B.4 Thermodynamics flashcards: IB Physics HL

All 75 B.4 Thermodynamics cards

1. DefinitionHL only

   State the first law of thermodynamics in the IB form and define every term including the sign conventions.

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   Q = ΔU + W: the thermal energy Q supplied TO a system equals the increase in internal energy ΔU plus the work W done BY the system on its surroundings. It is conservation of energy applied to thermal systems. Signs: Q > 0 heat into the gas, Q < 0 heat out; W > 0 the gas expands and does work on the surroundings, W < 0 the gas is compressed; ΔU > 0 means the temperature rises. All three are energies in J and are scalars. Exam tip: the commonest lost mark is treating W as work done ON the gas — the booklet form defines W as work done BY the gas, so every sign flips.

2. DefinitionHL only

   Define the internal energy of a system and state what it equals for an ideal gas.

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   Internal energy is the total of the random kinetic energies and the mutual potential energies of all the particles of the system, measured in the frame in which the system is at rest. Unit J, scalar. It is a state function: it depends only on the present state, never on the path taken to reach it. For an ideal gas the intermolecular potential energy is zero by assumption, so U is purely kinetic and U = 3/2 nRT = 3/2 Nk_BT for a monatomic gas — U depends only on T and the amount of gas. Exam tip: never write "the heat contained in the body"; heat is energy in transit, not a stored quantity.

3. DefinitionHL only

   Define the work done by an expanding gas and state when W = pΔV may be used.

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   Work done by a gas is the energy it transfers to its surroundings when it changes volume against an external pressure: W = pΔV. Derivation: the force on a piston of area A is F = pA, so for displacement Δx, W = FΔx = pAΔx = pΔV. Unit J, scalar. Valid only when the pressure is constant (isobaric) or changes negligibly; otherwise W is the area under the p–V curve. Exam tip: ΔV must be in m³ (1 dm³ = 1 × 10⁻³ m³) and p in Pa. W is positive for expansion and negative for compression, and W = 0 at constant volume however much the pressure changes.

4. DefinitionHL only

   Define an isothermal process and state the consequences for ΔU, Q and W.

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   An isothermal process takes place at constant temperature; for an ideal gas pV = constant (Boyle's law). Because U depends only on T, ΔU = 0, so the first law gives Q = W: all the thermal energy supplied becomes work done by the gas in expansion, and in compression all the work done on the gas is expelled as heat. Exam tip: the mark usually missed is the condition — the change must be carried out slowly and in good thermal contact with a large reservoir so the gas stays at the reservoir temperature. On a p–V diagram the path is a rectangular hyperbola; isotherms of higher T lie further from the origin and never cross.

5. DefinitionHL only

   Define an isobaric process and give expressions for W, ΔU and Q for a monatomic ideal gas.

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   An isobaric process occurs at constant pressure — for example a gas heated in a cylinder closed by a frictionless piston loaded only by the atmosphere and its own weight. Then V/T = constant (Charles's law), W = pΔV = nRΔT, ΔU = 3/2 nRΔT and hence Q = 5/2 nRΔT. Unit of each is J. Exam tip: students forget that in an isobaric expansion the temperature must rise, so Q > W and only part of the heat supplied appears as useful work. On a p–V diagram the path is a horizontal straight line and the work is the rectangular area beneath it, taken as negative if the gas is compressed.

6. DefinitionHL only

   Define an isovolumetric (isochoric) process and state its consequences.

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   An isovolumetric or isochoric process takes place at constant volume, for example a gas heated in a rigid sealed vessel. Since ΔV = 0 the gas does no work, W = 0, and the first law reduces to Q = ΔU: all the thermal energy supplied raises the internal energy and hence the temperature. For an ideal gas p/T = constant (the pressure law). Exam tip: the mark is for "no work is done because the volume does not change" — not "because the pressure is constant", which is the opposite process. On a p–V diagram the path is a vertical straight line with no area beneath it, so such a leg contributes no work to a cycle.

7. DefinitionHL only

   Define an adiabatic process and state two ways it can be achieved in practice.

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   An adiabatic process is one in which no thermal energy enters or leaves the system: Q = 0. The first law then gives ΔU = −W, so an adiabatic expansion (W > 0) cools the gas and an adiabatic compression (W < 0) heats it. It is achieved either by insulating the system thermally, or by carrying out the change so rapidly that there is no time for appreciable heat transfer — as in a sound wave, a bicycle pump, or a diesel compression stroke. For a monatomic ideal gas pV^(5/3) = constant. Exam tip: adiabatic does NOT mean constant temperature; confusing it with isothermal is the most frequent error in this sub-topic.

8. DefinitionHL only

   Define a cyclic process and state what is true of ΔU and the net work over one complete cycle.

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   A cyclic process is a sequence of changes that returns the system to its original state, with the same p, V and T. Because internal energy is a state function, ΔU = 0 over a complete cycle, so the first law gives Q_net = W_net: the net thermal energy absorbed equals the net work done by the gas. On a p–V diagram the net work equals the area enclosed by the loop — positive (a heat engine) if traversed clockwise, negative (a refrigerator or heat pump) if anticlockwise. Exam tip: state W_net = Q_H − Q_C, where Q_H and Q_C are the magnitudes of the heat absorbed and rejected; energy is exchanged in both directions during one loop.

9. DefinitionHL only

   Define a heat engine and describe the energy transfers involved in one cycle.

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   A heat engine is a device that operates in a cycle, absorbing thermal energy Q_H from a hot reservoir at temperature T_h, converting part of it into useful work W, and rejecting the remainder Q_C to a cold reservoir at T_c; energy conservation over a cycle gives W = Q_H − Q_C. Examples: petrol and diesel engines, steam turbines, jet engines. A reservoir is idealised as so large that its temperature does not change as energy is exchanged. Exam tip: the second law forbids Q_C = 0, so no heat engine can be 100 % efficient however perfectly it is built — the limit is not caused by friction or by poor insulation.

10. DefinitionHL only

    Define the thermal efficiency of a heat engine and state its possible range of values.

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    Thermal efficiency is the ratio of the useful work output to the thermal energy input per cycle: η = W/Q_H = (Q_H − Q_C)/Q_H = 1 − Q_C/Q_H. It is a dimensionless ratio, often quoted as a percentage, and 0 ≤ η < 1. Typical values: petrol engine ≈ 25–30 %, large steam turbine ≈ 40 %. Exam tip: the denominator must be the energy INPUT from the hot reservoir, not the net work and not the energy rejected — a very common slip. η can never reach 1 because Q_C = 0 would violate the Kelvin–Planck statement of the second law, and a real engine's η is always below the Carnot value for its two reservoir temperatures.

11. DefinitionHL only

    Describe the four stages of the Carnot cycle and state the significance of the Carnot efficiency.

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    The Carnot cycle consists of four reversible stages: isothermal expansion at T_h absorbing Q_H; adiabatic expansion cooling the gas from T_h to T_c; isothermal compression at T_c rejecting Q_C; adiabatic compression heating it back to T_h. Its efficiency η_C = 1 − T_c/T_h, with temperatures in kelvin, is the maximum possible efficiency of ANY engine working between those two reservoirs — Carnot's theorem. Exam tip: state that every stage is reversible and quasi-static, which is exactly why the cycle is unattainable in practice, and always convert °C to K before substituting. Over the reversible cycle ΔS = 0, so Q_H/T_h = Q_C/T_c.

12. DefinitionHL only

    State the Clausius statement of the second law of thermodynamics and give an everyday consequence.

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    Clausius statement: it is impossible for thermal energy to pass spontaneously — that is, with no other change and no work input — from a colder body to a hotter body. Consequence: a refrigerator or heat pump can move energy from cold to hot, but only because a compressor supplies work. Exam tip: the words "spontaneously" or "without work being done" are essential; omitting them makes the statement plainly false because refrigerators exist. Do not confuse this with the first law — heat flowing from cold to hot would conserve energy perfectly. It is forbidden because the total entropy of the universe would decrease.

13. DefinitionHL only

    State the Kelvin–Planck statement of the second law and explain what device it forbids.

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    Kelvin–Planck statement: it is impossible for an engine working in a cycle to convert thermal energy from a single reservoir completely into work with no other effect; some energy must always be rejected to a colder reservoir, so η < 1. It forbids a perpetual motion machine of the second kind, such as a ship propelled solely by cooling the ocean. Exam tip: the phrase "working in a cycle" earns a mark — in a single isothermal expansion Q does equal W and all the heat IS converted to work, but the gas does not return to its initial state, so it is not a cycle. The Clausius and Kelvin–Planck statements are logically equivalent.

14. DefinitionHL only

    Define entropy and state what the second law says about the entropy of an isolated system.

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    Entropy S is a state function measuring the degree of disorder of a system, equivalently the amount of its energy that is unavailable for doing useful work; a change is defined by ΔS = ΔQ/T for a reversible transfer of thermal energy ΔQ at absolute temperature T. Unit J K⁻¹, scalar. Second law: in an isolated system, and hence for the universe as a whole, entropy cannot decrease — ΔS > 0 for any real (irreversible) process and ΔS = 0 only for a reversible one. Exam tip: the entropy of a chosen system CAN decrease, as when water freezes, provided the surroundings gain more; always argue about the total entropy change.

15. DefinitionHL only

    Define macrostate, microstate and the number of microstates Ω, and relate them to entropy.

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    A macrostate is the state of a system described by its bulk measurable properties (p, V, T, total energy). A microstate is one specific arrangement of the individual particles — their positions and momenta — that produces that macrostate. Ω is the number of microstates corresponding to a given macrostate, a pure number. Boltzmann's relation is S = k_B ln Ω with k_B = 1.38 × 10⁻²³ J K⁻¹. Exam tip: the second law is statistical, not mechanical — a disordered macrostate has enormously more microstates, so a system evolves towards it with overwhelming probability rather than by necessity. The logarithm makes entropy additive when two systems are combined.

16. DefinitionHL only

    Distinguish between a reversible and an irreversible process, giving the entropy criterion for each.

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    A reversible process is an idealised change that can be exactly retraced so that both the system and its surroundings are returned to their initial states with no net change; it must be quasi-static (infinitely slow, always in equilibrium) and free of friction, turbulence and finite temperature differences, and for it ΔS_universe = 0. An irreversible process is any real process — heat flow across a finite temperature difference, free expansion, friction, mixing — and for it ΔS_universe > 0. Exam tip: reversible does not merely mean "the system can be brought back to its starting point"; a refrigerator does that, but only by increasing the entropy of its surroundings.

17. DefinitionHL only

    Describe how a refrigerator or a heat pump works in terms of the thermodynamic quantities involved.

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    A refrigerator or heat pump is a heat engine run in reverse — anticlockwise on a p–V diagram. Work W is supplied by a compressor, thermal energy Q_C is extracted from the cold space and Q_H = Q_C + W is delivered to the hot surroundings. A refrigerator's purpose is the removal of Q_C; a heat pump's is the delivery of Q_H, which is why a heat pump can deliver more energy to a house than the electrical energy it consumes. Exam tip: this breaks neither law — the extra energy comes from the cold outside air, and work is supplied, so the entropy gained by the hot reservoir exceeds that lost by the cold one.

18. EquationHL onlyData booklet: Yes

    State the first-law equation Q = ΔU + W, define each symbol with its SI unit and give the special cases.

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    Q = ΔU + W. Q = thermal energy transferred TO the system, J; ΔU = increase in internal energy, J; W = work done BY the system, J. Valid for any closed system undergoing any process. Rearranged: ΔU = Q − W. Special cases: isothermal ΔU = 0 so Q = W; adiabatic Q = 0 so ΔU = −W; isovolumetric W = 0 so Q = ΔU; complete cycle ΔU = 0 so Q_net = W_net. Common misuse: taking W as the work done ON the gas, which reverses the sign of every answer. Sanity check: supply 400 J to a gas that does 150 J of work and ΔU = +250 J, so its temperature rises.

19. EquationHL onlyData booklet: Yes

    State W = pΔV, define the symbols, give its validity condition and the general replacement for it.

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    W = pΔV. W = work done by the gas, J; p = the constant pressure, Pa; ΔV = change in volume, m³. Valid only at constant pressure (isobaric). In general W = ∫p dV = the area under the p–V curve, which is what must be used for isothermal or adiabatic paths, and for a closed cycle W_net = the enclosed area. Useful equivalent at constant p: pΔV = nRΔT. Common misuse: applying pΔV to an isothermal expansion, or leaving V in litres. Sanity check: 1.0 × 10⁵ Pa expanding by 2.0 × 10⁻³ m³ gives W = 2.0 × 10² J done by the gas.

20. EquationHL onlyData booklet: Yes

    State the internal energy of a monatomic ideal gas and the corresponding expression for ΔU.

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    U = 3/2 nRT = 3/2 Nk_BT, hence ΔU = 3/2 nRΔT = 3/2 (p₂V₂ − p₁V₁). Symbols: U internal energy, J; n amount of substance, mol; R = 8.31 J K⁻¹ mol⁻¹; T absolute temperature, K; N number of molecules; k_B = 1.38 × 10⁻²³ J K⁻¹. Valid for an ideal MONATOMIC gas only — three translational degrees of freedom and no molecular potential energy. Common misuse: applying it to air (diatomic), or substituting T in °C. Sanity check: 2.0 mol heated by 50 K gains ΔU = 1.5 × 2.0 × 8.31 × 50 ≈ 1.2 × 10³ J. Since U depends only on T, ΔU is the same for every path between two given states.

21. EquationHL onlyData booklet: Yes

    State the adiabatic relation for a monatomic ideal gas and show how to obtain the final temperature.

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    pV^(5/3) = constant, so p₁V₁^(5/3) = p₂V₂^(5/3); combining with pV = nRT gives TV^(2/3) = constant. Symbols: p in Pa, V in m³, T in K; the exponent 5/3 is γ = C_p/C_V for a monatomic gas. Valid for a reversible adiabatic (Q = 0) change of an ideal monatomic gas. Common misuse: using pV = constant (the isothermal law) instead, or assuming the temperature is unchanged. Sanity check: halving the volume gives p₂/p₁ = 2^(5/3) ≈ 3.2 and T₂/T₁ = 2^(2/3) ≈ 1.6, so an adiabatic compression heats the gas — exactly what makes a diesel engine ignite its fuel.

22. EquationHL onlyData booklet: Yes

    State the efficiency equation for a heat engine in all the forms you must be able to use.

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    η = useful work out / energy input = W/Q_H, and since W = Q_H − Q_C over a cycle, η = 1 − Q_C/Q_H. Symbols: W net work per cycle, J; Q_H energy taken from the hot reservoir per cycle, J; Q_C energy rejected to the cold reservoir per cycle, J. η is dimensionless with 0 ≤ η < 1. The same form works with powers: η = P_useful/P_input. Common misuse: dividing by Q_C, or by the net work instead of the input. Sanity check: Q_H = 800 J and Q_C = 560 J give W = 240 J and η = 240/800 = 0.30, i.e. 30 %.

23. EquationHL onlyData booklet: Yes

    State the Carnot efficiency equation, define its symbols and state precisely what it represents.

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    η_Carnot = 1 − T_c/T_h. T_c = absolute temperature of the cold reservoir, K; T_h = absolute temperature of the hot reservoir, K. It is the maximum theoretical efficiency of any engine operating between those two reservoirs, reached only by a fully reversible (Carnot) cycle. Common misuse: substituting Celsius temperatures — the single biggest source of lost marks here — or quoting it as a real engine's efficiency. Sanity check: steam at 500 °C (773 K) rejecting heat at 27 °C (300 K) gives η_C = 1 − 300/773 = 0.61, so a real turbine reaching 0.40 is plausible while a claimed 0.70 is impossible.

24. EquationHL onlyData booklet: Yes

    State the macroscopic entropy-change equation, define the symbols and give the conditions for its use.

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    ΔS = ΔQ/T. ΔS = change in entropy, J K⁻¹; ΔQ = thermal energy transferred reversibly, J; T = the absolute temperature at which the transfer occurs, K. Valid when T stays effectively constant during the transfer — a large reservoir, or a phase change. ΔS is positive for the body gaining energy and negative for the one losing it. Common misuse: using T in °C, or applying it when T changes appreciably. Sanity check: 500 J flowing from a reservoir at 400 K into one at 300 K gives ΔS = −500/400 + 500/300 = −1.25 + 1.67 = +0.42 J K⁻¹ > 0, consistent with the second law.

25. EquationHL onlyData booklet: Yes

    State Boltzmann's entropy equation, define the symbols and explain what it tells you.

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    S = k_B ln Ω. S = entropy of the macrostate, J K⁻¹; k_B = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹; Ω = the number of microstates corresponding to that macrostate, a pure number. It links the microscopic and macroscopic definitions of entropy and shows why disordered macrostates are favoured — they have vastly more microstates. Common misuse: using log₁₀ instead of ln, or treating Ω as a probability. Sanity check: doubling the volume available to N molecules doubles the positions open to each, so Ω → 2^N Ω and ΔS = k_B ln(2^N) = Nk_B ln 2 = nR ln 2 > 0, agreeing with the macroscopic result.

26. EquationHL onlyData booklet: No – derive

    Give the entropy change of an ideal gas in a reversible isothermal expansion and show where it comes from.

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    ΔS = nR ln(V₂/V₁). Derivation: isothermal so ΔU = 0 and Q = W = ∫p dV = ∫(nRT/V) dV = nRT ln(V₂/V₁); then ΔS = Q/T = nR ln(V₂/V₁). Symbols: n in mol, R = 8.31 J K⁻¹ mol⁻¹, V₂ and V₁ the final and initial volumes in the same unit; ΔS in J K⁻¹. Positive for expansion, negative for compression. Common misuse: forgetting that in the reversible case the reservoir loses exactly the same entropy, so ΔS_universe = 0, whereas in a free expansion no heat flows yet ΔS of the gas is identical and the universe gains. Sanity check: 1.0 mol doubling its volume gives ΔS = 8.31 × ln 2 = 5.8 J K⁻¹.

27. EquationHL onlyData booklet: No – derive

    State the work done by an ideal gas in a reversible isothermal change and the accompanying heat flow.

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    W = nRT ln(V₂/V₁) = p₁V₁ ln(V₂/V₁), obtained as the area under an isotherm using p = nRT/V. Symbols: W in J; n in mol; R = 8.31 J K⁻¹ mol⁻¹; T in K; the volumes in any consistent unit. Because ΔU = 0 for an isothermal change, Q = W, so exactly this much heat must be absorbed from the reservoir. Common misuse: using W = pΔV, which misjudges the area because p is not constant along an isotherm. Sanity check: 0.50 mol at 300 K expanding to three times its volume does W = 0.50 × 8.31 × 300 × ln 3 ≈ 1.4 × 10³ J, and absorbs 1.4 × 10³ J of heat.

28. EquationHL onlyData booklet: Yes

    State the ideal gas equation and show how it specialises to each thermodynamic process.

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    pV = nRT = Nk_BT: p in Pa, V in m³, n in mol, R = 8.31 J K⁻¹ mol⁻¹, T in K, N the number of molecules, k_B = 1.38 × 10⁻²³ J K⁻¹. For a fixed mass, p₁V₁/T₁ = p₂V₂/T₂. Isothermal: pV = constant. Isobaric: V/T = constant. Isovolumetric: p/T = constant. Adiabatic: pV/T and pV^(5/3) are both constant at once. Common misuse: using gauge instead of absolute pressure, or °C instead of K. Sanity check: 1.00 mol at 273 K and 1.013 × 10⁵ Pa occupies V = nRT/p = (8.31 × 273)/(1.013 × 10⁵) = 2.24 × 10⁻² m³ = 22.4 dm³, the molar volume at STP.

29. EquationHL onlyData booklet: No – derive

    Give Q, ΔU and W for each of the four standard processes of a monatomic ideal gas.

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    Isovolumetric: W = 0, ΔU = 3/2 nRΔT, Q = 3/2 nRΔT. Isobaric: W = pΔV = nRΔT, ΔU = 3/2 nRΔT, Q = 5/2 nRΔT. Isothermal: ΔU = 0, W = Q = nRT ln(V₂/V₁). Adiabatic: Q = 0, W = −ΔU = −3/2 nRΔT = 3/2 (p₁V₁ − p₂V₂). All energies in J and T in K. None of these is printed as such — build each from Q = ΔU + W. Common misuse: quoting Q = 3/2 nRΔT for an isobaric change and losing the pΔV term. Sanity check: for isobaric heating Q > ΔU because the gas must also push back the atmosphere as it expands.

30. Graph/diagramHL only

    Describe the p–V graph of an isothermal change of an ideal gas and state what can be extracted from it.

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    Axes: pressure p/Pa (often ×10⁵ Pa) on the y-axis against volume V/m³ (often ×10⁻³ m³) on the x-axis. Shape: a rectangular hyperbola, p = nRT/V, steep at small V and flattening at large V, approaching but never touching either axis. The area under the curve between V₁ and V₂ is the work done by the gas — count squares, or use W = nRT ln(V₂/V₁) — and because ΔU = 0 that area also equals the heat absorbed. The product pV read from any point equals nRT, giving T if n is known. Raising the temperature shifts the whole isotherm further from the origin; two isotherms can never cross.

31. Graph/diagramHL only

    Compare the adiabatic and the isothermal curves drawn on a p–V diagram from the same starting point.

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    Both fall as the gas expands, but the adiabat (pV^(5/3) = constant) is STEEPER than the isotherm (pV = constant), because in an adiabatic expansion the gas also cools, so the pressure drops for two reasons at once. Consequences: over the same volume change the area under the adiabat is smaller, so less work is done by the gas; an adiabatic expansion ends on a lower isotherm and an adiabatic compression on a higher one. Sketching tip: draw two isotherms labelled T₁ > T₂ and show the adiabat cutting across from one to the other. A larger γ makes the adiabat steeper; a diatomic gas has the smaller γ = 7/5, so its adiabat is slightly less steep than the monatomic 5/3 one.

32. Graph/diagramHL only

    Describe how isobaric and isovolumetric changes appear on a p–V diagram and how the work is found.

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    Axes: p/Pa against V/m³. An isobaric change is a horizontal straight line; the work done by the gas is the rectangular area beneath it, W = pΔV, positive when the arrow points right (expansion) and negative when it points left. An isovolumetric change is a vertical straight line with no area beneath it, so W = 0 and Q = ΔU. The temperature at any point is T = pV/nR, so points with a larger product pV are hotter: moving right along a horizontal line is heating, moving up a vertical line is heating. Exam tip: always mark the direction with an arrow — reversing a leg reverses the signs of W and Q.

33. Graph/diagramHL only

    Sketch and describe the Carnot cycle on a p–V diagram, identifying each leg.

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    Axes: p/Pa against V/m³, four legs traversed clockwise. A→B isothermal expansion along the hot isotherm T_h, absorbing Q_H. B→C adiabatic expansion, a steeper curve falling to the cold isotherm T_c. C→D isothermal compression along T_c, rejecting Q_C. D→A adiabatic compression back to A. The enclosed area is the net work per cycle, W = Q_H − Q_C, and η = 1 − T_c/T_h. Exam tip: the two adiabats must clearly be drawn steeper than the isotherms or the sketch loses a mark, and the arrows must be clockwise for an engine. Traversed anticlockwise the identical loop represents a refrigerator or heat pump.

34. Graph/diagramHL only

    Explain how to obtain the net work, net heat and efficiency of an engine from a closed loop on a p–V diagram.

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    Axes: p/Pa against V/m³. The net work done by the gas per cycle is the AREA ENCLOSED by the loop — positive (work output) if traversed clockwise, negative (work input, a refrigerator or heat pump) if anticlockwise. Method: count the squares inside the loop and multiply by the energy of one square, e.g. 0.5 × 10⁵ Pa × 0.5 × 10⁻³ m³ = 25 J per square. Since ΔU = 0 round a cycle, that area also equals Q_H − Q_C. Efficiency: η = enclosed area ÷ the sum of the heat inputs on the legs where Q is positive. Exam tip: never use the area under a single leg, and check the powers of ten on both axes before multiplying.

35. Graph/diagramHL only

    Describe the energy-flow (Sankey) diagram for a heat engine and for a heat pump, and how to read efficiency from it.

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    Heat engine: an arrow of width Q_H leaves the hot reservoir at T_h and splits into a narrower arrow W (useful work) and an arrow Q_C entering the cold reservoir at T_c, with Q_H = W + Q_C so the total width is conserved. Efficiency is the fraction of the original width that becomes work, η = W/Q_H, read directly from the widths. Heat pump: the arrows reverse — Q_C leaves the cold reservoir, W enters from the compressor, and Q_H = Q_C + W is delivered to the hot reservoir. Exam tip: widths must be to scale and must balance; a diagram showing work out with no rejected heat asserts a violation of the Kelvin–Planck statement.

36. Graph/diagramHL only

    Describe the linearised graph used to find the adiabatic index γ from p–V data, including gradient and intercept.

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    From pV^γ = constant, take natural logarithms: ln p = −γ ln V + ln(constant). Plot ln(p/Pa) on the y-axis against ln(V/m³) on the x-axis. Shape: a straight line of negative gradient, with gradient = −γ, so γ = −gradient and a monatomic gas should give 5/3 ≈ 1.67; the y-intercept is ln(constant). Draw maximum and minimum gradient lines through the error bars to obtain the uncertainty in γ. Exam tip: the argument of a logarithm must be a pure number, so divide by the unit and label the axis ln(p/Pa); systematic curvature indicates that the compression was not truly adiabatic because heat leaked out.

37. Graph/diagramHL only

    Describe the graph used to test whether a change is isothermal and how the temperature is obtained from it.

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    Plot pressure p/Pa on the y-axis against 1/V in m⁻³ on the x-axis. From pV = nRT, p = (nRT)(1/V), so an isothermal change of a fixed mass of ideal gas gives a straight line through the origin of gradient nRT; hence T = gradient/(nR) if n is known. A best-fit line that misses the origin signals a systematic error — usually a dead volume in the tubing and pressure sensor, or gauge rather than absolute pressure being plotted. Exam tip: state that T is held constant by compressing slowly in a water bath. An alternative is pV against p: horizontal for an ideal gas at constant T, curving for a real gas.

38. Concept/explainHL onlyData booklet: Yes

    Explain the first law of thermodynamics and state clearly the sign conventions used in the IB data booklet.

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    • Q = ΔU + W is conservation of energy applied to a gas
    • Q is thermal energy transferred INTO the system (Q < 0 when heat leaves)
    • ΔU is the increase in internal energy = total random kinetic + potential energy of the molecules; for an ideal gas intermolecular PE = 0 so U depends only on T
    • W is work done BY the gas on its surroundings (positive on expansion, negative on compression)
    • for a monatomic ideal gas ΔU = 3/2 nRΔT, so the sign of ΔU follows the sign of ΔT
    • U is a state function (path-independent) while Q and W depend on the path taken. Exam tip: the commonest error is treating W as work done on the gas; always state the convention and check ΔT.
39. Concept/explainHL only

    Explain what happens to Q, W and ΔU during an isothermal expansion of an ideal gas.

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    • Isothermal means the temperature is constant, so ΔT = 0 and ΔU = 0 for an ideal gas
    • the first law then gives Q = W: all the heat absorbed is converted into work done by the gas
    • the gas must be in thermal contact with a large reservoir and the change must be slow (quasi-static) so the gas stays in equilibrium at T
    • on a p–V diagram the path is an isotherm, pV = nRT = constant, a hyperbola
    • work done by the gas is the area under the isotherm, W = nRT ln(V₂/V₁)
    • the gas does work by pushing the piston, but the reservoir replaces exactly that energy. Exam tip: incomplete answers say no energy is transferred because ΔU = 0; energy is transferred, just in equal amounts as Q in and W out.
40. Concept/explainHL onlyData booklet: Yes

    Explain the features of an adiabatic compression of a monatomic ideal gas.

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    • Adiabatic means no thermal energy is transferred, Q = 0, achieved by insulation or by a change that is too rapid for heat flow
    • first law gives 0 = ΔU + W so ΔU = −W: work done ON the gas (W negative) raises the internal energy
    • therefore the temperature rises even though no heat is supplied
    • the relation pV^(5/3) = constant holds for a monatomic ideal gas (booklet), together with pV = nRT
    • examples: the fire syringe, a bicycle pump warming, rapid compression in a diesel engine, rising air parcels cooling adiabatically
    • the process is reversible only if it is slow and frictionless. Exam tip: students often claim no heat means no temperature change; Q = 0 does not mean ΔT = 0.
41. Concept/explainHL only

    Explain why an adiabatic curve is steeper than an isotherm through the same point on a p–V diagram.

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    • Along an isotherm pV = constant, so p ∝ V⁻¹
    • along an adiabatic pV^(5/3) = constant, so p ∝ V^(−5/3), a larger negative exponent
    • at the same (p, V) point the adiabatic gradient is 5/3 times the isothermal gradient in magnitude
    • physically, during an adiabatic expansion the gas both increases in volume AND falls in temperature because it does work with no heat input, so the pressure falls faster than in the isothermal case where T is held constant
    • hence an adiabatic drawn through a point cuts across isotherms towards lower temperature. Exam tip: sketch answers frequently draw the two curves crossing or the adiabatic shallower; label each curve and mark the direction of the change with an arrow.
42. Concept/explainHL onlyData booklet: Yes

    Explain the energy transfers in an isovolumetric (isochoric) process.

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    • Volume is constant so ΔV = 0 and the gas does no work, W = pΔV = 0
    • the area under a vertical line on a p–V diagram is zero
    • the first law reduces to Q = ΔU, so all the heat supplied raises the internal energy and hence the temperature
    • for a monatomic ideal gas Q = ΔU = 3/2 nRΔT
    • pressure and temperature are proportional, p/T = constant (Gay-Lussac), so the p–V path is a vertical line and the p–T path a straight line through the origin
    • real example: heating a gas in a sealed rigid container, or the constant-volume heat addition stage of the Otto cycle. Exam tip: students still write W = pΔV with the (unchanged) volume substituted, giving a large false work value.
43. Concept/explainHL onlyData booklet: Yes

    Explain the energy transfers in an isobaric expansion, including how the heat splits between work and internal energy.

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    • Pressure is constant, so the p–V path is a horizontal line and the work done by the gas is W = pΔV, the rectangular area under the line
    • the gas must be free to expand, e.g. against a frictionless piston held down by a constant weight and atmospheric pressure
    • V/T = constant (Charles), so expansion at constant p means T rises and ΔU is positive
    • for a monatomic ideal gas ΔU = 3/2 nRΔT = 3/2 pΔV, so Q = ΔU + W = 5/2 pΔV
    • hence 60% of the heat becomes internal energy and 40% becomes work in a monatomic gas. Exam tip: many students equate Q to W and forget that the gas also warms up.
44. Concept/explainHL only

    Explain how work done and net work are found from a p–V diagram, including cyclic processes.

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    • Work done by the gas equals the area between the curve and the volume axis, since W = ∫p dV and for a small step W = pΔV
    • the area must be read with the axis scales, giving joules from Pa × m³
    • expansion (moving right) gives positive work done BY the gas; compression (moving left) gives negative work, i.e. work done on the gas
    • for a closed cycle the net work per cycle is the area enclosed by the loop
    • a clockwise loop means net work done by the gas (a heat engine); anticlockwise means net work done on the gas (a refrigerator or heat pump)
    • around any complete cycle ΔU = 0, so Q_net = W_net. Exam tip: students count squares but forget to multiply by the value of one square in J.
45. Concept/explainHL onlyData booklet: Yes

    Explain how a heat engine works and why its efficiency can never be 100%.

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    • A heat engine takes thermal energy Q_H from a hot reservoir, converts part of it into useful work W, and rejects the remainder Q_C to a cold reservoir
    • energy conservation over one cycle: W = Q_H − Q_C, since ΔU = 0 for a complete cycle
    • efficiency η = W/Q_H = 1 − Q_C/Q_H
    • the Kelvin–Planck statement of the second law forbids an engine that converts heat entirely into work with no other effect, so Q_C > 0 always
    • the working substance must be returned to its initial state, which requires rejecting entropy to the cold reservoir
    • the ideal upper limit is η_Carnot = 1 − T_c/T_h. Exam tip: an answer of it loses energy to friction is incomplete; the limit is fundamental, not merely practical.
46. Concept/explainHL onlyData booklet: Yes

    Outline the four stages of the Carnot cycle and explain why it gives the maximum possible efficiency.

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    • Stage 1: isothermal expansion at T_h, gas absorbs Q_H from the hot reservoir and does work
    • Stage 2: adiabatic expansion, Q = 0, gas does work and cools from T_h to T_c
    • Stage 3: isothermal compression at T_c, work done on gas and Q_C rejected to the cold reservoir
    • Stage 4: adiabatic compression, work done on gas, temperature rises back to T_h
    • every stage is reversible (quasi-static, frictionless, no finite temperature difference during heat transfer), so the total entropy change of the universe is zero
    • hence η_Carnot = 1 − T_c/T_h, depending only on the reservoir temperatures. Exam tip: students muddle the order of the isothermal and adiabatic legs; on a p–V sketch the two adiabatics must be steeper than the two isotherms.
47. Concept/explainHL only

    State and explain the Clausius and Kelvin–Planck statements of the second law of thermodynamics.

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    • Clausius: thermal energy cannot spontaneously flow from a colder body to a hotter body; a refrigerator only does so because external work is supplied
    • Kelvin–Planck: no cyclic process can take heat from a single reservoir and convert it completely into work with no other effect, so some heat must always be rejected
    • the two statements are equivalent: violating one allows the construction of a device violating the other
    • the entropy form: the total entropy of an isolated system (the universe) never decreases, ΔS_total ≥ 0, with equality only for a reversible process
    • the second law gives time a direction (the arrow of time), unlike the first law which is satisfied by both directions. Exam tip: quoting one statement without applying it to the situation in the question scores no explanation marks.
48. Concept/explainHL onlyData booklet: Yes

    Explain what entropy measures and why the entropy of the universe increases in every real process.

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    • Entropy S is a measure of the disorder of a system, or equivalently of the number of microstates Ω consistent with the macrostate
    • a thermodynamic change is ΔS = ΔQ/T (booklet), in J K⁻¹, with T in kelvin
    • when heat Q flows from a hot body at T_h to a cold body at T_c the hot body loses Q/T_h and the cold body gains Q/T_c; since T_c < T_h the gain is larger and ΔS_total > 0
    • all real processes involve finite temperature differences, friction or free expansion, all of which are irreversible and generate entropy
    • a system's entropy can decrease (a freezer, a growing organism) only if the surroundings gain more. Exam tip: many answers assert entropy always increases without identifying which system.
49. Concept/explainHL onlyData booklet: Yes

    Explain the statistical (Boltzmann) interpretation of entropy, S = k_B ln Ω.

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    • Ω is the number of microstates (arrangements of individual molecular positions and momenta) that correspond to the same observed macrostate
    • k_B = 1.38 × 10⁻²³ J K⁻¹, so S is in J K⁻¹ and is an extensive quantity: the logarithm makes entropies of combined systems add
    • a macrostate with more microstates is overwhelmingly more probable, so a system evolves towards the macrostate of maximum Ω, i.e. maximum entropy
    • the second law is therefore statistical, not absolute: a spontaneous decrease is not impossible, merely so improbable for N ≈ 10²³ that it is never observed
    • a perfect crystal at 0 K has Ω = 1 and S = 0. Exam tip: describing entropy as messiness earns nothing; refer explicitly to numbers of microstates and probability.
50. Concept/explainHL only

    Distinguish between reversible and irreversible processes and give examples of each.

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    • A reversible process can be run backwards so that both the system AND the surroundings return exactly to their initial states, leaving no net change in the universe
    • it requires infinitesimally slow (quasi-static) change, no friction, and heat transfer only across infinitesimal temperature differences, so ΔS_universe = 0
    • an irreversible process increases the total entropy: heat flow across a finite temperature difference, free (unresisted) expansion of a gas into a vacuum, friction, mixing, and any rapid change
    • all real processes are irreversible; reversibility is an idealisation used to set performance limits such as the Carnot efficiency
    • in a free expansion into a vacuum W = 0 and Q = 0, so ΔU = 0 and T is unchanged, yet ΔS > 0. Exam tip: reversible does not mean the gas can simply be recompressed.
51. Concept/explainHL only

    Explain how a refrigerator or heat pump operates and how it is consistent with the second law.

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    • Work W is supplied by a compressor to drive a working fluid around a cycle in the anticlockwise sense on a p–V diagram
    • the fluid evaporates at low pressure inside the cold space, absorbing Q_C from it, and condenses at high pressure outside, releasing Q_H = Q_C + W to the surroundings
    • the room (surroundings) is therefore warmed by more than the fridge interior is cooled
    • the entropy decrease of the cold space is Q_C/T_c while the surroundings gain Q_H/T_h, and because Q_H = Q_C + W is larger and is delivered at the higher temperature the gain still exceeds the loss, so ΔS_universe > 0 (the work input itself carries no entropy)
    • a heat pump is the same device valued for Q_H rather than Q_C. Exam tip: saying a fridge makes cold is wrong; it transfers energy out, and it cannot cool a room with its door open.
52. Concept/explainHL onlyData booklet: Yes

    Explain why the internal energy of an ideal gas depends only on its temperature, and how this is used in thermodynamic problems.

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    • Internal energy is the sum of the random kinetic energies and the intermolecular potential energies of all the molecules
    • an ideal gas is defined to have no intermolecular forces, so the potential energy term is zero
    • the average translational kinetic energy per molecule is E_k = 3/2 k_BT, so for a monatomic gas U = 3/2 Nk_BT = 3/2 nRT and ΔU = 3/2 nRΔT = 3/2 Δ(pV)
    • U is therefore a state function fixed by T alone, whatever the path taken
    • consequences: ΔU = 0 for any isothermal change or any complete cycle; ΔU is the same for two different paths between the same two states, even though Q and W differ. Exam tip: students wrongly claim U depends on volume or pressure separately.
53. Worked problemHL onlyData booklet: Yes

    A gas absorbs 250 J of thermal energy and does 100 J of work on its surroundings. Determine the change in its internal energy.

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    First law: Q = ΔU + W with Q = heat into gas and W = work done by gas. Substituting: 250 J = ΔU + 100 J. Rearranging: ΔU = Q − W = 250 − 100 = +150 J. The internal energy increases by 1.5 × 10² J, so the temperature of the gas rises. Check/Trap: both quantities are positive here because heat goes IN and work is done BY the gas; if the gas had been compressed with 100 J of work done on it, W = −100 J and ΔU would be +350 J. In a Paper 1 item the distractor 350 J is offered precisely to catch the sign error.

54. Worked problemHL onlyData booklet: Yes

    0.50 mol of a monatomic ideal gas expands at a constant pressure of 1.0 × 10⁵ Pa from 8.0 × 10⁻³ m³ to 1.2 × 10⁻² m³. Determine W, ΔU and Q.

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    Work done by the gas (isobaric): W = pΔV = 1.0 × 10⁵ × (1.2 × 10⁻² − 8.0 × 10⁻³) = 1.0 × 10⁵ × 4.0 × 10⁻³ = 4.0 × 10² J. Internal energy: ΔU = 3/2 nRΔT = 3/2 Δ(pV) = 1.5 × 1.0 × 10⁵ × 4.0 × 10⁻³ = 6.0 × 10² J. (As a check, T₁ = pV/nR = 800/4.155 = 193 K and T₂ = 1200/4.155 = 289 K, so ΔT = 96 K and 1.5 × 0.50 × 8.31 × 96 = 6.0 × 10² J.) First law: Q = ΔU + W = 600 + 400 = 1.0 × 10³ J into the gas. Check/Trap: for a monatomic isobaric change Q = 5/2 pΔV always, so only 40% of the heat becomes useful work.

55. Worked problemHL onlyData booklet: No – derive

    2.0 mol of an ideal gas at 300 K expands isothermally until its volume has doubled. Determine the work done by the gas and the heat absorbed.

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    Isothermal, so ΔT = 0 and ΔU = 0. First law: Q = ΔU + W = W. Work is the area under the isotherm: W = nRT ln(V₂/V₁) = 2.0 × 8.31 × 300 × ln 2 = 4986 × 0.693 = 3.46 × 10³ J. So W = 3.5 × 10³ J done by the gas and Q = 3.5 × 10³ J absorbed from the reservoir (2 s.f., matching the data). Entropy change of the gas: ΔS = Q/T = 3455/300 = 11.5 J K⁻¹, which agrees with nR ln(V₂/V₁) = 2.0 × 8.31 × 0.693 = 11.5 J K⁻¹. Check/Trap: W = pΔV is invalid here because p is not constant; the logarithmic form is not printed in the booklet, so if the question supplies a p–V graph, count squares instead.

56. Worked problemHL onlyData booklet: Yes

    A monatomic ideal gas at 1.0 × 10⁵ Pa, 2.0 × 10⁻³ m³ and 290 K is compressed adiabatically to 1.0 × 10⁻³ m³. Determine the final pressure and temperature.

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    Adiabatic for a monatomic gas: pV^(5/3) = constant. So p₂ = p₁(V₁/V₂)^(5/3) = 1.0 × 10⁵ × 2^(5/3) = 1.0 × 10⁵ × 3.17 = 3.2 × 10⁵ Pa. Temperature from pV = nRT with n fixed: T₂/T₁ = p₂V₂/(p₁V₁) = (3.17 × 10⁵ × 1.0 × 10⁻³)/(1.0 × 10⁵ × 2.0 × 10⁻³) = 317/200 = 1.59. T₂ = 1.59 × 290 = 4.6 × 10² K. Energy: ΔU = 3/2 Δ(pV) = 1.5 × (317 − 200) = 1.8 × 10² J, and since Q = 0 this equals the work done ON the gas (W by the gas = −1.8 × 10² J). Check/Trap: the gas heats up with no heat supplied; using pV = constant would wrongly give 2.0 × 10⁵ Pa and no temperature change.

57. Worked problemHL onlyData booklet: Yes

    0.20 mol of a monatomic ideal gas in a sealed rigid container is heated from 300 K to 450 K. Determine the work done and the heat supplied.

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    Volume is constant, so ΔV = 0 and the work done by the gas is W = pΔV = 0 J. First law: Q = ΔU + W = ΔU. For a monatomic ideal gas ΔU = 3/2 nRΔT = 1.5 × 0.20 × 8.31 × (450 − 300) = 1.5 × 1.662 × 150 = 3.7 × 10² J. So 3.7 × 10² J of heat is supplied and all of it becomes internal energy. The pressure rises in proportion to T: p₂/p₁ = 450/300 = 1.5. Check/Trap: no work is done even though the pressure changes a lot, because the piston does not move; on a p–V diagram the path is a vertical line with zero area beneath it.

58. Worked problemHL onlyData booklet: Yes

    A monatomic ideal gas is taken clockwise round the rectangular cycle A(1.0 × 10⁻³ m³, 3.0 × 10⁵ Pa) → B(3.0 × 10⁻³ m³, 3.0 × 10⁵ Pa) → C(3.0 × 10⁻³ m³, 1.0 × 10⁵ Pa) → D(1.0 × 10⁻³ m³, 1.0 × 10⁵ Pa) → A. Determine the net work per cycle and the efficiency.

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    A→B isobaric: W = pΔV = 3.0 × 10⁵ × 2.0 × 10⁻³ = +600 J; ΔU = 3/2 Δ(pV) = 1.5(900 − 300) = +900 J; Q = +1500 J in. B→C isovolumetric: W = 0; ΔU = 1.5(300 − 900) = −900 J; Q = −900 J out. C→D isobaric: W = 1.0 × 10⁵ × (−2.0 × 10⁻³) = −200 J; ΔU = 1.5(100 − 300) = −300 J; Q = −500 J out. D→A isovolumetric: W = 0; ΔU = +300 J; Q = +300 J in. Net work = 600 − 200 = 4.0 × 10² J, equal to the enclosed area Δp × ΔV. Heat input Q_H = 1500 + 300 = 1800 J, so η = 400/1800 = 0.22 (22%). Check/Trap: ΔU sums to zero round the loop; divide by total heat IN only, not by net heat.

59. Worked problemHL onlyData booklet: Yes

    A heat engine absorbs 1.20 kJ per cycle from a hot reservoir and rejects 0.90 kJ to a cold reservoir, completing 20 cycles per second. Determine the efficiency and the useful power output.

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    Over one complete cycle ΔU = 0, so the useful work is W = Q_H − Q_C = 1200 − 900 = 300 J per cycle. Efficiency: η = W/Q_H = 300/1200 = 0.25, i.e. 25%. Power output: P = W × (cycles per second) = 300 × 20 = 6.0 × 10³ W = 6.0 kW. The waste power delivered to the cold reservoir is 900 × 20 = 18 kW. Check/Trap: efficiency has no unit and must not exceed 1; dividing W by Q_C (0.33) is a standard distractor. Also note η = 1 − Q_C/Q_H = 1 − 900/1200 gives the same answer, which is a quick check.

60. Worked problemHL onlyData booklet: Yes

    A steam turbine takes heat from steam at 550 K and rejects it to cooling water at 320 K. Its measured efficiency is 30%. Determine the maximum theoretical efficiency and comment.

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    Maximum (Carnot) efficiency: η_C = 1 − T_c/T_h with temperatures in kelvin. η_C = 1 − 320/550 = 1 − 0.582 = 0.418, i.e. 42%. The actual efficiency of 30% is 0.30/0.418 = 72% of the Carnot value. The shortfall arises because the real cycle is irreversible: heat is transferred across finite temperature differences, there is friction in bearings and turbine blades, and there are conduction losses from pipework — each generates entropy. Check/Trap: the temperatures must be absolute; using Celsius values (277 °C and 47 °C) would wrongly give 0.83. To raise η the engineer should raise T_h rather than lower T_c, since T_c is limited by the ambient environment.

61. Worked problemHL onlyData booklet: Yes

    0.50 kg of ice at 0 °C melts to water at 0 °C. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg⁻¹. Determine the entropy change of the ice.

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    Melting occurs at constant temperature T = 0 + 273 = 273 K. Heat absorbed: Q = mL = 0.50 × 3.3 × 10⁵ = 1.65 × 10⁵ J. Entropy change: ΔS = ΔQ/T = 1.65 × 10⁵ / 273 = 6.0 × 10² J K⁻¹, and it is positive because the heat flows into the ice. Physically the liquid has far more accessible microstates than the ordered crystal, so Ω and hence S increase. Check/Trap: T must be in kelvin, and ΔS = ΔQ/T is valid here only because the temperature is constant during the phase change; for a process with changing T you must either use a mean temperature for an estimate or an integral.

62. Worked problemHL onlyData booklet: Yes

    500 J of thermal energy is conducted through a metal bar from a reservoir at 600 K to a reservoir at 300 K. Determine the entropy change of the universe.

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    Each reservoir is large enough that its temperature does not change, so ΔS = ΔQ/T applies to each. Hot reservoir loses heat: ΔS_h = −500/600 = −0.833 J K⁻¹. Cold reservoir gains heat: ΔS_c = +500/300 = +1.667 J K⁻¹. The bar itself returns to its initial steady state, so ΔS_bar = 0. Total: ΔS_universe = −0.833 + 1.667 = +0.83 J K⁻¹. Check/Trap: the total is positive, as the second law requires for an irreversible process; heat flow across a finite temperature difference always generates entropy. If the two temperatures were made almost equal the process would approach reversibility and ΔS_universe → 0. Reversing the flow would give −0.83 J K⁻¹, which is why it never happens spontaneously.

63. Worked problemHL onlyData booklet: No – derive

    3.0 mol of an ideal gas expands isothermally to four times its original volume. Determine the entropy change of the gas.

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    For an isothermal expansion ΔU = 0 so Q = W = nRT ln(V₂/V₁), and ΔS = Q/T = nR ln(V₂/V₁). Substituting: ΔS = 3.0 × 8.31 × ln 4 = 24.93 × 1.386 = 34.6 J K⁻¹ ≈ 35 J K⁻¹, positive because the gas absorbs heat and each molecule has four times as much volume available, so Ω rises. Check/Trap: this result is the same for a free (unresisted) expansion into a vacuum, where Q = 0 and W = 0. Entropy is a state function, so ΔS depends only on the initial and final states, not on the path — but in the free expansion the entropy is generated internally and ΔS_universe = +35 J K⁻¹ rather than zero.

64. Worked problemHL onlyData booklet: Yes

    One mole of gas expands so that the number of microstates available to each of its 6.02 × 10²³ molecules doubles. Use S = k_B ln Ω to determine the entropy change.

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    If each molecule independently has twice as many available microstates, the total number for N molecules increases by a factor 2^N: Ω₂ = Ω₁ × 2^N. Then ΔS = k_B ln Ω₂ − k_B ln Ω₁ = k_B ln(Ω₂/Ω₁) = k_B ln(2^N) = N k_B ln 2. Substituting: ΔS = 6.02 × 10²³ × 1.38 × 10⁻²³ × 0.693 = 8.31 × 0.693 = 5.8 J K⁻¹. Check/Trap: this is exactly nR ln 2 = 8.31 × 0.693, confirming that the statistical and thermodynamic definitions of entropy agree. The logarithm is essential — it turns the multiplicative growth of microstates into an additive (extensive) entropy.

65. Worked problemHL onlyData booklet: Yes

    A gas is compressed isothermally and 250 J of work is done on it. Determine the heat transferred and state its direction.

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    Isothermal, so ΔT = 0 and for an ideal gas ΔU = 0. The booklet convention takes W as work done BY the gas, so compression gives W = −250 J. First law: Q = ΔU + W = 0 + (−250) = −250 J. The negative sign means 250 J of thermal energy is transferred OUT of the gas, into the reservoir. Check/Trap: the answer is not zero — a constant temperature does not mean no energy transfer. The magnitudes of Q and W are always equal for an isothermal change of an ideal gas, which is a quick way to spot the correct Paper 1 option.

66. Worked problemHL onlyData booklet: Yes

    A coal-fired power station releases 2.5 GW of thermal power and delivers 900 MW of electrical power. Steam enters the turbine at 800 K and cooling water is at 300 K. Determine the actual and Carnot efficiencies and the waste power.

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    Actual efficiency: η = useful power out / power in = 900 × 10⁶ / 2.5 × 10⁹ = 0.36, i.e. 36%. Carnot limit: η_C = 1 − T_c/T_h = 1 − 300/800 = 0.625, i.e. 63%. Waste thermal power: P_waste = 2.5 × 10⁹ − 900 × 10⁶ = 1.6 × 10⁹ W = 1.6 GW, discharged to the river or cooling towers. Check/Trap: efficiency may be computed with powers instead of energies provided both refer to the same time interval, since η = W/Q_H = P_out/P_in. The station reaches 36/62.5 = 58% of its theoretical limit; quoting an efficiency above η_C would immediately indicate an arithmetic or unit error.

67. Exam technique/trapHL onlyData booklet: Yes

    Explain the sign-convention trap in Q = ΔU + W and how to avoid losing marks on it.

    Show answer

    The trap: the booklet form defines Q as heat INTO the system and W as work done BY the gas, but many textbooks and students use ΔU = Q + W with W as work done ON the gas. Mixing the two flips the sign of the answer and loses the final mark. Why students fall for it: exam stems say the gas is compressed doing 200 J of work, and students enter +200. Correct approach: write down the convention explicitly, then decide the sign of each term physically — expansion means W positive, compression W negative; heat out means Q negative; a rise in T means ΔU positive. Finally sanity-check that the signs of ΔU and ΔT agree. If a question says work is done on the gas, put a minus sign in before substituting.

68. Exam technique/trapHL onlyData booklet: Yes

    Explain when W = pΔV may be used and what to do when it may not.

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    The trap: W = pΔV is valid ONLY at constant pressure (an isobaric change). Students apply it to isothermal, adiabatic and curved p–V paths, often substituting the initial pressure and one of the volumes. Correct approach: identify the process first. Isobaric → W = pΔV. Isovolumetric → W = 0. Any curved path → work equals the area under the p–V curve, found by counting squares (state the value of one square in J) or by splitting the region into rectangles and triangles. Isothermal → W = nRT ln(V₂/V₁) if no graph is supplied. A closed cycle → net work is the enclosed area, positive for clockwise. Command-term note: determine expects a numerical answer with working and a unit, whereas estimate allows sensible square-counting.

69. Exam technique/trapHL onlyData booklet: Yes

    Explain the common confusion between adiabatic and isothermal processes and how to distinguish them in an exam.

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    The trap: students assume Q = 0 implies ΔT = 0, or that constant temperature implies no energy transfer. Both are wrong. Adiabatic: Q = 0, so ΔU = −W and the temperature MUST change when work is done — compression heats, expansion cools. Isothermal: ΔT = 0 so ΔU = 0 and Q = W, so energy flows in and straight out again. Distinguishing cues in the stem: insulated, rapid, sudden or thermally isolated → adiabatic; in thermal contact with a reservoir, slow, or at constant temperature → isothermal. On a sketch the adiabatic is steeper (p ∝ V^(−5/3) versus p ∝ V⁻¹) and crosses isotherms. Only the adiabatic uses pV^(5/3) = constant, and only for a monatomic gas.

70. Exam technique/trapHL onlyData booklet: Yes

    Explain the traps in calculating and interpreting Carnot efficiency.

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    Trap 1: using Celsius. η_C = 1 − T_c/T_h requires absolute temperatures; converting first (T/K = θ/°C + 273) is the single most common lost mark in this sub-topic. Trap 2: quoting a real efficiency greater than η_C — always check your answer against the Carnot limit, since exceeding it means an arithmetic error, not a discovery. Trap 3: treating η_C as achievable; it applies only to a fully reversible cycle, which needs infinitely slow operation and would deliver zero power. Trap 4: dividing by Q_C rather than Q_H. Command-term note: suggest how the efficiency could be increased expects raise T_h (or lower T_c, limited by the environment) plus reduce friction and thermal losses, with a reason attached to each.

71. Exam technique/trapHL only

    Give command-term guidance for p–V diagram questions in thermodynamics.

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    Sketch a p–V diagram: axes labelled p/Pa and V/m³ with the origin shown, correct curvature (isotherm hyperbolic, adiabatic steeper, isobaric horizontal, isovolumetric vertical), an arrow showing the direction, and the states labelled A, B, C. Deduce the sign of Q: state ΔU from ΔT, state W from the area and direction, then apply Q = ΔU + W — the reasoning earns the marks, not the conclusion. Determine the work: read scales, give a numerical value with the unit J. Outline: short bullet points; explain: link cause to effect with because. Compare the isothermal and adiabatic curves: make paired statements about BOTH (steeper, lower final temperature, less work done), never a list about one alone. Describe requires detail but no reasons.

72. Exam technique/trapHL onlyData booklet: Yes

    Explain the trap in statements about entropy decreasing.

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    The trap: students write entropy always increases, then are contradicted by a question about water freezing, a refrigerator, or a gas being compressed — all of which lower the entropy of a system. Correct approach: always name the system. A LOCAL entropy decrease is allowed provided the surroundings gain at least as much, so that ΔS_universe ≥ 0; for a refrigerator the room warms by more than the cabinet cools. Equality holds only for an idealised reversible process. Second trap: assuming entropy is conserved like energy — it is not; irreversible processes create entropy. Third trap: writing ΔS = ΔQ/T for a process whose temperature changes; that form applies at constant T, such as a phase change or an isothermal expansion.

73. Exam technique/trapHL onlyData booklet: Yes

    Outline a practical investigation to determine the efficiency of a model heat engine, with variables, limitations and improvements.

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    Apparatus: a model Stirling or steam engine running between a beaker of boiling water (T_h) and an ice-water bath (T_c), lifting a known mass m through a height h via a pulley; thermometers or thermocouples in each bath; a stopwatch. Independent variable: hot reservoir temperature; dependent: efficiency; controlled: load mass, cold bath temperature, engine, lubrication. Useful work out = mgh per lift, or P_out = mgh × (lifts per second). Energy input estimated from the hot bath: Q_H = mcΔT of the water cooling in a measured time, using a lid to reduce evaporation. η = W/Q_H, then compare with 1 − T_c/T_h. Limitations: heat lost from the tubing, friction in bearings, Q_H overestimated by losses to the room. Improvements: lag the apparatus, repeat and average, use a data logger.

74. Exam technique/trapHL onlyData booklet: No – derive

    Outline an experiment to test whether a compression is adiabatic, including the graph used and its limitations.

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    Apparatus: a sealed gas syringe or a pressure-law flask connected to a pressure sensor and a fast-response temperature probe, with the volume read from the syringe scale, all logged with a data logger. Method: compress the plunger rapidly to different volumes, recording p and V at each. Test: for an adiabatic change pV^γ = constant, so plot ln(p/Pa) on the y-axis against ln(V/m³) on the x-axis; a straight line of gradient −γ confirms the relation. Expect γ = 5/3 for a monatomic gas such as argon, but about 1.4 for air, which is diatomic — a common source of an apparently wrong result. Controlled variables: mass of gas (no leaks), same syringe. Limitations: the compression is never fast enough to be perfectly adiabatic, so the measured γ falls between 1 and the true value; friction warms the plunger. Improvements: repeat, use max/min gradients on the log plot to find the uncertainty in γ.

75. Exam technique/trapHL only

    Explain how to handle uncertainties and significant figures in a thermodynamics efficiency calculation.

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    Efficiency is a ratio of measured energies, so combine FRACTIONAL uncertainties by addition: for η = W/Q_H, Δη/η = ΔW/W + ΔQ_H/Q_H, then convert back to an absolute uncertainty and round it to one significant figure, quoting η to the matching decimal place (e.g. η = 0.24 ± 0.03). If Q_H = mcΔT, remember that ΔT is a difference of two readings, so its absolute uncertainty is the SUM of the two thermometer uncertainties — this usually dominates. Report the final efficiency to no more significant figures than the least precise datum, typically 2 s.f. A systematic heat loss to the room makes every value of Q_H too large and so η too small, shifting all points the same way; random reading errors instead scatter points about the line and are reduced by repeating and averaging.

Practise this topic with exam-style questions: B.4 Thermodynamics questions (HL) · all flashcards

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