IB Physics · Theme B The particulate nature of matter
B.4 Thermodynamics HL only
Summary to follow. 13 syllabus statements · 29 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 13 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Closed system HL
Closed system
A system in which no mass can be transferred in or out, but energy can be transferred in both directions as heat or as work. A fixed quantity of gas in a cylinder sealed by a movable piston is a closed system: it can be heated or cooled, and it can do work on the piston or have work done on it, but no gas enters or leaves.
Internal energy (U)
The total energy of the particles of a system due to their random motion and their interactions: the sum of the random kinetic energies of the particles and the intermolecular potential energy. For an ideal gas there are no intermolecular forces, so the internal energy is only the total random kinetic energy of the particles, and it depends only on the temperature and the number of particles. Internal energy is a property of the state of the system. SI unit: joule (J).
Thermal energy transferred (Q)
Energy transferred between a system and its surroundings because of a temperature difference between them. Q is not something a system contains: it describes a transfer during a process, so it depends on the process and not only on the start and end states. In Clausius' sign convention Q is the resultant thermal energy supplied to the system: positive when the system gains energy as heat, negative when it loses energy as heat. SI unit: joule (J).
First law of thermodynamics
Q = ΔU + W: the resultant thermal energy Q supplied to a closed system equals the increase in its internal energy ΔU plus the resultant work W done by the system. It is the principle of conservation of energy applied to a closed system, in which energy can cross the boundary only as heat or as work. For a complete cycle ΔU = 0, so Q = W.
Clausius' sign convention
The convention used in IB problems for Q = ΔU + W. Q is the resultant thermal energy supplied TO the system (negative if the system loses thermal energy). W is the resultant work done BY the system: work done by the system is positive (the gas expands and pushes its surroundings), and work done ON the system is negative (the gas is compressed). ΔU is the increase in internal energy (negative for a decrease). Some chemistry texts write ΔU = Q + W with W the work done on the system; that is a different convention and must not be mixed with this one.
Students often think A closed system is sealed off from its surroundings completely, so neither matter nor energy can cross its boundary. In fact Yes. A closed system exchanges no mass with its surroundings, but energy can be transferred in both directions as heat or as work.
Students often think Heat is something a body contains: Q is the heat stored in the system, and supplying heat simply adds to this store. In fact No. Q is energy transferred to the system because of a temperature difference during a process. A system contains internal energy, not heat.
Work done by a gas (W = PΔV) HL
Work done by a gas (W = PΔV)
When the boundary of a closed system moves, the gas does work W = PΔV if its pressure P is constant, where ΔV is the change in volume. W is positive when the gas expands (ΔV > 0) and negative when it is compressed. When the pressure changes during the process, W equals the area under the P–V graph between the initial and final volumes, for example the area of a trapezium for a straight-line path. SI unit: joule (J), since 1 Pa × 1 m³ = 1 N m⁻² × 1 m³ = 1 N m = 1 J. Volumes must be in m³ (1 dm³ = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³).
Pressure–volume (P–V) diagram and work
A graph of pressure (vertical axis) against volume (horizontal axis) for a fixed quantity of gas, on which each state is a point and each process is a path. The work done by the gas in a process is the area under its path, and it therefore depends on the path taken, not only on the end points. For a cycle, the net work done by the gas is the area enclosed by the loop: positive when the loop is traversed clockwise (a heat engine), negative when anticlockwise.
Students often think W = PΔV can be used for any expansion by inserting one pressure, such as the final pressure, even if the pressure changes during the process. In fact No. W = PΔV applies only at constant pressure. When the pressure varies, W is the area under the P–V path.
Students often think PΔV means the change in PV, so the work done is found by subtracting P₁V₁ from P₂V₂. In fact No. PΔV is the pressure multiplied by the change in volume. The change in the product, Δ(PV) = P₂V₂ − P₁V₁, is a different quantity, equal to nRΔT for an ideal gas.
Change in internal energy of a monatomic ideal gas HL
Change in internal energy of a monatomic ideal gas
ΔU = (3/2)Nk_BΔT = (3/2)nRΔT, where N is the number of atoms, n the amount of substance and ΔT the change in absolute temperature. Because the internal energy of an ideal gas is its total random kinetic energy, ΔU depends only on ΔT: any two processes with the same temperature change give the same ΔU, whatever heat and work are involved. Using PV = nRT, the same result can be written ΔU = (3/2)Δ(PV). A temperature change is the same number in kelvin and in degrees Celsius.
Boltzmann constant (k_B) and molar gas constant (R)
k_B = 1.38 × 10⁻²³ J K⁻¹ is the constant that links the energy of individual particles to absolute temperature; R = 8.31 J mol⁻¹ K⁻¹ is the corresponding constant per mole. They are related by R = N_A k_B, so Nk_B = nR. In S = k_B ln Ω, k_B links the number of microstates to the macroscopic entropy.
Students often think The thermal energy supplied to a gas all goes into its internal energy, so ΔU equals Q in every process. In fact Only if the gas does no work. In general Q = ΔU + W; when a gas expands at constant pressure, part of Q leaves as work done by the gas.
Students often think The change in internal energy depends on the process: a gas that does work while it is heated ends up with a different ΔU from one that does not, even for the same temperature change. In fact No. ΔU = (3/2)nRΔT depends only on the temperature change. Any two processes with the same ΔT have the same ΔU.
Entropy (S) HL
Entropy (S)
A thermodynamic quantity that relates to the degree of disorder of the particles in a system: the more ways the particles' positions and energies can be arranged consistent with the observed macroscopic state, the greater the entropy. A gas has greater entropy than the same substance as a liquid, and a liquid greater than a solid. Entropy is a property of the state of the system; it is not a form of energy. SI unit: J K⁻¹.
Students often think Entropy means visible untidiness, so a messy room or a shuffled pack of cards has high entropy, whatever the particles are doing. In fact No. Thermodynamic entropy concerns the particles: the number of ways their positions and energies can be arranged consistent with the macrostate.
Students often think Entropy is the thermal energy of the particles, or is measured by the temperature, so it changes only when the temperature changes. In fact No. Entropy is measured in J K⁻¹, not J. It relates to the number of ways the particles can be arranged, not to how much energy they have.
Entropy change from heat and temperature (ΔS = ΔQ/T) HL
Entropy change from heat and temperature (ΔS = ΔQ/T)
When thermal energy ΔQ is transferred reversibly to a system at absolute temperature T, its entropy changes by ΔS = ΔQ/T. Entropy increases when thermal energy is supplied (ΔQ > 0) and decreases when it is removed. The relation applies directly when T stays constant, as in melting, boiling or a large reservoir. T must be in kelvin. SI unit of ΔS: J K⁻¹.
Microstate
One particular arrangement of the individual particles of a system, specifying the state of every particle. In the coin model each particular sequence of heads and tails, such as HTHT for four coins, is one microstate. All microstates of a system are equally probable.
Macrostate
The state of a system described by its macroscopic properties only, without specifying each particle. In the coin model the macrostate is the total number of heads. Different macrostates correspond to different numbers of microstates Ω: for four coins, 'four heads' has Ω = 1 and 'two heads' has Ω = 6. Because each microstate is equally probable, a macrostate with more microstates is more probable.
Statistical entropy (S = k_B ln Ω)
The entropy of a macrostate expressed in terms of the particles: S = k_B ln Ω, where Ω is the number of microstates that correspond to that macrostate and k_B = 1.38 × 10⁻²³ J K⁻¹. The logarithm is the natural logarithm. A macrostate with only one microstate has S = 0. The entropy change between two macrostates is ΔS = k_B ln(Ω₂/Ω₁).
Students often think Temperatures can be substituted in whichever unit they are given, since °C and K are both units of temperature. In fact No. These formulae need the absolute (kelvin) temperature: T/K = θ/°C + 273.
Students often think Each possible outcome, such as 0, 1, 2, 3 or 4 heads from four coins, is equally likely, because the coins are fair. In fact No. Each microstate is equally probable, and macrostates with more microstates are more probable: for four coins, 'two heads' (6 microstates) is six times as likely as 'four heads' (1 microstate).
Isolated system HL
Isolated system
A system in which neither mass nor energy can be transferred in or out. The total energy of an isolated system is constant, but energy can be redistributed within it, for example as heat flowing from a hotter to a colder part.
Second law of thermodynamics: Clausius form
It is impossible for thermal energy to flow spontaneously from a colder body to a hotter body. Thermal energy can be transferred from cold to hot (as in a refrigerator), but only if work is done, so the transfer is never the sole result of the process.
Second law of thermodynamics: Kelvin form
It is impossible for a device operating in a cycle to extract thermal energy from a single reservoir and convert it completely into work. Every heat engine must reject some thermal energy to a colder reservoir, so its efficiency is less than 1 even with no friction.
Second law of thermodynamics: entropy form
The entropy of an isolated system never decreases: it stays constant in a reversible process and increases in an irreversible process. The law sets constraints on which processes are possible and on the overall evolution of an isolated system, which moves towards macrostates of greater entropy. The Clausius, Kelvin and entropy forms are equivalent.
Students often think The second law says thermal energy cannot pass from a colder body to a hotter body by any means at all. In fact No. It forbids this happening spontaneously, as the sole result of a process. A refrigerator transfers thermal energy from cold to hot because work is done.
Students often think Thermal energy flows from the body that has more internal energy to the one that has less, until they have equal amounts. In fact No. Thermal energy flows spontaneously from higher to lower temperature, whatever the internal energies of the bodies.
Reversible and irreversible processes HL
Reversible and irreversible processes
A reversible process is an idealized process that can be run backwards, returning both the system and its surroundings to their original states; it would have to proceed infinitely slowly through equilibrium states with no friction and no heat flow across a finite temperature difference. An irreversible process cannot be reversed in this way. Real processes, such as heat flow from hot to cold, friction, mixing and free expansion, are almost always irreversible, so the entropy of a real isolated system increases.
Students often think Entropy changes only when thermal energy is transferred, so a process with Q = 0 has ΔS = 0. In fact No. ΔS = ΔQ/T gives the entropy change for a reversible transfer. In an irreversible process, such as the free expansion of a gas, entropy increases even though Q = 0.
Students often think A gas always does work when it expands, even into a vacuum, and so it always cools as it expands. In fact No. In a free expansion into a vacuum there is nothing for the gas to push against, so W = 0. For an ideal gas with Q = 0 as well, ΔU = 0 and the temperature is unchanged.
Local decrease of entropy HL
Local decrease of entropy
The entropy of a system that is not isolated can decrease, for example water freezing in a freezer or a living organism building ordered molecules. The second law is obeyed because the process transfers energy to the surroundings and the entropy of the surroundings increases by an equal or greater amount, so the total entropy of system plus surroundings does not decrease.
Students often think In any spontaneous process the entropy of each object involved increases, so the entropy of a system can never decrease. In fact No. The entropy of one part can decrease; the second law requires only that the total entropy of an isolated system does not decrease.
Students often think When a machine does work to cool or order something, it can lower the total entropy; the entropy of the surroundings need not be considered. In fact No. Work can lower the entropy of the system itself, but the energy transferred to the surroundings raises their entropy by an equal or greater amount.
Isovolumetric process HL
Isovolumetric process
A process at constant volume (ΔV = 0). The gas does no work (W = 0), so Q = ΔU. On a P–V diagram it is a vertical line.
Isobaric process
A process at constant pressure. The work done by the gas is W = PΔV, and for a monatomic ideal gas Q = ΔU + W = (3/2)nRΔT + nRΔT = (5/2)nRΔT. On a P–V diagram it is a horizontal line.
Isothermal process
A process at constant temperature. For an ideal gas ΔU = 0, so Q = W: thermal energy must be supplied as the gas expands and removed as it is compressed. PV = constant, so on a P–V diagram the path is a curve (a hyperbola).
Adiabatic process
A process in which no thermal energy is transferred between the gas and its surroundings (Q = 0), so ΔU = −W. In an adiabatic compression the work done on the gas raises its internal energy and temperature; in an adiabatic expansion the gas does work at the expense of its internal energy and cools. It is approached when the gas is well insulated or the change is rapid.
Students often think In an adiabatic process no heat enters or leaves the gas, so its temperature cannot change; adiabatic and isothermal processes are the same. In fact No. Adiabatic means Q = 0. Then ΔU = −W, so the temperature rises in an adiabatic compression and falls in an adiabatic expansion.
Students often think If the temperature of a gas stays constant, no thermal energy can have been transferred to or from it. In fact No. In an isothermal process of an ideal gas ΔU = 0, so Q = W: thermal energy must be supplied during an isothermal expansion and removed during a compression.
Adiabatic equation for a monatomic ideal gas HL
Adiabatic equation for a monatomic ideal gas
In an adiabatic process of a monatomic ideal gas, PV^(5/3) = constant, so P₁V₁^(5/3) = P₂V₂^(5/3). Because the exponent 5/3 is greater than 1, an adiabatic curve on a P–V diagram is steeper than the isothermal curve through the same point. Combined with PV = nRT, it gives the temperature change: compressing the gas adiabatically raises its temperature.
Students often think The exponent in the adiabatic relation for P can be taken as 3/5 as well as 5/3, since both numbers appear in the working. In fact No. PV^(5/3) = constant gives P₂ = P₁(V₁/V₂)^(5/3).
Students often think The new pressure is found by multiplying by the ratio of new volume to old volume, raised to the power 5/3. In fact No. P₁V₁^(5/3) = P₂V₂^(5/3) gives P₂ = P₁(V₁/V₂)^(5/3), which is larger than P₁ for a compression.
Cyclic process HL
Cyclic process
A sequence of processes that returns a gas to its initial state. Because internal energy is a property of the state, ΔU = 0 for a complete cycle, so the net thermal energy supplied equals the net work done by the gas (Q = W). Q and W are not zero over a cycle in general: the net work is the area enclosed by the loop on the P–V diagram.
Heat engine
A device that takes a working substance repeatedly round a cycle, absorbing thermal energy Q_h from a hot reservoir, doing useful work W and rejecting thermal energy Q_c to a cold reservoir, with W = Q_h − Q_c. A reservoir is a body so large that its temperature does not change as energy is exchanged with it.
Students often think Because the gas ends each cycle in the state it started in, the net work done by the gas in the cycle is zero, and likewise the net thermal energy supplied is zero. In fact No. The net work done by the gas in a cycle equals the area enclosed by the loop on the P–V diagram, which is not zero for a heat engine.
Students often think The work output of a cycle is the work done by the gas while it expands; the compression part of the cycle does not count. In fact No. The net work is the work done by the gas during expansion minus the work done on it during compression.
Efficiency of a heat engine (η) HL
Efficiency of a heat engine (η)
η = useful work / input energy = W/Q_h, where Q_h is the thermal energy supplied from the hot reservoir in one cycle. Since W = Q_h − Q_c, η = 1 − Q_c/Q_h. It is a ratio with no unit and is often quoted as a percentage. Different cycles between the same reservoirs can have different efficiencies.
Students often think The efficiency of an engine is Q_c/Q_h, the output to the cold reservoir divided by the input from the hot one. In fact No. That fraction, Q_c/Q_h, is the energy wasted. The efficiency is W/Q_h = 1 − Q_c/Q_h.
Students often think Efficiency is the work divided by the energy rejected to the cold reservoir, since that is the other energy that leaves the engine. In fact No. η = W/Q_h: the denominator is the energy supplied from the hot reservoir, not the energy rejected.
Carnot cycle HL
Carnot cycle
An idealized reversible cycle consisting of an isothermal expansion at the hot-reservoir temperature, an adiabatic expansion, an isothermal compression at the cold-reservoir temperature and an adiabatic compression. No engine working between two given reservoirs can be more efficient than a Carnot engine working between them.
Carnot efficiency
η_Carnot = 1 − T_c/T_h, where T_c and T_h are the absolute temperatures of the cold and hot reservoirs. It is the upper limit on the efficiency of any heat engine working between those reservoirs. It is less than 1 unless T_c = 0 K, and it rises as T_h is raised or T_c is lowered. Temperatures must be in kelvin.
Students often think The Carnot efficiency is the efficiency of one particular cycle, so an engine using a different cycle could be more efficient. In fact No. No engine of any design working between two reservoirs can exceed the efficiency of a Carnot engine between the same reservoirs.
Students often think The Carnot efficiency depends only on the temperature difference T_h − T_c, so raising both temperatures by the same amount leaves it unchanged. In fact No. η_Carnot = 1 − T_c/T_h = (T_h − T_c)/T_h depends on the ratio of the absolute temperatures. The same difference gives a lower efficiency at higher temperatures.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 The first law of thermodynamics, Q = ΔU + W, is applied to a closed system using Clausius' sign convention. Which statement about the law is correct? HL
Answer and reasoning
It applies to a system that no energy can enter or leave, because a closed system is sealed off from its surroundings. — A student who reads 'closed' in its everyday sense picks this. A closed system exchanges no MASS with its surroundings, but energy can cross its boundary as heat and as work; a system that exchanges neither mass nor energy is isolated, and for it Q = W = 0.
Q is the thermal energy that the system contains, and it forms one part of the system's total internal energy U. — A student who thinks of heat as something a body stores picks this. Q is energy transferred because of a temperature difference during a process; what the system contains is internal energy U, and the energy supplied as heat can leave again as work.
Q and W are properties of the state of the system, as U is, so each depends only on the initial and final states. — A student who treats heat and work as properties of the state picks this. Only U is a property of the state: two different processes between the same states have the same ΔU but can involve different amounts of heat and work.
It is conservation of energy: the thermal energy supplied equals the rise in internal energy plus the work done by the system. — Energy can cross the boundary of a closed system only as heat or as work. Conservation of energy then requires the thermal energy supplied, Q, to equal the increase in internal energy, ΔU, plus the energy that leaves as work done by the system, W.
2 Helium, a monatomic ideal gas, is kept at a constant pressure of 1.5 × 10⁵ Pa while it expands from 2.0 dm³ to 5.0 dm³. What is the work W, as it appears in Q = ΔU + W, for this process? HL
Answer and reasoning
−4.5 × 10² J — A student who treats W as the work done ON the gas gives the expansion a negative sign. In Clausius' convention W is the work done BY the gas, which is positive when the gas expands.
+4.5 × 10² J — At constant pressure W = PΔV. ΔV = 5.0 − 2.0 = 3.0 dm³ = 3.0 × 10⁻³ m³, so W = 1.5 × 10⁵ Pa × 3.0 × 10⁻³ m³ = 450 J. It is positive because the gas expands and does work on its surroundings.
+7.5 × 10² J — A student who multiplies the pressure by the final volume, 1.5 × 10⁵ × 5.0 × 10⁻³ = 750 J, has calculated PV, not the work. Work depends on the change in volume: W = PΔV = 1.5 × 10⁵ × 3.0 × 10⁻³ = 450 J.
+4.5 × 10⁵ J — A student who substitutes ΔV = 3.0 without converting dm³ to m³ gets an answer 1000 times too large. 1 dm³ = 10⁻³ m³, so ΔV = 3.0 × 10⁻³ m³ and W = 450 J.
Working Constant pressure, so W = PΔV. ΔV = 5.0 dm³ − 2.0 dm³ = 3.0 dm³ = 3.0 × 10⁻³ m³. W = 1.5 × 10⁵ Pa × 3.0 × 10⁻³ m³ = 4.5 × 10² J. The gas expands, so the work done by it is positive: W = +4.5 × 10² J.
3 A sample of 8.0 g of helium, a monatomic ideal gas of molar mass 4.0 g mol⁻¹, is heated at constant pressure from 20 °C to 80 °C. The thermal energy supplied is 2.49 × 10³ J. R = 8.31 J mol⁻¹ K⁻¹. What is the change in internal energy ΔU of the helium? HL
Answer and reasoning
2.5 × 10³ J — A student who thinks all the thermal energy supplied becomes internal energy picks the value of Q. At constant pressure the gas expands and does work, W = nRΔT = 997 J, so ΔU = Q − W = 1.5 × 10³ J.
1.5 × 10³ J — n = 8.0 g ÷ 4.0 g mol⁻¹ = 2.0 mol and ΔT = 60 K. ΔU = (3/2)nRΔT = 1.5 × 2.0 × 8.31 × 60 = 1.5 × 10³ J. The rest of the 2.5 × 10³ J supplied leaves as work done by the expanding gas.
8.3 × 10³ J — A student who adds 273 to the temperature change uses ΔT = 333 K. A change of 60 °C is a change of 60 K, so ΔU = 1.5 × 2.0 × 8.31 × 60 = 1.5 × 10³ J.
6.0 × 10³ J — A student who uses the mass, 8.0, as the amount of substance gets a value four times too large. n = m/M = 8.0 g ÷ 4.0 g mol⁻¹ = 2.0 mol.
Working n = m/M = 8.0 g ÷ 4.0 g mol⁻¹ = 2.0 mol. ΔT = 80 °C − 20 °C = 60 K. ΔU = (3/2)nRΔT = 1.5 × 2.0 mol × 8.31 J mol⁻¹ K⁻¹ × 60 K = 1496 J ≈ 1.5 × 10³ J. (Check: W = PΔV = nRΔT = 997 J, and ΔU + W = 2.5 × 10³ J = Q.)
4 Which statement correctly relates the entropy of a system to its particles? HL
Answer and reasoning
Entropy is the total kinetic energy of all the particles, so it is measured in joules. — A student who thinks entropy is a form of energy picks this. Entropy is measured in J K⁻¹, not J, and it can change with no change in energy, as in the free expansion of a gas.
Entropy is how untidy the objects in a system look, whatever their particles are doing. — A student who takes 'disorder' in its everyday sense picks this. Thermodynamic entropy concerns the arrangements of particles; rearranging large objects changes it negligibly.
The more ways the particles can be arranged, the greater their disorder and entropy. — Entropy relates to the degree of disorder of the particles: the more arrangements of positions and energies are consistent with the macrostate, the greater the entropy. This is why a gas has more entropy than the same substance as a liquid or solid.
Entropy, like energy, is conserved, so the particles keep the same disorder. — A student who assumes every thermodynamic quantity is conserved picks this. Entropy is not conserved: it increases in every irreversible process in an isolated system, and the disorder of the particles changes when, for example, a solid melts.
5 Which of the following is a correct statement of the second law of thermodynamics in its Clausius form? HL
Answer and reasoning
Thermal energy cannot be moved from a colder body to a hotter body by any means at all. — A student who drops the word 'spontaneously' picks this. A refrigerator moves thermal energy from its cold interior to the warmer room; this is allowed because work is done on it.
Thermal energy cannot flow spontaneously from a colder body to a hotter body. — This is the Clausius form. The word 'spontaneously' matters: thermal energy can be moved from cold to hot, as in a refrigerator, but only when work is done, so the transfer is not the sole result of the process.
Thermal energy flows spontaneously from the body with more internal energy to the one with less. — A student who confuses internal energy with temperature picks this. The direction of heat flow is set by temperature: a warm bath with a large internal energy still receives heat from a small, hotter cup of tea.
Cold flows spontaneously from a colder body into a hotter body until their temperatures are equal. — A student who thinks of cold as something that moves picks this. There is no substance called cold; cooling means thermal energy flowing out of the hotter body into the colder one.
6 A hot copper block and a cold copper block are placed in contact inside a sealed, perfectly insulated box, so that together they form an isolated system. They reach a common temperature. Which statement about entropy is correct? HL
Answer and reasoning
The total entropy is unchanged, because the entropy lost by the hot block is gained by the cold block. — A student who thinks entropy is conserved picks this. The cold block gains Q/T_c, which is larger than the Q/T_h lost by the hot block, so the total increases.
The total entropy increases, because heat flow across a temperature difference is irreversible. — Heat flow from a hotter to a colder body is irreversible, so the entropy of the isolated system increases. The hot block loses entropy Q/T_h, and the cold block gains more, Q/T_c, because T_c < T_h.
The entropy of each block increases, because thermal energy flows between the two blocks. — A student who thinks the entropy of every object must increase picks this. The hot block loses thermal energy and its entropy decreases; only the total entropy must increase.
The total entropy increases, because some of the energy is destroyed as the blocks come to equilibrium. — A student who thinks irreversible processes destroy energy picks this. The total energy of the isolated system is unchanged; the entropy increases because the process is irreversible, not because energy is lost.
7 Water placed in the freezer compartment of a refrigerator freezes, and the entropy of the water decreases. The refrigerator stands in a kitchen. Which statement explains why this is consistent with the second law of thermodynamics? HL
Answer and reasoning
The refrigerator transfers thermal energy to the kitchen, whose entropy rises by at least as much as the water's falls. — The water is not an isolated system, so its entropy can decrease locally. The refrigerator delivers to the kitchen the thermal energy removed from the water plus the work done on it, and the entropy of the kitchen increases by an equal or greater amount.
The entropy lost by the water is passed to the air inside the freezer, so the total entropy is unchanged. — A student who thinks entropy is conserved picks this. The freezer air is also being cooled, and its entropy falls too. The compensating increase is in the kitchen, and for a real, irreversible refrigerator it exceeds the decrease.
The water is not isolated; no thermal energy can pass to the warmer kitchen, whose entropy is unchanged. — A student who thinks thermal energy can never pass from a colder body to a hotter one, by any means, rules out any effect on the kitchen. A refrigerator uses work to transfer thermal energy from its cold interior to the warmer kitchen, and the kitchen's entropy rises by at least as much as the water's falls.
The work done by the refrigerator allows the total entropy to decrease, so no increase elsewhere is needed. — A student who thinks doing work can override the second law picks this. Work lets the water's entropy fall, but the total entropy of water plus surroundings cannot decrease: the kitchen's entropy rises to compensate.
8 Helium, a monatomic ideal gas, has a pressure of 1.0 × 10⁵ Pa and a volume of 3.0 × 10⁻³ m³. It is compressed adiabatically to a volume of 1.0 × 10⁻³ m³. For this process PV^(5/3) = constant. What is the final pressure of the helium? HL
Answer and reasoning
3.0 × 10⁵ Pa — A student who uses PV = constant treats the compression as isothermal. In an adiabatic compression the temperature rises, so the pressure rises by 3.0^(5/3) = 6.24, not 3.
1.9 × 10⁵ Pa — A student who uses the exponent 3/5 instead of 5/3 gets 3.0^(3/5) = 1.93. Solving PV^(5/3) = constant for P gives P₂ = P₁(V₁/V₂)^(5/3).
6.2 × 10⁵ Pa — P₂ = P₁(V₁/V₂)^(5/3) = 1.0 × 10⁵ × 3.0^(5/3) = 1.0 × 10⁵ × 6.24 = 6.2 × 10⁵ Pa. The pressure rises by more than the factor of 3 an isothermal compression would give, because the gas also warms.
1.6 × 10⁴ Pa — A student who writes the volume ratio as V₂/V₁ gets (1/3)^(5/3) = 0.160, a pressure that falls on compression. From P₁V₁^(5/3) = P₂V₂^(5/3), P₂ = P₁(V₁/V₂)^(5/3).
Working P₁V₁^(5/3) = P₂V₂^(5/3), so P₂ = P₁(V₁/V₂)^(5/3) = 1.0 × 10⁵ Pa × (3.0 × 10⁻³ / 1.0 × 10⁻³)^(5/3) = 1.0 × 10⁵ Pa × 3.0^(5/3) = 1.0 × 10⁵ Pa × 6.24 = 6.2 × 10⁵ Pa.
9 The working gas of a heat engine is taken round one complete cycle, returning to its initial state. Which statement about the cycle is correct? HL
Answer and reasoning
The net work done by the gas is zero, as the gas returns to the state it started in. — A student who treats work as a property of the state picks this. The net work done in a cycle is the area enclosed on the P–V diagram; for an engine it is positive, which is the whole purpose of the engine.
The net work done by the gas is the work it does while it expands, as compression is not output. — A student who counts only the expansion stroke picks this. The gas must be compressed again, and work is done on it then; the net work done by the gas is the expansion work minus the compression work, the area enclosed on the P–V diagram.
ΔU = 0, so the net thermal energy supplied to the gas equals the net work done by the gas. — Internal energy is a property of the state, and the gas returns to its initial state, so ΔU = 0 for the cycle. Q = ΔU + W then gives Q = W: the net heat supplied equals the net work done.
ΔU is positive, because thermal energy is supplied to the gas in every cycle of the engine. — A student who thinks supplied heat builds up in the gas picks this. Over a complete cycle the gas returns to its initial state, so ΔU = 0; the heat supplied leaves as work and as heat rejected.
Working U is a property of the state and the gas returns to its initial state, so ΔU = 0 over the cycle. Q = ΔU + W then gives net Q = net W; for an engine both are positive and equal to the area enclosed by the cycle on the P–V diagram.
10 A heat engine works between reservoirs at 600 K and 300 K. In each cycle it absorbs 2.5 kJ of thermal energy from the hot reservoir and rejects 1.8 kJ of thermal energy to the cold reservoir. What is the efficiency of the engine? HL
Answer and reasoning
72% — A student who divides the energy rejected by the energy supplied, 1.8 ÷ 2.5, has calculated the fraction wasted. Efficiency is the useful work divided by the input: 0.7 ÷ 2.5 = 28%.
39% — A student who divides the work by the energy rejected, 0.7 ÷ 1.8, has used the wrong denominator. The input energy is the 2.5 kJ absorbed from the hot reservoir.
50% — A student who uses 1 − T_c/T_h = 1 − 300/600 has found the Carnot efficiency, the maximum for these reservoirs. This engine's actual efficiency is W/Q_h = 28%.
28% — Useful work per cycle W = Q_h − Q_c = 2.5 − 1.8 = 0.7 kJ. η = W/Q_h = 0.7 ÷ 2.5 = 0.28 = 28%. This is below the Carnot limit of 50% for these reservoirs, as it must be.
Working W = Q_h − Q_c = 2.5 kJ − 1.8 kJ = 0.7 kJ. η = W/Q_h = 0.7 kJ ÷ 2.5 kJ = 0.28 = 28%. (Carnot limit 1 − 300/600 = 50%, so the engine is possible.)
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19 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Helium in a cylinder fitted with a piston is a closed system. During one process the piston does 600 J of work on the helium, and the helium loses 400 J of thermal energy to its surroundings. Using Clausius' sign convention, what is the change in internal energy ΔU of the helium? HL
Answer and reasoning
+2.0 × 10² J — Thermal energy is lost, so Q = −400 J; work is done on the gas, so W = −600 J. Q = ΔU + W gives ΔU = Q − W = −400 J − (−600 J) = +200 J. The work done on the gas adds more energy than the heat removes.
−2.0 × 10² J — A student who substitutes the sizes Q = 400 J and W = 600 J without signs gets ΔU = 400 − 600 = −200 J. The helium LOSES thermal energy (Q = −400 J) and has work done ON it (W = −600 J), so ΔU = −400 − (−600) = +200 J.
−1.0 × 10³ J — A student who takes the work done on the gas as W = +600 J, as in the chemistry form ΔU = Q + W, but uses Q = ΔU + W gets ΔU = −400 − 600 = −1000 J. In Clausius' convention work done on the system is negative, W = −600 J, which gives ΔU = +200 J.
−4.0 × 10² J — A student who thinks only heating and cooling change internal energy sets ΔU = Q = −400 J. The 600 J of work done by the piston also transfers energy to the helium, so ΔU = Q − W = −400 − (−600) = +200 J.
Working Clausius' sign convention: Q is the thermal energy supplied to the helium, so Q = −400 J (energy lost); W is the work done BY the helium, so W = −600 J (work done on it). Q = ΔU + W, so ΔU = Q − W = −400 J − (−600 J) = +200 J = +2.0 × 10² J.
2 A fixed mass of helium, a monatomic ideal gas, is taken from state X to state Y. On a P–V diagram (pressure on the vertical axis, volume on the horizontal axis) the process is a straight line from X (2.0 × 10⁻³ m³, 1.0 × 10⁵ Pa) to Y (6.0 × 10⁻³ m³, 3.0 × 10⁵ Pa), where each state is given as (volume, pressure). What is the work W, as it appears in Q = ΔU + W, for this process? HL
Answer and reasoning
−8.0 × 10² J — A student who treats W as the work done ON the gas gives the expansion a negative sign. In Clausius' convention W is the work done BY the gas, positive when the gas expands, as it does here.
+1.2 × 10³ J — A student who uses W = PΔV with the final pressure, 3.0 × 10⁵ × 4.0 × 10⁻³ = 1200 J, has treated the pressure as constant. It rises from 1.0 × 10⁵ Pa to 3.0 × 10⁵ Pa, so the work is the trapezium area under the line, 800 J.
+8.0 × 10² J — The pressure is not constant, so W is the area under the straight-line path, a trapezium: W = ½(1.0 × 10⁵ + 3.0 × 10⁵) Pa × 4.0 × 10⁻³ m³ = 800 J. It is positive because the gas expands.
+1.6 × 10³ J — A student who reads PΔV as the change in PV calculates P_YV_Y − P_XV_X = 1800 − 200 = 1600 J. That quantity equals nRΔT, not the work; the work is the area under the path, 800 J.
Working Pressure varies, so W = area under the P–V path. For a straight line this is a trapezium: W = ½(P_X + P_Y)(V_Y − V_X) = ½(1.0 × 10⁵ Pa + 3.0 × 10⁵ Pa) × (6.0 − 2.0) × 10⁻³ m³ = 2.0 × 10⁵ Pa × 4.0 × 10⁻³ m³ = 8.0 × 10² J. The gas expands, so W = +8.0 × 10² J.
3 Two identical samples, X and Y, each contain the same amount of helium, a monatomic ideal gas, at 300 K. X is heated at constant volume and Y is heated at constant pressure, and both finish at 350 K. Which comparison of the two processes is correct? HL
Answer and reasoning
ΔU is greater for Y, as more thermal energy must be supplied to Y than to X. — A student who thinks all the thermal energy supplied becomes internal energy picks this. Y does need more heat, but the extra leaves as work done by the expanding gas; ΔU depends only on ΔT, which is the same for X and Y.
Y has the smaller ΔU, since it loses some of its energy as work as it expands. — A student who thinks ΔU depends on the process picks this. For an ideal gas ΔU = (3/2)nRΔT, so equal temperature changes give equal ΔU. The work done by Y is extra energy supplied as heat, not energy taken from its internal energy.
ΔU is greater for Y, since its atoms move further apart and gain potential energy. — A student who includes intermolecular potential energy in the internal energy of an ideal gas picks this. The ideal gas model has no intermolecular forces, so U is kinetic energy only and depends only on temperature.
ΔU is the same for both, but more thermal energy must be supplied to Y than to X. — ΔU = (3/2)nRΔT depends only on the temperature change, which is the same for both. Y expands and does work W = nRΔT, so it needs Q = ΔU + W, more than X, for which W = 0 and Q = ΔU.
4 Helium, a monatomic ideal gas, expands reversibly and isothermally at 27 °C, doing 600 J of work on its surroundings. What is the entropy change of the helium? HL
Answer and reasoning
+2.0 J K⁻¹, as Q = W = 600 J is supplied to it at 300 K — At constant temperature ΔU = 0 for an ideal gas, so Q = ΔU + W = 600 J of thermal energy is supplied to the helium at T = 300 K. ΔS = ΔQ/T = 600 ÷ 300 = +2.0 J K⁻¹.
0 J K⁻¹, as no heat is transferred at a constant temperature — A student who thinks constant temperature means no heat transfer sets ΔQ = 0. The expanding gas does 600 J of work while its internal energy stays constant, so 600 J of thermal energy must be supplied, and its entropy rises.
−2.0 J K⁻¹, as the gas does work, so W = −600 J and Q = −600 J — A student who treats the work done by the gas as negative gets Q = ΔU + W = −600 J. In Clausius' convention work done BY the gas is positive, W = +600 J, so Q = +600 J and the entropy increases.
+22 J K⁻¹, as Q = 600 J is supplied at a temperature of 27 °C — A student who divides by 27, the temperature in °C, gets 22 J K⁻¹. ΔS = ΔQ/T needs the absolute temperature, 27 + 273 = 300 K.
Working Isothermal, ideal gas: ΔU = 0, so Q = ΔU + W = 0 + 600 J = 600 J (supplied to the gas). T = 27 + 273 = 300 K, constant and the process reversible, so ΔS = ΔQ/T = 600 J ÷ 300 K = +2.0 J K⁻¹.
5 Four identical fair coins are tossed. Each particular sequence of heads (H) and tails (T), such as HTHT, is a microstate, and all microstates are equally probable. The macrostate is the total number of heads. What is the probability of the macrostate 'two heads'? HL
Answer and reasoning
20.0% — A student who assumes the five macrostates (0, 1, 2, 3 or 4 heads) are equally likely gets 1/5. It is the 16 microstates that are equally probable; 'two heads' has 6 of them, 'four heads' only 1.
6.25% — A student who counts 'two heads' as the single arrangement HHTT gets 1/16. Six different sequences contain exactly two heads, so the probability is 6/16.
37.5% — There are 2⁴ = 16 equally probable microstates. Six of them have exactly two heads: HHTT, HTHT, HTTH, THHT, THTH and TTHH. P = 6/16 = 37.5%.
50.0% — A student who confuses the fraction of coins showing heads (2 of 4) with the probability of the outcome gets 1/2. The probability is the number of microstates in the macrostate divided by the total: 6/16.
Working Total microstates = 2⁴ = 16, all equally probable. Microstates with exactly two heads: HHTT, HTHT, HTTH, THHT, THTH, TTHH, so Ω = 6. P = 6/16 = 0.375 = 37.5%.
6 A set of four fair coins changes from the macrostate 'four heads' to the macrostate 'two heads'. Using S = k_B ln Ω, where Ω is the number of microstates of the macrostate and k_B = 1.38 × 10⁻²³ J K⁻¹, what is the change in entropy, expressed as a multiple of k_B and in J K⁻¹? HL
Answer and reasoning
5.00k_B = 6.90 × 10⁻²³ J K⁻¹ — A student who leaves out the logarithm calculates ΔS = k_B × 6 − k_B × 1 = 5k_B = 6.90 × 10⁻²³ J K⁻¹. Entropy is S = k_B ln Ω, so ΔS = k_B ln 6 − k_B ln 1 = k_B ln 6.
0.78k_B = 1.07 × 10⁻²³ J K⁻¹ — A student who presses the base-10 log key gets k_B log₁₀ 6 = 0.78k_B = 1.07 × 10⁻²³ J K⁻¹. The formula uses the natural logarithm: ln 6 = 1.79, not 0.778.
2.77k_B = 3.83 × 10⁻²³ J K⁻¹ — A student who uses Ω = 16, the total number of microstates of four coins, for the 'two heads' macrostate gets k_B ln 16 = 2.77k_B. Ω counts only the microstates of the macrostate in question: six for 'two heads'.
7 An inventor claims to have built a device that, operating in a cycle, extracts 1000 J of thermal energy from seawater at 290 K in each cycle and delivers 1000 J of work, with no other effect. Which judgement of this claim is correct? HL
Answer and reasoning
Impossible: although it conserves energy, it breaks the Kelvin form of the second law. — The Kelvin form says no device operating in a cycle can extract thermal energy from a single reservoir and convert it completely into work. Energy is conserved here, so the first law is satisfied, but the second law forbids the device.
Possible: the work output equals the thermal energy input, so no law of physics is broken. — A student who thinks energy conservation is the only constraint picks this. The first law is satisfied, but the second law (Kelvin form) forbids any cyclic device that turns heat from one reservoir entirely into work.
Impossible: seawater at 290 K is cold, so it contains no thermal energy to extract. — A student who thinks cold bodies contain no heat picks this. Seawater at 290 K has a very large internal energy. The claim fails because of the second law, not because the energy is absent.
Possible, but only if all friction and heat leaks in the device are removed. — A student who thinks friction is the only cause of inefficiency picks this. Even a frictionless device must reject some thermal energy to a colder reservoir; complete conversion from one reservoir is forbidden by the Kelvin form.
8 Which statement of the second law of thermodynamics in terms of entropy is correct for an isolated system? HL
Answer and reasoning
Its entropy increases in every process, whether the process is reversible or not. — A student who has learned the slogan 'entropy always increases' picks this. In a reversible process the entropy of an isolated system stays constant; it increases only in irreversible processes.
Its entropy is constant in every process, as entropy is conserved like energy. — A student who assumes entropy is conserved like energy picks this. Entropy is created in irreversible processes, so the entropy of an isolated system increases in any real process.
Its entropy increases only when its temperature rises and is otherwise fixed. — A student who thinks entropy is measured by temperature or thermal energy picks this. Entropy can increase at constant temperature, as when ice melts or a gas expands freely.
Its entropy stays constant in reversible processes and increases in irreversible processes. — This is the entropy form of the second law: the entropy of an isolated system never decreases. It is unchanged in an idealized reversible process and increases in an irreversible one.
9 A rigid, insulated container is divided by a partition. One half contains helium, a monatomic ideal gas; the other half is a vacuum. The partition is removed and the gas spreads freely through the whole container, which forms an isolated system. What happens to the entropy of the gas? HL
Answer and reasoning
It is unchanged, because no heat is transferred and so ΔS = ΔQ/T = 0. — A student who applies ΔS = ΔQ/T to every process picks this. That relation holds for reversible transfers. The free expansion is irreversible, and the entropy increases because the gas has more microstates in the larger volume.
It decreases, as the expanding gas does work and cools, so its entropy falls. — A student who thinks a gas does work whenever it expands picks this. Expanding into a vacuum, the gas pushes on nothing, so W = 0; with Q = 0 as well, ΔU = 0 and the temperature of the ideal gas is unchanged.
It increases, as the free expansion is irreversible although Q = 0. — The gas spreads into twice the volume, so many more microstates are available to it and S = k_B ln Ω increases. The expansion is irreversible: the gas does not return to one half by itself. ΔS = ΔQ/T does not apply to an irreversible process.
It is unchanged, since entropy, like energy, is conserved in an isolated system. — A student who thinks entropy is conserved like energy picks this. Energy is conserved in the isolated container, but entropy increases in every irreversible process, and free expansion is irreversible.
Working Isolated, so Q = 0; the gas pushes on nothing, so W = 0; hence ΔU = 0 and the temperature of the ideal gas is unchanged. The volume available to each atom doubles, so the number of microstates Ω increases and S = k_B ln Ω increases. The expansion is irreversible (the gas never returns to one half by itself), so ΔS > 0 even though Q = 0; ΔS = ΔQ/T applies only to reversible transfers.
10 In one minute a refrigerator removes 3000 J of thermal energy from its interior, which stays at 250 K, and delivers 4000 J of thermal energy to the kitchen, which stays at 300 K. What is the total entropy change of the interior and the kitchen in that minute? HL
Answer and reasoning
−12 J K⁻¹; the work done lets the total entropy decrease — A student who counts only the interior and thinks work overrides the second law gets −12 J K⁻¹. The 4000 J delivered to the kitchen raises its entropy by 13.3 J K⁻¹, so the total increases.
0 J K⁻¹; all the entropy that leaves the interior reaches the kitchen — A student who thinks entropy is conserved assumes the kitchen gains exactly the 12 J K⁻¹ the interior loses. The kitchen receives 4000 J at 300 K, which is 13.3 J K⁻¹, so the total rises by 1.3 J K⁻¹.
+25 J K⁻¹; the interior and the kitchen both gain entropy — A student who thinks every part gains entropy adds 12.0 and 13.3 J K⁻¹. The interior loses thermal energy, so its entropy decreases by 12.0 J K⁻¹; the total is 13.3 − 12.0 = +1.3 J K⁻¹.
+1.3 J K⁻¹; the kitchen gains more entropy than the interior loses — Interior: ΔS = −3000/250 = −12.0 J K⁻¹. Kitchen: ΔS = +4000/300 = +13.3 J K⁻¹. Total = +1.3 J K⁻¹. The interior's entropy decreases locally, but the kitchen's increases by more, as the second law requires.
Working Interior: ΔS = ΔQ/T = −3000 J ÷ 250 K = −12.0 J K⁻¹. Kitchen: ΔS = +4000 J ÷ 300 K = +13.3 J K⁻¹. Total ΔS = −12.0 + 13.3 = +1.3 J K⁻¹ (positive, consistent with the second law).
11 Which statement about a thermodynamic process of a fixed mass of monatomic ideal gas is correct? HL
Answer and reasoning
Adiabatic: the temperature is kept fixed, so ΔU = 0 and Q = W. — A student who confuses adiabatic with isothermal picks this. Adiabatic means no heat transfer (Q = 0); the temperature changes because work changes the internal energy.
Adiabatic: no thermal energy is transferred, so Q = 0 and W = −ΔU. — An adiabatic process is defined by Q = 0. The first law then gives 0 = ΔU + W, so W = −ΔU: work done by the gas lowers its internal energy, and work done on it raises it.
Isothermal: the temperature is kept fixed, so no thermal energy flows. — A student who thinks constant temperature means no heat transfer picks this. At constant temperature ΔU = 0, so Q = W: an expanding gas must be supplied with exactly as much heat as the work it does.
Isobaric: all of the thermal energy supplied becomes internal energy. — A student who thinks heat supplied all becomes internal energy picks this. At constant pressure the gas expands and does work PΔV, so Q = ΔU + PΔV = (5/2)nRΔT, of which only (3/2)nRΔT is ΔU.
Working Adiabatic: Q = 0 by definition, so Q = ΔU + W gives 0 = ΔU + W, i.e. W = −ΔU. Isothermal: ΔU = 0, so Q = W (not zero when the volume changes). Isobaric: Q = ΔU + PΔV, so Q ≠ ΔU when the gas expands.
12 Helium, a monatomic ideal gas, is compressed adiabatically. The piston does 450 J of work on the gas. Which statement correctly describes the change? HL
Answer and reasoning
ΔU = +450 J and T rises, as Q = 0 and W = −450 J — Adiabatic means Q = 0. Work is done on the gas, so W = −450 J. Q = ΔU + W gives ΔU = +450 J, and since ΔU = (3/2)nRΔT, the temperature rises.
ΔU = 0 and T is unchanged, as no heat enters the gas — A student who thinks adiabatic means constant temperature picks this. No heat enters, but 450 J of work is done on the gas, which raises its internal energy and hence its temperature.
ΔU = −450 J and T falls, because Q = 0 and W = +450 J — A student who takes work done on the gas as positive W gets ΔU = Q − W = −450 J. In Clausius' convention work done on the system is negative, so ΔU = +450 J.
ΔU = +450 J but T is unchanged; the energy is stored as PE — A student who thinks an ideal gas stores intermolecular potential energy picks this. An ideal gas has no intermolecular forces, so its internal energy is kinetic only and an increase in U means an increase in T.
Working Adiabatic: Q = 0. Work done on the gas: W = −450 J. Q = ΔU + W → 0 = ΔU − 450 J → ΔU = +450 J. For a monatomic ideal gas ΔU = (3/2)nRΔT, so ΔT > 0: the temperature rises.
13 Helium, a monatomic ideal gas, at 27 °C is compressed adiabatically to one-eighth of its initial volume. For this process PV^(5/3) = constant. What is the final temperature of the helium? HL
Answer and reasoning
3.0 × 10² K — A student who thinks adiabatic means constant temperature keeps T = 300 K. No heat is transferred, but the work done on the gas raises its internal energy, so its temperature rises to 1200 K.
1.2 × 10³ K — PV^(5/3) = constant gives P₂/P₁ = 8^(5/3) = 32. Then PV/T = constant gives T₂ = T₁ × (P₂/P₁) × (V₂/V₁) = 300 K × 32 × (1/8) = 1200 K.
9.6 × 10³ K — A student who multiplies the temperature by the pressure ratio, 32, has used P ∝ T, which holds only at constant volume. The volume falls by a factor of 8, so T₂ = 300 × 32 ÷ 8 = 1200 K.
3.8 × 10² K — A student who uses 27, the temperature in °C, gets 27 × 4 = 108 °C = 381 K. The gas laws need absolute temperature: T₁ = 300 K, so T₂ = 1200 K.
Working Initial T₁ = 27 + 273 = 300 K. Adiabatic: P₁V₁^(5/3) = P₂V₂^(5/3), so P₂/P₁ = (V₁/V₂)^(5/3) = 8^(5/3) = 32. Ideal gas: P₁V₁/T₁ = P₂V₂/T₂, so T₂ = T₁ × (P₂/P₁)(V₂/V₁) = 300 K × 32 × 1/8 = 1200 K = 1.2 × 10³ K. (Equivalently T₂ = T₁(V₁/V₂)^(2/3) = 300 × 4.)
14 Helium, a monatomic ideal gas, is the working substance of an engine. On a P–V diagram (pressure on the vertical axis, volume on the horizontal axis) each cycle is a rectangle with corners A (1.0 × 10⁻³ m³, 1.0 × 10⁵ Pa), B (1.0 × 10⁻³ m³, 3.0 × 10⁵ Pa), C (3.0 × 10⁻³ m³, 3.0 × 10⁵ Pa) and D (3.0 × 10⁻³ m³, 1.0 × 10⁵ Pa), each given as (volume, pressure). The gas goes round the cycle A → B → C → D → A. What is the net work W done by the gas in one cycle? HL
Answer and reasoning
+4.0 × 10² J — A → B and C → D are at constant volume, so W = 0. B → C: W = 3.0 × 10⁵ × 2.0 × 10⁻³ = +600 J. D → A: W = 1.0 × 10⁵ × (−2.0 × 10⁻³) = −200 J. Net W = +400 J, the area enclosed by the rectangle.
+6.0 × 10² J — A student who counts only the expansion B → C gets 600 J. Work of 200 J is done on the gas during the compression D → A, so the net work done by the gas is 600 − 200 = 400 J.
+8.0 × 10² J — A student who adds the sizes of the work in each stage, 600 + 200 J, has ignored signs. Work done on the gas during compression is negative, W = −200 J, so the net work is +400 J.
−4.0 × 10² J — A student who takes W as the work done on the gas gives the net work a negative sign. The cycle runs clockwise on the P–V diagram, so the gas does net positive work: W = +400 J in Clausius' convention.
Working A → B: V constant, W = 0. B → C: W = PΔV = 3.0 × 10⁵ Pa × (3.0 − 1.0) × 10⁻³ m³ = +600 J. C → D: V constant, W = 0. D → A: W = 1.0 × 10⁵ Pa × (1.0 − 3.0) × 10⁻³ m³ = −200 J. Net W = +600 − 200 = +4.0 × 10² J (= enclosed area ΔP × ΔV = 2.0 × 10⁵ Pa × 2.0 × 10⁻³ m³).
15 An engine takes helium, a monatomic ideal gas, round a rectangular cycle on a P–V diagram, with states given as (volume, pressure): A (1.0 × 10⁻³ m³, 1.0 × 10⁵ Pa) → B (1.0 × 10⁻³ m³, 3.0 × 10⁵ Pa) → C (3.0 × 10⁻³ m³, 3.0 × 10⁵ Pa) → D (3.0 × 10⁻³ m³, 1.0 × 10⁵ Pa) → A. The net work done by the gas per cycle is 400 J. Thermal energy is supplied to the gas only during A → B and B → C. What is the efficiency of the engine? HL
Answer and reasoning
22% — A → B: W = 0, Q = ΔU = (3/2)VΔP = 1.5 × 1.0 × 10⁻³ × 2.0 × 10⁵ = 300 J. B → C: Q = ΔU + W = (3/2)PΔV + PΔV = (5/2) × 3.0 × 10⁵ × 2.0 × 10⁻³ = 1500 J. Input = 1800 J, so η = 400/1800 = 22%.
33% — A student who takes the heat supplied during B → C to be only ΔU = (3/2)PΔV = 900 J gets an input of 1200 J. During B → C the gas also does 600 J of work, so Q = 1500 J, the input is 1800 J and η = 22%.
89% — A student who uses 1 − T_c/T_h with the lowest and highest temperatures of the cycle (T ∝ PV, a ratio of 9) gets 89%. That would be the Carnot limit between those temperatures; this cycle's efficiency is W/Q_in = 400/1800 = 22%.
29% — A student who divides the work by the heat rejected during C → D and D → A (900 + 500 = 1400 J) gets 29%. Efficiency divides by the energy supplied, 1800 J.
Working Using ΔU = (3/2)Δ(PV) and W = PΔV. A → B (isovolumetric): W = 0, Q_AB = (3/2)V(P_B − P_A) = 1.5 × 1.0 × 10⁻³ m³ × 2.0 × 10⁵ Pa = 300 J. B → C (isobaric): ΔU = (3/2)PΔV = 900 J, W = PΔV = 600 J, Q_BC = 1500 J. Input Q_h = 300 + 1500 = 1800 J. (Rejected: C → D 900 J, D → A 500 J; 1800 − 1400 = 400 J = W.) η = W/Q_h = 400 ÷ 1800 = 0.22 = 22%.
16 A Carnot engine works between a hot reservoir at 227 °C and a cold reservoir at 27 °C. In each cycle it absorbs 1200 J of thermal energy from the hot reservoir. How much work does it do in each cycle? HL
Answer and reasoning
1.1 × 10³ J — A student who substitutes the Celsius temperatures gets η = 1 − 27/227 = 0.88 and W = 1060 J. η_Carnot = 1 − T_c/T_h needs kelvin: 1 − 300/500 = 0.40.
4.8 × 10² J — T_h = 500 K and T_c = 300 K. η_Carnot = 1 − 300/500 = 0.40, so W = ηQ_h = 0.40 × 1200 J = 480 J.
7.2 × 10² J — A student who takes T_c/T_h = 0.60 as the efficiency has calculated the fraction of the input rejected to the cold reservoir. The efficiency is 1 − 0.60 = 0.40, so W = 480 J.
8.0 × 10² J — A student who divides the temperature difference by the cold temperature, 200/300, has used the wrong denominator. η_Carnot = (T_h − T_c)/T_h = 200/500 = 0.40.
17 Which statement about the Carnot cycle and its efficiency is correct? HL
Answer and reasoning
A Carnot engine is 100% efficient, since its cycle is reversible and has no friction. — A student who thinks friction is the only cause of inefficiency picks this. A Carnot engine must still reject thermal energy to the cold reservoir; its efficiency is 1 − T_c/T_h, which is less than 1.
The Carnot efficiency is unchanged if both temperatures are raised by the same number of kelvin. — A student who thinks the efficiency depends only on the temperature difference picks this. η = (T_h − T_c)/T_h: with the same difference but a larger T_h, the efficiency falls.
No engine working between two given reservoirs can be more efficient than a Carnot engine. — The Carnot cycle is reversible, and the second law shows that no engine working between the same two reservoirs can exceed its efficiency, 1 − T_c/T_h.
The Carnot efficiency is T_c/T_h, the ratio of the cold to the hot reservoir temperature. — A student who confuses efficiency with the fraction of energy rejected picks this. T_c/T_h is the fraction rejected to the cold reservoir; the efficiency is 1 − T_c/T_h.
18 An engineer claims that an engine working between reservoirs at 327 °C and 27 °C does 700 J of work for every 1200 J of thermal energy it absorbs from the hot reservoir. Which evaluation of the claim is correct? HL
Answer and reasoning
Possible: 58% is less than 100%, and the 500 J rejected balances the energy. — A student who thinks energy conservation is the only limit picks this. The first law is satisfied, but the second law limits any engine between these reservoirs to 1 − 300/600 = 50%.
Possible: the Carnot limit applies only to engines that run a Carnot cycle. — A student who thinks the Carnot limit belongs to one particular cycle picks this. It applies to every engine working between the two reservoirs, whatever its cycle.
Possible: 58% is below the Carnot limit of 92% for these two reservoirs. — A student who uses the Celsius temperatures gets 1 − 27/327 = 92%. With kelvin temperatures, 1 − 300/600 = 50%, which the claimed 58% exceeds.
Impossible: 58% is more than the Carnot limit of 50% for these two reservoirs. — The claimed efficiency is 700/1200 = 58%. With T_h = 600 K and T_c = 300 K, η_Carnot = 1 − 300/600 = 50%. No engine between these reservoirs can exceed that, so the claim violates the second law.
Working Claimed η = W/Q_h = 700 J ÷ 1200 J = 0.583 = 58%. T_h = 327 + 273 = 600 K, T_c = 27 + 273 = 300 K. η_Carnot = 1 − 300/600 = 0.50 = 50%. Since 58% > 50%, the claim is impossible.
19 A beaker of crushed ice at 0 °C is left in a warm room. The ice melts completely, and at the moment melting finishes the liquid water is still at 0 °C. Which statement correctly compares the entropy of the sample as liquid water with its entropy as crushed ice? HL
Answer and reasoning
The water has the greater entropy: its molecules are arranged far less regularly than in ice, although the temperature is unchanged. — Entropy relates to the degree of disorder of the particles. In ice the molecules sit in a regular crystal lattice; in liquid water they are free to move past one another and can be arranged in many more ways. The disorder of the particles, and so the entropy, increases on melting even though the temperature stays at 0 °C; thermodynamically, ΔS = ΔQ/T with ΔQ the latent heat absorbed.
The crushed ice has the greater entropy: a jumble of fragments is a more disordered arrangement than a smooth body of liquid. — A student who reads 'disorder' as visible untidiness picks this. Entropy concerns the arrangement and motion of the particles, not the shape of the lumps. Inside every fragment the water molecules are locked in a regular lattice; in the liquid they are far less ordered, so the water has the greater entropy.
The two are equal: entropy is fixed by the temperature, and the sample is at 0 °C both before and after it melts. — A student who thinks entropy is a measure of temperature picks this. Entropy depends on how the particles are arranged as well as on their energy: at the same temperature the liquid's molecules are far less ordered than the crystal's, so the entropy rises on melting. ΔS = ΔQ/T is non-zero here because latent heat ΔQ is absorbed at constant T.
The two are equal: entropy, like energy, is conserved, so the entropy of the ice is carried unchanged into the water. — A student who treats entropy as a conserved quantity picks this. Entropy is not conserved: it is a measure of the disorder of the particles, and it increases when a regular crystal becomes a liquid. The water's entropy exceeds the ice's by the latent heat absorbed divided by 273 K.
That was your twenty minutes. Real practice on B.4 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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