The big idea: 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 same gas can be taken from one state to another in different ways, and in each process we hold one thing fixed, which fixes how the first law ΔU = Q − W plays out.
The four standard processes are:
- 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 |
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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. 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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What a heat engine does: A heat engine runs in a cycle. Each cycle it:
1. takes in heat Qin from a hot reservoir, 2. does useful work W, and 3. rejects the leftover heat Qout to a cold reservoir.
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. 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)
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.
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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