The big idea: Leave a mug of tea on the desk and it slowly goes cold — the energy leaking away was stored inside it the whole time. That stored energy is the internal energy U of a gas: the total energy of all its particles added up:
- the random kinetic energy of the particles (how fast they jiggle and fly about), plus - the potential energy stored in the forces between them.
For an ideal gas the particles don't attract each other, so there is no potential energy — and then U depends on temperature alone. Hotter gas = more internal energy.
Internal energy ≠ heat: Internal energy is energy a gas already has inside it. Heat is energy on the move — energy flowing in or out because of a temperature difference. You change U by adding heat or by doing work.
Because an ideal gas's internal energy depends only on temperature, raising its temperature always raises U, and any process at constant temperature leaves U unchanged (ΔU = 0). This single fact decides the sign of ΔU in almost every exam question.
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Energy is just bookkeeping: The first law of thermodynamics is conservation of energy for a gas. The heat you add to a gas either raises its internal energy or gets spent doing work as the gas pushes outward — or some of each.
- heat ADDED to the gas (J)
- increase in internal energy of the gas (J)
- work done BY the gas on its surroundings (J)
Read it as a sentence: heat in = energy stored + work done out. Rearranged, the change in internal energy is ΔU = Q − W — the heat you put in minus the work the gas spends expanding.
500 J of heat is added to a gas. As it warms, the gas expands and does 200 J of work pushing back its surroundings. Find the change in internal energy.
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Get the signs right or lose every mark: In the booklet's convention, Q is heat ADDED to the gas and W is work done BY the gas. So:
- Heat added ⇒ Q is positive; heat removed ⇒ Q is negative. - Gas expands (does work on surroundings) ⇒ W is positive. - Gas is compressed (surroundings do work on it) ⇒ W is negative. - Temperature rises ⇒ ΔU is positive; temperature falls ⇒ ΔU is negative.
| Situation | Sign | Why |
|---|---|---|
| Heat added to the gas | Q > 0 | Energy flows in |
| Heat removed from the gas | Q < 0 | Energy flows out |
| Gas expands | W > 0 | Gas does work on its surroundings |
| Gas compressed | W < 0 | Surroundings do work on the gas |
| Temperature rises | ΔU > 0 | Particles move faster |
A trick that always works: Before you substitute, write the sign of each quantity in words next to it: 'heat removed → Q = −...', 'gas compressed → W = −...'. Then plug straight into ΔU = Q − W. The minus signs do the rest.
Pushing back the surroundings: When a gas expands it pushes its container (or a piston) outward, so it does work on the surroundings. At constant pressure that work is simply the pressure times the change in volume.
- work done BY the gas (J)
- pressure of the gas, held constant (Pa)
- change in volume (m³)
A gas expands at a constant pressure of P = 1.0×10⁵ Pa, increasing its volume by ΔV = 2.0×10⁻³ m³. Find the work done by the gas.
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Units must match: Use pascals (Pa) for pressure and cubic metres (m³) for volume, and W comes out in joules (J). A volume given in litres or cm³ must be converted to m³ first (1 L = 1×10⁻³ m³).
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How this is tested — the first law is HL only (B.4) and shows up at two levels:
Paper 1A
- A one-step 'find ΔU / Q / W' using the correct signs.
- 'What does the internal energy of an ideal gas depend on?' (temperature alone).
Paper 2
- A multi-stage process where you track Q, W and ΔU through each step.
- Often finishing with W = PΔV for the constant-pressure stage.
The classic trap: Getting the signs wrong. Q is heat added and W is work done by the gas — so heat removed makes Q negative and compression makes W negative. Decide each sign in words before substituting into ΔU = Q − W.
Three easy marks: (1) Quote the given first law Q = ΔU + W before substituting. (2) Decide the sign of every term in words first. (3) For an ideal gas, no temperature change ⇒ ΔU = 0.
A gas in a cylinder is compressed by a piston, which does 350 J of work on the gas. At the same time 500 J of heat is removed from the gas to the cooling surroundings. Determine the change in the internal energy of the gas, and state whether the gas warms or cools.
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