Ideal Gas Laws
Senior 2 · Physics
Ideal Gas Laws
Lesson 3 of 5
22 min
A balloon left in a hot car grows bigger. A football pumped hard in the cool morning feels harder at midday. An aerosol can thrown into a fire can explode. All three are about what heating does to a gas: at constant pressure it expands; in a rigid container its pressure rises. This lesson turns those two observations into laws you can calculate with.
Air is trapped in a thin glass tube by a small bead of liquid that can slide freely, so the pressure of the trapped air stays equal to air pressure. The tube is warmed in a water bath and the length (and so the volume) of the trapped air is measured at several temperatures.
Plotted against temperature in °C, the volume rises in a straight line. Extend that line backwards and it reaches zero volume at about −273 °C, absolute zero. Plotted against temperature in kelvin, the line passes through the origin: the volume is directly proportional to the kelvin temperature.
Charles's law: for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute (kelvin) temperature.
V/T = constant, so V₁/T₁ = V₂/T₂
Worked example. Umutoni's balloon holds 2.0 L of air at 27 °C. She leaves it in the sun, where it warms to 57 °C. Assuming the pressure stays constant, find its new volume.
Convert first: T₁ = 27 + 273 = 300 K; T₂ = 57 + 273 = 330 K.
V₂ = V₁ × T₂ ÷ T₁ = 2.0 × 330 ÷ 300 = 2.2 L.
Notice that the Celsius temperature more than doubled (27 → 57) but the volume grew by only 10%. Only kelvin temperatures are proportional to volume.
Now keep the volume fixed, in a rigid, sealed container with a pressure gauge, and heat it. The pressure rises in proportion to the kelvin temperature. (Guillaume Amontons first noticed this around 1700; Joseph Louis Gay-Lussac measured it precisely around 1800. The law is also called the pressure law or Amontons's law.)
Gay-Lussac's law: for a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute (kelvin) temperature.
P/T = constant, so P₁/T₁ = P₂/T₂
Worked example. The gas in a sealed aerosol can is at 300 kPa and 27 °C. The can is thrown onto a fire and heats to 327 °C. Find the new pressure.
T₁ = 300 K; T₂ = 327 + 273 = 600 K.
P₂ = P₁ × T₂ ÷ T₁ = 300 × 600 ÷ 300 = 600 kPa. The pressure doubles, and the can may burst. That is why the label says "Do not incinerate".
A note on tyres. A tyre gauge shows how much the pressure inside is above air pressure. The gas laws need the full (absolute) pressure. If the air in a tyre is at an absolute pressure of 290 kPa at 17 °C (290 K) and warms to 46 °C (319 K) on a hot tarmac road, its pressure rises to 290 × 319 ÷ 290 = 319 kPa.
Heating a gas makes its particles move faster. Faster particles hit the walls harder and more often.
In Gas Properties (Ideal screen), pump in some particles. Under "Hold Constant", choose Volume and heat the gas: watch the pressure rise. Then choose Pressure ↕V and heat again: the box widens instead, keeping the pressure fixed.
Tip
Kelvin or nothing. Write "T₁ = … + 273 = … K" as the first line of every Charles or Gay-Lussac calculation. Examiners give a mark for that line alone, and it prevents the most common error in this topic.
Which law? Look for the word "constant". Constant temperature → Boyle. Constant pressure → Charles. Constant volume (a rigid, sealed container) → Gay-Lussac.
A gas occupies 300 cm³ at 27 °C. At constant pressure, what is its volume at 127 °C?
Solution: T₁ = 300 K, T₂ = 400 K. V₂ = 300 × 400 ÷ 300 = 400 cm³.
At what temperature, in °C, will 300 cm³ of gas at 27 °C shrink to 150 cm³, at constant pressure?
Solution: Halving the volume means halving the kelvin temperature: T₂ = 300 × 150 ÷ 300 = 150 K = 150 − 273 = −123 °C.
A sealed gas cylinder in a store room is at 600 kPa and 20 °C. On a hot afternoon the room reaches 40 °C. Find the new pressure.
Solution: T₁ = 293 K, T₂ = 313 K. P₂ = 600 × 313 ÷ 293 ≈ 641 kPa.
Byiringiro says: "If I heat a gas from 20 °C to 40 °C at constant pressure, its volume doubles." Is he right?
Solution: No. Volume is proportional to the kelvin temperature: 293 K → 313 K, an increase of only about 7%. The volume would double only if the kelvin temperature doubled, from 293 K to 586 K (313 °C).
Explain, in terms of particles, why a football pumped in the cool morning feels harder at midday.
Solution: The sun warms the air inside the ball, so its particles move faster. They hit the inside of the ball harder and more often. The ball cannot stretch much (nearly constant volume), so the pressure rises and the ball feels harder.
Remember
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