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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 $\left(\gamma=\frac{5}{3}\right)$
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Verified Answer
The correct answer is:
$\frac{P}{64}$
For isothermal process: $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$
$$
\therefore \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\mathrm{P}_1}{2}
$$
For adiabatic process: $\mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma$
$$
\begin{aligned}
& \therefore \mathrm{P}_3=\mathrm{P}_2\left(\frac{\mathrm{V}_2}{\mathrm{~V}_3}\right)^\gamma=\mathrm{P}_2\left(\frac{2 \mathrm{~V}}{16 \mathrm{~V}}\right)^\gamma=\frac{\mathrm{P}_1}{2}\left(\frac{1}{8}\right)^{5 / 3} \\
& =\frac{\mathrm{P}}{2} \times \frac{1}{32}=\frac{\mathrm{P}}{64}
\end{aligned}
$$
$$
\therefore \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\mathrm{P}_1}{2}
$$
For adiabatic process: $\mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma$
$$
\begin{aligned}
& \therefore \mathrm{P}_3=\mathrm{P}_2\left(\frac{\mathrm{V}_2}{\mathrm{~V}_3}\right)^\gamma=\mathrm{P}_2\left(\frac{2 \mathrm{~V}}{16 \mathrm{~V}}\right)^\gamma=\frac{\mathrm{P}_1}{2}\left(\frac{1}{8}\right)^{5 / 3} \\
& =\frac{\mathrm{P}}{2} \times \frac{1}{32}=\frac{\mathrm{P}}{64}
\end{aligned}
$$
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