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.
U is the total energy of all the particles
- 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: For an ideal gas the particles don't attract each other, so there is no potential energy. 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.
- change in internal energy (J)
- number of particles, amount in moles
- change in temperature (K, or °C: the same size of step)
One mole of a monatomic ideal gas in an insulated cylinder is compressed by a piston: 60 J of work is done on the gas. Determine its rise in temperature.
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Melting and freezing change U at constant temperature: In a real substance U also holds the potential energy of the bonds between particles. When water freezes at 0 °C its temperature, and so the mean kinetic energy, stays the same, but energy (the latent heat) leaves and the potential energy falls. So U decreases. "Constant temperature means constant U" is true only for an ideal gas.
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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.
Where the first law comes from: It is conservation of energy applied to a closed system: no mass goes in or out, but energy can cross the boundary as heat or as work. Every joule of heat supplied is either stored as internal energy or leaves again as work.
An isolated system is stricter: neither mass nor energy crosses the boundary, so Q = 0, W = 0 and U stays constant.
- 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 the signs are
- 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).
Work is the area under the p–V graph
For a monatomic ideal gas at constant pressure, ΔU = 3/2 nRΔT = 3/2 pΔV (since pV = nRT). The heat in is then Q = ΔU + W = 3/2 pΔV + pΔV = 5/2 pΔV.
A monatomic ideal gas is compressed at a constant pressure of 30 kPa from 4.0 × 10⁻² m³ to 3.0 × 10⁻² m³. Show that about 750 J of thermal energy leaves the gas.
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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.
The words to know
| Word | What it means |
|---|---|
| Closed system | No mass in or out; energy crosses as heat or work |
| Isolated system | Neither mass nor energy crosses the boundary |
| Internal energy U | Total random kinetic energy + intermolecular potential energy of the particles |
| Work done by a gas W | pΔV at constant pressure; the area under the p–V graph in general |
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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