The pressure on one mole of gas at s.t.p. is doubled and the temperature is raised to 546 K. What is the final volume of the gas ? [one mole of a gas occupies a volume of 22.4 litres at stp.]

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 1 atm
V₁ = Initial volume of the gas = 22.4 litres
T₁ = Initial temperature of the gas = 273 K

Final conditions:

P₂ (Final pressure) = 2 atm
V₂ (Final volume) = ?
T₂ (Final temperature) = 546 K

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{1 \times 22.4}{273} = \dfrac{2\times \text{V}$2}{546}\\[0.5em] \text{V}2 = \dfrac{1 \times 22.4\times 546}{273\times 2} \\[0.5em] \text{V}_2 = 22.4 \text{ lit}

Therefore, final volume of the gas = 22.4 lit.