B.4 ThermodynamicsIB Physics HL: Revision notes

Section 1

The first law of thermodynamics

Applying conservation of energy to a closed system gives Q = ΔU + W, where Q is the energy transferred to the system as heat, ΔU is the change in internal energy, and W is the work done by the system.

When a gas changes volume at pressure P, W = PΔV (positive for expansion, negative for compression).

For a monatomic ideal gas, internal energy is all random kinetic energy, so ΔU = (3/2)Nk_BΔT = (3/2)nRΔT: ΔU depends only on the temperature change.

Key termsfirst law of thermodynamicsinternal energywork done by a gas

Common mistake

Sign conventions: Q is heat IN, W is work done BY the gas. Work done on the gas during compression makes W negative.

Section 2

The four special processes

Each keeps one quantity fixed:

Isovolumetric (constant V): W = 0, so Q = ΔU.
Isobaric (constant P): W = PΔV.
Isothermal (constant T): ΔU = 0, so Q = W; PV = constant. Must be slow.
Adiabatic (Q = 0): ΔU = −W; for a monatomic ideal gas PV^(5/3) = constant. Must be rapid or well insulated. Adiabatic compression heats the gas; expansion cools it.

Key termsisovolumetricisobaricisothermaladiabatic

Section 3

Entropy

Entropy S is a measure of the disorder of the particles in a system.

Macroscopic: ΔS = ΔQ/T, the energy transferred as heat divided by the kelvin temperature at which it is transferred.
Microscopic: S = k_B ln Ω, where Ω is the number of possible microstates (arrangements of the particles consistent with the same macroscopic state).

Melting, boiling and mixing all increase Ω and so increase S.

Key termsentropymicrostatesBoltzmann constant

Section 4

The second law of thermodynamics

The entropy of an isolated system never decreases. Real processes are irreversible, so the entropy of a real isolated system always increases. The entropy of a non-isolated system can decrease (a fridge cooling food, water freezing), but only if the surroundings gain at least as much entropy.

Consequences: heat flows spontaneously only from hot to cold, and no engine can convert heat completely into work.

Key termssecond law of thermodynamicsirreversible processisolated system

Exam tip

To test an impossible claim, add up ΔS = Q/T for every reservoir over one cycle. A negative total breaks the second law.

Section 5

Heat engines and efficiency

A heat engine takes a working gas round a cycle of processes, absorbing heat Q_H from a hot reservoir and releasing Q_C to a cold reservoir. Over a cycle ΔU = 0, so the useful work is W = Q_H − Q_C.

Efficiency η = useful work / input energy = W/Q_H.

Key termsheat engineefficiencycycle

Section 6

The Carnot cycle

The Carnot cycle (isothermal expansion at T_H, adiabatic expansion, isothermal compression at T_C, adiabatic compression) is the most efficient possible cycle between two reservoirs:

η_Carnot = 1 − T_C/T_H (temperatures in kelvin).

No real engine can exceed this; real engines fall short because of friction, heat leaks and other irreversible processes. A higher T_H or lower T_C raises the limit.

Key termsCarnot cycleCarnot efficiency

Common mistake

Always use kelvin in the Carnot efficiency: 1 − 27/627 is wrong; use 1 − 300/900.

That's the notes covered.

Carry on to the next subtopic.