B.4 Thermodynamics

UNIT SYLLABUS

HL 8 hours · HL only

Thermodynamics is the physics of what is possible. The first law ($Q = \Delta U + W$) is energy conservation for gases: heat in becomes internal energy plus work out. The second law is deeper — it introduces entropy, the measure of disorder, and declares that in an isolated system it never decreases. That single statement explains why heat flows hot-to-cold, why no engine can be perfectly efficient, and why time has a direction. You will trace the four named processes (isothermal, isobaric, isovolumetric, adiabatic) on p–V diagrams, assemble them into engine cycles, and prove with Carnot that even a perfect engine pays an entropy tax set by its reservoir temperatures.

Guiding Questions

? How do the laws of thermodynamics constrain what physical processes are possible?
? Why can thermal energy never be converted entirely into useful work in a cyclic process?

✓ What the IB expects you to master

Apply the first law of thermodynamics $Q = \Delta U + W$ as energy conservation for a closed system.
Calculate the work done when a gas changes volume: $W = P\Delta V$ (area under the p–V curve).
Relate internal energy change to temperature change: $\Delta U = \frac{3}{2}nR\Delta T$ for a monatomic ideal gas.
Interpret entropy both macroscopically ( $\Delta S = \Delta Q / T$ ) and microscopically ( $S = k_B\ln\Omega$ , counting microstates).
State the second law: the entropy of an isolated system never decreases; real processes are irreversible.
Explain how local entropy decreases are paid for by greater increases elsewhere.
Analyse isothermal ( $\Delta U = 0$ ), isobaric, isovolumetric ( $W = 0$ ) and adiabatic ( $Q = 0$ ) processes; use $PV^{5/3} = \text{constant}$ for adiabatic changes of a monatomic gas.
Analyse cyclic processes as heat engines with efficiency $\eta = \frac{\text{useful work}}{\text{input energy}}$ , bounded by Carnot: $\eta_{max} = 1 - \frac{T_c}{T_h}$ .

1 Key Formulas

First law of thermodynamics

Q = \Delta U + W

Work done by a gas

W = P\Delta V

Internal energy change

\Delta U = \tfrac{3}{2}nR\Delta T

Entropy change

\Delta S = \frac{\Delta Q}{T}

Entropy (statistical)

S = k_{B}\ln\Omega

Adiabatic (monatomic)

PV^{\frac{5}{3}} = \text{constant}

Engine efficiency

\eta = \frac{\text{useful work}}{\text{input energy}}

Carnot efficiency

\eta_{\text{max}} = 1 - \frac{T_{c}}{T_{h}}

2 Exam Preparation & Topic Explanations

Reading p–V diagrams like an examiner

Every process is a path on the p–V plane, and three facts decode any of them: work is the area under the path (positive for expansion); internal energy depends only on temperature, i.e. on $PV$ ; and the first law connects them. For cycles, net work is the enclosed area, clockwise = engine.

Tabulate $Q$ , $\Delta U$ and $W$ for each leg of a cycle — examiners award marks per cell, and the table catches sign errors before they propagate.

Pro Exam Strategy

Sign convention: in the IB, $W$ is work done BY the gas. Compression → $W < 0$ .
Adiabatic curves are steeper than isotherms through the same point — a standard "identify the process" question.
Isothermal: $\Delta U = 0$ . Isovolumetric: $W = 0$ . Adiabatic: $Q = 0$ . Memorise the three zeros.
Carnot efficiency needs kelvin temperatures — and real engines always achieve less.

3 MCQ Practice

Q1. An ideal gas expands isothermally, absorbing 400 J of heat. The work done by the gas is:

0 J
267 J
400 J
600 J

Q2. During an adiabatic compression of an ideal gas, its temperature:

Rises, because work is done on the gas and no heat escapes
Falls, because the gas does work
Stays constant, because no heat is exchanged
Falls, because the pressure increases

Q3. A heat engine operates between 500 K and 300 K. Its maximum possible efficiency is:

30%
40%
60%
100%

4 Short Answer Questions

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