Certain amount of a gas occupies a volume of 0.4 litre at 17°C. calculate the temperature at which the volume of gas is (a) doubled (b) reduced to half, pressure remaining constant?

(a) To find the temperature so that volume gets doubled :

V₁ = Initial volume of the gas = 0.4 litre
T₁ = Initial temperature of the gas = 17°C = 17 + 273 = 290 K
V₂ = Final volume of the gas = double of initial = 2 x 0.4 = 0.8 litre
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{\text{0.4}}{290} = \dfrac{\text{0.8}}{\text{T}$2} \\[1em] \text{T}2 = \dfrac{0.8 \times 290}{0.4} \\[1em] \text{T}_2 = 580 \text{ K}

∴ Final temperature in °C = 580 - 273 = 307°C

(b) To find the temperature so that volume gets reduced to half :

V₁ = Initial volume of the gas = 0.4 litre
T₁ = Initial temperature of the gas = 17°C = 17 + 273 = 290 K
V₂ = Final volume of the gas = reduced to half of initial = 0.2 litre
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{\text{0.4}}{290} = \dfrac{\text{0.2}}{\text{T}$2} \\[1em] \text{T}2 = \dfrac{0.2\times 290}{0.4} \\[1em] \text{T}_2 = 145

∴ Final temperature in °C = 145 - 273 = -128°C