1True or false
The total entropy of an isolated system never decreases.
Show answer
Answer: True
This is one form of the second law; it increases in real (irreversible) processes and stays constant only in ideal reversible ones.
!Common mistakeSome learners think a fridge lowers total entropy; the entropy of its inside falls, but the room's rises by more.
2True or false · ★ Challenge
In an isobaric expansion of an ideal gas, all the heat supplied is turned into work.
Show answer
Answer: False
At constant pressure the temperature rises as the gas expands, so part of the heat increases U; Q = ΔU + W.
!Common mistakeThat statement is true for isothermal expansion, not isobaric; check whether the temperature changes.
3True or false
In PV = nRT the temperature may be put in °C, as long as the same unit is used on both sides.
Show answer
Answer: False
T must be the absolute temperature in kelvin; at 0 °C a gas still has pressure and volume, so T cannot be zero there.
!Common mistakeThis mistake gives impossible results such as zero or negative volumes for gases below 0 °C.
4Multiple choice · ★ Challenge
Thermometer A is in thermal equilibrium with a water bath, and thermometer B is also in thermal equilibrium with the same bath. What can be concluded?
- AA and B hold exactly the same amount of internal energy as the water bath
- BA and B are at the same temperature, so no net heat flows between them
- CA and B must be made of the same liquid to agree
- DHeat flows from the bath into A and B all the time
Show answer
Answer: B. A and B are at the same temperature, so no net heat flows between them
This is the zeroth law: two bodies each in thermal equilibrium with a third are in equilibrium with each other, so they share a temperature.
!Common mistakeChoosing "same internal energy" confuses temperature with internal energy, which also depends on mass and material.
5Multiple choice · ★ Challenge
When a bottle of fizzy drink is opened, a little fog often appears at the neck. What is the best explanation?
- ACarbon dioxide reacts with oxygen in the air to form tiny droplets of water
- BThe bottle walls cool the gas by conduction as it leaves
- CThe gas expands suddenly, does work, cools, and water vapour condenses
- DPressure from outside squeezes the gas until it liquefies
Show answer
Answer: C. The gas expands suddenly, does work, cools, and water vapour condenses
The rapid expansion is adiabatic: Q ≈ 0, so ΔU = −W and the temperature falls enough for vapour to condense.
!Common mistakeChoosing a chemical reaction is wrong; the fog is condensed water vapour caused by cooling.
6True or false
A refrigerator breaks the second law of thermodynamics because it moves heat from cold to hot.
Show answer
Answer: False
The second law forbids this only without work; the fridge uses electrical work, so it obeys the law.
!Common mistakeLearners often forget the words "on its own" in the Clausius statement.
7Fill in the blank · ★ Challenge
A gas at 27 °C is heated at constant pressure until its volume doubles. Its new temperature is ______ °C.
Show answer
Answer: 327
V/T constant: T₂ = 2 × 300 K = 600 K = 600 − 273 = 327 °C.
!Common mistakeWriting 54 °C doubles the Celsius temperature; gas laws need kelvin.
8True or false
Heat and temperature are the same quantity: a hot object simply contains a large amount of heat.
Show answer
Answer: False
Temperature measures how hot a body is (average molecular kinetic energy); heat is energy transferred because of a temperature difference.
!Common mistakeEveryday speech ("this soup has a lot of heat") leads many learners to treat heat as something stored.
9Multiple choice · ★ Challenge
A gas goes from state X to state Y on a P–V diagram by two routes. Route 1 expands first at high pressure, then cools at constant volume; route 2 cools first, then expands at low pressure. Which statement is correct?
- ARoute 1 does more work, as it expands at the higher pressure
- BBoth routes do the same work, as they start and end together
- CRoute 2 does more work, as the gas is cooler while it expands
- DNeither does work, as the volume is the same at X and Y
Show answer
Answer: A. Route 1 does more work, as it expands at the higher pressure
Work equals the area under the path on a P–V diagram; route 1 expands at higher pressure, so its area is larger. Work depends on the path.
!Common mistakeChoosing "the same work" treats work as a state variable; only the change in internal energy is fixed by the start and end states.
10True or false
In a four-stroke engine the gas does useful work on the piston during only one of the four strokes.
Show answer
Answer: True
Only the power stroke delivers work; the flywheel's energy carries the engine through the other three.
!Common mistakeSome learners think every stroke pushes the car; intake, compression and exhaust actually take energy from the flywheel.
11Fill in the blank · ★ Challenge
An engine has a Carnot efficiency of 25% and its cold reservoir is at 330 K. Its hot reservoir must be at ______ K.
Show answer
Answer: 440
η = 1 − Tc/Th, so Th = Tc ÷ (1 − η) = 330 ÷ 0.75 = 440 K.
!Common mistakeWriting 412.5 K comes from Tc × 1.25; rearrange the formula properly: Th = Tc/(1 − η).
12True or false
Heat can be made to flow from a cold body to a hot body, but only if work is done by an outside agent.
Show answer
Answer: True
This is the Clausius statement of the second law; a fridge does it using electrical work.
!Common mistakeSome learners say heat can never flow from cold to hot; it cannot do so ON ITS OWN.
13Multiple choice · ★ Challenge
An ideal gas is compressed isothermally, with 450 J of work done on it. What is the heat Q for the gas?
- AQ = +450 J: it takes in 450 J of heat
- BQ = −450 J: it gives out 450 J of heat
- CQ = 0: its temperature does not change
- DQ = −900 J: it gives out twice the work
Show answer
Answer: B. Q = −450 J: it gives out 450 J of heat
Isothermal, so ΔU = 0 and Q = W; W = −450 J (work done on the gas), so Q = −450 J.
!Common mistakeChoosing Q = 0 confuses isothermal (constant T) with adiabatic (no heat).
14True or false
A pot of beans boiling without a lid is an open system, because both energy and matter (steam) cross its boundary.
Show answer
Answer: True
Heat enters from the stove and steam leaves, so both energy and matter cross the boundary.
!Common mistakeSome learners call it closed because the pot is solid; the open top lets steam (matter) out.
15Multiple choice · ★ Challenge
For a monatomic ideal gas Cv = 3R/2. What is the ratio γ = Cp/Cv?
- A1.67
- B1.40
- C1.33
- D0.60
Show answer
Answer: A. 1.67
Cp = Cv + R = 5R/2, so γ = (5R/2) ÷ (3R/2) = 5/3 ≈ 1.67.
!Common mistakeChoosing 0.60 inverts the ratio (Cv/Cp); γ is always greater than 1.
16Fill in the blank
The ratio of the largest volume of the cylinder to its smallest volume is called the ______ ratio.
Show answer
Answer: compression
A higher compression ratio gives a higher ideal efficiency in both Otto and Diesel cycles.
!Common mistakeWriting "pressure ratio" is a common slip; the ratio is defined using volumes.
17Multiple choice · ★ Challenge
A paddle wheel stirs a gas inside a rigid, insulated container, doing 200 J of work on it. Using ΔU = Q − W (W = work done BY the gas), what are Q, W and ΔU?
- AQ = +200 J, W = 0, ΔU = +200 J
- BQ = 0, W = +200 J, ΔU = −200 J
- CQ = 0, W = −200 J, ΔU = +200 J
- DQ = 0, W = 0, ΔU = 0
Show answer
Answer: C. Q = 0, W = −200 J, ΔU = +200 J
Insulated, so Q = 0; work is done on the gas, so W = −200 J; ΔU = 0 − (−200) = +200 J.
!Common mistakeChoosing Q = +200 J treats stirring as heating; energy given by a moving paddle is work, even though the gas warms up.
18Multiple choice · ★ Challenge
1000 J of heat flows from a body at 400 K to a body at 250 K. What is the total change in entropy (ΔS = Q/T)?
- A−1.5 J/K
- B+6.5 J/K
- C0 J/K
- D+1.5 J/K
Show answer
Answer: D. +1.5 J/K
Hot body: −1000 ÷ 400 = −2.5 J/K; cold body: +1000 ÷ 250 = +4.0 J/K; total = +1.5 J/K.
!Common mistakeChoosing 0 J/K assumes the entropy lost equals the entropy gained; the same heat at a lower temperature gives a larger entropy change.
19True or false
When a gas does work on its surroundings without receiving any heat, its internal energy decreases.
Show answer
Answer: True
ΔU = Q − W = 0 − W, which is negative when W > 0.
!Common mistakeSome learners think the internal energy stays constant if no heat flows; work done by the gas takes energy away too.
20Multiple choice · ★ Challenge
A rigid 2.0 L tank contains gas at 300 K and 1.0 × 10⁵ Pa. It is heated to 450 K. What is the new pressure?
- A6.7 × 10⁴ Pa
- B1.5 × 10⁵ Pa
- C2.5 × 10⁵ Pa
- D4.5 × 10⁵ Pa
Show answer
Answer: B. 1.5 × 10⁵ Pa
At constant volume P/T is constant: P₂ = 1.0 × 10⁵ × 450 ÷ 300 = 1.5 × 10⁵ Pa.
!Common mistakeChoosing 6.7 × 10⁴ Pa uses the ratio upside down; pressure rises when temperature rises at fixed volume.
21Multiple choice
Which of these processes is irreversible?
- AA drop of ink spreading through a glass of water
- BA very slow, frictionless compression of a gas
- CA frictionless pendulum swinging in a vacuum
- DGas expanding very slowly against a matched piston
Show answer
Answer: A. A drop of ink spreading through a glass of water
The ink never gathers back into a drop on its own: its spreading increases disorder (entropy), so it is irreversible.
!Common mistakeThe slow-compression options are tempting distractors, but ideal slow, frictionless changes are the textbook examples of reversible processes.
22Multiple choice · ★ Challenge
An engine works between 500 K and 300 K. To raise its Carnot efficiency, is it better to raise the hot temperature by 50 K or to lower the cold temperature by 50 K?
- ALower the cold temperature: 50% compared with 45%
- BRaise the hot temperature: 50% compared with 45%
- CEither way: both give the same 45% efficiency
- DNeither: the efficiency only depends on the fuel
Show answer
Answer: A. Lower the cold temperature: 50% compared with 45%
Lower Tc: 1 − 250/500 = 50%. Raise Th: 1 − 300/550 ≈ 45%. Lowering Tc helps more.
!Common mistakeChoosing "either way" assumes only the difference Th − Tc matters; the ratio Tc/Th is what counts.
23Multiple choice
Where does the heat removed from food in a refrigerator go?
- AIt is destroyed by the cooling gas in the pipes
- BInto the freezer compartment, where it is stored as ice
- CTo the kitchen air, through the hot grille at the back
- DInto the electricity supply, back to the grid
Show answer
Answer: C. To the kitchen air, through the hot grille at the back
The fridge pumps heat from the inside to the coils at the back, which give it (plus the compressor work) to the room.
!Common mistakeChoosing "destroyed" breaks conservation of energy; heat is moved, never destroyed.
24Short answer · ★ Challenge
Use the idea of entropy to explain why heat flows from a hot cup of tea to the cooler room, but never the other way on its own.
Show answer
Model answer: When heat Q leaves the tea at Thot, its entropy falls by Q/Thot; when the room gains Q at the lower Troom, its entropy rises by Q/Troom, which is larger. So the total entropy increases. The reverse flow would decrease total entropy, which the second law forbids for a natural process.
!Common mistakeSaying "heat rises" or "hot things push heat out" gives no reason; the direction is set by the total entropy, which must increase.
25Fill in the blank
For a refrigerator, the coefficient of performance is the heat removed from the cold space divided by the ______ done by the compressor.
Show answer
Answer: work
COP = Qc/W; a good fridge removes several joules of heat for each joule of work.
!Common mistakeWriting "heat given to the room" gives the heat-pump COP, which is a different ratio.
26Multiple choice · ★ Challenge
A heat pump puts 4.0 kW of heat into a building while using 1.0 kW of electrical power. Which is correct?
- ACOP = 3.0, and it takes 4.0 kW from outside
- BCOP = 0.25, and it takes 3.0 kW from outside
- CCOP = 4.0, and it takes 3.0 kW from outside
- DCOP = 4.0, and it takes 5.0 kW from outside
Show answer
Answer: C. COP = 4.0, and it takes 3.0 kW from outside
COP (heating) = heat delivered ÷ work = 4.0 ÷ 1.0 = 4.0; heat from outside = 4.0 − 1.0 = 3.0 kW.
!Common mistakeChoosing 5.0 kW adds the work to the heat delivered; the delivered heat already includes the work.
27Multiple choice
A gas is taken round a closed loop on a P–V diagram in the clockwise direction. What does this tell you?
- ANet work is done on the gas in each cycle, as happens in a fridge
- BThe gas does net positive work in each cycle, as in an engine
- CNo net work is done, as the gas returns to its start
- DThe internal energy of the gas rises after each cycle
Show answer
Answer: B. The gas does net positive work in each cycle, as in an engine
Clockwise, the expansion happens at higher pressure than the compression, so the net work done by the gas (the enclosed area) is positive.
!Common mistakeChoosing "no net work" confuses work with ΔU; over a cycle ΔU = 0 but the net work equals the enclosed area.
28Short answer · ★ Challenge
An inventor proposes a ship that takes heat from the warm sea water and turns all of it into work to drive the propeller. Use the second law to explain why this cannot work.
Show answer
Model answer: The Kelvin–Planck statement of the second law says no engine working in a cycle can take heat from one reservoir and turn it completely into work. The ship would need a colder reservoir to reject part of the heat to; with only the sea at one temperature, no net work can be obtained.
!Common mistakeAnswering "it breaks conservation of energy" is wrong: the energy books would balance; it is the second law, not the first, that forbids it.
29Multiple choice · ★ Challenge
The ideal Otto-cycle efficiency is η = 1 − 1/rγ−1, where r is the compression ratio. What is it for r = 8 and γ = 1.4?
- A87.5%
- B56%
- C44%
- D12.5%
Show answer
Answer: B. 56%
rγ−1 = 80.4 ≈ 2.30, so η = 1 − 1/2.30 ≈ 0.565 = 56%.
!Common mistakeChoosing 87.5% uses 1 − 1/r and forgets the power (γ − 1).
30Multiple choice
Hot tea is sealed in a perfect vacuum flask with a tight stopper. Which kind of system is this closest to?
- AA closed system
- BAn open system
- CAn isolated system
- DA cyclic system
Show answer
Answer: C. An isolated system
An ideal vacuum flask lets neither matter nor energy (heat) cross its boundary, so it is close to an isolated system.
!Common mistakeChoosing "closed" is tempting because no tea escapes, but a closed system can still exchange heat; the flask is designed to stop that too.
31Short answer · ★ Challenge
On one P–V diagram, describe the lines for an isobaric expansion, an isochoric heating and an isothermal expansion that all start from the same point.
Show answer
Model answer: Isobaric expansion: a horizontal line to the right (P constant, V grows). Isochoric heating: a vertical line upwards (V constant, P grows). Isothermal expansion: a curve sloping down to the right, part of a hyperbola PV = constant.
!Common mistakeDrawing the isothermal as a straight sloping line is a common error; PV = constant gives a curve that flattens out.
32Multiple choice
The first law of thermodynamics is a form of which principle?
- AThe conservation of mass
- BThe direction in which heat flows
- CThe conservation of energy
- DThe equality of temperature at equilibrium
Show answer
Answer: C. The conservation of energy
ΔU = Q − W says the energy gained by a system equals the heat in minus the work out: energy is conserved.
!Common mistakeChoosing "direction of heat flow" mixes the first law with the second law.
33Short answer · ★ Challenge
A student says: 'If I heat a gas, its temperature must rise.' Use the first law to describe a case where heat is added to an ideal gas but its temperature stays the same.
Show answer
Model answer: In an isothermal expansion the gas does work equal to the heat it absorbs: Q = W, so ΔU = 0 and the temperature stays constant. The heat supplied leaves the gas as work done on the surroundings.
!Common mistakeAnswering "when the gas is in a rigid container" is wrong: then W = 0 and all the heat raises U and the temperature.
34Fill in the blank
Pressure, volume and temperature are called state ______ because together they describe the condition of a gas at one moment.
Show answer
Answer: variables
State variables fix the state of a system; heat and work are not state variables because they depend on the process.
!Common mistakeWriting "functions of heat" mixes up properties of the state with energy transfers.
35Multiple choice · ★ Challenge
Moist air rising up the slopes of Mount Karisimbi cools and forms clouds. What is the best explanation of the cooling?
- AIt expands as the pressure falls, doing work at the expense of its internal energy
- BIt rises into colder air around it and loses its heat to that air by conduction
- CIt gets farther from the warm ground, which heats it less
- DIt mixes with snow on the peak, which absorbs its heat
Show answer
Answer: A. It expands as the pressure falls, doing work at the expense of its internal energy
Rising air expands almost adiabatically (air is a poor conductor); the work it does comes from its internal energy, so its temperature drops and vapour condenses.
!Common mistakeChoosing conduction is tempting, but air conducts heat very poorly; the cooling happens even with no heat transfer.
36Multiple choice
Why can a diesel engine use a much higher compression ratio than a petrol engine?
- AIts fuel does not burn, so it needs extra pressure to move
- BIts cylinders are made of thicker steel than petrol ones
- CIts spark plug is stronger, so it needs more compressed gas
- DIt compresses only air, so the fuel cannot ignite too early
Show answer
Answer: D. It compresses only air, so the fuel cannot ignite too early
In a petrol engine the fuel–air mixture would ignite early (knock) if compressed too much; a diesel compresses air alone and injects fuel at the end.
!Common mistakeChoosing the spark-plug option is wrong: a diesel engine has no spark plug at all.
37Multiple choice · ★ Challenge
A cup of water at 80 °C and a bucket of water at 60 °C are compared. Which has more internal energy?
- AThe cup, since its water is at the higher temperature of the two
- BBoth are equal, since both of them contain only pure liquid water
- CNeither has any, since neither is being heated now
- DThe bucket, since internal energy also depends on the mass of water
Show answer
Answer: D. The bucket, since internal energy also depends on the mass of water
Internal energy is the total kinetic and potential energy of all the molecules; the bucket has far more molecules.
!Common mistakeChoosing the cup treats temperature as the amount of energy; temperature measures the average kinetic energy per molecule.
38Short answer · ★ Challenge
Explain how the zeroth law makes it possible for a nurse to compare the temperatures of two patients using one clinical thermometer.
Show answer
Model answer: The thermometer is left in contact with the first patient until it is in thermal equilibrium with them, so it is at their temperature. It is then put in equilibrium with the second patient. By the zeroth law, if the readings are equal the two patients would be in equilibrium with each other, so they have the same temperature; if not, the higher reading shows the hotter patient.
!Common mistakeReading the thermometer before it reaches equilibrium is the practical error the zeroth law warns about; the reading only means something at equilibrium.
39Multiple choice
A sealed glass jar of air is left in a car in the sun and warms up. Which describes this process?
- AIsobaric: W = PΔV and ΔU = 0
- BIsothermal: ΔU = 0 and Q = W
- CAdiabatic: Q = 0 and ΔU = −W
- DIsochoric: W = 0 and ΔU = Q
Show answer
Answer: D. Isochoric: W = 0 and ΔU = Q
The rigid jar keeps the volume constant, so no work is done and all the heat raises the internal energy.
!Common mistakeChoosing isobaric assumes the pressure stays the same; in a sealed rigid jar the pressure rises as it warms.
40Fill in the blank
A gas has a molar heat capacity at constant pressure of 37.4 J mol⁻¹ K⁻¹ (R = 8.31 J mol⁻¹ K⁻¹). Its molar heat capacity at constant volume is ______ J mol⁻¹ K⁻¹.
Show answer
Answer: 29.1
Cv = Cp − R = 37.4 − 8.31 ≈ 29.1 J mol⁻¹ K⁻¹.
!Common mistakeAdding R instead of subtracting gives 45.7; Cv is the smaller one because no expansion work is done.
41Multiple choice · ★ Challenge
A car engine is 25% efficient and gives 30 kW of useful power. Petrol releases 34 MJ per litre. How much petrol does it burn in one hour?
- A3.2 L
- B9.5 L
- C0.0035 L
- D12.7 L
Show answer
Answer: D. 12.7 L
Heat input = 30 ÷ 0.25 = 120 kW; in one hour 120 000 × 3600 = 432 MJ; petrol = 432 ÷ 34 ≈ 12.7 L.
!Common mistakeChoosing 3.2 L uses only the useful 30 kW; the fuel must supply four times as much heat at 25% efficiency.
42Short answer · ★ Challenge
A fridge keeps its inside at 5 °C in a kitchen at 30 °C. Find the greatest possible coefficient of performance, COP = Tc/(Th − Tc), and explain why a real fridge has a lower COP.
Show answer
Model answer: Tc = 278 K, Th = 303 K. COPmax = 278 ÷ (303 − 278) = 278 ÷ 25 ≈ 11. A real fridge has friction in the compressor, heat leaking through the walls and door seals and temperature differences across its coils, so it needs more work for the same heat removed.
!Common mistakeUsing °C (5 ÷ 25 = 0.2) gives a meaningless COP; temperatures in this formula must be in kelvin.
43Short answer
Give two different ways of increasing the internal energy of the air trapped in a sealed syringe, and say for each whether it is heating or doing work.
Show answer
Model answer: Put the syringe in hot water: heat flows into the air (heating, Q > 0). Push the piston in quickly: work is done on the air (W done on the gas). Both raise the internal energy and temperature of the air.
!Common mistakeSaying "push the piston in slowly" is less convincing: in a slow compression heat can leak out, so the internal energy may hardly change.
44Short answer · ★ Challenge
For 1.0 mol of a gas, 250 J raises the temperature by 12 K at constant volume, but 350 J is needed for the same rise at constant pressure. Find Cv and Cp and check that Cp − Cv ≈ R.
Show answer
Model answer: Cv = 250 ÷ (1.0 × 12) ≈ 20.8 J mol⁻¹ K⁻¹. Cp = 350 ÷ 12 ≈ 29.2 J mol⁻¹ K⁻¹. Cp − Cv ≈ 8.3 J mol⁻¹ K⁻¹, which matches R = 8.31 J mol⁻¹ K⁻¹. The extra 100 J at constant pressure is the work done in expanding.
!Common mistakeDividing by the temperature only and forgetting the number of moles gives J/K, not a molar heat capacity; here n = 1.0 so the numbers agree.
45Short answer
Name the four strokes of a four-stroke petrol engine in order and say briefly what happens in each.
Show answer
Model answer: Intake: the piston moves down and the fuel–air mixture is drawn in. Compression: the piston moves up, compressing the mixture. Power: the spark ignites the mixture and the hot gas pushes the piston down. Exhaust: the piston moves up and pushes out the burnt gases.
!Common mistakePutting the spark at the start of the compression stroke is wrong; ignition happens at the END of compression, when the mixture is hottest and most compressed.
46Short answer · ★ Challenge
A fire piston is a tube with a tight piston. The air inside is at 300 K and is compressed so fast that no heat escapes, to one fifth of its volume (γ = 1.4). Using TVγ−1 = constant, find the final temperature and say whether a piece of cotton wool that ignites at about 500 K would catch fire.
Show answer
Model answer: T₂ = T₁ (V₁/V₂)γ−1 = 300 × 50.4 ≈ 300 × 1.90 ≈ 571 K. This is above 500 K, so the cotton wool catches fire.
!Common mistakeUsing PV/T = constant with the pressure unchanged gives T₂ = 60 K (cooling); in an adiabatic compression the pressure rises sharply and the gas heats up.
47Multiple choice
How many moles of gas occupy 0.025 m³ at 1.0 × 10⁵ Pa and 300 K (R = 8.31 J mol⁻¹ K⁻¹)?
- A0.10 mol
- B10 mol
- C0.0010 mol
- D1.0 mol
Show answer
Answer: D. 1.0 mol
n = PV/RT = 1.0 × 10⁵ × 0.025 ÷ (8.31 × 300) ≈ 1.0 mol.
!Common mistakeChoosing 0.0010 mol comes from entering the pressure in kPa (100) instead of pascals (1.0 × 10⁵); keep SI units throughout.
48Short answer · ★ Challenge
2.0 mol of ideal gas is heated at a constant pressure of 1.2 × 10⁵ Pa from 300 K to 400 K (R = 8.31 J mol⁻¹ K⁻¹). Use PV = nRT to find the work done by the gas and its increase in volume.
Show answer
Model answer: At constant pressure PΔV = nRΔT, so W = 2.0 × 8.31 × 100 = 1662 J ≈ 1.7 kJ. ΔV = W/P = 1662 ÷ 1.2 × 10⁵ ≈ 0.014 m³ (about 14 L).
!Common mistakeUsing the final temperature (400 K) instead of the change (100 K) gives the whole PV, not the work done in this expansion.
49Short answer
Air in a syringe is pushed in at a constant pressure of 1.0 × 10⁵ Pa from 60 cm³ to 20 cm³. How much work is done on the air?
Show answer
Model answer: ΔV = 20 − 60 = −40 cm³ = −40 × 10⁻⁶ m³. Work done by the gas W = PΔV = 1.0 × 10⁵ × (−4.0 × 10⁻⁵) = −4.0 J, so 4.0 J of work is done ON the air.
!Common mistakeUsing 40 instead of 40 × 10⁻⁶ m³ gives 4 000 000 J; 1 cm³ = 10⁻⁶ m³.
50Short answer · ★ Challenge
A gas goes round a triangular cycle: A (1.0 × 10⁵ Pa, 1.0 L) → B (4.0 × 10⁵ Pa, 1.0 L) → C (1.0 × 10⁵ Pa, 5.0 L) → A. Find the net work done by the gas per cycle and the net heat it absorbs.
Show answer
Model answer: Net work = area of the triangle = ½ × base × height = ½ × (5.0 − 1.0) × 10⁻³ m³ × (4.0 − 1.0) × 10⁵ Pa = 600 J. The cycle is clockwise (A→B up, B→C down to the right, C→A left), so the gas does +600 J. Over a cycle ΔU = 0, so net heat absorbed = 600 J.
!Common mistakeForgetting the ½ for a triangle gives 1200 J; check the shape before using an area formula.