MHT CET · Physics · Kinetic Theory of Gases
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 \(\gamma=5 / 3\) )
- A \(\mathrm{P} / 64\)
- B \(\mathrm{P} / 32\)
- C \(16 \mathrm{P}\)
- D \(32 \mathrm{P}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{P} / 64\)
Step-by-step Solution
Detailed explanation
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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