MHT CET · Physics · Thermodynamics
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 \(\left(\gamma=\frac{5}{3}\right)\)
- A \(\frac{P}{64}\)
- B \(\frac{\mathrm{P}}{128}\)
- C \(\frac{\mathrm{P}}{8}\)
- D \(\frac{P}{32}\)
Answer & Solution
Correct Answer
(A) \(\frac{P}{64}\)
Step-by-step Solution
Detailed explanation
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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