L.P.G. cylinder can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at 27°C. Due to a sudden fire in the building the temperature rises. At what temperature will the cylinder explode?

P₁ = Initial pressure of the gas = 12 atm

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

P₂ = Final pressure of the gas = 14.9 atm

T₂ = Final temperature of the gas = ?

V₁ = V₂ = V [constant 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{12\times\text{V}}{300} = \dfrac{14.9\times\text{V}}{\text{T}$2} \\[1em] \text{T}2 = \dfrac{14.9 \times 300}{12}\\[1em] \text{T}_2 = 372.5 \text {K} \\[1em]

∴ The temperature at which the cylinder will explode = 99.5°C.