed system as given by W = PΔV when its boundaries are changed can be described in terms of pressure and changes of volume of the system; that cyclic gas processes are used to run heat engines; that a heat engine can respond to different cycles and is characterized by its efficiency as given by η = useful work/input energy; that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/Th; A.3 — that power developed P is the rate of work done as given by P = ΔW/Δt = Fv; efficiency Command term: Determine
41B-2-16
First law of thermodynamics·B.4 Thermodynamics (HL)
Paper 2Hard10 marks
Short answer & extended response10 steps to full marksDetermine
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.
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
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Notes
Part (a)(i)
The piston is in equilibrium: pA = p0A + kx, where x is the compression of the spring
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)
Syllabus understandingB.4 — the first law Q = ΔU + W; that the work done by a closed system can be described in terms of pressure and changes of volume, including situations where pressure is not constant; ΔU = (3/2)nRΔT; A.2 — elastic restoring force FH = −kx and translational equilibrium; A.3 — elastic potential energy EH = ½k(Δx)² Command term: Determine
42B-2-17
Heat engines and the Carnot cycle·B.4 Thermodynamics (HL)
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
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.
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
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Notes
Part (a)(i)
P at t = 0 and at t = 40 years read as 2720 W and 1980 W
✓ 1
Accept ±20 W on each reading.
P ∝ e−λt, so λ = ln(2720/1980)/40 = 7.94 × 10−3 yr−1
✓ 1
An 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
Allow ECF from (a)(i). The decay constant must be in s−1.
N0 = A0/λ = 1.2 × 1025
✓ 1
Allow ECF from (b).
Mass = 1.2 × 1025 × 238 × 1.66 × 10−27 = 4.8 kg
✓ 1
Accept 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
✓ 1
Allow 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
✓ 1
Allow 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)
Syllabus understandingB.4 — that the Carnot cycle sets a limit for the efficiency of a heat engine as given by ηCarnot = 1 − Tc/Th; B.1 — Stefan–Boltzmann law; B.2 — emissivity; E.3 — the decay constant and the radioactive decay law N = N0e−λt; activity A = λN; T1/2 = ln 2/λCommand term: Determine
43B-2-26
Entropy and the Clausius form of the second law·B.4 Thermodynamics (HL)
Paper 2Hard11 marks
Short answer & extended response8 steps to full marksDetermine
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.
(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)
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Notes
Part (a)(i)
ΔQ/Δt = kAΔT/Δx = 390 × 4.0 × 10−4 × 100/0.250
✓ 1
= 62.4 W
✓ 1
Accept 62 W.
Part (a)(ii)
Q = 62.4 × 600 = 3.74 × 104 J; mass = 3.74 × 104/3.34 × 105 = 0.112 kg
✓ 1
Allow ECF.
Part (b)(i)
ΔShot = −3.74 × 104/373 = −100 J K−1
✓ 1
Negative sign required. ΔS = ΔQ/T applied only at constant temperature (phase change / reservoir).
ΔScold = +3.74 × 104/273 = +137 J K−1
✓ 1
ΔS = ΔQ/T applied only at constant temperature (phase change / reservoir).
Total = +37 J K−1
✓ 1
Accept 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
✓ 1
Accept an argument using the Carnot relation for a reversible machine.
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)
Syllabus understandingB.4 — entropy change ΔS = ΔQ/T; that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes (Clausius form); that the entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase of the entropy of the surroundings; B.1 — thermal conduction ΔQ/Δt = kAΔT/ΔxCommand term: Determine
44B-2-27
Internal energy and irreversible processes·B.4 Thermodynamics (HL)
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine
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.
(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)
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Notes
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
✓ 1
An 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
The constancy of V must be stated or used.
Part (d)(ii)
Δp = 2 × 212/(3 × 1.20 × 10−2) = 1.18 × 104 Pa
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
The 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)
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system; that the change in internal energy as given by ΔU = (3/2)nRΔT of a system is related to the change of its temperature; that entropy can be determined as given by ΔS = ΔQ/T; that processes in real isolated systems are almost always irreversible (Guidance: the second law in Clausius form and Kelvin form); A.3 — the principle of the conservation of energy; ΔEp = mgΔhCommand term: Determine
45B-2-28
A Stirling engine cycle·B.4 Thermodynamics (HL)
Paper 2Hard14 marks
Short answer & extended response10 steps to full marksDetermine
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
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Notes
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
✓ 1
Pressures: 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
✓ 1
Accept 750 J.
Part (b)(ii)
ΔU = 0 (isothermal), so Q = W = 0.150 × 8.31 × 720 × ln 2.5
✓ 1
Q = +822 J
✓ 1
Accept 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 %)
✓ 1
Accept 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
✓ 1
The 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)
Syllabus understandingB.4 — that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; the first law Q = ΔU + W; ΔU = (3/2)nRΔT; that cyclic gas processes are used to run heat engines; that a heat engine can respond to different cycles and is characterized by its efficiency η = useful work/input energy; that the Carnot cycle sets a limit for the efficiency of a heat engine as given by ηCarnot = 1 − Tc/Th; isothermal work W = nRT ln(Vf/Vi) given in the stem Command term: Determine
46B-2-29
Adiabatic processes·B.4 Thermodynamics (HL)
Paper 2Hard14 marks
Short answer & extended response9 steps to full marksDetermine
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
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Notes
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
Equal 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
✓ 1
Numerical 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)
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by the equation pV5/3 = constant; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; the first law Q = ΔU + W; ΔS = ΔQ/T; B.3 — changes of state of an ideal gas represented on pressure–volume diagrams Command term: Determine
47B-2-55
Isobaric heating of a balloon·B.4 Thermodynamics (HL)
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksDetermine
A scientific balloon contains 4.0 × 103 mol of helium, a monatomic ideal gas. The balloon is not completely filled, and an open duct at its base keeps the pressure of the helium equal to that of the surrounding air. During the night the balloon floats at a height where the air pressure is 12.0 kPa and the helium is at 220 K. After sunrise, sunlight warms the helium to 245 K while the balloon is at the same height. The density of the air at this height is 0.19 kg m−3.
(a)
(i)
Show that the volume of the helium at night is about 610 m³.
(1)
(ii)
Outline why the warming of the helium is an isobaric process.
(1)
(b)
(i)
Calculate the work done by the helium as it warms from 220 K to 245 K.
(2)
(ii)
Determine the thermal energy absorbed by the helium.
(2)
(iii)
Show that, for any isobaric expansion of a monatomic ideal gas, two-fifths of the thermal energy absorbed is used to do work.
(1)
(c)
(i)
Determine the increase in the upward buoyant force on the balloon as the helium warms.
(2)
(d)
(i)
At sunset the helium cools at constant pressure back to 220 K. State the sign of Q and the sign of W for the helium during the cooling, where Q = ΔU + W.
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
The pressure of the helium equals the pressure of the air at that height, which does not change while the balloon stays at the same height (the balloon is slack, so the helium can expand freely)
✓ 1
Reference to the unchanged air pressure is required.
W = pΔV = nRΔT and Q = (3/2)nRΔT + nRΔT = (5/2)nRΔT, so W/Q = 2/5
✓ 1
The use of pV = nRT at constant pressure must be seen. Numerical check alone (8.31/20.8) scores [0].
Part (c)(i)
Buoyant force = weight of displaced air: ΔFb = ρΔVg
✓ 1
= 0.19 × 69.2 × 9.81 = 129 N
✓ 1
Allow ECF from (b)(i). Accept 125–135 N.
Part (d)(i)
W negative (the helium is compressed, work is done on it) and Q negative (thermal energy leaves the helium)
✓ 1
Both signs are required.
Answers: (a)(i) 609 m³ · (b)(i) 8.31 × 105 J · (b)(ii) 2.08 × 106 J · (c)(i) 129 N (the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that the work done by or on a closed system as given by W = PΔV when its boundaries are changed can be described in terms of pressure and changes of volume of the system; that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔT of a system is related to the change of its temperature; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; B.3 — the ideal gas equation pV = nRT; A.2 — buoyancy Fb = ρVgCommand term: Determine
48B-2-56
Entropy changes in a freezer·B.4 Thermodynamics (HL)
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksDetermine
A tray holding 0.400 kg of water at 0 °C is placed in the freezer compartment of a refrigerator in a kitchen at 22 °C. The water freezes at 0 °C. While the water freezes, the electric motor of the compressor of the refrigerator works at a power of 80 W for 12.0 minutes. Assume that all the thermal energy removed from the freezer compartment during this time comes from the freezing water, and that the refrigerator transfers thermal energy to the kitchen, which acts as a reservoir at constant temperature.
Specific latent heat of fusion of water = 3.34 × 105 J kg−1.
(a)
(i)
Calculate the thermal energy that must be removed from the water to freeze it.
(1)
(b)
(i)
Determine the change in entropy of the water.
(2)
(c)
(i)
Calculate the energy transferred to the motor of the compressor.
(1)
(ii)
Determine the total change in entropy of the water and the kitchen while the water freezes.
(3)
(d)
(i)
The entropy of the water decreases. Explain, with reference to your answers, why this is consistent with the second law of thermodynamics.
(1)
(e)
(i)
State the Clausius form of the second law of thermodynamics. Hence explain why the refrigerator cannot freeze the water without work being done on it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Q = mL = 0.400 × 3.34 × 105 = 1.34 × 105 J
✓ 1
Part (b)(i)
The water stays at 273 K while it freezes, so ΔS = ΔQ/T with ΔQ = −1.34 × 105 J
✓ 1
Allow ECF from (a). Celsius temperature scores [0].
ΔS = −1.336 × 105/273 = −489 J K−1
✓ 1
The negative sign is required. Accept −489 to −491 J K−1 (rounding of (a)).
Part (c)(i)
W = Pt = 80 × 720 = 5.76 × 104 J
✓ 1
Unit conversion of 12.0 min to 720 s is required.
Part (c)(ii)
Energy transferred to the kitchen = 1.336 × 105 + 5.76 × 104 = 1.91 × 105 J
✓ 1
Allow ECF from (a) and (c)(i). The refrigerator works in a cycle, so all the energy entering it leaves to the kitchen.
ΔSkitchen = 1.91 × 105/295 = +648 J K−1
✓ 1
Kelvin temperature required.
Total = −489 + 648 = +159 J K−1
✓ 1
Allow ECF from (b). Accept 155–162 J K−1.
Part (d)(i)
The water is not an isolated system; its entropy decrease (−489 J K−1) is more than compensated by the increase in the entropy of the kitchen, so the total entropy increases (+159 J K−1), as the second law requires
✓ 1
Allow ECF from (b) and (c)(ii). "Entropy always increases" alone scores [0].
Part (e)(i)
Thermal energy cannot transfer spontaneously from a colder body to a hotter body
✓ 1
OWTTE; "without work being done" may be included.
The energy removed from the water at 273 K must be delivered to the warmer kitchen at 295 K; with no work, the kitchen would gain only 1.336 × 105/295 = 453 J K−1, less than the 489 J K−1 lost by the water, so the total entropy would decrease (by about 36 J K−1)
✓ 1
Allow ECF from (a) and (b). A qualitative argument (the transfer is from cold to hot, which the Clausius form forbids unless work is done) scores this mark.
Answers: (a)(i) 1.34 × 105 J · (b)(i) −489 J K−1 · (c)(i) 5.76 × 104 J · (c)(ii) +159 J K−1(the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that the entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase of the entropy of the surroundings; that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes and on the overall evolution of the system; that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; Guidance: the second law in Clausius form; B.1 — specific latent heat; A.3 — power as the rate of energy transfer Command term: Determine
49B-2-57
Adiabatic expansion in a cryocooler·B.4 Thermodynamics (HL)
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
A cryocooler keeps an infrared detector at a low temperature. In each cycle, a fixed mass of helium, a monatomic ideal gas, at a pressure of 2.00 MPa, a volume of 10.0 cm³ and a temperature of 120 K expands rapidly against a piston to a volume of 25.0 cm³. The graph shows the temperature T of the helium against its volume V during the expansion.
Temperature of the helium against its volume during the expansion (drawn to scale)
(a)
(i)
Calculate the amount of helium, in mol.
(1)
(ii)
Show that, for an adiabatic change of a monatomic ideal gas, TV2/3 = constant.
(2)
(b)
(i)
Use the graph to deduce that the expansion is adiabatic.
(2)
(ii)
State the temperature of the helium at the end of the expansion.
(1)
(c)
(i)
Determine the work done by the helium during the expansion.
(2)
(d)
(i)
Explain, using the first law of thermodynamics, why the temperature of the helium falls during the expansion although no thermal energy is removed from it.
(2)
(e)
(i)
After each expansion the piston is held fixed and the cold helium absorbs thermal energy from a metal shield around the detector, warming by 10 K at constant volume. The shield absorbs radiation from the surrounding housing at 300 K at a rate P = eσAT4, where T = 300 K, the emissivity e = 0.20 and the area A = 2.0 × 10−2 m². Determine the minimum number of cycles per second needed to remove this energy.
Substitute p = nRT/V into pV5/3 = constant: (nRT/V)V5/3 = constant
✓ 1
so TV2/3 = constant, since nR is constant
✓ 1
Any valid algebraic route.
Part (b)(i)
Three readings from the graph, e.g. (10.0 cm³, 120 K), (16.0 cm³, 88 K), (25.0 cm³, 65 K), and TV2/3 for each: 557, 559, 556 (K cm²)
✓ 1
Allow ECF from (a)(ii). Readings within ±1 K.
The products are constant (to within about 1 %), so the expansion follows TV2/3 = constant: it is adiabatic
✓ 1
A conclusion is required. Two points only: [1 max].
Part (b)(ii)
65 K
✓ 1
Accept 64–66 K, from the graph or from 120 × (10.0/25.0)2/3.
Part (c)(i)
Q = 0, so W = −ΔU = (3/2)nR(T1 − T2)
✓ 1
= 1.5 × 2.01 × 10−2 × 8.31 × (120 − 65) = 13.8 J
✓ 1
Allow ECF from (a)(i) and (b)(ii). Accept 13.5–14.0 J (13.7 J with unrounded values).
Part (d)(i)
The expansion is so rapid that Q = 0, and the helium does positive work on the piston (W > 0)
✓ 1
So ΔU = Q − W = −W < 0; the internal energy of a monatomic ideal gas is the kinetic energy of its molecules, so the mean kinetic energy and the temperature fall
✓ 1
The link from internal energy to temperature is required.
Part (e)(i)
Energy absorbed per cycle = (3/2)nRΔT = 1.5 × 2.01 × 10−2 × 8.31 × 10 = 2.51 J
✓ 1
Allow ECF from (a)(i). W = 0 at constant volume, so Q = ΔU.
P = 0.20 × 5.67 × 10−8 × 2.0 × 10−2 × 3004 = 1.84 W
✓ 1
Cycles per second = 1.84/2.51 = 0.73 s−1
✓ 1
Allow ECF from (e) first two marks. Accept 0.72–0.75 s−1.
Answers: (a)(i) 2.01 × 10−2 mol · (b)(i) TV2/3 ≈ 557 (K cm²), constant · (b)(ii) 65 K · (c)(i) 13.8 J · (e)(i) 0.73 s−1(the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that adiabatic processes in monatomic ideal gases can be modelled by the equation as given by pV5/3 = constant; that the first law of thermodynamics as given by Q = ΔU + W results from the application of conservation of energy to a closed system and relates the internal energy of a system to the transfer of energy as heat and as work; that the change in internal energy as given by ΔU = (3/2)NkBΔT = (3/2)nRΔT of a system is related to the change of its temperature; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; B.3 — the ideal gas equation; B.1/B.2 — Stefan–Boltzmann law and emissivity Command term: Determine
50B-2-58
Entropy and the erasure of information·B.4 Thermodynamics (HL)
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A memory register in a computer stores N bits. Each bit is a small cell that is in one of two states, 0 or 1. When the register holds random data, each of its possible arrangements of 0s and 1s (its microstates) is equally probable. Erasing the register resets every bit to 0, whatever its previous state.
In a simple model, the entropy of the register is S = kB ln Ω, where Ω is the number of microstates of the register that are consistent with what is known about it.
(a)
The register stores 8 bits of random data.
(i)
Show that the number of microstates of the register is 256.
(1)
(ii)
Determine the probability that the register contains exactly two 1s.
(2)
(b)
(i)
Determine the change in entropy of the register when the 8-bit register holding random data is erased.
(2)
(c)
(i)
Explain, with reference to the second law of thermodynamics, why erasing the register must transfer thermal energy to its surroundings.
(2)
(d)
(i)
The surroundings are at 300 K. Show that the minimum thermal energy transferred to the surroundings for each bit erased is about 3 × 10−21 J.
(2)
(e)
(i)
A processor operates from a supply of 1.10 V and draws a current of 15.0 A. It erases 2.0 × 1017 bits per second, and all the electrical energy it uses becomes thermal energy. Determine, for each bit erased, the ratio of the thermal energy produced by the processor to the minimum value in (d).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Each bit has 2 possible states, independently of the others: Ω = 28 = 256
✓ 1
Multiplication (not addition) of the possibilities must be clear.
Part (a)(ii)
Number of microstates with two 1s = number of ways of choosing 2 of the 8 bits = 8 × 7/2 = 28
✓ 1
Listing or a combinatorial argument.
Probability = 28/256 = 0.109
✓ 1
Allow ECF from (a)(i). Accept 0.11.
Part (b)(i)
After erasure only one microstate is possible (all 0s): Ω = 1, so S = 0
Allow ECF from (a)(i). The negative sign is required.
Part (c)(i)
The entropy of the register decreases; it is not isolated, and the second law requires the total entropy of the register and its surroundings not to decrease
✓ 1
So the entropy of the surroundings must increase by at least as much; at constant temperature this requires thermal energy ΔQ = TΔS to be transferred to them
✓ 1
"Energy is conserved" scores [0]: the argument must use entropy.
Part (d)(i)
Per bit Ω falls by a factor of 2, so ΔS = −kB ln 2 (= ΔS from (b) ÷ 8); minimum Q = kBT ln 2
Answers: (a)(i) 256 · (a)(ii) 0.109 · (b)(i) −7.65 × 10−23 J K−1 · (d)(i) 2.87 × 10−21 J · (e)(i) 2.9 × 104(the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; that entropy S is a thermodynamic quantity that relates to the degree of disorder of the particles in a system; that the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes and on the overall evolution of the system; that the entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase of the entropy of the surroundings; Guidance: the microstates of a system are equally probable; simple combinatorial models; B.5 — electrical power P = IVCommand term: Determine
51B-2-59
A Carnot cycle analysed with entropy·B.4 Thermodynamics (HL)
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDetermine
A reversible engine uses 0.100 mol of helium, a monatomic ideal gas. Its cycle ABCDA has four stages:
A → B: isothermal expansion at Th = 500 K from VA = 1.00 × 10−3 m³ to VB = 2.00 × 10−3 m³; B → C: adiabatic expansion until the temperature is Tc = 300 K; C → D: isothermal compression at 300 K; D → A: adiabatic compression back to A.
The work done by an ideal gas in an isothermal change from Vi to Vf at temperature T is nRT ln(Vf/Vi). For an adiabatic change of a monatomic ideal gas, TV2/3 = constant.
(a)
(i)
Determine the volume at C.
(2)
(ii)
Deduce that VC/VD = VB/VA.
(2)
(b)
(i)
Determine the thermal energy Qh taken in by the helium during A → B, and the change in entropy of the helium during A → B.
(2)
(ii)
Explain why the entropy of the helium does not change during the stages B → C and D → A, and hence determine the thermal energy Qc given out by the helium during C → D.
(2)
(iii)
Hence show that the efficiency of the engine is 1 − Tc/Th.
(1)
(c)
(i)
The engine completes 20 cycles per second and drives a generator of efficiency 0.90 connected to a resistor of resistance 24 Ω. Determine the current in the resistor.
(2)
(d)
(i)
In a real engine of this type the hot reservoir is at 550 K, and the 288 J of (b)(i) passes into the helium while the helium is at 500 K. Determine the total change in entropy of the hot reservoir and the helium during A → B, and explain what your answer shows about the process.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ThVB2/3 = TcVC2/3, so VC = VB(Th/Tc)3/2
✓ 1
VC = 2.00 × 10−3 × (500/300)1.5 = 4.30 × 10−3 m³
✓ 1
Accept 4.3 × 10−3 m³.
Part (a)(ii)
Both adiabatic stages connect the temperatures Th and Tc, so VC/VB = (Th/Tc)3/2 and VD/VA = (Th/Tc)3/2
✓ 1
Allow ECF from (a)(i). The use of the same temperature ratio for D → A is the key step.
Hence VC/VB = VD/VA, which rearranges to VC/VD = VB/VA (= 2.00)
✓ 1
Part (b)(i)
ΔU = 0, so Qh = W = 0.100 × 8.31 × 500 × ln 2 = 288 J
✓ 1
ΔS = Qh/Th = 288/500 = 0.576 J K−1
✓ 1
Allow ECF from the first marking point.
Part (b)(ii)
In the reversible adiabatic stages Q = 0, so ΔS = 0; over a whole cycle the helium returns to its initial state, so the entropy it loses in C → D equals the 0.576 J K−1 it gained in A → B
✓ 1
Allow ECF from (b)(i).
Qc = TcΔS = 300 × 0.576 = 173 J
✓ 1
Allow ECF from (b)(i). ALT: nRTc ln(VC/VD) with the ratio 2.00 from (a)(ii).
Part (b)(iii)
η = (Qh − Qc)/Qh and Qc/Qh = Tc/Th from (b)(ii), so η = 1 − Tc/Th (= 0.40 = 115/288)
✓ 1
Allow ECF from (b)(ii). The ratio Qc/Qh = Tc/Th must come from the equal entropy changes.
Part (c)(i)
Electrical power = 0.90 × (288 − 173) × 20 = 0.90 × 115 × 20 = 2070 W
✓ 1
Allow ECF from (b)(i) and (b)(ii).
I = √(P/R) = √(2070/24) = 9.3 A
✓ 1
Accept 9.2–9.4 A. Award [1 max] for 9.8 A (generator efficiency ignored).
Part (d)(i)
Hot reservoir: ΔS = −288/550 = −0.524 J K−1; helium: +0.576 J K−1 (as in (b)(i))
✓ 1
Allow ECF from (b)(i). The negative sign for the reservoir is required.
Total = −0.524 + 0.576 = +0.052 J K−1
✓ 1
Accept +0.05 J K−1. Allow ECF from (b)(i).
The total entropy increases, so the transfer of thermal energy across a finite temperature difference is irreversible; the engine is no longer reversible and its efficiency is below the Carnot value for 550 K and 300 K
✓ 1
A link to irreversibility is required; "entropy increases" alone scores [0].
Answers: (a)(i) 4.30 × 10−3 m³ · (b)(i) 288 J; 0.576 J K−1 · (b)(ii) 173 J · (c)(i) 9.3 A · (d)(i) +0.052 J K−1(the remaining parts are explanations — see the table above)
Syllabus understandingB.4 — that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by ηCarnot = 1 − Tc/Th; that entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature as given by ΔS = ΔQ/T and also in terms of the properties of individual particles of the system as given by S = kB ln Ω; that adiabatic processes in monatomic ideal gases can be modelled by the equation as given by pV5/3 = constant; that isovolumetric, isobaric, isothermal and adiabatic processes are obtained by keeping one variable fixed; that cyclic gas processes are used to run heat engines; that a heat engine can respond to different cycles and is characterized by its efficiency as given by η = useful work/input energy; that processes in real isolated systems are almost always irreversible and consequently the entropy of a real isolated system always increases; isothermal work nRT ln(Vf/Vi) given in the stem; B.5 — electrical power P = I2RCommand term: Determine
52B-2-66
Gas springs and oscillations·B.4 Thermodynamics (HL)
Paper 2Hard20 marks
Short answer & extended response20 steps to full marksDetermine
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.
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 point
Mark
Notes
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
✓ 1
The hidden step (four supports) must be seen.
p = 1.01 × 105 + 3.92 × 104 = 1.40 × 105 Pa
✓ 1
Atmospheric pressure must be added. Answer to at least 3 s.f.
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)
✓ 1
Measuring over several oscillations is required for this mark.
T = 0.93/4 = 0.23 s
✓ 1
Accept 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
Dependent 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
✓ 1
Allow 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
✓ 1
ΔS = 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)
Syllabus understandingB.3 — pressure p = F/A; the ideal gas equation and Boyle's law; B.4 — the first law Q = ΔU + W; adiabatic processes in monatomic ideal gases modelled by pV5/3 = constant; entropy change ΔS = ΔQ/T; A.2 — translational equilibrium; C.1 — conditions for simple harmonic motion, T = 2π√(m/k) and energy in SHM; C.4 — natural frequency, driving frequency and resonance Command term: Determine
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