The big idea: A gas is made of tiny particles in constant, random motion, bouncing off the walls of their container. Each collision pushes on the wall — together they create the pressure.
Three things describe a gas sample: its pressure P, its volume V and its temperature T (always in kelvin). Change one and the others respond. The gas laws describe exactly how.
What 'ideal' means: The ideal gas model makes two key assumptions:
- the particles themselves have no volume (they are point-like), and - there are no forces between them (no attraction or repulsion).
Real gases behave most like an ideal gas at high temperature and low pressure — when the particles are far apart and moving fast. They deviate most at low temperature and high pressure.
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Keep the temperature and the amount of gas fixed, and pressure and volume are inversely proportional — this is Boyle's law. Squeeze the gas into half the volume and the pressure doubles.
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A quick sanity check: P and V multiply to a constant (PV = k). So if V is doubled at constant T, P must be halved to keep the product the same. You can often answer a Boyle's-law MCQ just by this reasoning — no calculator needed.
Worked example — squeezing a gas
A sealed syringe holds 60.0 cm³ of gas at 100 kPa. The plunger is pushed in to 24.0 cm³ at the same temperature. Calculate the new pressure.
Solution
- Temperature is constant, so use Boyle's law — formula first:
- Rearrange for the new pressure:
- Substitute (volumes in the same unit, so cm³ is fine):
- Work it out — keep the unit:
Final answer
P₂ = 250 kPa (volume fell to 0.4×, so pressure rose to 2.5×).
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Heating a gas at constant volume makes the particles move faster and hit the walls harder, so the pressure rises. Pressure is directly proportional to the kelvin temperature.
Animated graph
Watch the graph build step by step in study mode.
Always work in kelvin: Convert every temperature with T(K) = θ(°C) + 273.
Using °C breaks the gas laws — only the kelvin scale starts at true zero (0 K), so only kelvin gives the right proportion.
When two or three of P, V and T change at once, use the combined gas law, which is given in the data booklet.
- initial and final pressure (any unit, kept the same on both sides)
- initial and final volume (same unit on both sides)
- initial and final absolute temperature, in kelvin (K)
Worked example — heating in a fixed container
A rigid 2.0 dm³ container holds gas at 101 kPa and 27 °C. It is heated to 127 °C. Calculate the new pressure.
Solution
- Convert both temperatures to kelvin first:
- Volume is fixed (V1 = V2), so the combined law simplifies — formula first:
- Rearrange for the new pressure:
- Substitute the kelvin values:
- Work it out:
Final answer
P₂ = 135 kPa (the temperature rose by a factor of 400/300, so the pressure did too).
How this is tested: S1.5.1 shows up most as a quick Paper 1A MCQ.
- A Boyle's-law ratio question — 'the volume is doubled at constant temperature, what happens to the pressure?' — answerable by reasoning alone. - An ideal-gas question — which gas/conditions deviate least from ideal? (answer: high temperature, low pressure). - A combined-gas-law calculation when two quantities change at once.
The trap that costs the mark every year: leaving the temperature in °C instead of converting to kelvin.
Score it cleanly: (1) Convert every T to kelvin before anything else. (2) Cancel whatever is constant (fixed V or fixed T) to shrink the equation. (3) Write the formula, then substitute — carry the unit.
IB-style question — combined gas law
A 0.500 dm³ balloon of gas is at 96 kPa and 17 °C. It is taken to a warmer, lower-pressure place where it expands to 0.620 dm³ at 87 kPa. Calculate the final temperature, in °C. [3]
How to score the marks
- Mark 1 — convert to kelvin. T1 = 17 + 273 = 290 K (everything else can stay in its given units, kept the same on both sides).
- Mark 2 — rearrange the given combined gas law for T2 and substitute:
- Mark 3 — evaluate, then convert back to °C as asked:
Final answer
T₂ ≈ 326 K = 53 °C.