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A monoatomic gas at pressure $p_1$ and $V_1$ is compressed adiabatically to $\frac{1}{8}$ th its original volume. What is the final pressure of the gas?
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Verified Answer
The correct answer is:
$32 \mathrm{p}_1$
$\quad \mathrm{p}_1 \mathrm{~V}_1^{5 / 3}=\mathrm{p}^{\prime}\left(\frac{\mathrm{V}_1}{8}\right)^{5 / 3}$
$$
\begin{aligned}
p^{\prime} & =p_1(8)^{5 / 3} \\
& =p_1 \times 2^5 \\
p^{\prime} & =32 p_1
\end{aligned}
$$
$$
\begin{aligned}
p^{\prime} & =p_1(8)^{5 / 3} \\
& =p_1 \times 2^5 \\
p^{\prime} & =32 p_1
\end{aligned}
$$
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