The big idea: A gas can be taken from one state to another in different ways. In each process we hold one thing fixed, and that 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 |
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)
Worked example — work in an isobaric expansion
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.
Solution
- Write the given formula first:
- Find ΔV, then substitute:
- Work it out — keep the unit:
Final answer
W = 600 J (work done by the gas, since it expands).
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What a heat engine does: A heat engine runs in a cycle. Each cycle it:
1. takes in heat Q_in from a hot reservoir, 2. does useful work W, and 3. rejects the leftover heat Q_out to a cold reservoir.
Energy is conserved each cycle, so W = Q_in − Q_out. 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)
Worked example — efficiency of an engine
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.
Solution
- Write the given formula first:
- Substitute the heat values:
- Work it out:
Final answer
η = 0.25, i.e. 25%. (Only 200 J of the 800 J becomes useful work.)
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)
Worked example — the maximum efficiency
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.
Solution
- Write the given formula first:
- Substitute the kelvin temperatures:
- Work it out:
Final answer
ηCarnot = 0.40, i.e. 40% — the best any real engine between these temperatures could reach.
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Where it shows up: Heat engines and the four processes are HL only (B.4):
- Paper 1A — 'which process has W = 0?', 'which has ΔU = 0?', or a one-step η = 1 − Q_out/Q_in. - Paper 2 — determine an efficiency, compare it with the Carnot value, or read work = area off a p–V cycle.
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.
IB-style question — comparing real and Carnot efficiency
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.
Solution
- (a) Actual efficiency = useful work ÷ energy input:
- (b) Carnot efficiency from the kelvin temperatures:
- (c) The actual value (0.40) is below the Carnot ceiling (0.50) because of friction and unwanted heat loss.
Final answer
(a) η = 0.40 (40%). (b) ηCarnot = 0.50 (50%). (c) Real losses (friction, turbulence, heat leaks) always push the actual efficiency below the Carnot maximum.