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A monoatomic gas of presuure 'P' having volume 'V' expands isothermally to a volume '2V' and then adiabatically to a volume '16V'. The final pressure of the gas is (ratio of specific heats $=\frac{5}{3}$)
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Verified Answer
The correct answer is:
$\frac{\mathrm{P}}{64}$
$\gamma=\frac{5}{3}$
Case I: $\quad P_{1} V_{1}=P_{2} V_{2}$
$\mathrm{PV}=\mathrm{P}_{2} \times 2 \mathrm{~V}$
$\therefore \quad P_{2}=\frac{P}{2}$
Case II : $\quad P_{2} V_{2}^{\gamma}=P_{3} V_{3}^{\gamma}$
$$
\begin{array}{l}
\left(\frac{P}{2}\right)(2 V)^{\gamma}=P_{3}(16 V)^{\gamma} \\
P_{3}=\frac{P}{2} \frac{(2 V)^{\gamma}}{(16 V)^{\gamma}}=\frac{P}{2}\left(\frac{1}{8}\right)^{\gamma} \\
=\frac{P}{2}\left(\frac{1}{2^{3}}\right)^{5 / 3}=\frac{P}{2}\left(\frac{1}{2}\right)^{5} \\
=\frac{P}{2 \times 32}=\frac{P}{64}
\end{array}
$$
Case I: $\quad P_{1} V_{1}=P_{2} V_{2}$
$\mathrm{PV}=\mathrm{P}_{2} \times 2 \mathrm{~V}$
$\therefore \quad P_{2}=\frac{P}{2}$
Case II : $\quad P_{2} V_{2}^{\gamma}=P_{3} V_{3}^{\gamma}$
$$
\begin{array}{l}
\left(\frac{P}{2}\right)(2 V)^{\gamma}=P_{3}(16 V)^{\gamma} \\
P_{3}=\frac{P}{2} \frac{(2 V)^{\gamma}}{(16 V)^{\gamma}}=\frac{P}{2}\left(\frac{1}{8}\right)^{\gamma} \\
=\frac{P}{2}\left(\frac{1}{2^{3}}\right)^{5 / 3}=\frac{P}{2}\left(\frac{1}{2}\right)^{5} \\
=\frac{P}{2 \times 32}=\frac{P}{64}
\end{array}
$$
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