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IB Physics B.4: Thermodynamics worked solutions

Worked solutions to every practice question in B.4: Thermodynamics of Photon Physics: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

Type A1 · A1 Work direction[1 mark]

An ideal gas is taken round a cycle in three stages. Stage 1: the pressure rises at constant volume. Stage 2: the gas expands while its pressure falls steadily. Stage 3: the gas is compressed at constant pressure back to its starting state. In which stages do the surroundings do work on the gas?

A. 3 only

B. 1 and 3

C. 2 only

D. 1, 2 and 3

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Work is done on the gas only while its volume falls. Stage 1 has no volume change, so no work is done at all; B counts the pressure rise as work. Stage 2 is an expansion, so the gas does work (C). Stage 3 is the only compression.

Final answer

Question A2

Type A2 · A3 First-law signs[1 mark]

350\,\mathrm{J} of thermal energy is removed from an ideal gas while 120\,\mathrm{J} of work is done on the gas. What is the change in internal energy of the gas?

A. -470\,\mathrm{J} B. -230\,\mathrm{J}

C. +230\,\mathrm{J} D. +470\,\mathrm{J}

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Signs first: Q=-350\,\mathrm{J} (removed) and W=-120\,\mathrm{J} (work done on the gas). \Delta U=Q-W=-350-(-120)=-230\,\mathrm{J}. A treats the 120\,\mathrm{J} as work done by the gas; C and D give the heat the wrong sign.

Final answer

Question B1

Type B1 · B1 Name the process[1 mark]

In which process does an ideal gas convert all the thermal energy supplied to it into work done by the gas?

A. isovolumetric B. isobaric

C. isothermal D. adiabatic

Show solutionHide solution

At constant temperature \Delta U=0, so Q=W. At constant volume W=0 and all of Q raises U (A). At constant pressure three fifths of Q goes into U for a monatomic gas (B). An adiabatic process has Q=0 (D).

Final answer

Question B2

Type B2 · B2 Insulated piston[1 mark]

The outlet of a bicycle pump is sealed and the plunger is pushed in quickly, so that no thermal energy is transferred. W is the work done by the air. Which row describes the air in the pump?

A. Q=0, W<0, \Delta U>0

B. Q=0, W>0, \Delta U<0

C. Q<0, W<0, \Delta U=0

D. Q=0, W<0, \Delta U=0

Show solutionHide solution

Q=0 and the air is compressed, so W<0; then \Delta U=Q-W=-W>0 and the air warms. B describes an adiabatic expansion, C an isothermal compression. D forgets that work changes the internal energy when no heat flows.

Final answer

Question E2

Type E2 · E3 Count microstates[1 mark]

Four distinguishable particles share a total energy of 2E. Each particle can hold only whole multiples of E. How many microstates are possible?

A. 2 B. 4

C. 6 D. 10

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Either one particle holds 2E (4 choices of particle) or two particles hold E each (4\times3/2=6 choices of pair): 4+6=10. A counts the two ways of splitting the energy, not the arrangements. B and C count only one of the two kinds.

Final answer

Question C1

Type C1 · C1 Cycle area & U[1 mark]

An ideal gas is taken once round a cycle that runs clockwise on a pressure–volume diagram. Which statement is correct for the whole cycle?

A. \Delta U=0, and Q_\text{net}=W_\text{net}>0.

B. \Delta U=0, and W_\text{net}=0.

C. \Delta U>0, because thermal energy was supplied.

D. \Delta U=0, and no net thermal energy is supplied.

Show solutionHide solution

Internal energy depends only on the state, and the gas ends where it started, so \Delta U=0. Then Q_\text{net}=W_\text{net}, the enclosed area, positive for a clockwise loop. B mixes up U and W; C ignores that U is a state quantity; D would make the loop's area zero.

Final answer

Question C3

Type C3 · C2 Engine balance[1 mark]

In each cycle a heat engine does 150\,\mathrm{J} of useful work and rejects 450\,\mathrm{J} of thermal energy to its cold reservoir. What is its efficiency?

A. 0.25 B. 0.33

C. 0.50 D. 0.75

Show solutionHide solution

Energy in per cycle =150+450=600\,\mathrm{J}, so \eta=150/600=0.25. B divides by the rejected heat, D is the fraction rejected, and C divides by 450-150.

Final answer

Question D1

Type D1 · D1 Carnot diagram[1 mark]

A Carnot engine runs between a hot and a cold reservoir. During which stage of its cycle is thermal energy transferred to the cold reservoir?

A. isothermal expansion B. adiabatic expansion

C. isothermal compression D. adiabatic compression

Show solutionHide solution

A Carnot cycle is two isotherms and two adiabats, and heat crosses only on the isotherms. Thermal energy enters during the expansion at T_h (A). The gas cools during the adiabatic expansion but exchanges no heat (B is the trap). During compression at T_c, \Delta U=0 and W<0, so Q=W<0: heat leaves to the cold reservoir.

Final answer

Question D2

Type D2 · D3 Second-law limit[1 mark]

A refrigerator runs with its door open in a closed, thermally insulated kitchen. What happens to the temperature of the air in the kitchen over several hours?

A. It falls steadily. B. It stays constant.

C. It rises. D. It falls, then returns to its starting value.

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The refrigerator moves thermal energy from the air in front of it to the coils at its back, and both are in the same kitchen, so that transfer cancels. The electrical energy supplied also ends up as thermal energy in the kitchen, which nothing removes. To cool a room, heat must be rejected outside it. An open oven warms the kitchen for the same reason.

Final answer

Question E1

Type E1 · E1 Entropy language[1 mark]

Which statement about entropy is correct?

A. The entropy of any system can never decrease.

B. Entropy is conserved in every process.

C. A macrostate with more microstates has greater entropy.

D. Entropy is a form of energy, measured in joules.

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S=k_B\ln\Omega grows with the number of microstates. A applies the isolated-system rule to every system; a gas that loses heat loses entropy. B is false for every irreversible process. Entropy is measured in \mathrm{J\,K^{-1}} and is not an energy (D).

Final answer

Tutorial practice

Question A3

Type A3 · A2 Work area[1 mark]

An ideal gas expands from 1.0\times10^{-3}\,\mathrm{m^3} at 3.0\times10^{5}\,\mathrm{Pa} to 4.0\times10^{-3}\,\mathrm{m^3} at 1.0\times10^{5}\,\mathrm{Pa}. On a pressure–volume diagram its path is a straight line. What is the work done by the gas?

A. 300\,\mathrm{J} B. 600\,\mathrm{J}

C. 900\,\mathrm{J} D. 1200\,\mathrm{J}

Show solutionHide solution

The pressure changes, so use the area under the line: a trapezium, \tfrac12(3.0+1.0)\times10^{5}\times3.0\times10^{-3}=600\,\mathrm{J}. A multiplies by the final pressure, C by the initial pressure, and D adds those two rectangles.

Final answer

Question D3

Type D3 · D2 Carnot ratio[1 mark]

A Carnot engine has an efficiency of 0.25. The temperature of the hot reservoir stays fixed. By what factor must the kelvin temperature of the cold reservoir be multiplied to raise the efficiency to 0.40?

A. 0.63 B. 0.80

C. 1.25 D. 1.60

Show solutionHide solution

T_c/T_h=1-\eta goes from 0.75 to 0.60, so the factor is 0.60/0.75=0.80. A divides the efficiencies, D is their inverse ratio, and C inverts the correct ratio.

Final answer

Question A4

Type A4 · A4 Constant-p heat; A3 First-law signs[6 marks]

A cylinder holds 0.30\,\mathrm{mol} of an ideal monatomic gas under a piston. The gas is cooled at a constant pressure of 2.0\times10^{5}\,\mathrm{Pa}, and its volume falls by 2.0\times10^{-3}\,\mathrm{m^3} from an initial 5.0\times10^{-3}\,\mathrm{m^3}.

(a) Show that the gas starts at a temperature of about 400\,\mathrm{K}. [1]

(b) Calculate W for this change. [1]

(d) State which way the thermal energy flows. [1]

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(a) T=PV/nR=(2.0\times10^{5})(5.0\times10^{-3})/(0.30\times8.31)=401\,\mathrm{K} (A1). (b) W=P\Delta V=2.0\times10^{5}\times(-2.0\times10^{-3})=-400\,\mathrm{J}, so 400\,\mathrm{J} of work is done on the gas (A1). (c) For a monatomic gas \Delta U=\tfrac32nR\Delta T=\tfrac32P\Delta V at constant pressure (M1), so \Delta U=-600\,\mathrm{J} (A1); the final temperature is 241\,\mathrm{K}. Q=\Delta U+W=-600+(-400)=-1000\,\mathrm{J} (A1). (d) Q is negative, so 1000\,\mathrm{J} flows from the gas to the surroundings (A1). The gas loses more heat than the work put into it.

Final answer

(a) 401\,\mathrm{K}. (b) -400\,\mathrm{J}. (c) \Delta U=-600\,\mathrm{J}; Q=-1000\,\mathrm{J}. (d) Out of the gas: 1000\,\mathrm{J} leaves.

Question A5

Type A5 · A5 Same heat, paths; A4 Constant-p heat[5 marks]

Two identical samples, X and Y, each contain 0.40\,\mathrm{mol} of an ideal monatomic gas at 300\,\mathrm{K}. X is sealed in a rigid flask. Y is under a freely moving piston, so its pressure stays constant. Each sample then receives 500\,\mathrm{J} of thermal energy.

(a) Calculate the temperature rise of X. [1]

(b) Show that the temperature rise of Y is about 60\,\mathrm{K}. [2]

(d) Explain why the temperature rise of Y is smaller, although both samples received the same energy. [1]

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(a) X cannot expand, so W=0 and \Delta U=Q: \Delta T=500/(1.5\times0.40\times8.31)=100\,\mathrm{K} (A1). (b) At constant pressure W=P\Delta V=nR\Delta T, so Q=\tfrac32nR\Delta T+nR\Delta T=\tfrac52nR\Delta T (M1); \Delta T=500/(2.5\times0.40\times8.31)=60.2\,\mathrm{K} (A1). (c) W=nR\Delta T=0.40\times8.31\times60.2=200\,\mathrm{J}, two fifths of Q (A1). (d) Internal energy depends only on temperature. Y keeps 300\,\mathrm{J} as internal energy because 200\,\mathrm{J} leaves as work; X keeps all 500\,\mathrm{J} (A1).

Final answer

(a) 100\,\mathrm{K}. (b) 60.2\,\mathrm{K}. (c) 200\,\mathrm{J}. (d) Y uses part of Q to do expansion work; only \Delta U raises the temperature.

Question B3

Type B3 · B3 Adiabatic state; B4 Adiabatic work[7 marks]

A sample of 0.050\,\mathrm{mol} of an ideal monatomic gas at 1.0\times10^{5}\,\mathrm{Pa} and 300\,\mathrm{K} is compressed rapidly and reversibly to one quarter of its volume. No thermal energy is transferred.

(a) Show that the initial volume is about 1.25\times10^{-3}\,\mathrm{m^3}. [1]

(b) Calculate the final pressure. [2]

(d) Determine W for this compression, with its sign. [2]

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(a) V=nRT/P=0.050\times8.31\times300/1.0\times10^{5}=1.25\times10^{-3}\,\mathrm{m^3} (A1). (b) PV^{5/3} is constant (M1): P_2=1.0\times10^{5}\times4^{5/3}=1.01\times10^{6}\,\mathrm{Pa} (A1). (c) V_2=3.12\times10^{-4}\,\mathrm{m^3}; T_2=P_2V_2/nR (M1) =756\,\mathrm{K} (A1). The same follows from T_2=300\times4^{2/3}. (d) Q=0, so W=-\Delta U (M1). \Delta U=\tfrac32nR\Delta T=1.5\times0.050\times8.31\times456=284\,\mathrm{J}, so W=-284\,\mathrm{J} (A1). Writing +284\,\mathrm{J} as the work done by the gas loses this mark.

Final answer

(a) 1.25\times10^{-3}\,\mathrm{m^3}. (b) 1.01\times10^{6}\,\mathrm{Pa}. (c) 756\,\mathrm{K}. (d) -284\,\mathrm{J} (284\,\mathrm{J} of work is done on the gas).

Question C2

Type C2 · C1 Cycle area & U[1 mark]

An ideal gas is taken round the cycle X\toY\toZ\toX. At X the volume is 1.0\times10^{-3}\,\mathrm{m^3} and the pressure 1.0\times10^{5}\,\mathrm{Pa}. X\toY: the pressure rises at constant volume to 4.0\times10^{5}\,\mathrm{Pa}. Y\toZ: a straight-line expansion to 3.0\times10^{-3}\,\mathrm{m^3} and 1.0\times10^{5}\,\mathrm{Pa}. Z\toX: compression at constant pressure. What is W for one complete cycle?

A. -300\,\mathrm{J} B. +200\,\mathrm{J}

C. +300\,\mathrm{J} D. +500\,\mathrm{J}

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Net work is the enclosed triangle: \tfrac12(2.0\times10^{-3})(3.0\times10^{5})=300\,\mathrm{J}. The loop goes up, then down to the right, then left: clockwise, so the sign is positive. A has the wrong sign. B is the size of the work done on the gas in Z\toX; D is the area under Y\toZ alone.

Final answer

Question D4

Type D4 · D3 Second-law limit[1 mark]

Which process is ruled out by the second law of thermodynamics?

A. An ideal gas expands isothermally and does work equal to the thermal energy it absorbs.

B. A cyclic engine takes thermal energy from one reservoir and converts all of it into work.

C. A refrigerator transfers thermal energy from a cold region to a hot region while work is done on it.

D. The entropy of a gas decreases while the gas is compressed isothermally.

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B breaks the Kelvin form: no cyclic device turns all the heat it takes in into work. A is allowed because it is a single process, not a cycle: the gas ends with a larger volume. C is the Clausius form with work supplied. In D the surroundings gain at least as much entropy as the gas loses.

Final answer

Question C5

Type C5 · C4 Reverse cycle; C1 Cycle area & U[6 marks]

A heat pump warms a building. Its working substance, an ideal monatomic gas, is modelled by a rectangular cycle on a pressure–volume diagram with corners A (1.0\times10^{-3}\,\mathrm{m^3}, 1.0\times10^{5}\,\mathrm{Pa}), B (3.0\times10^{-3}\,\mathrm{m^3}, 1.0\times10^{5}\,\mathrm{Pa}), C (3.0\times10^{-3}\,\mathrm{m^3}, 2.0\times10^{5}\,\mathrm{Pa}) and D (1.0\times10^{-3}\,\mathrm{m^3}, 2.0\times10^{5}\,\mathrm{Pa}), taken in the order A\toB\toC\toD\toA.

(a) Explain why the cycle needs a net input of work. [1]

(b) Calculate the net work in one cycle. [1]

(d) Hence state the thermal energy delivered to the building in each cycle. [1]

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(a) The compression C\toD happens at a higher pressure than the expansion A\toB, so more work is done on the gas than by it; the loop runs anticlockwise (A1). (b) Enclosed area =(1.0\times10^{5})(2.0\times10^{-3})=200\,\mathrm{J}, so W_\text{net}=-200\,\mathrm{J} (A1). (c) C\toD: W=2.0\times10^{5}\times(-2.0\times10^{-3})=-400\,\mathrm{J}; \Delta U=1.5(200-600)=-600\,\mathrm{J} (M1); Q=-1000\,\mathrm{J} (A1). D\toA: W=0, Q=\Delta U=1.5(100-200)=-150\,\mathrm{J} (A1). (d) Heat leaves the gas only in these two stages: 1000+150=1150\,\mathrm{J} (A1). Check: the gas takes in 500+450=950\,\mathrm{J} on A\toB and B\toC, and 950+200=1150\,\mathrm{J}. Add the two magnitudes; do not subtract them.

Final answer

(a) The loop is anticlockwise: the compression happens at a higher pressure than the expansion. (b) 200\,\mathrm{J} done on the gas (W=-200\,\mathrm{J}). (c) C\toD: -1000\,\mathrm{J}; D\toA: -150\,\mathrm{J}. (d) 1150\,\mathrm{J}.

Question D5

Type D5 · C2 Engine balance; D2 Carnot ratio[7 marks]

A heat engine takes in thermal energy from a reservoir at 800\,\mathrm{K} and rejects thermal energy to the air at 320\,\mathrm{K}. Its useful mechanical power output is 84\,\mathrm{kW} and its efficiency is 0.42.

(a) Calculate the greatest possible efficiency of an engine working between these temperatures. [1]

(b) Calculate the rate at which the engine takes in thermal energy, and the rate at which it rejects thermal energy. [2]

(d) Suggest two reasons why the efficiency of the real engine is below the Carnot value. [2]

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(a) \eta_\text{Carnot}=1-320/800=0.60 (A1). (b) Input power =84/0.42=200\,\mathrm{kW} (A1); rejected =200-84=116\,\mathrm{kW} (A1). (c) Carnot output =0.60\times200=120\,\mathrm{kW} (M1), so it rejects 80\,\mathrm{kW} (A1). (d) Each real irreversibility generates entropy and wastes work (A1 each): friction between moving parts; thermal energy crossing a finite temperature difference into and out of the gas; strokes too fast for the gas to pass through equilibrium states; heat leaking through the cylinder walls.

Final answer

(a) 0.60. (b) 200\,\mathrm{kW} in; 116\,\mathrm{kW} rejected. (c) 80\,\mathrm{kW}. (d) Any two of: friction in moving parts; heat transfer across a finite temperature difference; processes too fast to be reversible; heat leaking through the engine walls.

Question E3

Type E3 · E4 Invert Boltzmann[1 mark]

A system with \Omega possible microstates has entropy S. A second system has entropy 3S. How many microstates does the second system have?

A. 3\Omega B. \Omega+3

C. \Omega^{3} D. \Omega^{1/3}

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3S=3k_B\ln\Omega=k_B\ln\Omega^{3}. Entropy is a logarithm of \Omega, so tripling S cubes \Omega. A scales \Omega as if S were proportional to it; D is the result for S/3.

Final answer

Question E4

Type E4 · E2 Two entropy signs[1 mark]

A tray of water at 0\,{}^\circ\mathrm{C} freezes inside a freezer that is running in a kitchen. What are the entropy changes of the water and of its surroundings (the freezer and the kitchen)?

A. water decreases; surroundings increase by more

B. water decreases; surroundings decrease

C. water increases; surroundings decrease by more

D. water unchanged, because its temperature is constant

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Freezing releases thermal energy at a fixed temperature, so the water's entropy falls by Q/T. The water, the freezer and the kitchen together form a real, irreversible system, so their total entropy rises: the surroundings gain more than the water loses. D confuses constant temperature with zero entropy change.

Final answer

Question E5

Type E5 · E5 Freezing entropy; E2 Two entropy signs[6 marks]

A tray holds 0.20\,\mathrm{kg} of water at 0\,{}^\circ\mathrm{C}. It is placed on the cooling plate of a freezer, which stays at -18\,{}^\circ\mathrm{C}, and the water freezes completely at 0\,{}^\circ\mathrm{C}. Specific latent heat of fusion of water =3.34\times10^{5}\,\mathrm{J\,kg^{-1}}.

(a) Calculate the thermal energy removed from the water. [1]

(b) Calculate the entropy change of the water. [2]

(d) The water stays at 0\,{}^\circ\mathrm{C} while it freezes. Explain why its internal energy still falls. [1]

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(a) Q=mL=0.20\times3.34\times10^{5}=6.68\times10^{4}\,\mathrm{J} (A1). (b) \Delta S=\Delta Q/T with T=273\,\mathrm{K} (M1): -6.68\times10^{4}/273=-245\,\mathrm{J\,K^{-1}} (A1). A denominator of 0 or a Celsius value earns nothing. (c) The plate is at 255\,\mathrm{K}: +6.68\times10^{4}/255=+262\,\mathrm{J\,K^{-1}} (A1). The total is +17\,\mathrm{J\,K^{-1}}: heat crossing a temperature difference is irreversible, so total entropy rises, as the second law requires (A1). (d) The molecules move into fixed positions, bound more tightly; their potential energy falls while their random kinetic energy, set by the temperature, stays the same (A1).

Final answer

(a) 6.68\times10^{4}\,\mathrm{J}. (b) -245\,\mathrm{J\,K^{-1}}. (c) +262\,\mathrm{J\,K^{-1}}; total +17\,\mathrm{J\,K^{-1}}, an increase. (d) The intermolecular potential energy falls as the molecules lock into the solid; their mean kinetic energy is unchanged.

Advanced practice

Question B4

Type B4 · B5 Compare paths; B3 Adiabatic state; A2 Work area[6 marks]

An ideal monatomic gas starts at 3.0\times10^{5}\,\mathrm{Pa} and 1.0\times10^{-3}\,\mathrm{m^3}. In three separate experiments its volume is doubled: at constant pressure, isothermally, and adiabatically (reversibly).

(a) Calculate the final pressure in each experiment. [3]

(b) Calculate W for the constant-pressure experiment and for the adiabatic one. [2]

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(a) Constant pressure: unchanged, 3.0\times10^{5}\,\mathrm{Pa} (A1). Isothermal: PV constant, 1.5\times10^{5}\,\mathrm{Pa} (A1). Adiabatic: P_2=3.0\times10^{5}\times2^{-5/3}=9.45\times10^{4}\,\mathrm{Pa} (A1). (b) W=P\Delta V=3.0\times10^{5}\times1.0\times10^{-3}=300\,\mathrm{J} (A1). Adiabatic: W=-\Delta U=\tfrac32(P_1V_1-P_2V_2)=1.5(300-189)=167\,\mathrm{J} (A1). (c) All three start at the same state and cover the same volume range. At each volume the isobar is highest and the adiabat, which is steeper, is lowest; the isotherm lies between, so its area does too (A1).

Final answer

(a) 3.0\times10^{5}\,\mathrm{Pa}; 1.5\times10^{5}\,\mathrm{Pa}; 9.45\times10^{4}\,\mathrm{Pa}. (b) 300\,\mathrm{J}; 167\,\mathrm{J}. (c) At every volume the isotherm lies between the isobar and the adiabat, so the area under it does too.

Question C4

Type C4 · C3 Cycle heat ledger; B3 Adiabatic state; C2 Engine balance[8 marks]

An ideal monatomic gas is the working substance of an engine. Each cycle has three stages. A\toB: the gas is heated at a constant volume of 2.0\times10^{-3}\,\mathrm{m^3} from 1.0\times10^{5}\,\mathrm{Pa} to 4.0\times10^{5}\,\mathrm{Pa}. B\toC: reversible adiabatic expansion until the pressure is 1.0\times10^{5}\,\mathrm{Pa}. C\toA: compression at constant pressure back to A.

(a) Use the adiabatic law to show that the gas occupies about 4.6\times10^{-3}\,\mathrm{m^3} at C. [2]

(b) Copy and complete a table of Q, \Delta U and W for each stage. [4]

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(a) P_BV_B^{5/3}=P_CV_C^{5/3} (M1), so V_C=2.0\times10^{-3}\times4^{3/5}=4.59\times10^{-3}\,\mathrm{m^3} (A1). (b) For a monatomic gas \Delta U=\tfrac32\Delta(PV) on every stage (M1). A\toB: W=0, \Delta U=1.5(800-200)=900\,\mathrm{J}=Q (A1). B\toC: Q=0, \Delta U=1.5(459-800)=-511\,\mathrm{J}, W=+511\,\mathrm{J} (A1). C\toA: W=1.0\times10^{5}(2.0-4.59)\times10^{-3}=-259\,\mathrm{J}, \Delta U=1.5(200-459)=-389\,\mathrm{J}, Q=-649\,\mathrm{J} (A1). Check: the \Delta U column adds to zero. (c) W_\text{net}=511-259=251\,\mathrm{J} (M1); \eta=251/900=0.28 (A1). The denominator is the heat taken in, 900\,\mathrm{J}, not the net heat.

Final answer

(a) 4.59\times10^{-3}\,\mathrm{m^3}. (b) A\toB: Q=900\,\mathrm{J}, \Delta U=900\,\mathrm{J}, W=0; B\toC: Q=0, \Delta U=-511\,\mathrm{J}, W=+511\,\mathrm{J}; C\toA: Q=-649\,\mathrm{J}, \Delta U=-389\,\mathrm{J}, W=-259\,\mathrm{J}. (c) 0.28.

Question E6

Type E6 · E6 Reservoir entropy; D2 Carnot ratio; Predicted type[8 marks]

An inventor claims that in every cycle her engine takes 5000\,\mathrm{J} of thermal energy from a reservoir at 500\,\mathrm{K}, does 3200\,\mathrm{J} of work, and rejects 1800\,\mathrm{J} to a reservoir at 300\,\mathrm{K}.

(a) Show that the claim obeys the first law of thermodynamics. [1]

(b) Calculate the entropy change of each reservoir in one cycle. [2]

(d) Determine the greatest work per cycle that an engine taking 5000\,\mathrm{J} from the hot reservoir could do, and show that the total entropy change is then zero. [3]

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(a) Over a cycle the working gas returns to its starting state, so \Delta U=0 and Q_\text{net}=W: 5000-1800=3200\,\mathrm{J} (A1). (b) \Delta S_\text{hot}=-5000/500=-10.0\,\mathrm{J\,K^{-1}} (A1); \Delta S_\text{cold}=+1800/300=+6.0\,\mathrm{J\,K^{-1}} (A1). (c) The engine's own entropy returns to its starting value each cycle (M1), so the isolated system of engine and reservoirs changes by -4.0\,\mathrm{J\,K^{-1}}, a decrease the second law forbids (A1). Equivalently, the claimed efficiency 0.64 beats the Carnot limit 0.40. (d) \eta_\text{Carnot}=1-300/500=0.40, so W_\text{max}=0.40\times5000=2000\,\mathrm{J} (M1 A1). It then rejects 3000\,\mathrm{J}: \Delta S=-10.0+3000/300=0 (A1). A reversible engine generates no entropy.

Final answer

(a) 3200+1800=5000\,\mathrm{J}, with \Delta U=0 over a cycle. (b) Hot: -10.0\,\mathrm{J\,K^{-1}}; cold: +6.0\,\mathrm{J\,K^{-1}}. (c) Total -4.0\,\mathrm{J\,K^{-1}}: the entropy of an isolated system would fall. (d) 2000\,\mathrm{J}; it rejects 3000\,\mathrm{J}, and -10.0+10.0=0.

Question E7

Type E7 · E7 Free expansion; E3 Count microstates; Predicted type[5 marks]

A rigid, thermally insulated box is divided into two equal halves by a thin partition. One half holds 0.10\,\mathrm{mol} of an ideal monatomic gas at 300\,\mathrm{K}; the other half is a vacuum. The partition is removed and the gas fills the box.

(a) State the values of Q, W and \Delta U for the gas, and hence its final temperature. [2]

(b) Explain why PV^{5/3}=\text{constant} cannot be used to find the final pressure. [1]

(c) Each molecule can now be found in either half of the box, so the number of microstates is multiplied by 2^{N}, where N is the number of molecules. Calculate the entropy change of the gas. [2]

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(a) The box is insulated, so Q=0; the gas pushes against nothing, so W=0 (A1). Then \Delta U=0, and for an ideal gas U depends only on T, so T stays 300\,\mathrm{K} (A1). (b) The gas rushes into the vacuum without passing through equilibrium states, so the process is irreversible and does not follow the reversible adiabat; use PV=nRT instead, which gives half the initial pressure (A1). (c) N=0.10\times6.02\times10^{23}=6.02\times10^{22}; \Delta S=k_B\ln2^{N}=Nk_B\ln2 (M1) =6.02\times10^{22}\times1.38\times10^{-23}\times0.693=0.576\,\mathrm{J\,K^{-1}} (A1). No heat flowed, yet the entropy rose: adiabatic does not mean constant entropy unless the change is reversible.

Final answer

(a) Q=0, W=0, \Delta U=0; the temperature stays 300\,\mathrm{K}. (b) The expansion is irreversible; the adiabatic law holds only for a reversible adiabatic change. (c) +0.576\,\mathrm{J\,K^{-1}}.

Connected questions

Question X1

Type X1 · B.4 + B.3 + B.5[9 marks]

A vertical cylinder with a freely moving piston of area 5.0\times10^{-3}\,\mathrm{m^2} holds 0.10\,\mathrm{mol} of an ideal monatomic gas at 300\,\mathrm{K} and a constant pressure of 1.0\times10^{5}\,\mathrm{Pa}. A small heater inside the cylinder is connected to a 6.0\,\mathrm{V} supply and carries a current of 0.50\,\mathrm{A} for 40\,\mathrm{s}. Assume that all the energy from the heater is transferred to the gas.

(a) Calculate the energy transferred by the heater. [1]

(b) Show that the initial volume of the gas is about 2.5\times10^{-3}\,\mathrm{m^3}. [1]

(d) Calculate the distance moved by the piston. [2]

(e) Calculate the increase in the mean kinetic energy of a gas atom. [2]

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(a) E=VIt=6.0\times0.50\times40=120\,\mathrm{J} (A1). (b) V=nRT/P=0.10\times8.31\times300/1.0\times10^{5}=2.49\times10^{-3}\,\mathrm{m^3} (A1). (c) At constant pressure W=P\Delta V=nR\Delta T (M1), so Q=\tfrac32nR\Delta T+nR\Delta T=\tfrac52nR\Delta T (M1); \Delta T=120/(2.5\times0.10\times8.31)=57.8\,\mathrm{K} (A1). (d) \Delta V=nR\Delta T/P=4.80\times10^{-4}\,\mathrm{m^3} (M1); distance =\Delta V/A=0.096\,\mathrm{m} (A1). (e) Mean kinetic energy =\tfrac32k_BT (M1), so the increase is 1.5\times1.38\times10^{-23}\times57.8=1.20\times10^{-21}\,\mathrm{J} (A1).

Final answer

(a) 120\,\mathrm{J}. (b) 2.49\times10^{-3}\,\mathrm{m^3}. (c) 57.8\,\mathrm{K}. (d) 0.096\,\mathrm{m}. (e) 1.20\times10^{-21}\,\mathrm{J}.

Question X2

Type X2 · B.4 + A.3[8 marks]

An engine takes in 4.0\,\mathrm{kJ} of thermal energy from a hot reservoir at 650\,\mathrm{K} in each cycle, and completes 25 cycles every second. Its efficiency is 0.30. The engine drives an electrical generator of efficiency 0.90.

(a) Calculate the mechanical power output of the engine. [2]

(b) Calculate the rate at which the engine rejects thermal energy. [1]

(d) Calculate the electrical power output and the overall efficiency from thermal input to electrical output. [2]

(e) State what happens to the energy that the generator does not convert to electrical energy. [1]

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(a) Work per cycle =0.30\times4.0=1.2\,\mathrm{kJ} (M1); power =1.2\times25=30\,\mathrm{kW} (A1). (b) (4.0-1.2)\times25=70\,\mathrm{kW} (A1). (c) No engine beats Carnot, so 0.30\le1-T_c/650 (M1), giving T_c\le0.70\times650=455\,\mathrm{K} (A1). (d) 0.90\times30=27\,\mathrm{kW} (A1); overall efficiency =27/100=0.27, the product of the two efficiencies (A1). (e) Friction and resistive heating in the generator dissipate it as thermal energy in the surroundings, raising their entropy (A1).

Final answer

(a) 30\,\mathrm{kW}. (b) 70\,\mathrm{kW}. (c) 455\,\mathrm{K}. (d) 27\,\mathrm{kW}; 0.27. (e) It is dissipated as thermal energy in the surroundings.

Question X3

Type X3 · B.4 + B.1[8 marks]

A 0.30\,\mathrm{kg} block of ice at -10\,{}^\circ\mathrm{C} is dropped into a large lake at 12\,{}^\circ\mathrm{C}, whose temperature does not change. The ice melts and the meltwater warms to 12\,{}^\circ\mathrm{C}. Use c_\text{ice}=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}, c_\text{water}=4200\,\mathrm{J\,kg^{-1}\,K^{-1}} and L_f=3.34\times10^{5}\,\mathrm{J\,kg^{-1}}.

(a) Show that the lake supplies about 1.2\times10^{5}\,\mathrm{J}. [3]

(b) Calculate the entropy change of the lake. [1]

(d) Explain, using the second law, why the total entropy gained by the ice and meltwater over the whole process must exceed 427\,\mathrm{J\,K^{-1}}. [2]

(e) Outline, in terms of microstates, why melting increases the entropy of the water molecules. [1]

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(a) Warm the ice: 0.30\times2100\times10=6300\,\mathrm{J} (M1). Melt it: 0.30\times3.34\times10^{5}=1.002\times10^{5}\,\mathrm{J}. Warm the meltwater: 0.30\times4200\times12=15\,120\,\mathrm{J} (M1). Total 1.22\times10^{5}\,\mathrm{J} (A1). (b) The lake stays at 285\,\mathrm{K}: \Delta S=-1.22\times10^{5}/285=-427\,\mathrm{J\,K^{-1}} (A1). (c) +1.002\times10^{5}/273=+367\,\mathrm{J\,K^{-1}} (A1). (d) Ice and lake together are isolated, and heat flowing across a temperature difference is irreversible, so their total entropy must increase (M1). The lake loses 427\,\mathrm{J\,K^{-1}}, so the ice and meltwater must gain more than that (A1). (e) In the solid each molecule is held near one site; in the liquid the molecules can rearrange, so the number of accessible microstates rises (A1).

Final answer

(a) 1.22\times10^{5}\,\mathrm{J}. (b) -427\,\mathrm{J\,K^{-1}}. (c) +367\,\mathrm{J\,K^{-1}}. (d) Lake and ice form an isolated system and the process is irreversible, so the total change is positive. (e) Molecules in the liquid can be arranged in far more ways, so \Omega and S=k_B\ln\Omega increase.

Question X4

Type X4 · B.4 + A.2 + A.3[8 marks]

A vertical cylinder is closed by a piston of mass 12\,\mathrm{kg} and area 4.0\times10^{-3}\,\mathrm{m^2} that slides without friction. The air outside is at 1.0\times10^{5}\,\mathrm{Pa}. The cylinder holds an ideal monatomic gas.

(a) Show that the pressure of the gas is about 1.3\times10^{5}\,\mathrm{Pa} when the piston is at rest. [2]

(b) The gas is heated slowly and receives 150\,\mathrm{J}. Show that the gas does 60\,\mathrm{J} of work, and calculate the distance the piston rises. [3]

(c) Calculate the gain in gravitational potential energy of the piston, and state where the rest of the 60\,\mathrm{J} goes. [2]

(d) The piston is then clamped and the gas receives another 150\,\mathrm{J}. State, with a reason, whether its temperature rise is larger or smaller than in (b). [1]

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(a) The piston is in equilibrium: PA=P_\text{air}A+mg (M1), so P=1.0\times10^{5}+117.6/4.0\times10^{-3}=1.29\times10^{5}\,\mathrm{Pa} (A1). (b) Slow heating keeps the pressure constant, so W=nR\Delta T and Q=\tfrac52nR\Delta T (M1), giving W=\tfrac25\times150=60\,\mathrm{J} (A1). Rise =W/(PA)=60/(1.294\times10^{5}\times4.0\times10^{-3})=0.116\,\mathrm{m} (A1). (c) mgh=12\times9.8\times0.116=13.6\,\mathrm{J} (A1). The other 46.4\,\mathrm{J} is work done pushing back the outside air, P_\text{air}Ah (A1). (d) Larger, by a factor 5/3: with the piston clamped W=0, so all 150\,\mathrm{J} becomes internal energy instead of 90\,\mathrm{J} (A1).

Final answer

(a) 1.29\times10^{5}\,\mathrm{Pa}. (b) W=\tfrac25Q=60\,\mathrm{J}; 0.116\,\mathrm{m}. (c) 13.6\,\mathrm{J}; the other 46.4\,\mathrm{J} pushes back the outside air. (d) Larger: no work is done, so all 150\,\mathrm{J} raises the internal energy.

From the book

Photon Physics: Study Guide for the IB Diploma (SL & HL)

Thermodynamics from the core ideas to full exam questions, with question types, worked examples and graded practice.

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