The big idea: The same gas can be taken from one state to another in different ways. In each process we hold one thing fixed, which fixes how the first law ΔU = Q − W plays out.
You have seen this: Cover the nozzle of a bicycle pump and push down: the trapped air heats up. Open a fizzy drink and the escaping gas cools.
The four standard processes
- Isothermal: temperature T constant
- Isobaric: pressure P constant
- Isovolumetric (isochoric): volume V constant
- Adiabatic: no heat flows (Q = 0)
First law, signs: ΔU = Q − W, where W is the work done by the gas. Internal energy U depends only on temperature, so ΔU = 0 whenever T does not change. That single fact unlocks most of this topic.
| Process | What's constant | Consequence |
|---|---|---|
| Isothermal | T | ΔU = 0, so Q = W |
| Isobaric | P | W = PΔV (work = pressure × volume change) |
| Isovolumetric | V | W = 0, so Q = ΔU |
| Adiabatic | Q = 0 (no heat) | ΔU = −W |
The adiabatic law
- pressure (Pa)
- volume (m³)
The adiabat is steeper than the isotherm: Through the same point on a p–V diagram, an adiabatic curve falls more steeply than an isothermal one. In an adiabatic expansion the gas also cools, so its pressure drops for two reasons: more volume and a lower temperature.
0.20 mol of a monatomic ideal gas at 2.0 × 10⁶ Pa and 1.0 × 10⁻³ m³ expands adiabatically to 8.0 × 10⁻³ m³. Determine its final pressure and temperature.
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Entropy in an isothermal compression: Compressed at constant temperature, the gas gives out heat (Q = W < 0), so its entropy decreases while the surroundings' entropy increases.
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Plot pressure P against volume V and you get a p–V diagram. The work done by the gas is the area under the curve between the start and end volumes.
The sign of the work
- If the gas expands (V grows) it does positive work.
- If it is compressed the work is negative.
At constant pressure: When the pressure stays fixed, the area under the curve is just a rectangle, so the work is W = PΔV — pressure times the change in volume. This is the one work formula given in the data booklet.
- work done by the gas (J)
- pressure of the gas (Pa)
- change in volume (m³)
A gas is held at a constant pressure of 2.0 × 10⁵ Pa while it expands from 1.0 × 10⁻³ m³ to 4.0 × 10⁻³ m³. Find the work done by the gas.
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Cycles
Taking a gas round a cycle
- The gas ends where it started, so ΔU = 0 for the whole cycle and the net heat in equals the net work out.
- The net work done by the gas is the area inside the loop on the p–V diagram.
- A loop run clockwise does more work expanding than is done on it compressing: net work comes out, so it can run a heat engine.
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What a heat engine does: A heat engine runs in a cycle. Each cycle has three stages.
One cycle
Take in heat
It takes in heat Qin from a hot reservoir.
Do work
It does useful work W.
Reject heat
It rejects the leftover heat Qout to a cold reservoir.
Heat in, work out, leftover heat away.
Energy is conserved: Energy is conserved each cycle, so W = Qin − Qout. You can never turn all of Qin into work.
- efficiency (no unit; a fraction or %)
- heat taken in from the hot reservoir per cycle (J)
- heat rejected to the cold reservoir per cycle (J)
A heat engine takes in 800 J of energy from its hot source each cycle and rejects 600 J to the surroundings. Find its efficiency.
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There is a ceiling: No engine working between a hot and a cold reservoir can beat the Carnot efficiency.
About the Carnot efficiency
- It depends only on the two absolute temperatures (in kelvin).
- A real engine always falls below this value because of friction, turbulence and unwanted heat loss.
- maximum possible efficiency (no unit)
- absolute temperature of the cold reservoir (K)
- absolute temperature of the hot reservoir (K)
The Carnot cycle
Where U is greatest, and why the ceiling exists: U depends on temperature alone, so the gas has its greatest internal energy all along the hot isotherm (A → B) and its least along the cold one.
The second law (Kelvin form) says some heat must always go to the cold reservoir, so no engine reaches 100%. A steam turbine between 823 K and 303 K can never beat 1 − 303/823 = 0.63.
An engine works between a hot reservoir at 500 K and a cold reservoir at 300 K. Find the maximum (Carnot) efficiency it could possibly have.
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How this is tested — heat engines and the four processes are HL only (B.4) and appear at two levels:
Paper 1A
- 'Which process has W = 0?' or 'which has ΔU = 0?'
- A one-step η = 1 − Qout/Qin.
Paper 2
- Determine an efficiency and compare it with the Carnot value.
- Read work = area off a p–V cycle.
The classic trap: Leaving reservoir temperatures in °C for the Carnot formula — they must be in kelvin. And if your real efficiency comes out above the Carnot value you have slipped: a real engine is always below its Carnot ceiling.
Three easy marks: (1) For Carnot, put both temperatures in kelvin before dividing. (2) Efficiency is a fraction with no unit — multiply by 100 for a percentage. (3) A real efficiency that comes out above the Carnot value means you slipped — it must be lower.
The words to know
| Word | What it means |
|---|---|
| Adiabatic | No heat in or out (Q = 0); pV5/3 = constant for a monatomic ideal gas |
| Isothermal | Temperature constant, so ΔU = 0 for an ideal gas |
| Cycle | A set of processes that returns the gas to its starting state, so ΔU = 0 |
| Carnot cycle | Two isothermal and two adiabatic processes; the most efficient possible cycle |
A power station boiler runs at 600 K and rejects waste heat to a river at 300 K. Each second it takes in 5.0 × 10⁸ J and delivers 2.0 × 10⁸ J of useful work. (a) Calculate the actual efficiency. (b) Calculate the Carnot efficiency. (c) State why the actual value is lower.
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