Study of Gas Laws — Question 29

Initial conditions :

P₁ = 700 mm Hg

V₁ = Initial volume of the gas = 20 litres

T₁ = Initial temperature of the gas = 27°C + 273 = 300 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 :

$\dfrac{700 \times 20 }{300} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{7 \times 2 \times 273}{3\times 76} \\[1em] \text{V}_2 = 16.76 \space l \\[1em]$

22.4 l of gas at S.T.P. weighs 70 g

∴ 16.76 l of gas at S.T.P. will weigh

= $\dfrac{70 \times 16.76}{22.4}$ = 52.38 g

∴ Weight of gas = 52.38 g