Gas Laws: Boyle's, Charles's, and Pressure Law

Gases behave in predictable, quantifiable ways when you change their pressure, volume, or temperature. The three classical gas laws — Boyle's law, Charles's law, and the pressure law — describe these relationships. Together, they form the experimental foundation for the ideal gas equation.

The Three Gas Laws at a Glance

graph TD
    A["Gas Laws"] --> B["Boyle's Law\np ∝ 1/V\n(constant T)"]
    A --> C["Charles's Law\nV ∝ T\n(constant p)"]
    A --> D["Pressure Law\np ∝ T\n(constant V)"]
    B --> E["p₁V₁ = p₂V₂"]
    C --> F["V₁/T₁ = V₂/T₂"]
    D --> G["p₁/T₁ = p₂/T₂"]
    E --> H["Combined Gas Law\np₁V₁/T₁ = p₂V₂/T₂"]
    F --> H
    G --> H
    H --> I["Ideal Gas Equation\npV = nRT"]

Boyle's Law (Pressure and Volume)

Boyle's law states that for a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume:

pV = constant (at constant T)

Equivalently: p₁V₁ = p₂V₂

If you compress a gas into a smaller volume (without changing the temperature), the pressure increases. This makes sense from a molecular perspective: the same number of molecules occupy a smaller space, so they hit the walls of the container more frequently, creating more pressure.

A graph of p against V gives a curve (a rectangular hyperbola). A graph of p against 1/V gives a straight line through the origin, confirming the inverse proportionality.

Experimental Verification

Boyle's law can be verified using a sealed gas syringe or a Boyle's law apparatus (a column of gas trapped by oil in a glass tube connected to a pressure gauge):

Record the pressure and volume of a fixed mass of gas at room temperature.
Gradually change the volume (e.g., using a pump or adding weights to a piston) and record the corresponding pressure.
Plot p against 1/V — a straight line through the origin confirms Boyle's law.
The temperature must remain constant throughout. Allow time for the gas to return to room temperature between readings.

Charles's Law (Volume and Temperature)

Charles's law states that for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute temperature (in kelvin):

V/T = constant (at constant p)

Equivalently: V₁/T₁ = V₂/T₂

As you heat a gas (keeping the pressure constant, e.g., in a container with a movable piston), it expands. The molecules move faster and push the piston outward until the pressure equalises.

A graph of V against T (in kelvin) gives a straight line through the origin. If plotted in °C, the line intercepts the temperature axis at approximately −273 °C.

The Pressure Law (Pressure and Temperature)

The pressure law (sometimes called Gay-Lussac's law) states that for a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature:

p/T = constant (at constant V)

Equivalently: p₁/T₁ = p₂/T₂

Heating a gas in a sealed, rigid container increases the speed of the molecules. They strike the walls more frequently and with greater force, so the pressure rises.

A graph of p against T (in kelvin) gives a straight line through the origin. If plotted in °C, the line extrapolates to zero pressure at approximately −273 °C.

Absolute Zero and the Kelvin Scale

Both Charles's law and the pressure law extrapolate to zero volume or zero pressure at approximately −273.15 °C. This temperature is called absolute zero — the lowest possible temperature, at which the particles have minimum kinetic energy (they cannot be completely stationary due to quantum mechanical zero-point energy, but this is beyond A-level).

The Kelvin scale is defined so that:

T (K) = θ (°C) + 273

where 0 K = −273 °C = absolute zero.

The gas laws only work with temperatures in kelvin. Using Celsius in the proportional relationships will give incorrect answers, because the gas laws require a true zero point (absolute zero) as the baseline.

Molecular Explanations

Each gas law can be explained using kinetic theory:

Law	What is held constant	Molecular explanation
Boyle's	Temperature (= constant average KE)	Reducing V means molecules travel less distance between walls → collide with walls more often → pressure increases
Charles's	Pressure	Higher T means greater average KE → molecules hit walls harder → to maintain same pressure, V must increase so collision rate per unit area stays the same
Pressure	Volume	Higher T means greater average KE → molecules hit walls harder AND more often → pressure increases

The Combined Gas Law

When more than one variable changes simultaneously, the three individual laws combine into:

p₁V₁/T₁ = p₂V₂/T₂

This is extremely useful for problems where pressure, volume, and temperature all change. Each individual gas law is a special case:

Constant T: p₁V₁ = p₂V₂ (Boyle's)
Constant p: V₁/T₁ = V₂/T₂ (Charles's)
Constant V: p₁/T₁ = p₂/T₂ (Pressure)

Worked Example 1: Boyle's Law

A gas occupies 500 cm³ at a pressure of 1.0 × 10⁵ Pa. The gas is compressed at constant temperature to a volume of 200 cm³. What is the new pressure?

p₁V₁ = p₂V₂

1.0 × 10⁵ × 500 = p₂ × 200

p₂ = (1.0 × 10⁵ × 500) / 200 = 2.5 × 10⁵ Pa

Worked Example 2: Charles's Law

A balloon contains 2.0 litres of air at 27 °C. It is heated to 127 °C at constant pressure. What is the new volume?

First, convert to kelvin: T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K.

V₁/T₁ = V₂/T₂

2.0/300 = V₂/400

V₂ = 2.0 × 400/300 = 2.67 litres (3 s.f.)

Worked Example 3: Pressure Law

A sealed container of gas has a pressure of 1.5 × 10⁵ Pa at 300 K. The container is heated to 450 K. What is the new pressure?

p₁/T₁ = p₂/T₂

1.5 × 10⁵ / 300 = p₂ / 450

p₂ = 1.5 × 10⁵ × 450 / 300 = 2.25 × 10⁵ Pa

Worked Example 4: Combined Gas Law

A weather balloon at sea level (p = 1.01 × 10⁵ Pa, T = 20 °C) has a volume of 1.5 m³. It rises to an altitude where the pressure is 0.40 × 10⁵ Pa and the temperature is −40 °C. What is the new volume?

T₁ = 20 + 273 = 293 K, T₂ = −40 + 273 = 233 K

p₁V₁/T₁ = p₂V₂/T₂

V₂ = V₁ × (p₁/p₂) × (T₂/T₁)

V₂ = 1.5 × (1.01 × 10⁵ / 0.40 × 10⁵) × (233/293)

V₂ = 1.5 × 2.525 × 0.7952 = 3.01 m³

The balloon roughly doubles in volume as it rises, despite the drop in temperature, because the pressure decrease is the dominant effect.

Worked Example 5: Car Tyre Problem

A car tyre is inflated to 2.2 × 10⁵ Pa at 15 °C. After a motorway journey, the tyre temperature rises to 45 °C. Assuming the volume of the tyre does not change, what is the new pressure?

T₁ = 15 + 273 = 288 K, T₂ = 45 + 273 = 318 K

p₂ = p₁ × T₂/T₁ = 2.2 × 10⁵ × 318/288 = 2.2 × 10⁵ × 1.104 = 2.43 × 10⁵ Pa

The pressure increases by about 10%. This is why tyre pressure checks should be done when the tyres are cold — the recommended pressure assumes cold conditions.

Graphical Analysis

Graph	Shape	What it shows
p vs V (constant T)	Rectangular hyperbola	Boyle's law: p ∝ 1/V
p vs 1/V (constant T)	Straight line through origin	Confirms inverse proportionality
V vs T in K (constant p)	Straight line through origin	Charles's law: V ∝ T
V vs T in °C (constant p)	Straight line, x-intercept at −273 °C	Extrapolates to absolute zero
p vs T in K (constant V)	Straight line through origin	Pressure law: p ∝ T
p vs T in °C (constant V)	Straight line, x-intercept at −273 °C	Extrapolates to absolute zero
pV vs p (ideal gas, constant T)	Horizontal line	Confirms pV = constant

Common Exam Mistakes

Mistake	How to avoid it
Using °C instead of K	ALWAYS convert: T(K) = θ(°C) + 273
Forgetting which variable is constant	Read the question carefully; identify the process type
Using the wrong law when two variables change	Use the combined gas law: p₁V₁/T₁ = p₂V₂/T₂
Not recognising a Boyle's law graph	p vs V is a hyperbola; p vs 1/V is linear through origin
Confusing "constant pressure" with "constant volume"	Constant pressure = isobaric (piston moves); constant volume = isochoric (rigid container)

Summary

Boyle's law: pV = constant (constant T). Pressure and volume are inversely proportional.
Charles's law: V/T = constant (constant p). Volume and temperature are directly proportional.
Pressure law: p/T = constant (constant V). Pressure and temperature are directly proportional.
Combined gas law: p₁V₁/T₁ = p₂V₂/T₂ for changes involving all three variables.
All temperatures must be in kelvin (T = θ + 273).
Absolute zero (0 K = −273 °C) is where the extrapolated lines reach zero volume or zero pressure.
The gas laws have molecular explanations rooted in kinetic theory.

A-Level Deep Dive: Gas Laws

Spec mapping

Edexcel 9PH0 specification Topic 9 — Thermodynamics covers Boyle's law, Charles's law and the pressure law, the combined gas law, the ideal gas equation $pV = nRT$ (with molar form $pV = NkT$ ), absolute zero on the Kelvin scale, and the experimental relationships between $p$ , $V$ and $T$ for a fixed mass of gas (refer to the official specification document for exact wording). Gas-law content is examined in 9PH0 Paper 2 (Topics 6–9) but is heavily synoptic with Topic 5 (kinetic theory) for the link $pV = \tfrac{1}{3}Nm\langle c^2 \rangle$ and Topic 14 (astrophysics) for stellar atmospheres in option questions. The Edexcel formula booklet provides $pV = nRT$ and $pV = NkT$ but does not restate Boyle's, Charles's or the pressure law — these proportionalities and their constant-of-each-process conditions must be memorised.

Worked example with full mark scheme

Question (8 marks):

A diving cylinder of fixed internal volume $V = 1.20 \times 10^{-2}\;\text{m}^3$ contains air at a gauge pressure of $2.00 \times 10^{7}\;\text{Pa}$ when the cylinder is on the surface at a temperature of $15\,°\text{C}$ . Treat the air as an ideal gas. Take the molar gas constant $R = 8.31\;\text{J K}^{-1}\,\text{mol}^{-1}$ and atmospheric pressure as $1.01 \times 10^{5}\;\text{Pa}$ .

(a) Calculate the amount in moles of air in the cylinder. (4)

(b) The diver descends; the cylinder warms slightly so the air inside reaches a temperature of $24\,°\text{C}$ . Show, using the appropriate gas law, that the new absolute pressure of the gas in the cylinder is approximately $2.07 \times 10^{7}\;\text{Pa}$ . State the assumption you have used. (4)

Solution with mark scheme:

(a) Step 1 — convert to absolute pressure and absolute temperature.

Absolute pressure inside the cylinder $p = 2.00 \times 10^{7} + 1.01 \times 10^{5} \approx 2.010 \times 10^{7}\;\text{Pa}$ .

Temperature $T = 15 + 273 = 288\;\text{K}$ .

M1 — converting Celsius to Kelvin and recognising that gauge pressure must be added to atmospheric pressure to obtain the absolute pressure used in $pV = nRT$ . Common error: ignoring the atmospheric correction (small here, but flagged in mark schemes when omitted) or leaving $T$ in Celsius (catastrophic).

Step 2 — rearrange $pV = nRT$ for $n$ .

$n = \dfrac{pV}{RT}$

M1 — correct rearrangement of the ideal gas equation.

Step 3 — substitute and evaluate.

$n = \dfrac{(2.010 \times 10^{7}) \times (1.20 \times 10^{-2})}{8.31 \times 288} = \dfrac{2.412 \times 10^{5}}{2393} \approx 100.8\;\text{mol}$

A1 — value in the range $100\;\text{mol}$ to $101\;\text{mol}$ .

A1 — answer quoted to an appropriate number of significant figures (3 s.f.) with the unit "mol".

(b) Step 1 — identify the appropriate gas law.

Volume of the steel cylinder is fixed (isochoric process). Apply the pressure law:

$\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$

M1 — selection of the pressure law (or equivalently $pV = nRT$ with $V$ and $n$ constant).

Step 2 — convert temperatures.

$T_1 = 288\;\text{K}$ , $T_2 = 24 + 273 = 297\;\text{K}$ .

M1 — both temperatures correctly in Kelvin.

Step 3 — solve for $p_2$ .

$p_2 = p_1 \times \dfrac{T_2}{T_1} = (2.010 \times 10^{7}) \times \dfrac{297}{288} \approx 2.073 \times 10^{7}\;\text{Pa}$

A1 — answer rounding to $2.07 \times 10^{7}\;\text{Pa}$ as printed.

B1 — explicit assumption stated, e.g. "the cylinder is rigid so the volume of the gas is constant" or "no air leaks from the cylinder so $n$ is constant" or "air behaves as an ideal gas".

Total: 8 marks (M4 A3 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A weather balloon containing helium has a volume of $0.85\;\text{m}^3$ at ground level, where the pressure is $1.01 \times 10^{5}\;\text{Pa}$ and the temperature is $17\,°\text{C}$ . The balloon rises to an altitude where the pressure is $2.5 \times 10^{4}\;\text{Pa}$ and the temperature is $-43\,°\text{C}$ .

(a) Calculate the volume of the helium at altitude. (3)

(b) The balloon's elastic envelope can stretch to a maximum volume of $3.0\;\text{m}^3$ before bursting. State, with a reason, whether the balloon will burst at this altitude. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO2.2) — applying the combined gas law $\dfrac{p_1 V_1}{T_1} = \dfrac{p_2 V_2}{T_2}$ with both temperatures converted to Kelvin ( $T_1 = 290\;\text{K}$ , $T_2 = 230\;\text{K}$ ).
M1 (AO1.1) — correct rearrangement $V_2 = V_1 \dfrac{p_1}{p_2} \dfrac{T_2}{T_1}$ .
A1 (AO1.1) — $V_2 = 0.85 \times \dfrac{1.01 \times 10^{5}}{2.5 \times 10^{4}} \times \dfrac{230}{290} \approx 2.72\;\text{m}^3$ (accept $2.7\;\text{m}^3$ ).

(b)

M1 (AO3.2a) — comparing computed volume with the $3.0\;\text{m}^3$ burst threshold.
A1 (AO3.2a) — concluding "the balloon will not burst because $2.72\;\text{m}^3 < 3.0\;\text{m}^3$ ".

(c)

B1 (AO1.2) — any one of: helium behaves as an ideal gas; no helium escapes (so $n$ constant); the gas is in thermal equilibrium with its surroundings at each altitude.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. This question type — combined-gas-law calculation followed by an "evaluate against a threshold" judgement and an assumption statement — is a routine 9PH0 Paper 2 structure for Topic 9.

Synoptic links

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