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

A tyre pumped to a pressure of 6 atmosphere suddenly burst. Room termp

1.	A tyre pumped to a pressure of 6 atmosphere suddenly burst. Room termperature is 25^(@)C. Calculate the temperature of escaping air.
Answer» Solution :From `P_(1)^(1-gamma) T_(1)^(gamma)= P_(2)^(1-gamma) T_(2)^(gamma)` , Here, `P_(1)=6 " atm " 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)=[(298)^(1.4)/(6^(0.4))]^(1/1.4)=(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 `202533=178.7` `:. T_(2)=178.7 K rArr t_(2)=178.7-273=-94.3^(@) C` .