The pressure inside a tyre is 4 atm at `27^(@)C`.If the tyre bursts suddenly, its final temperature will be

Options

A. `300(4)^(7//2)`
B. `300(4)^(2//7)`
C. `300(2)^(7//2)`
D. `300(4)^(-2//7)`

Answer

D

Explanation

In an adiabatic process,
`p_(2)^((1-gamma))T_(2)^(gamma)=p_(1)^(1-gamma)T_(1)^(gamma)`
Final temperature `T_(2)=T_(1)((p_(1))/(p_(2)))^(1-gamma//y)`
= `300((4)/(1))^((1-7//5))/(7//5)=300(4)^(-2//7)`