An LPG cylinder can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at `27^@C` . Because of a sudden fire in the building , the temperature rises. At what temperature will the cylinder explode ?

Explanation

Given, `P_(1)=15 atm`, `T_(2)=?`
`P_(2)=12 atm`, `T_(2)=300K(273+27)`
Since the volume of the cylinder remains constant,
`(P_(1))/(T_(1))=(P_(2))/(T_(2))`
`:. (15)/(T_(1))=(12)/(300)`
`T_(1)=375 K`
i.e., the cylinder will explode above `375 K` or `102^(@)C`.