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A monoatomic gas at pressure ' $\mathrm{P}$ ', having volume ' $\mathrm{V}$ ' expands isothermally to a volume ' $2 \mathrm{~V}$ ' and then adiabatically to a volume ' $16 \mathrm{~V}$ '. The final pressure of the gas is (Take $\gamma=5 / 3$ )
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Verified Answer
The correct answer is:
$\mathrm{P} / 64$
After isothermal expansion:
$$
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& P_2=P_1 \frac{V_1}{V_2} \\
& P_2=P_1 \frac{V}{2 V} \\
& P_2=\frac{P}{2}
\end{aligned}
$$
After adiabatic expansion:
$$
\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma \\
& \mathrm{P}_3=\mathrm{P}_2\left(\frac{\mathrm{V}_2}{\mathrm{~V}_3}\right)^\gamma \\
& \mathrm{P}_3=\frac{\mathrm{P}}{2}\left(\frac{2 \mathrm{~V}}{16 \mathrm{~V}}\right)^{5 / 3} \\
& \mathrm{P}_3=\frac{\mathrm{P}}{2}\left(\frac{1}{8}\right)^{5 / 3} \\
& \mathrm{P}_3=\frac{\mathrm{P}}{64}
\end{aligned}
$$
$$
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& P_2=P_1 \frac{V_1}{V_2} \\
& P_2=P_1 \frac{V}{2 V} \\
& P_2=\frac{P}{2}
\end{aligned}
$$
After adiabatic expansion:
$$
\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma \\
& \mathrm{P}_3=\mathrm{P}_2\left(\frac{\mathrm{V}_2}{\mathrm{~V}_3}\right)^\gamma \\
& \mathrm{P}_3=\frac{\mathrm{P}}{2}\left(\frac{2 \mathrm{~V}}{16 \mathrm{~V}}\right)^{5 / 3} \\
& \mathrm{P}_3=\frac{\mathrm{P}}{2}\left(\frac{1}{8}\right)^{5 / 3} \\
& \mathrm{P}_3=\frac{\mathrm{P}}{64}
\end{aligned}
$$
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