Air `(gamma = 1.4 )` is pumped at 20atm pressure in a motor tyre at `20^@C`. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmoshpheric air.

Explanation

`gamma=1.4`. `T_1 = 20^@C=293K`,
`P_1=2atm`,`P_2=1atm`
We know for adiabatic process
`P_1^(1-gamma) xx T_1^(1-gamma)=P^(1-gamma) xx T_2^gamma`
or `(2)^(1-1.4) xx (293)^1.4 = (1)^(1-1.4) xx T_2^(1.4)`
implies `(2)^(-0.4) xx (293)^1.4 = T_2^1.4`
implies `2153.78 = T_2^1.4`
`T_2=(2153.78)^(1//1.4)`
`=240.3K`.