Study of Gas Laws — Question 13

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 760 mm Hg
V₁ = Initial volume of the gas = 100 cm³
T₁ = Initial temperature of the gas = 0°C = 273 K

Final conditions :

P₂ (Final pressure) = increased one and a half times of P₁

= (1 + $\dfrac{1}{2}$) of 760

= $\dfrac{3}{2}$ x 760

= 3 x 380

= 1140 mm Hg

V₂ (Final volume) = ?

T₂ (Final temperature) = increased by one-fifth of 273 K

= $\Big(1 + \dfrac{1}{5}\Big)$ of 273

= $\dfrac{6}{5}$ x 273 = $\dfrac{1638}{5}$

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 :

$\dfrac{760 \times 100 }{273} = \dfrac{1140 \times \text{V}_2}{\dfrac{1638}{5}} \\[1em] \dfrac{760 \times 100 }{273} = \dfrac{5 \times 1140 \times \text{V}_2}{1638} \\[1em] \text{V}_2 = \dfrac{760 \times 100 \times 1638}{273 \times 5 \times 1140} \\[1em] \text{V}_2 = 80 \text{ cm}^3$

∴ Final volume of the gas = 80 cm³