# Thermodynamics

> Physics Higher Level · IB DP Physics 2025+
> Source: https://www.owlsprep.com/study/ib-physics-hl-u2-thermodynamics/

This module covers core thermodynamic laws, key gas processes, PV diagram analysis, and second law constraints for closed systems, aligned to IB Physics HL assessment criteria.

**Prerequisites:** [Ideal gas law and kinetic model assumptions](https://www.owlsprep.com/study/ib-physics-hl-u2-ideal-gases/); [Energy transfer via work and heat](https://www.owlsprep.com/study/ib-physics-sl-u2-energy-transfer/)

## Learning objectives

- Distinguish between macroscopic thermodynamic state variables and process variables
- Apply the first law of thermodynamics to isochoric, isobaric, isothermal, and adiabatic processes
- Calculate work done from pressure-volume graphs for closed ideal gas systems
- Evaluate second of thermodynamics constraints on maximum heat engine efficiency

## State vs Process Variables

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.

**Internal Energy** — 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.

*Notation:* U

**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.

**Check your understanding**

Test your understanding of variable types:

1. Which of the following is a process variable?

   - Temperature
   - Heat transferred
   - Pressure
   - Entropy

   *Why:* Heat transfer depends entirely on the path taken between two states, so it is a process variable.

## First Law of Thermodynamics

> **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.

## Pressure-Volume Graph Analysis

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.

> **tip**
>
> For non-rectangular cycles, split the enclosed area into simple geometric shapes (triangles, rectangles) to calculate total area, no integration required for IB HL questions.

## Second Law and Entropy

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.

> **mnemonic**
>
> Remember the four process rules: Isochoric = Volume constant, Isobaric = Pressure constant, Isothermal = Temperature constant, Adiabatic = No heat transfer, using the first letter of each keyword to match the constant variable.

## Common pitfalls

- **Wrong:** Mixing up sign conventions for work done on vs by the system
  - Why it fails: Many introductory courses use W as work done on the system, which flips the first law equation
  - Correct: Always use IB's standard ΔU = Q - W, where W is work done BY the gas on surroundings
- **Wrong:** Claiming ΔU = 0 for isothermal processes for all gases
  - Why it fails: This rule only applies to ideal gases, where internal energy depends solely on temperature
  - Correct: Explicitly state the ideal gas assumption before setting ΔU = 0 for isothermal processes
- **Wrong:** Calculating work for non-isobaric processes as PΔV
  - Why it fails: PΔV only holds when pressure is constant across the entire process
  - Correct: For variable pressure, calculate the area under the PV curve to find total work
- **Wrong:** Assigning non-zero heat transfer to adiabatic processes
  - Why it fails: Students often forget adiabatic processes are defined by zero net heat exchange
  - Correct: Set Q = 0 for all adiabatic processes, and use ΔU = -W to relate variables
- **Wrong:** Suggesting a heat engine can reach 100% efficiency
  - Why it fails: The second law forbids this unless the cold reservoir is at absolute 0 K, which is physically unachievable
  - Correct: Always reference the Carnot maximum efficiency limit for any real heat engine analysis

## 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 |

## 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.

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