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An ideal gas at pressure ' $p$ ' is adiabatically compressed so that its density becomes twice that of the initial. If $\gamma=\frac{c_p}{c_v}=\frac{7}{5}$, then final pressure of the gas is
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The correct answer is:
$2.63 p$
Since density become twice, its volume will become half.
$$
\therefore \frac{\mathrm{V}_1}{\mathrm{~V}_2}=2
$$
$\therefore$ For adiabatic process we have
$$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(2)^{7 / 5}=(2)^{1.4}=2.63
\end{aligned}
$$
$$
\therefore \frac{\mathrm{V}_1}{\mathrm{~V}_2}=2
$$
$\therefore$ For adiabatic process we have
$$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(2)^{7 / 5}=(2)^{1.4}=2.63
\end{aligned}
$$
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