# First Law of Thermodynamics

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u20-first-law-of-thermodynamics/

This module covers the first law of thermodynamics, CIE's standard sign convention, and applications to common thermal processes and p-V diagrams. You will learn to calculate energy changes and avoid common sign errors that cost exam marks.

**Prerequisites:** [Internal energy of ideal gases](https://www.owlsprep.com/study/cie-9702-u20-internal-energy/); [Work done by expanding gases](https://www.owlsprep.com/study/cie-9702-u20-work-done-gas/)

## Learning objectives

- State the first law of thermodynamics and interpret each term per CIE convention
- Apply the first law to common thermal processes including adiabatic, isothermal and constant volume processes
- Calculate changes in internal energy from heat transfer and work done
- Interpret the first law for processes shown on pressure-volume (p-V) diagrams

## Core Definition and CIE Sign Convention

**First Law of Thermodynamics** — A restatement of the principle of conservation of energy for thermal processes, relating the change in a system's internal energy to heat added to the system and work done by the system.

*Notation:* \Delta U = Q - W

*Example:* Positive values indicate energy added to the system (Q) or work done by the system (W) per CIE 9702 convention.

> **warning**
>
> Different textbooks use different conventions. CIE 9702 always uses the convention where W = work done *by* the system, not work done *on* the system. Using the wrong convention will always give the wrong sign for ΔU.

**Worked example:** A gas absorbs 250 J of heat from its surroundings and does 130 J of work on the surroundings. Calculate the change in internal energy of the gas.

1. Identify values with CIE convention:
2. Heat enters the system, so $Q = +250$ J. Work is done by the system, so $W = +130$ J.
3. Substitute into the first law equation:
4. $$\Delta U = Q - W = 250 - 130$$
5. Calculate the result: $\Delta U = +120$ J

*Calculator:* forbidden

## Application to Common Thermal Processes

The first law simplifies for different standard processes that are commonly tested in CIE exams. We can fix one variable to get a simpler relationship between the remaining terms.

**Adiabatic Process** — A thermal process where no heat is exchanged between the system and surroundings. This occurs for rapid expansions/compressions where there is no time for heat transfer.

*Example:* Adiabatic processes always have $Q = 0$ by definition.

| Process Name | Key Condition | Simplified First Law |
| --- | --- | --- |
| Constant Volume (Isochoric) | $W = 0$ (no work done) | $\Delta U = Q$ |
| Adiabatic | $Q = 0$ (no heat transfer) | $\Delta U = -W$ |
| Isothermal (ideal gas) | $\Delta U = 0$ (constant temperature) | $Q = W$ |
| Free Expansion | $Q = 0, W = 0$ | $\Delta U = 0$ |

**Worked example:** An ideal gas undergoes adiabatic compression. 400 J of work is done *on* the gas by the surroundings. Calculate the change in internal energy.

1. Adiabatic process means no heat transfer: $Q = 0$
2. Work is done on the gas, so work done *by* the gas is negative: $W = -400$ J
3. Substitute into first law:
4. $$\Delta U = Q - W = 0 - (-400) = +400$$
5. Conclusion: Internal energy of the gas increases by 400 J, so its temperature rises.

> **Exam tip:** Always check if the question tells you work is done on or by the gas, and adjust the sign of W accordingly.

*Calculator:* forbidden

## First Law on p-V Diagrams

CIE frequently asks questions about processes plotted on pressure-volume (p-V) diagrams. The magnitude of work done by the gas equals the area under the process curve on a p-V diagram.

If the process moves from left to right (volume increases), the gas expands, so W is positive. If it moves right to left (volume decreases), the gas is compressed, so W is negative.

> **tip**
>
> For any full cyclic process that starts and ends at the same point on the p-V diagram, the total change in internal energy $\Delta U = 0$, so total heat added equals total work done: $Q = W$.

**Worked example:** A gas expands at constant pressure of 2 Pa from $V = 1 \text{ m}^3$ to $V = 4 \text{ m}^3$. 10 J of heat is added to the gas during the expansion. Calculate the change in internal energy.

1. Calculate work done by the gas for constant pressure: $W = p\Delta V$
2. $$W = 2 \times (4 - 1) = 6 \text{ J}$$
3. We know $Q = +10$ J (heat added)
4. Apply first law:
5. $$\Delta U = Q - W = 10 - 6 = +4 \text{ J}$$
6. Final answer: internal energy increases by 4 J

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the 'work done on the system' convention instead of CIE's convention.
  - Why it fails: Different sources use different conventions, but CIE 9700 strictly uses work done by the system.
  - Correct: Always use $\Delta U = Q - W$ where W is positive when work is done by the system.
- **Wrong:** Assuming $\Delta U = 0$ for any expansion process.
  - Why it fails: Internal energy only stays constant for isothermal expansion of an ideal gas, not adiabatic expansion.
  - Correct: Only set $\Delta U = 0$ if the process is explicitly stated to be isothermal.
- **Wrong:** Using $W = p\Delta V$ for non-constant pressure processes.
  - Why it fails: This formula only holds when pressure is constant. For changing pressure, work equals area under the p-V curve.
  - Correct: Calculate the area under the process curve to find W before applying the first law.
- **Wrong:** Taking Q as positive when heat leaves the system.
  - Why it fails: Sign conventions for Q are often mixed up with W.
  - Correct: Q is positive when heat enters the system, negative when heat leaves the system.

## Cheatsheet

| Concept | CIE Convention | Equation |
| --- | --- | --- |
| First Law | ΔU = change in internal energy, Q = heat added, W = work done BY system | $\Delta U = Q - W$ |
| Constant Volume | No work done | $\Delta U = Q$ |
| Adiabatic | No heat exchange | $Q = 0 \rightarrow \Delta U = -W$ |
| Isothermal (ideal gas) | Constant internal energy | $\Delta U = 0 \rightarrow Q = W$ |
| p-V Work | Expansion (left→right) = W positive | Area under curve = \|W\| |
| Full Cycle | Returns to start state | $\Delta U = 0 \rightarrow Q = W$ |

## What's next

Mastering the first law of thermodynamics and its sign convention is critical for all higher thermal physics topics in CIE A-Level Physics. The principles you have learned here are applied to analyze engine efficiency, adiabatic and isothermal processes, and entropy changes. Consistent practice with sign conventions will help you avoid losing easy marks in both multiple choice and structured questions. Next, you will explore specific types of thermal processes in more detail before moving on to the second law of thermodynamics.

- [Electric fields (A2)](https://www.owlsprep.com/study/cie-9702-u21-overview/)
- [Electric field concepts](https://www.owlsprep.com/study/cie-9702-u21-electric-field-concepts/)

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