A motor-car tyre has a pressure of 2,atm at 2 | vedclass

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Question

$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$

$\mathbf{o r}$

$\mathrm{TP}^{(1-\gamma) /\gamma}=$ constant

$\mathr

Vedclass · Accepted Answer

$-27\,^oC$

Vedclass · Answer

$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$

$\mathbf{o r}$

$\mathrm{TP}^{(1-\gamma) /\gamma}=$ constant

$\mathrm{T}_{1} \mathrm{P}_{1}^{(1-\gamma) / \gamma}=\mathrm{T}_{2} \mathrm{P}_{2}^{(1-\gamma) / \gamma}$

$\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}ight)^{\frac{1-\gamma}{\gamma}}=300 	imes\left(\frac{2}{1}ight)^{\frac{1-1.4}{1.4}}$

$=\frac{300}{(2)^{2/7}}=246 \mathrm{K}=-27^{\circ} \mathrm{C}$

A motor-car tyre has a pressure of $2\,atm$ at $27\,^oC$. It suddenly bursts. If $\left( {\frac{{{C_p}}}{{{C_v}}} = 1.4} \right)$ for air, find resulting temp

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