# Gas laws

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u19-gas-laws/

This module covers the three empirical gas laws relating pressure, volume and absolute temperature for a fixed mass of ideal gas, their combination into the ideal gas equation, and common problem-solving approaches for CIE A-Level exams.

**Prerequisites:** Basic kinetic theory of matter; SI unit conversion for temperature and pressure

## Learning objectives

- State and apply Boyle's, Charles' and the Pressure Law for ideal gases
- Combine individual gas laws to form the general and ideal gas equations
- Convert units correctly for all gas law calculations
- Solve exam problems for changing gas conditions with fixed mass

## Boyle's Law

**Boyle's Law** — For a fixed mass of gas at constant absolute temperature, pressure is inversely proportional to volume: $p \propto \frac{1}{V}$

*Example:* If a sealed gas is halved in volume at constant temperature, its pressure doubles

This law comes from experimental observation, and aligns with kinetic theory: reducing volume increases the number of gas molecules per unit volume, leading to more frequent collisions with container walls, hence higher pressure at the same temperature.

**Worked example:** A sealed syringe contains 50 cm³ of gas at 1.0 × 10⁵ Pa pressure. The plunger compresses the gas to 25 cm³ with no temperature change. Calculate the new pressure.

1. State Boyle's Law for constant temperature:
2. $$p_1 V_1 = p_2 V_2$$
3. List known values: $p_1 = 1.0 \times 10^5$ Pa, $V_1 = 50$ cm³, $V_2 = 25$ cm³
4. Rearrange to solve for $p_2$ and substitute values:
5. $$p_2 = \frac{p_1 V_1}{V_2} = \frac{(1.0 \times 10^5)(50)}{25} = 2.0 \times 10^5 \text{ Pa}$$

> **Exam tip:** Always confirm temperature is constant before applying Boyle's Law; CIE questions often test if you check this core condition.

## Charles' Law and the Pressure Law

**Absolute zero** — The lowest theoretically possible temperature, where an ideal gas has zero volume and zero pressure, equal to 0 K or -273.15 °C. All gas law calculations require temperature in Kelvin, not Celsius.

Charles' Law states that for a fixed mass of gas at constant pressure, volume is directly proportional to absolute temperature: $V \propto T$, giving $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. The Pressure Law (Gay-Lussac's Law) states that for fixed mass at constant volume, pressure is directly proportional to absolute temperature: $p \propto T$, giving $\frac{p_1}{T_1} = \frac{p_2}{T_2}$.

**Worked example:** A sealed rigid container holds gas at 27 °C and 1.2 × 10⁵ Pa. Calculate the pressure when heated to 127 °C, with no volume change.

1. Convert temperatures from Celsius to Kelvin by adding 273:
2. $$T_1 = 27 + 273 = 300 \text{ K}, \quad T_2 = 127 + 273 = 400 \text{ K}$$
3. Volume is constant, so apply the Pressure Law:
4. $$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$
5. Rearrange and substitute to find $p_2$:
6. $$p_2 = \frac{p_1 T_2}{T_1} = \frac{(1.2 \times 10^5)(400)}{300} = 1.6 \times 10^5 \text{ Pa}$$

> **Exam tip:** CIE markers explicitly penalise answers that use Celsius directly in gas law calculations. Always convert to Kelvin first.

## Combining Gas Laws: The Ideal Gas Equation

The three individual gas laws can be combined into a single general equation that works for any change in conditions for a fixed amount of ideal gas. For n moles of gas, this becomes the full ideal gas equation of state.

**Ideal Gas Equation** — The equation of state for an ideal gas, relating all four state variables. For a fixed mass of gas (constant n), this simplifies to the combined gas law: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

*Notation:* pV = nRT

*Example:* 1 mole of ideal gas at STP (273 K, 1.0 × 10⁵ Pa) occupies 22.4 dm³, which satisfies $pV = nRT$

**Check your understanding**

Check your unit conversion before proceeding: What is 0.5 dm³ converted to m³ (to match R's units)?

1. What is 0.5 dm³ in m³?

   - 5 × 10⁻⁴ m³
   - 5 × 10⁻³ m³
   - 0.5 × 10⁻² m³

   *Answer:* 5 × 10⁻⁴ m³

   *Why:* Correct. 1 dm = 0.1 m, so 1 dm³ = (0.1 m)³ = 10⁻³ m³. 0.5 dm³ = 0.5 × 10⁻³ = 5 × 10⁻⁴ m³. R in J mol⁻¹ K⁻¹ always requires p in Pa and V in m³.

**Worked example:** Calculate the volume occupied by 0.25 moles of oxygen at 27 °C and 1.0 × 10⁵ Pa, given $R = 8.31$ J mol⁻¹ K⁻¹.

1. Convert temperature to Kelvin: $T = 27 + 273 = 300$ K
2. Rearrange the ideal gas equation for volume:
3. $$V = \frac{nRT}{p}$$
4. Substitute all values in correct units:
5. $$V = \frac{0.25 \times 8.31 \times 300}{1.0 \times 10^5} = 6.23 \times 10^{-3} \text{ m}^3 = 6.23 \text{ dm}^3$$

## Solving Changing Condition Problems

Most CIE exam questions on gas laws describe a process where a fixed mass of gas changes from one set of conditions to another. For these problems, the combined gas law avoids calculating moles twice, and works for any change in p, V or T.

**Worked example:** A weather balloon contains 2.0 m³ of helium at ground level (pressure 1.0 × 10⁵ Pa, temperature 17 °C). At altitude, pressure is 5.0 × 10⁴ Pa and temperature is -13 °C. Calculate the new volume.

1. Convert all temperatures to Kelvin:
2. $$T_1 = 17 + 273 = 290 \text{ K}, \quad T_2 = -13 + 273 = 260 \text{ K}$$
3. Write the combined gas law and rearrange for $V_2$:
4. $$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} \implies V_2 = \frac{p_1 V_1 T_2}{p_2 T_1}$$
5. Substitute values and calculate:
6. $$V_2 = \frac{(1.0 \times 10^5)(2.0)(260)}{(5.0 \times 10^4)(290)} \approx 3.6 \text{ m}^3$$

**Exam command terms**

CIE uses specific command terms for gas law problems, with different expectations:

- **Show that** — You must show all steps, including the gas law you use and unit conversion *(Work to more significant figures than the final answer to demonstrate the result)*

- **Calculate** — You must show full working and give a numerical answer with correct units

## Common pitfalls

- **Wrong:** Using Celsius instead of Kelvin in gas law calculations
  - Why it fails: All gas law proportionalities rely on absolute temperature, so Celsius gives incorrect results
  - Correct: Always add 273 to Celsius to convert to Kelvin before starting any calculation
- **Wrong:** Using cm³/dm³ directly in $pV = nRT$ without unit conversion
  - Why it fails: R in J mol⁻¹ K⁻¹ requires volume in m³, so unmatched units give wrong orders of magnitude
  - Correct: Convert volume: 1 cm³ = 10⁻⁶ m³, 1 dm³ = 10⁻³ m³, before substituting into the ideal gas equation
- **Wrong:** Applying Boyle's Law when temperature changes
  - Why it fails: Boyle's Law only holds for constant temperature, which is a core requirement
  - Correct: Use the combined gas law $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$ for problems where temperature changes
- **Wrong:** Using the combined gas law for leaking containers with changing mass
  - Why it fails: All gas laws assume fixed mass of gas, so escaping gas changes n
  - Correct: Calculate initial and final n separately using $pV = nRT$ for containers that leak or gain gas
- **Wrong:** Inverting proportionality in Boyle's Law (smaller volume gives lower pressure)
  - Why it fails: Inverse proportionality is easy to mix up when rearranging
  - Correct: Always check your answer makes physical sense: smaller volume = higher pressure, lower pressure = larger volume

## Cheatsheet

| Law | Relationship | Conditions | Formula |
| --- | --- | --- | --- |
| Boyle's Law | $p \propto 1/V$ | Fixed mass, constant $T$ | $p_1 V_1 = p_2 V_2$ |
| Charles' Law | $V \propto T$ | Fixed mass, constant $p$ | $V_1/T_1 = V_2/T_2$ |
| Pressure Law | $p \propto T$ | Fixed mass, constant $V$ | $p_1/T_1 = p_2/T_2$ |
| Combined Gas Law | $pV/T = \text{constant}$ | Fixed mass, any change | $p_1 V_1 / T_1 = p_2 V_2 / T_2$ |
| Ideal Gas Equation | $pV = nRT$ | Any ideal gas | $pV = nRT$ |

## What's next

Gas laws are the foundation for the entire topic of ideal gases, and underpin key concepts in thermodynamics and kinetic theory that you will encounter in later topics. Understanding the relationship between pressure, volume and temperature is critical for solving problems on internal energy, heat transfer and the kinetic model of an ideal gas, which are commonly tested in both Paper 1 and Paper 2 of CIE A-Level Physics. The assumptions of an ideal gas that we use to derive the gas laws also connect directly to the kinetic theory of gases, where we derive pressure in terms of average molecular kinetic energy.

- [Ideal gas equation of state](https://www.owlsprep.com/study/cie-9702-u19-ideal-gas-equation-of-state/)
- [Kinetic theory of gases](https://www.owlsprep.com/study/cie-9702-u19-kinetic-theory-of-gases/)
- [Root mean square speed](https://www.owlsprep.com/study/cie-9702-u19-root-mean-square-speed/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-9702-u19-gas-laws/