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

Under adiabatic change
`T_(2)/T_(1) = (P_(1)/P_(2))^((1- gamma)/gamma)` or `T_(2) = T_(1) (P_(1)//P_(2))^((1-gamma)/gamma)`
`therefore T_(2) = 300 (4//1)^((1-(7//5))/(7//5)` , `gamma =1.4 = 7//5` for air or `T_(2) =300 (4)^(-2//7)`