Gas laws

CIE A-Level Physics· Unit 19: Ideal gases· 7 min read

1. Boyle's Law★★☆☆☆⏱ 10 min

📘 Definition

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: $p_{1} = 1.0 \times 1 0^{5}$ 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.

2. Charles' Law and the Pressure Law★★☆☆☆⏱ 12 min

📘 Definition

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.

3. Combining Gas Laws: The Ideal Gas Equation★★★☆☆⏱ 15 min

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.

📘 Definition

Ideal Gas Equation

pV = nRT

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: $\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}$

Example:

1 mole of ideal gas at STP (273 K, 1.0 × 10⁵ Pa) occupies 22.4 dm³, which satisfies $p V = n R T$

✓ Quick check

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

What is 0.5 dm³ in m³?
- 5 × 10⁻⁴ m³
- 5 × 10⁻³ m³
- 0.5 × 10⁻² 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
$V = \frac{0.25 \times 8.31 \times 300}{1.0 \times 10^5} = 6.23 \times 10^{-3} \text{ m}^3 = 6.23 \text{ dm}^3$

4. Solving Changing Condition Problems★★★☆☆⏱ 15 min

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
$V_2 = \frac{(1.0 \times 10^5)(2.0)(260)}{(5.0 \times 10^4)(290)} \approx 3.6 \text{ m}^3$

5. Common Pitfalls

Wrong move:

Using Celsius instead of Kelvin in gas law calculations

Why:

All gas law proportionalities rely on absolute temperature, so Celsius gives incorrect results

Correct move:

Always add 273 to Celsius to convert to Kelvin before starting any calculation

Wrong move:

Using cm³/dm³ directly in $p V = n R T$ without unit conversion

Why:

R in J mol⁻¹ K⁻¹ requires volume in m³, so unmatched units give wrong orders of magnitude

Correct move:

Convert volume: 1 cm³ = 10⁻⁶ m³, 1 dm³ = 10⁻³ m³, before substituting into the ideal gas equation

Wrong move:

Applying Boyle's Law when temperature changes

Why:

Boyle's Law only holds for constant temperature, which is a core requirement

Correct move:

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 move:

Using the combined gas law for leaking containers with changing mass

Why:

All gas laws assume fixed mass of gas, so escaping gas changes n

Correct move:

Calculate initial and final n separately using $p V = n R T$ for containers that leak or gain gas

Wrong move:

Inverting proportionality in Boyle's Law (smaller volume gives lower pressure)

Why:

Inverse proportionality is easy to mix up when rearranging

Correct move:

Always check your answer makes physical sense: smaller volume = higher pressure, lower pressure = larger volume

6. Quick Reference 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	$p V / T = constant$	Fixed mass, any change	$p_{1} V_{1} / T_{1} = p_{2} V_{2} / T_{2}$
Ideal Gas Equation	$p V = n R T$	Any ideal gas	$p V = n R T$

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.

2022 · 1
MCQ: Volume change of sealed gas
2023 · 2
Calculation of ideal gas volume
2021 · 1
Constant volume pressure change

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.