A tyre pumped to a pressure of 6 atmosphere suddenly bursts. Room temperature is 25^(@)C. Calculate the temperature of escaping air. (gamma=1.4)

A tyre pumped to a pressure of 6 atmosphere suddenly bursts. Room temp

1.	A tyre pumped to a pressure of 6 atmosphere suddenly bursts. Room temperature is 25^(@)C. Calculate the temperature of escaping air. (gamma=1.4)
Answer» SOLUTION :From `P_(1)^(1-gamma) T_(1)^(gamma)=P_(2)^(1-gamma) T_(2)^(gamma)` Here, `P_(1)=6atm P_(2) =1 ATM` `T_(1)=273+25=298K, gamma=1.4` `(6)^(1-1.4) (298)^(1.4)=(1)^(1-1.4)T_(2)^(1.4)` `T_(2)^(1.4) =(298)^(1.4)6^(-0.4)=((298)^((1.4)/(1.4)))/((6)^((0.4)/(1.4)))=(298)/((6)^((2)/(7)))` using logarithms `log T_(2)=2.4742-(2)/(7) (0.7782)` `=2.4742 -0.2209=2.2533`. Anti log of `2.2533=178.7` `:.T_(2)=178.7K rArr t_(2)=178.7-273 = -94.3^(@)C` .