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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :From `P_(1)^(1-gamma) T_(1)^(gamma)=P_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)^(1-gamma) T_(2)^(gamma)` <br/> Here, `P_(1)=6atm P_(2) =1 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>` <br/> `T_(1)=273+25=298K, gamma=1.4` <br/> `(6)^(1-1.4) (298)^(1.4)=(1)^(1-1.4)T_(2)^(1.4)` <br/> `T_(2)^(1.4) =(298)^(1.4)6^(-0.4)=((298)^((1.4)/(1.4)))/((6)^((0.4)/(1.4)))=(298)/((6)^((2)/(7)))` <br/> using logarithms `log T_(2)=2.4742-(2)/(7) (0.7782)` <br/> `=2.4742 -0.2209=2.2533`. <br/> Anti log of `2.2533=178.7` <br/> `:.T_(2)=178.7K rArr t_(2)=178.7-273 = -94.3^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` .</body></html>