Key Idea: A kettle turns electrical energy into hot water, a car engine burns fuel to push pistons, a fridge pumps heat out of cold food — all of it is thermodynamics: the accounting of heat, work and internal energy in a gas. The first law is energy conservation (Q = ΔU + W); the second law says disorder (entropy) never decreases, which is why heat only flows hot → cold and why no engine is perfect. It is HL only (B.4).
📐 The formulas you're given
- heat ADDED to the gas (J)
- increase in internal energy (J) — for an ideal gas, depends only on T
- work done BY the gas (J); W = PΔV at constant pressure
- entropy change (J K⁻¹) — T in kelvin
- efficiency of a heat engine
- the maximum possible efficiency between two temperatures
🔁 The four processes
| Process | What stays constant | Consequence |
|---|---|---|
| Isothermal | temperature T | ΔU = 0, so Q = W |
| Isobaric | pressure P | W = PΔV |
| Isovolumetric | volume V | W = 0, so Q = ΔU |
| Adiabatic | no heat flow (Q = 0) | ΔU = −W |
On a p–V diagram, the work done by the gas is the area under the line. The second law adds the arrow of time: in any real (irreversible) change, the total entropy of an isolated system increases.
✏️ IB-style worked examples (one per micro)
500 J of heat is added to a gas, and the gas does 200 J of work as it expands. Determine the change in its internal energy.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
1200 J of heat flows from a hot body at 400 K to a cold body at 300 K. Determine the net entropy change of the isolated system.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A heat engine takes in 800 J per cycle and rejects 600 J to the surroundings. Determine its efficiency, and state the maximum possible efficiency if it works between 500 K and 300 K.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Important: 1. Get the signs right in Q = ΔU + W: Q is heat added, W is work done by the gas. Heat removed → Q negative; gas compressed → W negative. 2. Entropy uses kelvin, never °C. 3. No real engine beats the Carnot efficiency — and you can never reach η = 1. 4. Identify the process first: isovolumetric → W = 0; isothermal → ΔU = 0.
Tap each card to reveal the answer.
State the first law of thermodynamics. Q = ΔU + W — heat added = increase in internal energy + work done by the gas (energy conservation).
Which process has W = 0? Isovolumetric (constant volume) — no area under a vertical line on a p–V diagram.
State the second law of thermodynamics. The entropy of an isolated system never decreases (it increases for any irreversible process).
Why does heat flow hot → cold, never the reverse? Because that direction increases total entropy; the reverse would decrease it, which the second law forbids.
Carnot efficiency between 600 K and 300 K? 0.50 (50%) — η = 1 − 300/600.
On a p–V diagram, what does the area under the curve give? The work done by the gas.
Exam tips
- Write Q = ΔU + W with signs first, then substitute — most marks are lost on signs.
- Keep all temperatures in kelvin for entropy and Carnot efficiency.
- Identify the process (isothermal / isobaric / isovolumetric / adiabatic) to know what's zero.
- Real efficiency < Carnot efficiency, always; η = 1 is impossible.
- Total entropy change of a real, isolated process is positive.