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IB Diploma Physics HL · Theme B.4 Thermodynamics (HL)
Mini-Lesson

Thermodynamics

This HL-only mini-lesson covers Theme B.4 — Thermodynamics: internal energy, the first law Q = ΔU + W, work done by a gas, the four processes, entropy and the second law, and heat-engine efficiency.

Q = ΔU + W processes entropy & efficiency

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Watch for the HL flag on higher-level extensions. Press Start when you're ready.

B.4 · first law

Internal energy & the first law

The internal energy U of an ideal gas depends only on its temperature (U = (3/2)nRT for a monatomic gas). The first law is conservation of energy for a gas:

Q = ΔU + WQ = heat added TO the gas; W = work done BY the gas; ΔU = change in internal energy

Sign care: in this IB convention, W is the work the gas does on its surroundings. If the gas is compressed, W is negative (work is done on it).

Quick check

Quick check

?Heat Q = 500 J is added to a gas, which does W = 200 J of work pushing back a piston. What is the change in internal energy?
Calculate

Calculate

#Heat 500 J is supplied to a gas that does 200 J of work. Find the change in internal energy ΔU.
J
Hint: first law — ΔU = Q − W = 500 − 200.
B.4 · work

Work done by a gas

When a gas expands at constant pressure it does work on its surroundings equal to the area under the p–V graph:

W = pΔVconstant-pressure (isobaric) work, in joules
Worked example — an expanding gas

A gas at 2.0 × 10⁵ Pa expands by ΔV = 0.0030 m³ at constant pressure.

W = pΔV = 2.0 × 10⁵ × 0.0030 = 600 J

Calculate

Calculate

#A gas at constant pressure 2.0 × 10⁵ Pa expands by 0.0030 m³. Find the work it does.
J
Hint: W = pΔV = 2.0e5 × 0.0030.
Sort it

Which thermodynamic process?

Tap an item, then tap the group it belongs to.

📦 Isovolumetric

🎚️ Isobaric

🌡️ Isothermal

B.4 · second law

Entropy & the second law

Entropy S measures the disorder (number of accessible microstates) of a system. The second law states that the entropy of an isolated system never decreases — it stays the same for a reversible process and increases for any real (irreversible) one.

This gives time its direction: heat flows spontaneously from hot to cold, never the reverse, because that increases total entropy. No heat engine can be 100% efficient.

Quick check

Quick check

?Why can no real heat engine convert all its input heat into useful work?
B.4 · efficiency

Heat-engine efficiency

A heat engine's efficiency is the useful work out per unit heat in. The maximum possible (Carnot) efficiency depends only on the reservoir temperatures (in kelvin):

η = W ÷ Q_H · η_carnot = 1 − T_c ÷ T_hthe Carnot value is an unreachable ideal ceiling
Worked example — Carnot limit

An engine works between T_h = 500 K and T_c = 300 K.

η = 1 − 300 ÷ 500 = 0.40 = 40%

Calculate

Calculate

#Find the maximum (Carnot) efficiency of an engine operating between 500 K and 300 K. Give a percentage.
%
Hint: η = (1 − T_c ÷ T_h) × 100 = (1 − 300 ÷ 500) × 100.
Calculate

Calculate

#A heat engine takes in 1000 J of heat and produces 300 J of work. Find its efficiency as a percentage.
%
Hint: η = W ÷ Q_H × 100 = 300 ÷ 1000 × 100.
Match it

Match statement to law/quantity

Tap a statement on the left, then its match on the right.

Statement
Answer
Recap

The big ideas to know

First law: Q = ΔU + W (heat in = internal-energy rise + work done by gas)

Work: W = pΔV (area under a p–V graph)

Processes: isovolumetric (W=0) · isobaric · isothermal (ΔU=0) · adiabatic (Q=0)

Second law: entropy of an isolated system never decreases

Efficiency: η = W/Q_H; Carnot ceiling η = 1 − T_c/T_h

That completes Thermodynamics for IB Diploma Physics HL. Press Finish to see your score.

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