• # question_answer 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 $(\gamma =1.4)$ A) $300\,{{(4)}^{1.4/0.4}}$ B) $300\,{{\left( \frac{1}{4} \right)}^{-0.4/1.4}}$ C) $300\,{{(2)}^{-0.4/1.4}}$ D) $300\,{{(4)}^{-0.4/1.4}}$

 [d] For adiabatic process $\frac{{{T}^{\gamma }}}{{{P}^{\gamma -1}}}=$constant Þ$\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{P}_{1}}}{{{P}_{2}}} \right)}^{\frac{1-\gamma }{\gamma }}}$Þ$\frac{{{T}_{2}}}{300}={{\left( \frac{4}{1} \right)}^{\frac{(1-1.4)}{1.4}}}$ Þ${{T}_{2}}=300{{(4)}^{-\frac{0.4}{1.4}}}$