A motor-car tyre has a pressure of 2,atm at 27,^oC . It suddenly bursts. If left( {frac{{{Cₚ

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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

$27\, K$

$27\,^oC$

$-27\,^oC$

$246\,^oC$

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

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An ideal gas heat engine operates in Carnot's cycle between $227\,^oC$ and $127\,^oC$ . It absorbs $6.0 \times 10^4\,cal$ at higher temperature. The amount of heat converted into work is equal to

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

Solution

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