Thermodynamics (HL) — IB Physics HL HL Study Guide

For: IB Physics HL candidates sitting IB Physics HL.

Covers: first law of thermodynamics, entropy and the second law of thermodynamics, heat engine efficiency, and PV diagrams and thermodynamic cycles, with exam-focused explanations and worked examples.

You should already know: IGCSE Physics, basic algebra and calculus.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Thermodynamics (HL)?

Thermodynamics is the branch of physics that quantifies the relationships between heat, work, temperature, and energy transfer in macroscopic systems, with HL content extending beyond SL to include quantitative entropy calculations, cyclic process analysis, and heat engine efficiency limits. It is assessed across IB Physics HL Paper 1 (multiple choice), Paper 2 (structured response), and is relevant to Topic 8 (Energy Production) for real-world application. Standard notation used across the topic includes $U$ for internal energy, $Q$ for heat transfer, $W$ for work done, $S$ for entropy, and $p,V,T$ for pressure, volume and absolute temperature respectively.

2. First law of thermodynamics

The first law of thermodynamics is a restatement of the principle of energy conservation, adapted for thermal systems where energy is transferred as both heat and work. First, define all core terms to avoid sign convention errors (a top exam mark loser):

• Internal energy ($U$): The sum of the random kinetic and intermolecular potential energies of all particles in a system, measured in joules (J). For an ideal gas, internal energy depends only on temperature, as intermolecular potential energy is zero.
• Heat transfer ($Q$): Energy transferred between the system and surroundings due to a temperature difference. $Q$ is positive if heat flows into the system, negative if heat flows out of the system.
• Work done ($W$): Energy transferred via a force acting over a distance. The IB uses the convention where $W$ is positive if work is done by the system on the surroundings, and negative if work is done on the system by the surroundings.

The formal statement of the first law is: $$\Delta U=Q-W$$

Key Special Cases (Exam-Frequent)

1. Isochoric (constant volume): No expansion or compression, so $W=0$, $\Delta U=Q$
2. Isobaric (constant pressure): Work done is $W=p\Delta V$, so $\Delta U=Q-p\Delta V$
3. Isothermal (constant temperature): For ideal gas $\Delta U=0$, so $Q=W$
4. Adiabatic (no heat transfer): $Q=0$, so $\Delta U=-W$

Worked Example

A fixed mass of ideal gas in a piston absorbs 320 J of heat from the surroundings, and expands to do 180 J of work pushing the piston out. Calculate the change in internal energy of the gas.

1. Identify values: $Q=+320$ J (heat enters system), $W=+180$ J (work done by system)
2. Substitute into first law: $\Delta U=320-180=140$ J The internal energy of the gas increases by 140 J, consistent with the gas retaining more heat energy than it uses to do work.

3. Entropy and second law

The first law confirms energy is conserved in all processes, but it does not explain why some processes occur spontaneously while others do not. This is the role of the second law of thermodynamics, which relies on the state function entropy.

• Entropy ($S$): A measure of the disorder or randomness of a system, measured in J K⁻¹. Higher entropy corresponds to more dispersed energy and more microstates available to the system's particles. For a reversible process occurring at constant temperature, the change in entropy is given by: $$\Delta S=\frac{Q_{\text{rev}}}{T}$$ where $T$ is absolute temperature in Kelvin, and $Q_{\text{rev}}$ is the heat transferred reversibly to the system.

Second Law Statements (Both Exam-Required)

1. Clausius form: Heat cannot spontaneously transfer from a colder body to a hotter body without external work being input.
2. Entropy form: The total entropy of an isolated system (the system plus its surroundings) can never decrease over time: $$\Delta S_{\text{total}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}\ge 0$$ The equals sign applies only to ideal reversible processes; all real spontaneous processes have $\Delta S_{\text{total}}>0$.

Worked Example

A 100 g sample of ice at 0°C (273 K) melts completely to water at 0°C, absorbing 33400 J of heat from a room held at constant 27°C (300 K). Calculate the total entropy change of the universe for this process.

1. Calculate system entropy change: $\Delta S_{\text{system}}=\frac{33400}{273}\approx 122.3$ J K⁻¹
2. Calculate surroundings entropy change: The room loses 33400 J, so $\Delta S_{\text{surroundings}}=\frac{-33400}{300}\approx -111.3$ J K⁻¹
3. Total entropy change: $\Delta S_{\text{total}}=122.3-111.3=11.0$ J K⁻¹ > 0, consistent with the second law, as melting ice is spontaneous at room temperature.

4. Heat engines and efficiency

A heat engine is a device that converts thermal energy into useful mechanical work by operating between a high-temperature hot reservoir ($T_h$) and low-temperature cold reservoir ($T_c$). All heat engines follow the same core energy flow:

1. Heat $Q_h$ is absorbed from the hot reservoir
2. A portion of the energy is converted to useful work $W$
3. Waste heat $Q_c$ is rejected to the cold reservoir

For a full cycle of the engine, the system returns to its initial state, so $\Delta U=0$ by the first law, giving $W=Q_h-Q_c$.

Thermal Efficiency

The thermal efficiency $\eta$ of a heat engine is the ratio of useful work output to total heat input, expressed as a decimal or percentage: $$\eta =\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}$$ Per the second law, efficiency can never equal 1 (100%), as some waste heat must always be rejected to the cold reservoir.

Carnot Efficiency (Maximum Possible Efficiency)

The Carnot cycle is an ideal, fully reversible cycle that gives the upper limit of efficiency for any heat engine operating between $T_h$ and $T_c$: $${\eta }_{\text{Carnot}}=1-\frac{T_c}{T_h}$$ Note that $T_c$ and $T_h$ must be in Kelvin for this calculation; using Celsius will give impossible efficiency values over 100%.

Worked Example

A coal-fired power plant operates as a heat engine between a boiler temperature of 550°C (823 K) and a cooling water temperature of 15°C (288 K). It absorbs 2.4 x 10⁹ J of heat from the boiler per hour. Calculate (a) its maximum possible efficiency, (b) the maximum useful work output per hour.

1. Maximum efficiency: ${\eta }_{\text{Carnot}}=1-\frac{288}{823}\approx 1-0.35=0.65$, or 65%
2. Maximum work output: $W=\eta Q_h=0.65\times 2.4\times 10^9=1.56\times 10^9$ J per hour Real power plants operate at ~30-40% efficiency due to friction, heat loss, and irreversible processes.

5. PV diagrams and cycles

A Pressure-Volume (PV) diagram plots the pressure of a system against its volume, and is a core tool for analysing thermodynamic processes and cycles. Two critical rules apply to all PV diagrams:

1. The area under the curve of a single process equals the work done by the system during that process.
2. For a full thermodynamic cycle (a sequence of processes that returns the system to its initial state), the area enclosed by the cycle equals the net work done by the system over the cycle.

Common Process Shapes on PV Diagrams

Clockwise cycles correspond to heat engines (net work done by the system), while counterclockwise cycles correspond to refrigerators or heat pumps (net work done on the system).

Worked Example

A rectangular cycle on a PV diagram has vertices at (0.001 m³, 1x10⁵ Pa), (0.003 m³, 1x10⁵ Pa), (0.003 m³, 3x10⁵ Pa), (0.001 m³, 3x10⁵ Pa), traversed clockwise. Calculate the net work done per cycle.

1. The area enclosed by the rectangle is width × height: $\Delta V=0.003-0.001=0.002$ m³, $\Delta p=3x10^5-1x10^5=2x10^5$ Pa
2. Net work = $0.002\times 2x10^5=400$ J, which matches the sum of work done across all four processes in the cycle.

6. Common Pitfalls (and how to avoid them)

• Wrong first law sign convention: You mix up $W$ as work done on vs by the system, using non-IB textbook conventions. Why? Multiple conventions exist across learning resources. Correct move: Explicitly write the IB sign convention ($\Delta U=Q-W$, $Q+$ = heat in, $W+$ = work out) next to your working in exams to avoid mark deductions, even if you think you applied it correctly.
• Using Celsius in entropy/Carnot calculations: You plug in Celsius temperature values instead of Kelvin, leading to efficiency values over 100% or negative entropy changes that violate the second law. Why? You skip unit conversion in a rush. Correct move: Circle every temperature value in a question and convert to Kelvin before substituting into any formula.
• Only calculating system entropy for second law questions: You conclude a process is non-spontaneous because the system entropy decreases, ignoring the surroundings. Why? You forget the second law applies to the total isolated system. Correct move: Always calculate $\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}$ to test spontaneity; a system can decrease in entropy if the surroundings' entropy increases more.
• Miscounting work from PV cycles: You only calculate the area under the expansion curve, forgetting to subtract work done on the system during compression. Why? You don't associate negative work with compression. Correct move: For any cycle, net work is exactly the area enclosed by the cycle; clockwise cycles give positive work, counterclockwise give negative.
• Confusing actual and Carnot efficiency: You use the Carnot temperature formula for real engine questions that give $Q_h$ and $Q_c$ values. Why? You assume all engines are ideal Carnot engines. Correct move: Use $\eta =1-Q_c/Q_h$ for actual efficiency, only use the Carnot formula when asked for maximum possible efficiency or for a fully reversible Carnot engine.

7. Practice Questions (IB Physics HL Style)

Question 1 (5 marks)

A fixed mass of ideal gas is compressed adiabatically, with 420 J of work done on the gas during the process. (a) State the value of $Q$ for this process. (1 mark) (b) Calculate the change in internal energy of the gas, including sign. (2 marks) (c) Explain why the temperature of the gas increases during this compression. (2 marks)

Solution

(a) Adiabatic processes involve no heat transfer, so $Q=0$ J. (1 mark) (b) Use the first law: $\Delta U=Q-W$. Work is done on the system, so $W=-420$ J. Substitute: $\Delta U=0-(-420)=+420$ J. (2 marks for correct substitution and sign) (c) For an ideal gas, internal energy is directly proportional to absolute temperature. The positive change in internal energy means the total kinetic energy of the gas particles increases, which corresponds to a rise in temperature. (2 marks for linking internal energy to temperature)

Question 2 (5 marks)

A heat engine operates between a hot reservoir at 700 K and a cold reservoir at 350 K. Per cycle, it absorbs 900 J of heat from the hot reservoir and rejects 540 J of heat to the cold reservoir. (a) Calculate the actual thermal efficiency of the engine, as a percentage. (2 marks) (b) Calculate the maximum possible efficiency for an engine operating between these two temperatures. (2 marks) (c) State one reason why the actual efficiency is lower than the maximum. (1 mark)

Solution

(a) $\eta =1-\frac{Q_c}{Q_h}=1-\frac{540}{900}=0.4$, so 40%. (2 marks for correct calculation and percentage conversion) (b) ${\eta }_{\text{Carnot}}=1-\frac{350}{700}=0.5$, so 50%. (2 marks for correct use of Kelvin temperature and calculation) (c) Any valid answer: friction between moving parts, heat loss to the surroundings, irreversible mixing of gas particles during the cycle. (1 mark)

Question 3 (3 marks)

A thermodynamic cycle for an ideal gas is plotted on a PV diagram, with the cycle traversed counterclockwise. (a) State whether the net work is done on or by the system. (1 mark) (b) State the change in entropy of the system over one full cycle, justifying your answer. (2 marks)

Solution

(a) Counterclockwise cycles have negative net work, so work is done on the system. (1 mark) (b) $\Delta S_{\text{system}}=0$ J K⁻¹. Entropy is a state function, and the system returns to its exact initial state after one full cycle, so there is no change in entropy. (2 marks for correct value and state function justification)

8. Quick Reference Cheatsheet

9. What's Next

The thermodynamics concepts you mastered in this guide are foundational to multiple later IB Physics HL topics, including Topic 3 (Ideal Gases, where you will apply first law calculations to gas expansions and compressions) and Topic 8 (Energy Production, where you will use heat engine efficiency to evaluate the performance of fossil fuel, nuclear, and solar thermal power plants). If you select the Thermal Physics option for Paper 3, you will extend these concepts to heat transfer mechanisms, refrigeration cycles, and statistical entropy calculations. Thermodynamics is also a core prerequisite for university-level STEM courses including mechanical engineering, chemistry, and environmental science.

As you practise official past paper questions on thermodynamics, you may run into confusing sign convention scenarios, complex PV cycle analysis, or multi-part entropy problems that trip you up. You can ask Ollie, our AI tutor, for step-by-step walkthroughs of any problem, custom practice questions tailored to your specific weak spots, or a 10-minute revision quiz to test your knowledge right before your mock or final exams.