The big idea: Pump up a bike tyre and the pump gets warm; squeeze a gas into less space and its pressure climbs. Those links between a gas's pressure P, volume V and temperature T are the gas laws.
What pressure is: Pressure is the force acting perpendicular to a surface per unit area: p = F/A, in pascals (1 Pa = 1 N m⁻²).
A gas pushing with 2.0 × 10³ N on a piston of area 0.020 m² exerts p = 2.0 × 10³ ÷ 0.020 = 1.0 × 10⁵ Pa, about atmospheric pressure.
Three laws, one for each fixed quantity: Hold one of them fixed and the other two are linked in a simple way — there are three of these links, one for each thing you keep fixed.
| Law | What is held fixed | The link | In words |
|---|---|---|---|
| Boyle's | temperature T | P V = constant | squeeze it (V down) → pressure up |
| Charles' | pressure P | V ÷ T = constant | heat it → it expands |
| Gay-Lussac's | volume V | P ÷ T = constant | heat a sealed can → pressure up |
New word — absolute temperature: Absolute temperature means temperature measured in kelvin (K), counted from absolute zero (the coldest possible, −273 °C).
Convert: T (K) = θ (°C) + 273. So 27 °C = 300 K.
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Animated graph
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Temperature is ALWAYS in kelvin: Charles' and Gay-Lussac's laws only work with temperature in kelvin, not °C.
Always do T (K) = °C + 273 before you put a temperature into a gas-law formula.
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The three laws are really one rule: the combined gas law. It is given in the data booklet.
What stays the same: For a fixed amount of gas the quantity P V ÷ T stays the same — so its value at one moment equals its value at another.
- pressure of the gas (Pa)
- volume of the gas (m³)
- absolute temperature (K) — always in kelvin
How to use it for two states: Compare a 'before' (state 1) and an 'after' (state 2) of the same gas:
P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂.
If one quantity is held fixed it cancels: fixed T → P₁V₁ = P₂V₂ (Boyle); fixed P → V₁ ÷ T₁ = V₂ ÷ T₂ (Charles).
Each law is a special case
- Fixed T → P V = constant (Boyle)
- Fixed P → V ÷ T = constant (Charles)
- Fixed V → P ÷ T = constant (Gay-Lussac)
Always, before you start
- Temperature in kelvin (°C + 273)
- Same units on both sides
- Pressure and volume units just have to match
A gas has a volume of 6.0 m³ at a pressure of 100 kPa. At the same temperature it is compressed to 2.0 m³. Find the new pressure.
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pV is an energy: Pa × m³ = N m⁻² × m³ = N m = J. So a constant such as K = pV (the gradient of a p against 1/V graph) has the unit J.
Showing changes on a p–V diagram
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How this is tested — gas-law questions almost always involve a graph or a before/after pair:
Paper 1B (data)
- Plot P against 1/V at fixed temperature — a straight line through the origin.
- Read its slope = the constant K and state K's SI unit.
Paper 2
- Use P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to find a percentage change in pressure, volume or amount of gas.
The classic trap: Leaving the temperature in °C — every gas-law T must be in kelvin (add 273).
Why P against 1/V is a straight line: Boyle says P V = K. Divide by V: P = K × (1/V).
That is the form y = (slope) × x, so a graph of P (y-axis) against 1/V (x-axis) is a straight line through the origin whose slope is K.
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The words to know
| Word | What it means |
|---|---|
| Pressure | Perpendicular force per unit area, p = F/A (Pa) |
| Isotherm | A line of constant temperature on a p–V diagram |
| Absolute zero | 0 K = −273 °C, where an ideal gas would have zero pressure |
| Combined gas law | pV/T = constant for a fixed amount of gas |
At constant temperature a student plots pressure P (in Pa) against 1/V (in m⁻³) for a fixed mass of gas. The points lie on a straight line through the origin passing through (0.10, 3.0) and (0.40, 12). Find the constant K and state its SI unit.
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Why the graph is a straight line through the origin: Rearrange the gas law so the plotted quantity is the subject, then name what is constant. At constant T, pV = constant gives 1/p = V ÷ constant: 1/p ∝ V. At constant p, V = (constant ÷ p) × T: V ∝ T, but only with T in kelvin.