Study of Gas Laws — Question 24

1.2 lit N₂ at 25°C and 748 mm Hg

Initial conditions :

P₁ = Initial pressure of the gas = 748 mm Hg

V₁ = Initial volume of the gas = 1.2 lit

T₁ = Initial temperature of the gas = 25°C = 25 + 273 = 298 K

Final conditions :

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 :

$\dfrac{748 \times 1.2}{298} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{748 \times 1.2\times 273}{298 \times 760} \\[1em] \text{V}_2 = 1.081 \space \text{lit} \\[1em]$

∴ Volume of N₂ at S.T.P. = 1.081 lit and volume of O₂ at S.T.P. is 1.25 lit. Hence, volume of O₂ at S.T.P. is greater than N₂ at S.T.P.