Physics › Engineering physics › The first law of thermodynamics
The first law of thermodynamics
For a fixed mass of gas, the first law relates heat transfer, change in internal energy and work done: heat in equals internal energy gained plus work done by the gas. Three of the four standard processes each set one term to zero, constant pressure instead makes the work simply p times the volume change, and a curve on the p-V diagram turns work into an area you can measure.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Ideal gases and the gas laws and Molecular kinetic theory.
IN THIS TOPIC
- Apply Q = ΔU + W with AQA's sign convention, W being work done by the gas.
- Calculate work from pΔV at constant pressure, and read work as area on a p-V diagram.
- Handle the four non-flow processes, isothermal, adiabatic, constant pressure and constant volume, knowing which term each one reduces to zero.
- Use pV = constant and pV to the gamma = constant for isothermal and reversible adiabatic changes of an ideal gas.
COMMON MISCONCEPTION
Heating a gas always makes it hotter.
The energy ledger
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, and the gas can do work. The first law of thermodynamics is the ledger line connecting them:
Read it with AQA's convention nailed down. Q is the heat supplied to the gas. W is the work done by the gas as it pushes its surroundings back. Heat in, then, either stays as internal energy or leaves as work; energy is conserved, itemised.
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. Where does W come from? A gas at pressure p pushing a piston of area A exerts force pA, and moving it a distance moves volume ΔV = A × distance, so at constant pressure
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. Compress a gas 150 J 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
Exam questions run gases through four named non-flow processes, and each one sets a term of the equation to zero. At constant volume nothing moves, so no work is done, W = 0, and every joule of heat lands in internal energy. At constant pressure the gas expands as it is heated, and the heat splits, Q = ΔU + pΔV.
Isothermal means constant temperature. The gas stays on one isotherm of the ideal gas equation, so
and since U depends only on temperature, ΔU = 0 and Q = W. An isothermally expanding gas turns heat into work continuously, its temperature never rising. You can pour heat into a gas without warming it at all, provided it does work at the same rate. In practice isothermal means slow, in good thermal contact, so the temperature can equalise throughout.
Adiabatic is the opposite case, no heat at all, Q = 0, in practice fast or well insulated. That is the whole of the definition, and on its own it gives only ΔU = −W. The power law below is a narrower claim needing two further conditions: the gas must be ideal, with constant heat capacities, and the change must be reversible, meaning slow and smooth enough that the gas holds one pressure and one temperature throughout. Granted those,
where γ, about 1.4 for air, is the ratio of the gas's two principal specific heat capacities. Q = 0 alone will not do it: a gas rushing into a vacuum takes in no heat and does no work on anything, so ΔU = 0 while pVγ does not stay constant. With Q = 0 the equation reads ΔU = −W, so a gas doing work adiabatically does that work at the expense of its internal energy and cools; a gas compressed adiabatically heats. A bicycle pump warming as you pump, and a diesel engine igniting fuel with no spark plug, are both adiabatic compression doing exactly this.
| Process | Held fixed | Q | ΔU | W |
|---|---|---|---|---|
| constant volume | V | = ΔU | = Q | zero |
| constant pressure | p | ΔU + pΔV | Q − W | pΔV |
| isothermal | T (pV = constant) | = W | zero | = Q |
| adiabatic | no heat flow (reversible: pVγ = constant) | zero | = −W | = −ΔU |
GUIDED PRACTICE
The diesel squeeze
Air at 1.0 × 105 Pa fills 8.0 × 10−4 m3 of a cylinder. It is compressed adiabatically to 1.0 × 10−4 m3. Take γ = 1.4. Find the final pressure, and use pV/T to find the factor the absolute temperature rises by.
Show the working
p2 = p1(V1/V2)γ = 1.0 × 105 × 8.01.4 = 1.8 × 106 Pa.
pV/T is constant, so T2/T1 = p2V2/p1V1 = 18.4 × (1/8.0) = 2.3. Room-temperature air at 290 K reaches about 670 K, far past diesel's ignition point, with not a joule of heat supplied.
Work as area
Plot any process on a p-V diagram, pressure against volume, and the work done by the gas appears as geometry. Each small expansion δV does work p δV, a thin strip under the curve, so the total work is the area under the curve between the two volumes. Constant pressure gives a rectangle, p × ΔV. Curved processes give areas you estimate by counting squares, and the examiners set exactly that.
Direction matters. Rightward along the curve, expansion, the gas does positive work. Leftward, compression, the surroundings do the work and W is negative. On the isotherm-and-adiabat figure above, the adiabat's steeper fall means less area beneath it, so an adiabatic expansion between two volumes does less work than an isothermal one, because no heat enters to replace the internal energy converted into work.
INDEPENDENT PRACTICE
A slow isothermal squeeze
A gas at 1.2 × 105 Pa occupies 5.0 × 10−3 m3. It is compressed slowly at constant temperature to 2.0 × 10−3 m3. Find the final pressure, and account for the heat flow during the process.
Show the working
pV constant: p2 = 1.2 × 105 × 5.0/2.0 = 3.0 × 105 Pa.
Isothermal, so ΔU = 0 and Q = W. The gas is compressed, W is negative, so Q is negative too. Every joule of work done on the gas flows straight out as heat, and slowness is what gives it time to escape, holding the temperature steady.
ASSESSMENT FOCUS
- Open every first-law answer by fixing the signs. Q is heat into the gas, W is work done by the gas. Most dropped marks in 3.11.2 are sign slips, not physics slips.
- Name the process, then set its zero term. Constant volume gives W = 0, isothermal gives ΔU = 0, adiabatic gives Q = 0. Writing the full equation and crossing one term out is working that earns marks.
- pΔV only serves at constant pressure. Anywhere else, work is the area under the p-V curve, and counting squares is a legitimate, expected method.
- Adiabatic calculations run on p1V1γ = p2V2γ, which assumes an ideal gas taken reversibly, as every adiabatic question set at this level is. Temperature questions then go through pV/T = constant. Do not reach for pV = constant when the process is adiabatic; that line belongs to the isotherm.
- Explain answers in molecules when invited. Adiabatic compression: the piston does work on the molecules, their mean kinetic energy rises, so temperature rises. That chain scores where a bare formula does not.
CHECK YOURSELF
A gas in a rigid sealed cylinder receives 900 J of heat. It is then allowed to expand at constant temperature, doing 300 J of work. For each stage, state Q, ΔU and W, and the overall change in internal energy.
Show a hint
Rigid means constant volume; constant temperature means ΔU = 0. Take the equation one stage at a time.
Show the answer
Stage 1, constant volume: W = 0, so Q = ΔU = +900 J. All the heat becomes internal energy and the gas warms.
Stage 2, isothermal: ΔU = 0, so Q = W = +300 J. The gas draws another 300 J of heat and spends it entirely on work.
Overall ΔU = 900 + 0 = +900 J, while the total heat supplied was 1200 J. The missing 300 J left as work, and the totals balance.
Q = ΔU + W, with W the work done by the gas: heat in becomes internal energy or work out.
Constant volume gives W = 0, isothermal gives ΔU = 0, adiabatic gives Q = 0.
On a p-V diagram, work is the area under the curve; pV is constant on an isotherm, pV^γ on a reversible adiabat of an ideal gas.
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 of thermodynamics questions page.
WHERE TO GO NEXT
- Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- Apply Q = ΔU + W with AQA's sign convention, W being work done by the gas.
- Calculate work from pΔV at constant pressure, and read work as area on a p-V diagram.
- Handle the four non-flow processes, isothermal, adiabatic, constant pressure and constant volume, knowing which term each one reduces to zero.
- Use pV = constant and pV to the gamma = constant for isothermal and reversible adiabatic changes of an ideal gas.
Open the full revision checklist to track your progress across the whole unit.