The pressure in the tyre of a car is four times the atmospheric pressure at $300 K$. If this tyre suddenly bursts, its new temperature will be $(\gamma = 1.4)$
Medium
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(d) For adiabatic process $\frac{{{T^\gamma }}}{{{P^{\gamma - 1}}}} = $ constant
==>$\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 - \gamma }}{\gamma }}}$
==>$\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 - \gamma }}{\gamma }}}$
==>$\frac{{{T_2}}}{{300}} = {\left( {\frac{4}{1}} \right)^{\frac{{(1 - 1.4)}}{{1.4}}}}$
==>${T_2} = 300{(4)^{ - \frac{{0.4}}{{1.4}}}}$
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