A gas occupies 500 cm³ at S.T.P. At what temperature will the volume of the gas be reduced by 20% of its original volume, pressure being constant?

V₁ = Initial volume of the gas = 500 cm³
T₁ = Initial temperature of the gas = 273 K
V₂ = Final volume of the gas = reduced by 20% of its original = 500 - ($\dfrac{20}{100}$ x 500) = 500 - 100 = 400 cm³
T₂ = Final temperature of the gas = ?

By Charles' Law:

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

Substituting the values :

$\dfrac{500}{273} = \dfrac{400}{\text{T}$2} \\[1em] \text{T}2 = \dfrac{400 \times 273}{500} \\[1em] \text{T}_2 = 218.4 \text{ K}

Final temperature = 218.4 K