A tyre pumped to a pressure of 3 atmospheres suddenly bursts. Calculate the fall in temperature due to adiabatic expansion. The temperature of air before expansion is `27^(@)C` and value of `gamma = 1.4`.

Explanation

We know that `T_(2)^(gamma) P_(2)^(1-gamma) = T_(1)^(gamma) P_(1)^(1-gamma) implies [(T_(2))/(T_(1))]^(gamma) = [(P_(1))/(P_(2))]^(1-gamma) implies [(T_(2))/(300)]^(1.4) = [(3)/(1)]^(1-1.4) `
`implies [(T_(2))/(300)]^(1.4) = [(1)/(3)]^(0.4) implies T_(2) = 219.2 K implies T_(1)- T_(2) = (300-219.2)K = 80.8K`