Stretch: Stretch: harder material that separates the top grades
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This topic is about bookkeeping: energy enters a gas by heating or by work, and the first law tracks where it goes. For Cambridge (Paper 4), Eduqas (Component 1) and WJEC (Unit 3) the first law and p–V work are core content. For AQA it is half of the Engineering physics option (Paper 3 Section B), where it extends to adiabatic changes, engine indicator diagrams, efficiencies and heat pumps. The marks go to learners who use one sign convention consistently and read p–V diagrams accurately.
By the end you’ll be able to
Apply the first law of thermodynamics with the correct board sign convention (Cambridge: ΔU = q + w; AQA/Eduqas/WJEC: Q = ΔU + W)
Calculate work done as pΔV or the area under/enclosed by a p–V curve
Distinguish isothermal, adiabatic (pV^γ = constant), isobaric and isovolumetric processes
Analyse four-stroke petrol and diesel indicator diagrams and calculate indicated and brake power (AQA engineering)
Use maximum theoretical efficiency (T_H − T_C)/T_H and coefficients of performance for heat pumps and refrigerators
What the exam asks
All four boards: internal energy, the first law, W=pΔV at constant pressure, and work as the area under a p–V graph.
Tables round a cycle (a favourite of Cambridge, Eduqas and WJEC): fill in Q, and for each stage, using the fact that for a complete cycle.
vii.Check your understanding
3 questions on first law of thermodynamics, p–V diagrams and heat engines. Every option is explained once you answer.
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PromptCard 1 of 3
State the first law in each board’s convention.
W
ΔU
ΔU=0
Eduqas/WJEC:U=23nRT for a monatomic ideal gas, so ΔU can be calculated directly from Δ(pV).
AQA Engineering option:
isothermal, adiabatic (pVγ=constant), constant-pressure and constant-volume changes;
petrol and diesel cycles and their indicator diagrams;
indicated, brake and friction power, and overall, thermal and mechanical efficiency;
the second law and the maximum efficiency THTH−TC;
coefficients of performance of refrigerators and heat pumps.
AQA 6-markers: comparing real and theoretical indicator diagrams, or explaining why practical engines fall short of the theoretical efficiency.
Core ideas
Internal energy
Internal energy U is the sum of the randomly distributed kinetic and potential energies of the molecules. For an ideal gas there is no potential energy, so U depends only on temperature: U=23nRT=23pV for a monatomic gas. Two consequences:
If T is unchanged (isothermal), ΔU=0.
Around any complete cycle, the gas returns to its starting state, so ΔUcycle=0.
The first law and its sign conventions
Energy is conserved: the increase in internal energy equals the energy supplied by heating plus the work done on the gas. Boards write this in two equivalent ways.
Board
Equation
W means
Expansion
AQA (Engineering)
Q=ΔU+W
work done by the gas
W>0
Eduqas / WJEC
ΔU=Q−W
work done by the gas
W>0
Cambridge 9702
ΔU=q+W
work done on the gas
W<0
In every convention, Q (or q) is positive when energy enters by heating and ΔU is positive when the internal energy rises.
Work done by a gas
At constant pressure, W=pΔV. In general, the work done by the gas is the area under the p–V curve between the two volumes. It is positive for expansion and negative for compression.
For a cycle, the net work done by the gas equals the area enclosed by the loop. A loop traversed clockwise does net work on the surroundings (a heat engine). An anticlockwise loop needs net work input (a heat pump or refrigerator).
The four non-flow processes
Process
Condition
First law (AQA form, Q=ΔU+W)
Isothermal
T constant, pV=constant
ΔU=0, so Q=W
Adiabatic
no heat transfer, pVγ=constant
Q=0, so ΔU
Constant pressure (isobaric)
W=pΔV
Q=ΔU+pΔV
Constant volume (isovolumetric)
W=0
Q=ΔU
Adiabatic compression heats a gas. The work done on it has nowhere to go but into internal energy. This is how a diesel engine ignites its fuel without a spark.
On a p–V diagram an adiabatic curve is steeper than an isothermal curve through the same point.
γ is the ratio of the principal molar heat capacities: 1.67 for a monatomic gas, 1.4 for air. After using pVγ=constant to find a pressure, use TpV=constant to find the temperature. Only AQA tests pVγ.
Heat engines and the second law (AQA)
A heat engine takes in energy QH from a hot source, does work W and must reject energy QC to a cold sink. The second law says no engine can turn all of QH into work.
Real engines fall well short because of friction, energy lost by heating through the cylinder walls, processes that happen too quickly to be reversible, and incomplete combustion. Combined heat and power schemes improve the overall use of fuel by using the rejected QC to heat buildings.
Engine cycles and indicator diagrams (AQA)
A four-stroke engine has induction, compression, power (expansion) and exhaust strokes, so there is one cycle every two revolutions.
Theoretical petrol (Otto) cycle: adiabatic compression; heat input at constant volume (instant spark ignition); adiabatic expansion; heat rejection at constant volume.
Theoretical diesel cycle: as above, but heat is input at constant pressure while the fuel is injected and burns. The compression ratio is higher.
Real indicator diagrams have rounded corners (valves and combustion take time), a lower peak pressure and a small anticlockwise pumping loop at low pressure for induction and exhaust. So the useful area is smaller than the theoretical one.
Power chain (all on the AQA data sheet):
input power=calorific value×fuel flow rate
indicated power=(area of p–V loop)×(cycles per second)×(number of cylinders)
brake power=Tωfriction power=indicated power−brake power
These are reversed heat engines: work W moves energy from cold to hot.
COPref=WQC=QH−QCQCCWQH=QH−QCQH
Theoretical maxima: COPref=TH−TCTC and COPhp=TH. The temperature forms are not on the data sheet, so learn them. Note that COPhp=COPref+1, and both are larger when the temperature difference is small.
Worked examples
Exam technique
Table questions: fill in the easy cells first. W=0 at constant volume; ΔU=0 for isothermal stages; Q=0 for adiabatic stages. Then use the fact that the ΔU values sum to zero around the cycle, and apply the first law row by row.
Areas: check the axis units. kPa×litres=J, and 105Pa×10 J per grid square.
Four-stroke trap: divide rev/s by 2 to get cycles per second. Two-stroke engines have one cycle per revolution, and the question will say so.
Kelvin always in THTH−T and the COP formulas.
Plausibility check: a real efficiency above the Carnot limit or a COP below 1 for a heat pump means you have made an error.
Common mistakes
Quick recap
First law: AQA Q=ΔU+W and Eduqas/WJEC ΔU=Q−W (W by the gas); Cambridge ΔU=q+W (W on the gas).
Work = area under the p–V graph; net work per cycle = area of the loop; W=pΔV at constant pressure.
Isothermal: ΔU=0. Adiabatic: Q=0 and pVγ constant. Constant volume: .
Ideal gas: U depends only on T (23nRT for a monatomic gas), and around any cycle.
Engines: efficiency =QHW, with a maximum of ; there is one cycle per two revolutions in a four-stroke engine.
Indicated power = loop area × cycles/s × cylinders; brake power =Tω; friction power = indicated − brake.