Calculate the temperature at which the volume of a gas is doubled if its pressure at the same time increases from 70 cm of Hg to 80 cm of Hg. Consider the gas was initially at ice point.

Initial conditions :

P₁ = Initial pressure of the gas = 70 cm of Hg

V₁ = Initial volume of the gas = V

T₁ = Initial temperature of the gas = 273 K

Final conditions:

P₂ (Final pressure) = 80 cm of Hg

V₂ (Final volume) = 2V

T₂ (Final temperature) = ?

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{70 \times \text{V}}{273} = \dfrac{80 \times \text{2V}}{\text{T}$2} \\[1em] \text{T}2 = \dfrac{80 \times \text{2V} \times 273}{70 \times \text{V}} \\[1em] \text{T}2 = \dfrac{16 \times 273}{7} \\[1em] \text{T}2 = 624 \text{ K} \\[1em]

∴ Final temperature of the gas = 624 - 273 = 351°C