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A monatomic gas of volume ' $\mathrm{V}$ ' and pressure ' $\mathrm{P}$ ' expands isothermally to a volume ' $27 \mathrm{~V}$ ' and then compressed adiabatically to a volume ' $\mathrm{V}$ '. The final pressure of the gas is
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Verified Answer
The correct answer is:
$9 P$
$P_1=P: V_1=V ; V_2=27 V$
For isothermally
$\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& P_2=\frac{P_1 V_1}{V_2} \\
& =\frac{P \times V}{27 V}=\frac{P}{27}
\end{aligned}$
For adiabatically
$\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma \\
& \frac{\mathrm{P}}{27} \times(27 \mathrm{~V})^{\frac{5}{3}}=\mathrm{P}_3 \times \mathrm{V}^{\frac{5}{3}} \\
& \frac{\mathrm{P}}{27} \times(3)^{3 \times \frac{5}{3}}=\mathrm{P}_3
\end{aligned}$
The final pressure of the gas $\mathrm{P}_3=9 \mathrm{P}$
For isothermally
$\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& P_2=\frac{P_1 V_1}{V_2} \\
& =\frac{P \times V}{27 V}=\frac{P}{27}
\end{aligned}$
For adiabatically
$\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma \\
& \frac{\mathrm{P}}{27} \times(27 \mathrm{~V})^{\frac{5}{3}}=\mathrm{P}_3 \times \mathrm{V}^{\frac{5}{3}} \\
& \frac{\mathrm{P}}{27} \times(3)^{3 \times \frac{5}{3}}=\mathrm{P}_3
\end{aligned}$
The final pressure of the gas $\mathrm{P}_3=9 \mathrm{P}$
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