Thermodynamics

Physics Higher Level· B.5· 25 min read

1. State vs Process Variables★★★☆☆⏱ 6 min

Thermodynamic state variables describe the current condition of a system, and their values depend only on the current state, not the path taken to reach it. Common state variables include pressure, volume, temperature, internal energy, and entropy. Process variables describe energy transfers that occur during a change between two states, and their values depend entirely on the path taken between the initial and final states.

📘 Definition

Internal Energy

U

The total sum of all microscopic kinetic and potential energies of all particles in the system. For ideal gases, there are no inter-particle potential energies, so U depends only on temperature.

📐 Worked Example

A fixed mass of ideal gas changes from state A (P=100 kPa, V=2 m³) to state B (P=200 kPa, V=1 m³). Compare the change in internal energy for a path that goes via an isobaric step then isochoric step, versus a path that goes via an isochoric step then isobaric step.

1
First, note that the initial and final temperature for both paths are identical, calculated via the ideal gas law PV = nRT.
2
Since internal energy U for an ideal gas depends only on temperature, ΔU is identical for both paths.
3
The total work done and total heat transferred will be different for the two paths, as these are process variables.

✓ Quick check

Test your understanding of variable types:

Which of the following is a process variable?
- Temperature
- Heat transferred
- Pressure
- Entropy
Reveal answer
Heat transferred —
Heat transfer depends entirely on the path taken between two states, so it is a process variable.

2. First Law of Thermodynamics★★★★☆⏱ 8 min

Exam tip:

IB strictly uses the convention where W is work done BY the system, not on the system. Never use the alternate university convention ΔU = Q + W unless explicitly stated.

3. Pressure-Volume Graph Analysis★★★★☆⏱ 6 min

The work done by a gas during expansion is equal to the area under the pressure-volume curve for that process. For a cyclic process that returns to the initial state, the net work done by the system is equal to the area enclosed by the cycle path. Clockwise cycles produce net work output, while counter-clockwise cycles require net work input.

📐 Worked Example

Calculate the net work done by 1 mole of ideal gas following a rectangular cycle defined by points (V=1 m³, P=100 kPa), (V=3 m³, P=100 kPa), (V=3 m³, P=300 kPa), (V=1 m³, P=300 kPa), returning to the start.

1
The enclosed shape is a rectangle, so area = ΔP × ΔV.
2
$\Delta P = 300 \times 10^3 - 100 \times 10^3 = 200 \times 10^3 \text{ Pa}$
3
$\Delta V = 3 - 1 = 2 \text{ m}^3$
4
Net work = 200,000 Pa × 2 m³ = 400,000 J = 400 kJ.

4. Second Law and Entropy★★★★★⏱ 5 min

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. This sets a hard upper limit on the maximum possible efficiency of any heat engine, known as the Carnot efficiency, which depends only on the absolute temperatures of the hot and cold heat reservoirs.

📐 Worked Example

Calculate the maximum possible efficiency of a heat engine operating between a hot reservoir at 500 K and a cold reservoir at 300 K.

1
$\eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}$
2
Substitute values: η = 1 - 300/500 = 1 - 0.6 = 0.4
3
Maximum efficiency = 40%, no real engine can exceed this value.

5. Common Pitfalls

Wrong move:

Mixing up sign conventions for work done on vs by the system

Why:

Many introductory courses use W as work done on the system, which flips the first law equation

Correct move:

Always use IB's standard ΔU = Q - W, where W is work done BY the gas on surroundings

Wrong move:

Claiming ΔU = 0 for isothermal processes for all gases

Why:

This rule only applies to ideal gases, where internal energy depends solely on temperature

Correct move:

Explicitly state the ideal gas assumption before setting ΔU = 0 for isothermal processes

Wrong move:

Calculating work for non-isobaric processes as PΔV

Why:

PΔV only holds when pressure is constant across the entire process

Correct move:

For variable pressure, calculate the area under the PV curve to find total work

Wrong move:

Assigning non-zero heat transfer to adiabatic processes

Why:

Students often forget adiabatic processes are defined by zero net heat exchange

Correct move:

Set Q = 0 for all adiabatic processes, and use ΔU = -W to relate variables

Wrong move:

Suggesting a heat engine can reach 100% efficiency

Why:

The second law forbids this unless the cold reservoir is at absolute 0 K, which is physically unachievable

Correct move:

Always reference the Carnot maximum efficiency limit for any real heat engine analysis

6. Quick Reference Cheatsheet

Process Name	Constant Variable	Heat Q	Work W	ΔU
Isochoric	Volume	Q = nCvΔT	0	ΔU = Q
Isobaric	Pressure	Q = nCpΔT	W = PΔV	ΔU = Q - PΔV
Isothermal	Temperature	Q = W	W = nRT ln(V₂/V₁)	0
Adiabatic	No heat transfer	0	W = -ΔU	ΔU = nCvΔT

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Paper 2
Heat engine efficiency calculation
2023 · Paper 3
Adiabatic process derivation
2022 · Paper 2
First law applied to isobaric system

What's Next

Mastering thermodynamics is the foundation for solving complex cyclic heat engine problems, which are a frequent 6-8 mark extended response on IB Physics HL Paper 2. You will next apply these core rules to analyse real-world heat pumps, refrigerators, and Carnot cycles, which are heavily weighted in Topic B.5 assessments. You will also connect thermodynamic entropy concepts to the broader study of statistical mechanics, which links macroscopic system properties to microscopic particle behaviour. Ensure you can quickly identify process types from PV graphs before progressing, as this skill is required for all follow-up problem sets.