Ideal Gas Laws
Senior 2 · Physics
Ideal Gas Laws
Lesson 2 of 5
20 min
Put your finger over the nozzle of a bicycle pump and push the handle: the harder you push, the smaller the air becomes and the harder it pushes back. Every breath you take works the same way: your diaphragm moves down, your lungs get bigger, the pressure inside them drops, and air flows in. The link between the pressure and the volume of a gas was first measured by Robert Boyle in 1662.
Air is trapped in a sealed syringe connected to a pressure gauge. The piston is pushed in slowly, so the temperature of the air does not change. The results are:
| Volume V (cm³) | 60 | 40 | 30 | 20 | 15 |
|---|---|---|---|---|---|
| Pressure P (kPa) | 100 | 150 | 200 | 300 | 400 |
| P × V (kPa·cm³) | 6000 | 6000 | 6000 | 6000 | 6000 |
When the volume is halved (from 60 to 30 cm³) the pressure doubles (from 100 to 200 kPa). The product P × V stays the same.
For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume.
PV = constant, so P₁V₁ = P₂V₂
"Fixed mass" means no gas gets in or out. "Constant temperature" means the change is slow enough for any heat to escape.
Halve the volume and the same number of particles is packed into half the space. Each particle, moving at the same average speed (the temperature has not changed), reaches a wall twice as often. Twice as many collisions per second on each square metre of wall means twice the pressure.
A bicycle pump holds 120 cm³ of air at 100 kPa. Mugisha blocks the outlet and pushes the piston in slowly until the volume is 40 cm³. Find the new pressure.
P₁V₁ = P₂V₂, so P₂ = P₁V₁ ÷ V₂ = 100 × 120 ÷ 40 = 300 kPa.
Check the direction: the volume went down to one third, so the pressure should go up three times. It did.
Lake Kivu holds dissolved gases, and bubbles sometimes rise from its depths. A bubble of volume 2.0 cm³ is 20 m below the surface. At that depth the pressure is the air pressure (about 85 kPa at the lake's height) plus the pressure of the water: 85 000 + 1000 × 9.8 × 20 = 281 000 Pa = 281 kPa. Assuming the temperature stays the same, what is its volume when it reaches the surface?
V₂ = P₁V₁ ÷ P₂ = 281 × 2.0 ÷ 85 ≈ 6.6 cm³. The bubble grows to more than three times its size as it rises.
Tip
Units need only match. In P₁V₁ = P₂V₂ you can use kPa and cm³ directly, as long as both sides use the same units. You do not have to convert to Pa and m³.
Sanity check: squeezing a gas (smaller V) must give a bigger P. If your answer moved the same way as the volume, you have divided the wrong way round.
In each question the temperature and the mass of gas are constant.
A gas occupies 500 cm³ at 80 kPa. It is compressed to 200 cm³. Find the new pressure.
Solution: P₂ = 80 × 500 ÷ 200 = 200 kPa.
A gas at 150 kPa occupies 2.0 L. To what volume must it expand for its pressure to fall to 100 kPa?
Solution: V₂ = P₁V₁ ÷ P₂ = 150 × 2.0 ÷ 100 = 3.0 L.
Do these results obey Boyle's law? P (kPa): 100, 125, 200. V (cm³): 50, 40, 25.
Solution: PV = 100 × 50 = 5000; 125 × 40 = 5000; 200 × 25 = 5000. PV is constant, so yes.
A diver breathing compressed air deep in a lake must never hold her breath while swimming up to the surface. Use Boyle's law to explain why.
Solution: As she rises, the water pressure on her falls. By Boyle's law, the air in her lungs expands as the pressure falls. If she holds her breath, the expanding air can over-stretch and damage her lungs. Breathing out steadily lets the extra volume escape.
Remember
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