Calculate the volume of dry air at S.T.P. that occupies 28 cm³ at 14°C and 750 mm Hg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mm Hg.

Initial conditions :

P₁ = 750 - water vapour pressure = 750 - 12 = 738 mm Hg

V₁ = Initial volume of the gas = 28 cm³

T₁ = Initial temperature of the gas = 14°C + 273 = 287 K

Final conditions [S.T.P.] :

P₂ (Final pressure) = 760 mm of Hg

T₂ (Final temperature) = 273 K

V₂ (Final volume) = ?

By Gas Law:

$\dfrac{\text{P}$1\times\text{V}1}{\text{T}1} = \dfrac{\text{P}2\times\text{V}2}{\text{T}2}

Substituting the values :

$\dfrac{738 \times 28 }{287} = \dfrac{760 \times \text{V}$2}{273} \\[1em] \text{V}2 = \dfrac{738 \times 28 \times 273}{287 \times 760} \\[1em] \text{V}_2 = 25.86 \text{ cm}^3 \\[1em]

∴ Volume of dry gas = 25.9 cm³