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11.Thermodynamics
normal
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
A
$27\, K$
B
$27\,^oC$
C
$-27\,^oC$
D
$246\,^oC$
Solution
$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$
$\mathbf{o r}$
$\mathrm{TP}^{(1-\gamma) /\gamma}=$ constant
$\mathrm{T}_{1} \mathrm{P}_{1}^{(1-\gamma) / \gamma}=\mathrm{T}_{2} \mathrm{P}_{2}^{(1-\gamma) / \gamma}$
$\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{\frac{1-\gamma}{\gamma}}=300 \times\left(\frac{2}{1}\right)^{\frac{1-1.4}{1.4}}$
$=\frac{300}{(2)^{2/7}}=246 \mathrm{K}=-27^{\circ} \mathrm{C}$
Standard 11
Physics
