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A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is ( $\gamma=5 / 3$ )
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Verified Answer
The correct answer is:
32
$$
\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{8}, \gamma=\frac{5}{3}
$$
For adiabatic process,
$$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{5 / 3}=(2)^5=32
\end{aligned}
$$
\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{8}, \gamma=\frac{5}{3}
$$
For adiabatic process,
$$
\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=(8)^{5 / 3}=(2)^5=32
\end{aligned}
$$
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