A motor-car tyre has a pressure of $2\, atm$ at $27\,^oC$. It suddenly burst's. If $\left( {\frac{{{C_p}}}{{{C_v}}}} \right) = 1.4$ for air, find the resulting temperatures (Given $4^{1/7} = 1.219$)
Medium
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$\mathrm{PV}^{\gamma}=$ constant (Adiabatic suddenly)
$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$
or $\quad \mathrm{T} \mathrm{P}^{(1-\gamma) / \gamma}=$ constant
$\mathrm{T}_{1} \mathrm{P}_{1}^{(1-\mathrm{\gamma}) / \gamma}=\mathrm{T}_{2} \mathrm{P}_{2}^{(1-\gamma) / \gamma}$
$\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{(\mathrm{1}-\mathrm{y}) / \gamma}=\mathrm{T}_{1}(2)^{\frac{-0.4}{1.4}}=\frac{300}{(4)^{1 / 7}}$
$\Rightarrow \mathrm{T}_{2}=246 \mathrm{K}=-27^{\circ} \mathrm{C}$
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