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