Question

A monoatomic gas of pressure \( P \) having volume \( V \) expands isothermally to a volume \( 2V \) and then adiabatically to a volume \( 16V \). The final pressure of the gas is
\textit{(Ratio of specific heats = \( \frac{5}{3} \))}

• A. \( \frac{P}{16} \)
• B. \( \frac{P}{8} \)
• C. \( \frac{P}{32} \)
• D. \( \frac{P}{64} \)

Hint:

In problems involving isothermal and adiabatic processes, remember to use Boyle’s Law for isothermal and the adiabatic relation \( P V^{\gamma} = \text{constant} \) for adiabatic expansion.

Answer 1

The Correct Option is D

Step 1: Isothermal Expansion.
For isothermal expansion, the pressure and volume follow Boyle's Law, where \( P_1 V_1 = P_2 V_2 \). So after the isothermal expansion, the pressure is reduced to \( \frac{P}{2} \).
Step 2: Adiabatic Expansion.
For adiabatic expansion, the pressure and volume follow the equation \( P V^{\gamma} = \text{constant} \), where \( \gamma = \frac{5}{3} \). Using this relation, we can find the final pressure: \[ P_2 V_2^{\gamma} = P_3 V_3^{\gamma} \] Substituting the values and solving gives: \[ P_3 = \frac{P}{64} \]
Step 3: Conclusion.
Thus, the correct answer is (D) \( \frac{P}{64} \).