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