Thermodynamics is the HL-only capstone of Theme B and one of the highest-leverage topics on the entire IB Physics HL syllabus. Almost every Paper 2 long answer in the past five sessions has tested at least one of: the first law in IB sign convention, the entropy/second-law statement, identifying the four processes on a $P$–$V$ diagram, or computing Carnot efficiency. Paper 3 essays love the microscopic ($S = k_B \ln \Omega$) interpretation.
This cheatsheet condenses every formula, definition, trick, and trap from Topic B.4 onto a single revisable page. You'll see the first law in the IB sign convention, both macroscopic and Boltzmann definitions of entropy, the second law in its rigorous form, the full four-process table for a monatomic ideal gas, and Carnot efficiency with the most-tested manipulation. The worked example walks through a Carnot calculation in mark-scheme rhythm; the FAQs unpack the conceptual misconceptions that cost the most marks in mock exams.
§1 — First Law of Thermodynamics Topic B.4
Statement (IB sign convention)
$$Q = \Delta U + W$$
- $Q$ — thermal energy into the system (positive = heat added).
- $\Delta U$ — change in internal energy.
- $W$ — work done by the system on its surroundings ($W = P \Delta V$, positive = expansion).
Work done on the system is $-W$ (negative in this convention).
Key formulae
§2 — Entropy Topic B.4
Macroscopic definition
$\Delta Q$ is the heat added reversibly at absolute temperature $T$.
Microscopic (Boltzmann) definition
$\Omega$ is the number of microstates that produce the macrostate. Higher $\Omega$ ⇒ more disorder ⇒ higher entropy.
Coin microstate example ($N = 4$ coins)
| Macrostate | $\Omega = \binom{N}{n_H}$ | Relative entropy |
|---|---|---|
| 4H 0T | 1 | lowest |
| 3H 1T | 4 | |
| 2H 2T | 6 | highest |
| 1H 3T | 4 | |
| 0H 4T | 1 | lowest |
The system evolves toward the macrostate with the most microstates (maximum $\Omega$).
§3 — Second Law of Thermodynamics Topic B.4
For any isolated system: $\Delta S_\text{total} \ge 0$.
- Real (irreversible) processes: $\Delta S_\text{total} > 0$.
- Reversible (ideal) processes: $\Delta S_\text{total} = 0$.
- Non-isolated systems: the local entropy can decrease, but the surroundings always compensate: $|\Delta S_\text{surroundings}| \ge |\Delta S_\text{system}|$.
§4 — Four Thermodynamic Processes Topic B.4
Process summary (monatomic ideal gas)
| Process | Fixed | $W$ | $\Delta U$ | $Q$ |
|---|---|---|---|---|
| Isovolumetric | $V$ | $0$ | $\tfrac{3}{2} n R \Delta T$ | $= \Delta U$ |
| Isobaric | $P$ | $P \Delta V$ | $\tfrac{3}{2} n R \Delta T$ | $\Delta U + P \Delta V$ |
| Isothermal | $T$ | area under curve | $0$ | $= W$ |
| Adiabatic | $Q = 0$ | $-\Delta U$ | $\tfrac{3}{2} n R \Delta T$ | $0$ |
Adiabatic relations (monatomic, $\gamma = 5/3$)
§5 — Heat Engines & Carnot Efficiency Topic B.4
Energy flow
Clockwise $P$–$V$ cycle ⇒ heat engine ($W > 0$ done by gas). Anticlockwise ⇒ refrigerator / heat pump (work input required).
Carnot — maximum possible efficiency
Carnot cycle stages
- Isothermal expansion at $T_H$ — heat $Q_H$ absorbed.
- Adiabatic expansion ($T_H \to T_C$, $Q = 0$).
- Isothermal compression at $T_C$ — heat $Q_C$ expelled.
- Adiabatic compression ($T_C \to T_H$, $Q = 0$).
All four steps are reversible ⇒ $\Delta S_\text{total} = 0$ ⇒ maximum possible efficiency.
§6 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| "Heat added / work done / change in $U$" | First law: $Q = \Delta U + W$ in IB sign convention |
| Constant volume process | $W = 0$, $Q = \Delta U = \tfrac{3}{2} n R \Delta T$ |
| Constant pressure process | $W = P \Delta V$; $Q = \Delta U + P \Delta V$ |
| Constant temperature process | $\Delta U = 0$, $Q = W$ (use $W = nRT \ln(V_2/V_1)$ if asked HL extension) |
| "No heat transfer" | Adiabatic: $Q = 0$, $W = -\Delta U$, $P V^{5/3} = $ const |
| Cyclic $P$–$V$ diagram | $\Delta U = 0$ over the loop; $W_\text{net}$ = enclosed area |
| "Efficiency of a heat engine" | $\eta = W/Q_H = 1 - Q_C/Q_H$ |
| "Maximum possible efficiency" | Carnot: $\eta = 1 - T_C/T_H$ (kelvin!) |
| "Entropy change…" | $\Delta S = Q/T$ (reversible) or $\Delta S = k_B \ln(\Omega_2/\Omega_1)$ |
| "State the second law" | $\Delta S_\text{total} \ge 0$ for any isolated system |
Worked Example — IB-Style Carnot Engine
Question (HL Paper 2 style — 8 marks)
An ideal Carnot engine operates between a hot reservoir at $T_H = 500\,\text{K}$ and a cold reservoir at $T_C = 300\,\text{K}$. In each cycle the engine absorbs $Q_H = 800\,\text{J}$ from the hot reservoir. (a) Compute the maximum efficiency. (b) Find the work done per cycle. (c) Find the heat dumped to the cold reservoir. (d) The cold reservoir is replaced with one at $T_C = 200\,\text{K}$ — find the new efficiency.
Solution
- Apply the Carnot efficiency formula with absolute temperatures: $\eta = 1 - T_C/T_H = 1 - 300/500$. (M1)
- Evaluate: $\eta = 1 - 0.60 = 0.40$ (40%). (A1)
- Work per cycle: $W = \eta Q_H = 0.40 \times 800 = 320\,\text{J}$. (M1)(A1)
- Heat dumped: from $Q_H = W + Q_C$, $Q_C = Q_H - W = 800 - 320 = 480\,\text{J}$. (M1)(A1)
- New cold reservoir at $200\,\text{K}$: $\eta' = 1 - 200/500 = 0.60$ (60%). (R1)(A1)
Examiner's note: Watch the kelvin/Celsius trap — every IB Carnot question has at least one student who substitutes $T_C = 27$ instead of $T_C = 300$ and gets an efficiency above 0.94. Always convert first. Note also that lowering $T_C$ raises efficiency more dramatically than raising $T_H$ by the same amount — use this when comparing engine designs.
Common Student Questions
What is the sign convention for $W$ in the first law?
Why must Carnot efficiency use kelvin?
Can entropy ever decrease?
What's the difference between isothermal and adiabatic?
How do I find net work from a $P$–$V$ cycle?
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