Ideal Gas Laws
Senior 2 · Physics
Ideal Gas Laws
Lesson 4 of 5
22 min
Real problems rarely keep one quantity fixed. A weather balloon launched from Kigali rises into air that is both thinner and colder. A hospital oxygen cylinder must be labelled with how much gas it holds. To handle these, the three laws you have met are combined into one equation that works for any ideal gas under any conditions.
Boyle (PV constant), Charles (V/T constant) and Gay-Lussac (P/T constant) all fit inside one statement. For a fixed mass of gas:
P₁V₁ / T₁ = P₂V₂ / T₂
If one quantity is constant, it cancels and you get back one of the three laws. For example, if T₁ = T₂ you are left with P₁V₁ = P₂V₂, which is Boyle's law.
Worked example. A weather balloon holds 5.0 m³ of helium at 85 kPa and 27 °C. It rises to a height where the pressure is 50 kPa and the temperature is −13 °C. Find its new volume.
T₁ = 27 + 273 = 300 K; T₂ = −13 + 273 = 260 K.
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 5.0 × (8550) × (260300) = 5.0 × 1.7 × 0.867 ≈ 7.4 m³.
Check the directions: lower pressure makes it bigger (× 1.7); lower temperature makes it smaller (× 0.867). Overall it grows.
Experiments also show that, at the same pressure and temperature, the volume of a gas is proportional to the number of moles, n (twice as many particles need twice as much space). Putting all of this together gives the ideal gas equation:
PV = nRT
where P is in pascals (Pa), V in cubic metres (m³), n in moles (mol), T in kelvin (K), and R = 8.31 J/(mol·K) is the molar gas constant, the same for every ideal gas.
Unlike the ratio laws, PV = nRT requires SI units: Pa, not kPa; m³, not cm³ or litres.
What volume does 1 mol of an ideal gas occupy at 0 °C (273 K) and 101 325 Pa?
V = nRT ÷ P = 1 × 8.31 × 273 ÷ 101 325 ≈ 0.0224 m³ = 22.4 L. Every ideal gas, whatever its particles, has this molar volume under these conditions.
A classroom in Kigali measures 10 m × 8.0 m × 3.0 m. The air is at 85 kPa and 20 °C. How many moles of air does it hold?
V = 10 × 8.0 × 3.0 = 240 m³; P = 85 000 Pa; T = 293 K.
n = PV ÷ (RT) = 85 000 × 240 ÷ (8.31 × 293) ≈ 8380 mol.
Since one mole of air has a mass of about 29 g, that is about 8380 × 0.029 ≈ 240 kg of air, about the mass of four adults, in a room that looks empty!
A steel cylinder of volume 0.020 m³ contains 4.0 mol of gas at 300 K. Find the pressure.
P = nRT ÷ V = 4.0 × 8.31 × 300 ÷ 0.020 = 498 600 Pa ≈ 499 kPa, about five times air pressure at sea level.
Tip
Two kinds of gas-law question, two unit rules
Use R = 8.31 J/(mol·K).
A gas occupies 400 cm³ at 100 kPa and 27 °C. Find its volume at 200 kPa and 327 °C.
Solution: T₁ = 300 K, T₂ = 600 K. V₂ = 400 × (100200) × (600300) = 400 × 0.5 × 2 = 400 cm³. Doubling the pressure halved the volume, and doubling the kelvin temperature doubled it back.
How many moles of gas are there in 0.024 m³ at 100 kPa and 300 K?
Solution: n = PV ÷ (RT) = 100 000 × 0.024 ÷ (8.31 × 300) ≈ 0.963 mol.
Find the temperature of 2.0 mol of gas held in 0.050 m³ at 100 kPa.
Solution: T = PV ÷ (nR) = 100 000 × 0.050 ÷ (2.0 × 8.31) ≈ 301 K (about 28 °C).
A fixed mass of gas has both its kelvin temperature and its volume doubled. What happens to its pressure?
Solution: P = nRT ÷ V. Doubling T doubles P; doubling V halves it. The two effects cancel: the pressure is unchanged.
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