Question
A car tyre contains air at a pressure of $4$ atm and its temperature is $27°C$. The tyre suddenly bursts. Calculate the resulting temperature. $(\gamma=1.4)$

Answer

P_1 = 4atm, P_2 = 1atm, $T_1 -27^\circ C = 300K$ and $\gamma=1.4$ The sudden burst of tyre is an adiabatic process, in which , $\text{P}^{1-\gamma}_1\text{T}^\gamma_2=\text{P}^{1-\gamma}_2\text{T}^{\gamma}_2$
$\therefore\text{T}_2=\text{T}_1\Big(\frac{\text{P}_1}{\text{P}_2}\Big)^{\frac{1-\gamma}{\gamma}}=\text{T}_1\Big(\frac{\text{P}_2}{\text{P}_1}\Big)^{\frac{\gamma-1}{\gamma}}$
$=300\Big(\frac{4}{1}\Big)^{\frac{1.4-1}{1.4}}=201.9$
$=202\text{K}$ or $-71^\circ\text{C}.$

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