School / IB / PHYSICS HL / Thermodynamics Question practice
Thermodynamics About this practice The first law of thermodynamics, entropy and thermodynamic processes.
Physics: Higher Level Thermodynamics
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1 State the first law of thermodynamics, defining each term used. Easy 3 marks No calculator + 2 Marking analysis: A learner attempts the following task: “State the first law of thermodynamics, defining each term used.” Their response addresses only this point: “States the first law: ΔU = Q + W (or ΔU = Q − W depending on sign convention, provided stated consistently).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 3 A gas absorbs 500 J of thermal energy and does 200 J of work on its surroundings as it expands. Calculate the change in internal energy of the gas. Easy 3 marks Calculator + 4 Marking analysis: A learner attempts the following task: “A gas absorbs 500 J of thermal energy and does 200 J of work on its surroundings as it expands. Calculate the change in internal energy of the gas.” Their response addresses only this point: “Uses the first law ΔU = Q − W, where W is work done by the gas.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks Calculator + 5 Distinguish between an isothermal process and an adiabatic process for an ideal gas. Easy 3 marks No calculator + 6 Marking analysis: A learner attempts the following task: “Distinguish between an isothermal process and an adiabatic process for an ideal gas.” Their response addresses only this point: “States that an isothermal process occurs at constant temperature, so the internal energy of an ideal gas does not change (ΔU = 0).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 7 Define entropy in simple qualitative terms, and state the second law of thermodynamics. Easy 2 marks No calculator + 8 Marking analysis: A learner attempts the following task: “Define entropy in simple qualitative terms, and state the second law of thermodynamics.” Their response addresses only this point: “Defines entropy as a measure of the disorder, or the number of possible microscopic arrangements (microstates), of a system.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 2 marks No calculator + 9 Explain, in terms of entropy, why thermal energy spontaneously flows from a hot object to a cold object and never the reverse. Easy 3 marks No calculator + 10 Marking analysis: A learner attempts the following task: “Explain, in terms of entropy, why thermal energy spontaneously flows from a hot object to a cold object and never the reverse.” Their response addresses only this point: “States that when thermal energy transfers from a hot object to a cold object, the entropy decrease of the hot object is smaller in magnitude than the entropy increase of the cold object (since entropy change for a given heat transfer is larger at lower temperature).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 11 A heat engine absorbs 800 J of thermal energy from a hot reservoir and does 300 J of useful work. Calculate the thermal efficiency of the engine. Easy 3 marks Calculator + 12 Marking analysis: A learner attempts the following task: “A heat engine absorbs 800 J of thermal energy from a hot reservoir and does 300 J of useful work. Calculate the thermal efficiency of the engine.” Their response addresses only this point: “Uses efficiency = useful work output / thermal energy input.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks Calculator + 13 State the Carnot theorem, and explain why no real heat engine operating between two given temperatures can be more efficient than a Carnot engine. Easy 2 marks No calculator + 14 Marking analysis: A learner attempts the following task: “State the Carnot theorem, and explain why no real heat engine operating between two given temperatures can be more efficient than a Carnot engine.” Their response addresses only this point: “States that the Carnot engine is a theoretical, idealized (reversible) heat engine that achieves the maximum possible efficiency between two given temperatures.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 2 marks No calculator + 15 A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate its maximum possible (Carnot) efficiency. Easy 3 marks Calculator + 16 Marking analysis: A learner attempts the following task: “A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate its maximum possible (Carnot) efficiency.” Their response addresses only this point: “Uses Carnot efficiency = 1 − T_cold/T_hot, with temperatures in kelvin.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks Calculator + 17 Explain why increasing the temperature of the hot reservoir, or decreasing the temperature of the cold reservoir, increases the maximum possible efficiency of a heat engine. Easy 3 marks No calculator + 18 Marking analysis: A learner attempts the following task: “Explain why increasing the temperature of the hot reservoir, or decreasing the temperature of the cold reservoir, increases the maximum possible efficiency of a heat engine.” Their response addresses only this point: “States the Carnot efficiency formula: efficiency = 1 − T_cold/T_hot.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 19 Outline why a refrigerator requires external work input to transfer thermal energy from a cold interior to a warmer exterior, in terms of the second law of thermodynamics. Easy 3 marks No calculator + 20 Marking analysis: A learner attempts the following task: “Outline why a refrigerator requires external work input to transfer thermal energy from a cold interior to a warmer exterior, in terms of the second law of thermodynamics.” Their response addresses only this point: “States that heat naturally (spontaneously) flows only from hot to cold, never from cold to hot, without external influence.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 21 A gas expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 0.020 m³ to 0.050 m³. Calculate the work done by the gas. Easy 3 marks Calculator + 22 Marking analysis: A learner attempts the following task: “A gas expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 0.020 m³ to 0.050 m³. Calculate the work done by the gas.” Their response addresses only this point: “States that for a constant-pressure (isobaric) process, work done by the gas = PΔV.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks Calculator + 23 Explain why no work is done by or on a gas during an isochoric (constant volume) process, and state what happens to any thermal energy transferred to the gas in this case. Easy 3 marks No calculator + 24 Marking analysis: A learner attempts the following task: “Explain why no work is done by or on a gas during an isochoric (constant volume) process, and state what happens to any thermal energy transferred to the gas in this case.” Their response addresses only this point: “States that work done by a gas = PΔV, and since volume does not change (ΔV = 0) in an isochoric process, no work is done.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 25 1.0 mol of an ideal gas undergoes an isothermal expansion at 300 K from a volume of 0.010 m³ to 0.020 m³. Calculate the work done by the gas, using W = nRT ln(V₂/V₁). (R = 8.31 J K⁻¹ mol⁻¹) Medium 4 marks Calculator + 26 Marking analysis: A learner attempts the following task: “1.0 mol of an ideal gas undergoes an isothermal expansion at 300 K from a volume of 0.010 m³ to 0.020 m³. Calculate the work done by the gas, using W = nRT ln(V₂/V₁). (R = 8.31 J K⁻¹ mol⁻¹)” Their response addresses only this point: “States W = nRT ln(V₂/V₁).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks Calculator + 27 0.500 kg of ice at 0°C melts completely and reversibly at constant temperature, absorbing 1.67 × 10⁵ J of thermal energy. Calculate the change in entropy of the ice as it melts. Easy 3 marks Calculator + 28 Marking analysis: A learner attempts the following task: “0.500 kg of ice at 0°C melts completely and reversibly at constant temperature, absorbing 1.67 × 10⁵ J of thermal energy. Calculate the change in entropy of the ice as it melts.” Their response addresses only this point: “States ΔS = Q/T for a reversible process at constant temperature.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks Calculator + 29 State the Kelvin-Planck statement of the second law of thermodynamics, and explain what it implies about the possibility of a heat engine with 100% efficiency. Easy 3 marks No calculator + 30 Marking analysis: A learner attempts the following task: “State the Kelvin-Planck statement of the second law of thermodynamics, and explain what it implies about the possibility of a heat engine with 100% efficiency.” Their response addresses only this point: “States that it is impossible to construct a heat engine that, operating in a cycle, converts thermal energy completely into work with no other effect (i.e. without exhausting some heat to a cold reservoir).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + Self-assessed Not marked yet
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