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The first law and its sign convention
Heat supplied to a gas raises its internal energy, leaves as work the gas does, or both: this is the first law of thermodynamics, Q = ΔU + W. Compression makes W negative and heat lost makes Q negative, so the signs settle whether the gas warmed or cooled. At constant pressure the work is pΔV.
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The first law of thermodynamics, part 1 of 2. Part 2 is Non-flow processes and p-V diagrams.
Builds on Ideal gases and the gas laws and The kinetic theory model and its derivation and What temperature means for a gas.
IN THIS TOPIC
- Apply Q = ΔU + W with AQA's sign convention, W being work done by the gas.
COMMON MISCONCEPTION
Compressing a gas cannot change its temperature, because no heat is supplied.
Work done on the gas raises its internal energy: with Q = 0, ΔU = −W and W is negative for a compression, so U and the temperature rise. A bicycle pump warms for exactly this reason.
The first law and its sign convention
A fixed mass of gas holds internal energy U, the kinetic energy of its molecules (for an ideal gas there is no potential term, so U depends on temperature alone). Two things can change U: heat can flow in or out, and work can be done by or on the gas. The first law of thermodynamics relates the three:
For the convention used here, Q = ΔU + W, where Q is energy transferred to the gas by heating and W is work done by the gas. Energy supplied by heating therefore either stays in the gas as internal energy or leaves as work done on the surroundings.
Describe the drawing
A gas trapped in a cylinder by a piston. A wide coral arrow of 600 joules of heat enters from the left and divides: an amber arrow of 380 joules raising the internal energy of the gas, and a narrower cyan arrow of 220 joules leaving as work on the piston. The arrow widths are drawn in proportion, so the ledger visibly balances.
Signs carry the physics. A gas compressed has work done on it, so W is negative. A gas losing heat has negative Q. Feed the signed numbers in and the sign of ΔU shows whether the gas warmed or cooled. To find W, take a gas at pressure p pushing a piston of area A: it exerts force pA, and moving the piston a distance sweeps out volume ΔV = A × distance, so at constant pressure
Describe the drawing
A gas in a cylinder with the piston on the right. A coral arrow labelled Q equals plus 500 joules points into the gas, a cyan arrow labelled W equals plus 200 joules leaves past the outward-moving piston, and the amber label delta U equals plus 300 joules sits inside the gas between them. A ledger beside the cylinder adds the 300 joules kept to the 200 joules out and totals the 500 joules in, so every sign in Q equals delta U plus W is positive and visible.
WORKED EXAMPLE
First-law practice
A gas absorbs 600 J of heat while expanding, doing 220 J of work on the piston. Find the change in its internal energy, and state what happens to its temperature.
ΔU = Q − W = 600 − 220 = +380 J.
Internal energy rose, and for an ideal gas U tracks temperature, so the gas got hotter, but by less than 600 J of heating alone would suggest. The other 220 J left as work.
Now reverse it. Do 150 J of work compressing a gas while it loses 150 J of heat: ΔU = −150 − (−150) = 0. Squeezed and cooled in balance, its temperature never moved.
Four processes, four zero terms has moved to the next part of this lesson.
Work as area has moved to the next part of this lesson.
ASSESSMENT FOCUS
- Open every first-law answer by fixing the signs. Q is heat into the gas, W is work done by the gas: a sign attached the wrong way round reverses the sign of ΔU and the conclusion about temperature with it.
- ΔU answers come from the rearrangement ΔU = Q − W. Write the law, rearrange, then substitute with the signs already attached; substituting first is where the sign slips happen.
- For an ideal gas the internal energy is kinetic only and tracks the absolute temperature, so the sign of ΔU is the whole answer to any warmer-or-colder question.
- W = pΔV applies only while the pressure is constant, and ΔV is the change in volume, never the final volume. For any other process the work is the area under the p-V curve.
CHECK YOURSELF
A gas loses 240 J of heat to its surroundings while it is being compressed, 380 J of work being done on it. Find the change in its internal energy, and state what happens to the temperature of the gas.
Show a hint
Fix both signs before touching the equation: heat out and work done on the gas are both negative in AQA's convention.
Show the answer
Signs first: heat leaves, so Q = −240 J, and the gas is compressed, so the work done by the gas is W = −380 J.
ΔU = Q − W = −240 − (−380) = +140 J.
Internal energy rose, and for an ideal gas U tracks temperature, so the gas got hotter: the compression put in more energy than the cooling carried away.
Q = ΔU + W, with W the work done by the gas: heat in becomes internal energy or work out.
Signs first: compression makes W negative, heat lost makes Q negative, and the sign of ΔU then says warmer or cooler.
At constant pressure the work is pΔV, the gas's pressure times the volume it sweeps.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the first law and its sign convention questions page.
WHERE TO GO NEXT
- Gradients and areas under graphs is the maths this lesson leans on, worked through step by step.
USED LATER IN
CHECK YOUR PROGRESS
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- Apply Q = ΔU + W with AQA's sign convention, W being work done by the gas.
Open the full revision checklist to track your progress across the whole unit.