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A gas at N.T.P. is suddenly compressed to $\left(\frac{1}{4}\right)^{\mathrm{th}}$ of its original volume. The final pressure in (Given $\gamma=$ ratio of sp. heats $=\frac{3}{2}$ ) atmosphere is ( $\mathrm{P}=$ original pressure $)$
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The correct answer is:
$8\ P$
In Adiabatic compression, $\mathrm{PV}^\psi=$ constant
Given $\mathrm{V}_{\text {new }}=\frac{1}{4} \mathrm{~V}$ and $\gamma=\frac{3}{2}$
$\begin{aligned}
& \therefore \quad \frac{\mathrm{P}_{\mathrm{new}}}{\mathrm{P}}=\left(\frac{\mathrm{V}}{\mathrm{V}_{\text {new }}}\right)^7=\left(\frac{\mathrm{V}}{\frac{1}{4} \mathrm{~V}}\right)^{3 / 2} \\
& \therefore \quad \mathrm{P}_{\mathrm{new}}=8 \mathrm{P}
\end{aligned}$
Given $\mathrm{V}_{\text {new }}=\frac{1}{4} \mathrm{~V}$ and $\gamma=\frac{3}{2}$
$\begin{aligned}
& \therefore \quad \frac{\mathrm{P}_{\mathrm{new}}}{\mathrm{P}}=\left(\frac{\mathrm{V}}{\mathrm{V}_{\text {new }}}\right)^7=\left(\frac{\mathrm{V}}{\frac{1}{4} \mathrm{~V}}\right)^{3 / 2} \\
& \therefore \quad \mathrm{P}_{\mathrm{new}}=8 \mathrm{P}
\end{aligned}$
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