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A tyre pumped to a pressure of 6 atmosph...

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)`

Text Solution

Verified by Experts

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` .
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