A Level Physics 9702 · A Level · Topic 16 · ~4 min read
Thermodynamics
Internal energy as the sum of random kinetic and potential energies of molecules, work done on or by a gas W = pΔV, and the first law of thermodynamics ΔU = q + W with its sign convention.
Internal energyRandom kinetic & potential energyTemperature and internal energyWork done W = pΔVFirst law ΔU = q + WSign conventionGas processes
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In 30 seconds
- Internal energy is the sum of the random kinetic and potential energies of all the molecules. For an ideal gas there's no potential energy, so U depends only on T.
- Work done on a gas when compressed at constant pressure: W = pΔV (with ΔV the decrease in volume).
- First law: ΔU = q + W. ΔU is the increase in internal energy, q the energy supplied by heating, W the work done on the system.
- Signs: heat in → q positive; gas compressed → W positive; gas expands → W negative.
By the end you can
- Understand internal energy as the sum of a random distribution of molecular kinetic and potential energies.
- Relate a rise in temperature to an increase in the kinetic energy of the molecules.
- Use W = pΔV for work done when the volume of a gas changes at constant pressure.
- Recall and use the first law of thermodynamics, ΔU = q + W.
1Internal energy
What it is
Internal energy — the sum of the random kinetic energies and potential energies of the molecules in a system.
- Raising the temperature increases the mean kinetic energy of the molecules, so U increases.
- Changing state at constant temperature changes the potential energy (bonds), so U changes even though T doesn't.
- Ideal gas: no intermolecular forces, so U is all kinetic: U = (3/2)NkT. ΔU = 0 whenever T is unchanged.
2The first law
ΔU = q + W
W = pΔV ; ΔU = q + W DATA & FORMULAE
W = work done ON the systemWorked example
A gas at a constant 2.0 × 10⁵ Pa expands from 1.5 × 10⁻³ m³ to 2.5 × 10⁻³ m³, while 500 J is supplied by heating. Find the change in internal energy.
- work done by gas = pΔV = 2.0 × 10⁵ × 1.0 × 10⁻³ = 200 J, so W = −200 J C1 A1
- ΔU = q + W = 500 − 200 = +300 J A1
Worked example
A gas is compressed quickly (no time for heat transfer) with 150 J of work done on it. Find ΔU and say what happens to its temperature.
- q = 0, W = +150 J C1
- ΔU = +150 J, so the temperature rises A1
| Process | Condition | First law |
|---|---|---|
| Constant volume | W = 0 | ΔU = q |
| Isothermal (ideal gas) | ΔU = 0 | q = −W |
| No heat transfer (adiabatic) | q = 0 | ΔU = W |
| Full cycle | back to start | ΔU = 0 |
Common mistakeSign errors with W. In this syllabus W is work done ON the gas. An expanding gas does work, so W is negative. The 2024 Paper 4 report flagged this every year.
Linking questionIdeal gases and U = (3/2)NkT are in Ideal gases; specific heat capacity is in Temperature; and work = force × distance is from Work, energy and power.
Test yourself — cover the answers first
Quick check
- Define internal energy.
- A gas expands. Is W positive or negative?
- What is ΔU for an ideal gas at constant temperature?
- A gas is heated at constant volume with 80 J. Find ΔU.
- Why does the internal energy of ice increase as it melts at 0 °C?
Exam-style
- A gas is compressed at constant pressure 1.0 × 10⁵ Pa from 4.0 × 10⁻⁴ m³ to 1.5 × 10⁻⁴ m³ and loses 40 J by heating. Find ΔU.
- An ideal gas expands isothermally, doing 120 J of work. How much energy was supplied by heating?
- Explain why a bicycle pump gets warm when used quickly.
Show answers
- the sum of the random kinetic and potential energies of the molecules
- negative (work is done by the gas)
- zero
- 80 J
- the molecular potential energy increases
- W = +25 J, q = −40 J ⇒ ΔU = −15 J
- ΔU = 0, W = −120 J ⇒ q = +120 J
- work is done on the air (W > 0) with little time for heat loss (q ≈ 0), so ΔU > 0 and T rises
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