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Learn7408Paper 3B · Option: Engineering physicsThermodynamics and enginesThe first law of thermodynamics
Thermodynamics and engines · Subtopic 1 of 2

The First Law of Thermodynamics

The first law is conservation of energy for a system: ΔU=q+W\Delta U=q+W, where qq is the energy transferred to the system by heating and WW is the work done on the system. When a gas expands at constant pressure, it does work pΔVp\Delta V on its surroundings, so W=−pΔVW=-p\Delta V. On a pp–VV graph, work is the area under the curve.

Cambridge 9702 Paper 4AQA 7408 Paper 3B≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use W=pΔVW=p\Delta V and distinguish work done on and by a gas.
  • State and apply ΔU=q+W\Delta U=q+W with the 9702 sign convention.
  • Find work from the area under a pp–VV graph.
  • Apply the first law to isothermal, constant-volume and adiabatic changes.
Pause and think

Pumping up a bicycle tyre quickly makes the pump barrel warm. No heat was supplied. Where did the energy come from?

Section 2 of 14

Core idea

Start hereA bicycle pump gets hot

Pump up a tyre quickly and the pump barrel warms. You did work on the air, squeezing it, and that work went into its internal energy. Energy can enter a gas as heat or as work, and the first law keeps the books.

The big idea

ΔU=q+W\Delta U=q+W: qq heat to the gas; WW work done on it. Work by a gas at constant pressure: pΔVp\Delta V (area under pp–VV).

1. Work by a gas

Work done by an expanding gas is the area under the p–V graph
123456501001502002500volume (×10⁻³ m³)pressure (kPa)constant pressurework = pΔV = 600 J
200×103×3×10−3=600200\times10^3\times3\times10^{-3}=600 J done by the gas on the surroundings.
W=pΔVW=p\Delta V
  1. F=pAF=pA
    Force of gas on a piston.
  2. W=FΔx=pAΔxW=F\Delta x=pA\Delta x
    Piston moves Δx\Delta x.
  3. =pΔV=p\Delta V
    AΔx=ΔVA\Delta x=\Delta V.

2. Applying the law

Worked exampleExpansion with heating

2.0 L → 5.0 L at 200 kPa; absorbs 1500 J.

Goal
ΔU\Delta U.
Strategy
Work done on gas is negative when it expands.
  1. pΔV=200×103×3.0×10−3=600 Jp\Delta V=200\times10^3\times3.0\times10^{-3}=600\text{ J}
    Work by gas.
  2. W=−600 JW=-600\text{ J}
    On the gas.
  3. ΔU=1500−600=900 J\Delta U=1500-600=900\text{ J}
    First law.
Answer+900+900 J: temperature rises.
✓ CheckUnits: Pa × m³ = J ✓.
Worked exampleRapid compression

No time for heat flow.

Goal
Sign of ΔU\Delta U.
Strategy
q=0q=0.
  1. ΔU=0+W\Delta U=0+W
    Adiabatic.
  2. W>0⇒ΔU>0W>0\Rightarrow\Delta U>0
    Work done on the gas.
AnswerGas heats up: the warm pump.
✓ CheckMatches the hook ✓.
A cycle on a p–V diagram: net work = enclosed area
123456123450Vp
Going clockwise, the gas does more work expanding (at high pressure) than is done on it compressing (at low pressure). Engines run on such cycles. Over a full cycle ΔU=0\Delta U=0.
Common misconceptionSign convention Cambridge uses WW = work done on the gas; some books use work by it. Check which.
Pause and think

Isothermal expansion of an ideal gas: sign of qq?

Summary

  • ΔU=q+W\Delta U=q+W (WW on gas).
  • W=pΔVW=p\Delta V = area.
Key words
Isothermal
Constant temperature.
Adiabatic
No heat transfer.
HistoryClausius and the two laws. Rudolf Clausius stated the first law in modern form in 18501850, combining Joule's experiments with the idea of internal energy. In 18651865 he summed up thermodynamics: "The energy of the universe is constant. The entropy of the universe tends to a maximum."
Fun factWhy spray cans get cold. When you hold down an aerosol nozzle, the gas inside expands rapidly, doing work on its surroundings. There's little time for heat to flow in, so q≈0q\approx0 and ΔU=W<0\Delta U=W<0. The internal energy drops and the can gets noticeably cold.
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