A tyre pumped at a pressure of 3.375 atm and at `27^(@)C` suddenly bursts. What is the final temperature? `(gamma=1.5)`

Explanation

We know that `(T_(2))/(T_(1))=((P_(2))/(P_(1)))^(((gamma-1)/(gamma)))`
where `P_(1)=3.375` atm, `P_(2)=1` atm, `gamma=1.5,T_(1)=300K`
`:.T_(2)=300xx((1)/(3.375))^((1.5-1)/(1.5))=300xx((1000)/(3375))^((1)/(3))=300xx((10)/(15))`
`T_(2)=200Kor-73^(@)C`