IB Physics HL B.4 Thermodynamics Questions

SLSL HLHL

Practise IB Physics HL B.4 by reading p–V cycles, tracking heat, work and internal energy, then using entropy and reservoir temperatures to assess heat-engine limits.

83 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Use Q = ΔU + W for a closed system with Q positive into it and W positive for work done by the gas; translate words into signed quantities first.
Apply the process constraint before balancing energy: Q = 0 for adiabatic, W = 0 for isochoric and ΔU = 0 for isothermal ideal-gas changes.
At constant pressure, use W = p(Vf − Vi), converting graph axes and kPa or MPa consistently before combining work with an internal-energy change.
For a variable-pressure path, calculate work from the whole area under the p–V curve between the stated volumes.
Include the rectangular baseline below a non-zero pressure-axis origin when estimating work under an open curve.
For a complete cycle, use the enclosed p–V area for net work and compare expansion and compression areas to justify whether work is supplied or required.
Estimate curved cycle work by counting grid squares and multiplying by the energy represented by both axis scales.
For two paths between the same ideal-gas states, keep ΔU fixed while using their different work areas to compare the required heat.
For a monatomic gas at constant pressure, combine ΔU = 3pΔV/2 with work to obtain Q = 5pΔV/2.
At constant volume, obtain monatomic ideal-gas ΔU from 3VΔp/2 or 3nRΔT/2, and identify heat transfer with ΔU.
Use the temperature change rather than the final temperature in ΔU; a Celsius interval has the same numerical value in kelvin.
For adiabatic work, find the actual endpoint temperatures and use W = −ΔU; explain expansion cooling and compression heating through work transfer.
Complete energy or sign tables for each process, and check that ΔU sums to zero over a full cycle so net Q equals net W.
Distinguish heat removed as a positive amount from negative Q for the gas, retaining consistent signs in the first-law calculation.
Recognise constant-volume and constant-pressure legs from vertical and horizontal p–V paths, and identify zero boundary work from ΔV = 0.
Define isothermal by constant temperature and adiabatic by absent thermal-energy transfer; a fast change or good insulation can approximate the latter.
Verify isothermal behaviour with numerical pV products at two graph points, then use pV = constant for pressure or volume ratios.
Compare expansion paths from the same initial state: the adiabat is steeper and cooler, while an isotherm gives greater work to the same final volume.
Sketch the complete cycle with its specified horizontal or vertical legs and correctly curved isotherms or adiabats, locating the required endpoints.
Use both process shape and the stated work direction to identify an isothermal leg and infer cycle direction.
For a monatomic ideal-gas adiabat, calculate pressure using p₁V₁^(5/3) = p₂V₂^(5/3), keeping the volume ratio in the correct order.
Solve an adiabatic volume ratio with V₂/V₁ = (p₁/p₂)^(3/5), choosing the states connected by the adiabatic leg.
Derive TV^(2/3) = constant from pV^(5/3) = constant and pV = nRT, then use it to calculate temperature or a volume ratio.
Convert absolute temperatures to kelvin before adiabatic ratios, and calculate any missing initial volume from nRT/p before applying the invariant.
Identify the state at the start of an adiabat after any preceding fixed-volume pressure rise, rather than reusing the original pressure.
Explain entropy through particle disorder and possible energy arrangements, including greater multiplicity on vaporization.
Compare entropy at fixed volume through temperature and disorder; explain isothermal expansion entropy through increased volume at unchanged temperature.
Locate the largest gas entropy on a cycle by explaining the changes on its legs, and distinguish the gas entropy from the surroundings during phase changes.
After a gas completes a cycle, explain zero net gas entropy change through return to the same state, even though its surroundings may gain entropy.
For reversible transfer at a constant temperature, calculate ΔS = Q/T in J/K; obtain Q from latent heat or isothermal work where needed.
Explain zero entropy change on an ideal reversible adiabat using both reversibility and Q = 0; neither condition alone suffices for the general claim.
During changing-temperature cooling, explain the entropy sign through heat loss or reduced disorder without using a finite Q divided by an arbitrary single T.
Use S = kB ln Ω to compare multiplicities: halving entropy gives √Ω, and energy assignments to distinguishable particles determine microstate counts.
Apply the second law to the combined isolated system, allowing a local entropy decrease only when it is at least compensated by the surroundings.
State the Clausius restriction on spontaneous cold-to-hot heat flow, and the Kelvin restriction on complete heat-to-work conversion in a cycle.
Separate the first-law cycle balance Qin − Qout = W from the second-law requirement to reject heat; energy conservation alone does not set Qout > 0.
For a small heat transfer from a warm region to colder surroundings, compare Q/Th lost with the larger Q/Tc gained.
For an open cooling room, include entropy carried by incoming air; for a powered refrigerator in an insulated room, include electrical work input.
Explain the more likely direction of change in an isolated particle model through increased multiplicity and entropy, retaining the statistical interpretation.
Explain rejected heat’s degradation through its dispersal into the surroundings and reduced ability to supply useful work relative to the ambient environment.
Calculate actual engine efficiency from net cycle work and positive heat input, or from 1 − Qc/Qh; distinguish gross input from net heat.
Use isothermal ideal-gas work as the heat input only for that leg; find net cycle work separately from heat rejection or enclosed p–V area.
Rearrange efficiency to calculate rejected heat, and multiply a stated fraction of Carnot efficiency by the Carnot limit to obtain actual efficiency.
From useful output power, calculate input and waste powers using efficiency, then multiply waste power by elapsed seconds for wasted energy.
Calculate the Carnot bound with kelvin temperatures or the fixed-amount gas-state ratio Tc/Th = (pcVc)/(phVh), and compare actual efficiency with it.
When one reservoir temperature changes, update Tc/Th rather than scaling efficiency by the same factor; at fixed Th, halving Tc changes η = 0.4 to 0.7.
Explain real-engine efficiency below the Carnot bound through irreversibility, and link efficient design to less wasted input and greater useful output.
Assess the stated theoretical engine cycle through thermal-transfer demands: slow isothermal operation versus fast or insulated adiabatic operation.
For show-that results, give the governing relation and full substitution or enough extra precision to verify the supplied approximation.

Covered syllabus points

B.4.1 (HL)—First law of thermodynamics B.4.2 (HL)—Gas work B.4.3 (HL)—Internal energy change B.4.4 (HL)—Entropy and disorder B.4.5 (HL)—Entropy equations B.4.6 (HL)—Second law of thermodynamics B.4.7 (HL)—Irreversibility B.4.8 (HL)—Local entropy decrease B.4.9 (HL)—Gas processes B.4.10 (HL)—Adiabatic ideal gas B.4.11 (HL)—Heat engine cycles B.4.12 (HL)—Heat engine efficiency B.4.13 (HL)—Carnot efficiency limit

Exam points

Apply Q = ΔU + W with Q positive into the gas and W positive for work done by the gas; heat removed and compression work have negative signs.
For adiabatic expansion, set Q = 0 and explain that positive work by the gas lowers internal energy and temperature.
For adiabatic compression, use W = −ΔU: work done on the gas increases internal energy rather than transferring heat into it.
For an isothermal ideal-gas change, use ΔU = 0 so Q = W; justify the heat-flow direction from expansion or compression.
For an isochoric change, use W = 0 and Q = ΔU; connect a pressure rise to higher temperature, internal energy and inward heat transfer.
At constant pressure, calculate work with p(Vf − Vi), then combine it with ΔU to find heat supplied or removed.
For the monatomic ideal-gas model, combine ΔU = 3nRΔT/2 with W; at constant pressure this gives Q = 5pΔV/2.
Explain isobaric expansion heating through both positive work and a rise in internal energy, then invoke the first law to show Q > 0.
For two routes between the same ideal-gas states, keep ΔU the same and compare area under each p–V path to decide which requires more heat.
For a complete cycle, use net ΔU = 0 so net Q equals net W; distinguish net heat from the gross thermal input on one leg.
Use an anticlockwise p–V cycle to infer negative net work by the gas and hence net thermal energy released over the cycle.
Complete process tables by combining each leg’s constraint with Q = ΔU + W, and check that internal-energy changes sum to zero over a cycle.
For adiabatic work calculations, find the endpoint temperatures from the gas states and use W = −3nR(Tf − Ti)/2 for a monatomic gas.
Distinguish signed Q for the gas from the positive amount described as heat removed, while retaining consistent signs in the first-law working.
For a show-that heat result, show the first-law substitution and sufficient working or extra precision instead of repeating the target.

[Maximum number: 4]

This question is in two parts

With reference to the first law of thermodynamics, explain for the change of state A to B, why energy is transferred from the surroundings to the gas.

Exam points

At constant pressure, calculate W = p(Vf − Vi) using the volume change and compatible pressure and volume units.
Identify expansion as work done by the gas, compression as work done on it, and a constant-volume leg as zero boundary work.
For a varying-pressure change, use the entire area under the p–V path between the initial and final volumes.
For one complete cycle, use the enclosed p–V area for net work and state whether it is work done by or on the gas.
Compare the areas under expansion and compression paths to justify the sign of net work, including cycles requiring net work input.
Estimate curved cycle areas by counting grid squares and converting each square with both axis scales before giving work in joules.
For work under a curve with a non-zero pressure-axis origin, include the rectangular baseline below the plotted range.
Check the native graph’s axis labels and powers of ten before multiplying pressure by a volume change or a counted area.

[Maximum number: 3]

This question is in two parts

Estimate the total work done in the cycle.

Exam points

For a monatomic ideal gas at constant pressure, combine ΔU = 3nRΔT/2 with pV = nRT to use ΔU = 3pΔV/2.
At fixed volume, use ΔU = 3VΔp/2 for the monatomic ideal-gas model, retaining a negative sign when the pressure falls.
For isochoric cooling, set W = 0 so heat removed equals the decrease in internal energy; distinguish the removal magnitude from signed ΔU.
Use the temperature difference in ΔU = 3nRΔT/2; a Celsius temperature interval equals the same interval in kelvin.
Explain adiabatic expansion cooling through work done by the gas: internal energy and temperature decrease even though Q = 0.

[Maximum number: 2]

The p V diagram shows a heat engine cycle consisting of adiabatic, isothermal and isovolumetric parts. The working substance of the engine is an ideal gas.

The following data are available:

\begin{aligned} & p_{\mathrm{A}}=5.00 \times 10^{5} \mathrm{~Pa} \\ & V_{\mathrm{A}}=2.00 \times 10^{-3} \mathrm{~m}^{3} \\ & T_{\mathrm{A}}=602 \mathrm{~K} \\ & p_{\mathrm{B}}=3.00 \times 10^{4} \mathrm{~Pa} \\ & p_{\mathrm{C}}=4.60 \times 10^{3} \mathrm{~Pa} \end{aligned}

Calculate the thermal energy (heat) taken out of the gas from B to C.

Alternative 1:
$Q=\Delta U=\frac{3}{2}V_C\Delta P$ .
$Q=\frac{3}{2}\times3.33\times10^{-2}\times(3.00\times10^4-4.60\times10^3)=1268.7\approx1270\ \mathrm{J}$ .

Alternative 2:
$Rn=\frac{5\times10^5\times2\times10^{-3}}{602}=1.66$ OR $T_C=\frac{4.6\times10^3\times3.33\times10^{-2}}{1.66}=92.2\ \mathrm{K}$ .
$\Delta U=\frac{3}{2}\times1.661\times(602-92.21)=1270\ \mathrm{J}$ .

Award [2] for BCA. Accept negative values.
Award MP1 if $T_C=92$ is taken from (e).

Exam points

Describe entropy as related to microscopic disorder or the number of possible particle arrangements.
Explain positive entropy change on vaporization through greater particle disorder or more accessible arrangements in steam.
For melting at constant temperature, distinguish increased entropy of the water molecules from decreased entropy of the surroundings losing heat.
Compare gas entropy at the same volume by linking higher temperature to greater disorder, and isochoric cooling to less disorder.
Explain entropy increase in an isothermal gas expansion through increased volume and greater disorder despite unchanged temperature.
Locate maximum entropy on a gas cycle by following its changes on each leg; justify the state using expansion and subsequent cooling.

[Maximum number: 2]

The p V diagram shows a heat engine cycle consisting of adiabatic, isothermal and isovolumetric parts. The working substance of the engine is an ideal gas.

The following data are available:

\begin{aligned} & p_{\mathrm{A}}=5.00 \times 10^{5} \mathrm{~Pa} \\ & V_{\mathrm{A}}=2.00 \times 10^{-3} \mathrm{~m}^{3} \\ & T_{\mathrm{A}}=602 \mathrm{~K} \\ & p_{\mathrm{B}}=3.00 \times 10^{4} \mathrm{~Pa} \\ & p_{\mathrm{C}}=4.60 \times 10^{3} \mathrm{~Pa} \end{aligned}

Suggest, for the change $A \Rightarrow B$ , whether the entropy of the gas is increasing, decreasing or constant.

Increasing because thermal energy/heat is being provided to the gas « and temperature is constant, $\Delta S=\frac{\Delta Q}{T} \gg$

Exam points

For reversible heat transfer at constant temperature, calculate ΔS = Q/T using absolute temperature and report the result in J/K.
For vaporization at the boiling point, calculate Q = mL before dividing by the kelvin temperature; convert grams to kilograms.
For an isothermal ideal-gas expansion with work given, use ΔU = 0 so Q = W before calculating the gas’s entropy change.
For an ideal reversible adiabatic change, use Q = 0 and conclude ΔS = 0; both reversibility and the adiabatic condition matter.
Explain entropy decrease during reversible cooling by the negative heat transfer, using the stated change of temperature and volume.
Use S = kB ln Ω to compare microstate numbers: halving entropy makes the new multiplicity √Ω, rather than Ω/2.
Compare microstate numbers by counting possible assignments of energy to distinguishable particles, rather than comparing total energy alone.

[Maximum number: 2]

A student is investigating the emf and internal resistance of a cell, using the circuit shown. The ammeter and the voltmeter are ideal.

The graph shows the variation of the voltmeter reading V with the ammeter reading I.

The student then replaces the variable resistor with an electric heating element in order to investigate the thermal output of the system. The heating element is used to raise the temperature of the water to the boiling point.

After the boiling point is reached, 12 g of the water are vaporized. The specific latent heat of vaporization of water is $2.3 \times 10^{6} \mathrm{Jkg}^{-1}$ .

Calculate the change in entropy of the vaporized water.

Use of $\Delta S=\frac{\Delta Q}{T}$ $=\frac{(0.012 \mathrm{~kg})\left(2.3 \times 10^{6}\right)}{100+273}=74 \mathrm{~J} \mathrm{~K}^{-1}$

Exam points

Apply the second law to the combined isolated system: total entropy does not decrease, while an individual non-isolated part can lose entropy.
State the Clausius restriction: thermal energy does not flow spontaneously from a colder body to a hotter one.
Use the Kelvin form for a cyclic engine: some heat must be rejected, so a cycle cannot convert all thermal input into work.
Distinguish cyclic operation from a single process, where an isothermal expansion can convert supplied heat entirely into work.
Separate first-law energy accounting from the second-law restriction: Qin − Qout = W alone does not require a non-zero Qout.
When a cooling gas loses entropy, show that the surroundings gain enough entropy for the combined change to be zero or positive.
For an open refrigerator in an insulated room, include its electrical work input when explaining why room temperature and total entropy rise.
For spontaneous melting, distinguish increased total entropy from the entropy change of the surroundings that supply the latent heat.
Explain a preferred transition in an isolated particle model through greater microstate multiplicity and higher entropy of the final configuration.
For a gas returning to its original state, use its zero net entropy change and the heat rejected to infer an entropy increase of the surroundings.

[Maximum number: 2]

This question is in two parts. Part 1 is about simple harmonic motion (SHM). Part 2 is about gas in an engine.

Part 1 Simple harmonic motion (SHM)
An object is placed on a frictionless surface. The object is attached by a spring fixed at one end and oscillates at the end of the spring with simple harmonic motion (SHM).

The tension F in the spring is given by F=k x where x is the extension of the spring and k is a constant.

The shaded area WXYZ is 610 J . The total thermal energy transferred out of the gas in one cycle is 1.3 kJ .

Explain why the total thermal energy transferred out of the gas is degraded.

Exam points

For isothermal gas compression, identify decreased gas entropy and increased surroundings entropy as heat leaves the gas.
Identify the entropy-decreasing leg of a gas cycle from its process: in the given cycle, isobaric compression and cooling B → C lowers gas entropy.
For a small heat transfer from warmer room to colder surroundings, compare the entropy lost Q/Th with the larger entropy gained Q/Tc.
In an open cooling room, include entropy carried by incoming air; falling temperature alone does not prove that the room’s total entropy decreases.

[Maximum number: 1]

A gas is compressed at constant temperature.

What are the changes in entropy of the gas and its surroundings during the compression?

Entropy of the gas

Entropy of surroundings

Increases

Decreases

Increases

Decreases

Exam points

Define isothermal as constant temperature and adiabatic as no transfer of thermal energy between gas and surroundings.
Identify isochoric paths as vertical and isobaric paths as horizontal on a p–V graph, reading the actual volume and pressure axes.
Explain rapid compression as approximately adiabatic because little time is available for heat transfer; thermal insulation offers another route.
For an isothermal change of a fixed ideal-gas sample, use pV = constant to calculate a new pressure or the requested volume ratio.
Test whether a gas path is isothermal by calculating pV at two actual graph points and comparing the numerical products.
Distinguish an expansion adiabat from an isotherm starting at the same state: the adiabat is steeper and reaches lower pressure and temperature.
For expansions from one state to the same final volume, identify zero isothermal ΔU and greater isothermal work than adiabatic work.
For an adiabatic change, set Q = 0 and use ΔU = −W; compression raises internal energy and temperature when work is done on the gas.
For an isothermal ideal-gas change, use ΔU = 0 and Q = W; explain heat rejection during compression through work done on the gas.
Complete a cycle using the specified fixed-volume and fixed-pressure lines, joined by correctly curved adiabatic or isothermal paths.
Sketch adiabatic compression from the end of an isothermal expansion to a state above the initial point, then return vertically at fixed volume.
Use both process shape and the stated work direction to identify an isothermal leg and the clockwise or anticlockwise direction of a cycle.

[Maximum number: 1]

This question is in two parts

A gas undergoes a thermodynamic cycle. The P-V diagram for the cycle is shown below.

In the changes of state B to C and D to A , the gas behaves as an ideal gas and the changes in state are adiabatic.

State what is meant by an adiabatic change of state.

Exam points

For a fixed monatomic ideal gas on an adiabat, use p₁V₁^(5/3) = p₂V₂^(5/3) and solve pressure from the correctly ordered volume ratio.
When finding volume from a pressure ratio, apply the reciprocal exponent: V₂/V₁ = (p₁/p₂)^(3/5).
Derive TV^(2/3) = constant by substituting p = nRT/V into pV^(5/3) = constant and cancelling the fixed amount nR.
Use T₂/T₁ = (V₁/V₂)^(2/3) on a monatomic adiabat, or find the endpoint pressure first and then apply pV = nRT.
For an adiabatic temperature-to-volume comparison, use V₂/V₁ = (T₁/T₂)^(3/2) with both temperatures in kelvin.
Convert Celsius to kelvin before an adiabatic temperature ratio, then convert the final temperature back only if Celsius is requested.
Identify the states joined by the adiabat before taking ratios; a preceding isochoric pressure rise changes the pressure at its start.
If the initial gas volume is missing, calculate it with V = nRT/p before applying the adiabatic pressure–volume relation.
For a show-that adiabatic result, display the equation and full substitution or enough extra significant figures to verify the stated approximation.

[Maximum number: 2]

The p V diagram shows a heat engine cycle consisting of adiabatic, isothermal and isovolumetric parts. The working substance of the engine is an ideal gas.

The following data are available:

\begin{aligned} & p_{\mathrm{A}}=5.00 \times 10^{5} \mathrm{~Pa} \\ & V_{\mathrm{A}}=2.00 \times 10^{-3} \mathrm{~m}^{3} \\ & T_{\mathrm{A}}=602 \mathrm{~K} \\ & p_{\mathrm{B}}=3.00 \times 10^{4} \mathrm{~Pa} \\ & p_{\mathrm{C}}=4.60 \times 10^{3} \mathrm{~Pa} \end{aligned}

Show that the volume at C is $3.33 \times 10^{-2} \mathrm{~m}^{3}$ .

Alternative 1:
Use the adiabatic formula $p_A V_A^{5/3}=p_C V_C^{5/3}$ , so $V_C=\left(\frac{p_A}{p_C}\right)^{3/5}V_A$ .
$V_C=\left(\frac{5.00\times10^5}{4.60\times10^3}\right)^{3/5}\times2.00\times10^{-3}=3.333\times10^{-2}\ \mathrm{m^3}$ .

Alternative 2:
$V_C=V_B$ and $p_AV_A=p_BV_B$ .
$V_C=\frac{5\times10^5\times2\times10^{-3}}{3\times10^4}$ .

Alternative 3:
$V_C=V_B$ and $n=0.2\ \mathrm{mol}$ .
$V_C=\frac{0.2\times8.31\times602}{3\times10^4}$ .

For MP2, working or answer to at least 4 significant figures must be seen.

Exam points

Explain heat-engine operation through positive net work by the gas: expansion work exceeds the work required for compression over one cycle.
When a cycle path changes, compare the enclosed p–V areas to predict the change in net work, and link the conclusion to the area explicitly.
Recognise a Carnot p–V cycle as two isothermal curves joined by steeper adiabatic paths, rather than horizontal or vertical process legs.
Critique the stated ideal cycle using thermal-transfer requirements: slow isothermal operation versus rapid or thermally insulated adiabatic operation.

[Maximum number: 1]

The pressure-volume diagram shows a 3-stage cyclic process of 0.20 moles of an ideal gas in a closed container. A-B is an isothermal process. C-A is an adiabatic process.

The following data are available:

\begin{aligned} \text { Pressure at } A & =2.0 \times 10^{6} \mathrm{~Pa} \\ \text { Volume at } A & =1.0 \times 10^{-3} \mathrm{~m}^{3} \\ \text { Pressure at } B & =2.0 \times 10^{5} \mathrm{~Pa} \\ \text { Volume at } B & =1.0 \times 10^{-2} \mathrm{~m}^{3} \end{aligned}

Outline why this cycle can be used for a heat engine.

Exam points

Calculate engine efficiency as net work divided by total heat input, or η = 1 − Qc/Qh using positive heat-transfer magnitudes.
For an isothermal ideal-gas input leg, use ΔU = 0 so its heat input equals its expansion work before finding cycle efficiency.
In a process table, divide net cycle work by the positive heat supplied on the input legs, rather than by net heat over the cycle.
Use the enclosed p–V cycle area for net work in η = Wnet/Qh, and keep graph-estimation tolerance distinct from the supplied heat input.
Rearrange efficiency to find rejected heat as Qc = (1 − η)Qh, identifying the correct heat-input process.
If an engine runs at a stated fraction of Carnot efficiency, multiply that fraction by ηCarnot before calculating actual input or waste power.
From useful output power, find input power Pout/η and waste power Pout(1/η − 1); multiply by the elapsed seconds to obtain waste energy.
Explain efficient engine design through less rejected energy and more useful output per unit input, even though total energy remains conserved.

[Maximum number: 2]

The tension F in the spring is given by F=k x where x is the extension of the spring and k is a constant.

The shaded area WXYZ is 610 J . The total thermal energy transferred out of the gas in one cycle is 1.3 kJ .

Determine the efficiency of cycle WXYZ.

efficiency $=\frac{610}{1300}$ ;
0.47 or 47 %;

Marking guidance:

Award [2] for a bald correct answer.
Award [1] for $\frac{610}{(610+1300)}=0.32$ .
Award [0] for any fraction that would exceed 1 even if later fudged.
Award [0] for efficiency $=\frac{T_{C}-T_{H}}{T_{H}}=38 \%$ , question does not imply that this is a Carnot cycle.

Exam points

Calculate the Carnot limit ηmax = 1 − Tc/Th using reservoir temperatures in kelvin, converting both Celsius values before taking their ratio.
Compare a stated engine efficiency with the calculated Carnot limit, showing the calculation before deciding whether it respects the second law.
Explain why a real engine falls below the Carnot limit by linking the reversible ideal model to irreversibility or dissipative losses.
For fixed hot-reservoir temperature, update the ratio Tc/Th when the cold temperature changes; do not scale efficiency by the same factor.
For a fixed gas amount, use Tc/Th = (pcVc)/(phVh) to obtain Carnot efficiency from the relevant cold and hot cycle states.

[Maximum number: 2]

The p V diagram shows a heat engine cycle consisting of adiabatic, isothermal and isovolumetric parts. The working substance of the engine is an ideal gas.

The following data are available:

\begin{aligned} & p_{\mathrm{A}}=5.00 \times 10^{5} \mathrm{~Pa} \\ & V_{\mathrm{A}}=2.00 \times 10^{-3} \mathrm{~m}^{3} \\ & T_{\mathrm{A}}=602 \mathrm{~K} \\ & p_{\mathrm{B}}=3.00 \times 10^{4} \mathrm{~Pa} \\ & p_{\mathrm{C}}=4.60 \times 10^{3} \mathrm{~Pa} \end{aligned}

The highest and lowest temperatures of the gas during the cycle are 602 K and 92 K .

The efficiency of this engine is about 0.6 . Outline how these data are consistent with the second law of thermodynamics.

$e_{\mathrm{C}}=1-\frac{92}{602}=0.847$

this engine has $e<e_{\mathrm{c}}$ as it should

Marking guidance:

Award [0] if no calculation shown.

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