Lesson 7: The General Gas Law

08 Lesson 7: The General Gas Law

7.1 Combining the three laws

For a fixed mass of gas:

Boyle: V ∝ 1/P (T constant)
Charles: V ∝ T (P constant) Combining: V ∝ T/P → PV/T = constant.

For two states of the same mass of gas:

P₁V₁/T₁ = P₂V₂/T₂ (T in kelvin; P in any consistent unit; V in any consistent unit)

Special cases: T constant → Boyle's law; P constant → Charles's law; V constant → pressure law.

7.2 Enrichment: the ideal gas equation

In general, PV = nRT, where

n = number of moles of the gas,
P in pascals, V in m³, T in kelvin,
R = 8.31 J/(mol·K) the universal gas constant. A mole is the amount of a substance whose mass in grams equals its molecular mass: one mole of oxygen (O₂) has a mass of 32 g; nitrogen (N₂) 28 g; carbon dioxide (CO₂) 44 g.

Example: the volume of one mole of any gas at STP is V = nRT/P = 1 × 8.31 × 273/1.013 × 10⁵ = 0.0224 m³ = 22.4 litres.

7.3 Standard temperature and pressure (STP)

STP: 0 °C (273 K) and 76 cm Hg (1 atm = 1.013 × 10⁵ Pa).

7.4 Worked examples

Example 1 (to STP). A gas has volume 450 cm³ at 27 °C and 60 cm Hg. Find its volume at STP. P₁V₁/T₁ = P₂V₂/T₂: 60 × 450/300 = 76 × V₂/273 → V₂ = 60 × 450 × 273/(300 × 76) = 323 cm³.

Example 2 (find T). A gas of 3 L at 2 atm and 27 °C changes to 6 L at 1.5 atm. T₂ = T₁ P₂V₂/(P₁V₁) = 300 × (1.5 × 6)/(2 × 3) = 300 × 1.5 = 450 K = 177 °C.

Example 3 (find P). A fixed mass of gas at 76 cm Hg, 300 K, and 500 cm³ is compressed to 250 cm³ and heated to 360 K: P₂ = 76 × 500 × 360/(300 × 250) = 182.4 cm Hg.

Example 4 (gas that escapes). An open flask of air is heated from 7 °C (280 K) to 147 °C (420 K) at constant pressure. The volume of the flask is constant, so the amount of gas ∝ 1/T. Fraction left = 280/420 = 2/3; fraction that escaped = 1/3 = 33.3 % of the original amount.

✍️ Worked examples

Example 5 (mixture of two gases — partial pressures).

Gas A: 6 L at 20 cm Hg.
Gas B: 4 L at 30 cm Hg.
Both are transferred (at constant temperature) to a 3-L container.

A: P_A V_A = P'_A × 3 → P'_A = 20 × 6/3 = 40 cm Hg.
B: P'_B = 30 × 4/3 = 40 cm Hg. The pressure of a mixture equals the sum of the pressures each gas would produce alone: P = 40 + 40 = 80 cm Hg.

Example 6 (ideal gas equation). Two moles of a gas at 300 K in a 0.01 m³ container: P = nRT/V = 2 × 8.31 × 300/0.01 = 4.99 × 10⁵ Pa (about 4.9 atm).

Example 7 (helium balloon). A helium balloon rises in the air. As it rises the surrounding pressure falls (and the temperature falls less than the pressure), so by PV/T = constant its volume increases, until it may burst.