Question

A monoatomic gas having $ \gamma = \frac{5}{3} $ is stored in a thermally insulated container and the gas is suddenly compressed to $ \left( \frac{1}{8} \right)^{\text{th}} $ of its initial volume. The ratio of final pressure and initial pressure is:

• A. 28
• B. 32
• C. 40
• D. 16

Hint:

In adiabatic processes, the relationship between pressure and volume follows the equation \( P_1 V_1^\gamma = P_2 V_2^\gamma \). When the volume changes, you can find the new pressure by applying this equation with the given values for \( V_1 \), \( V_2 \), and \( \gamma \).

Answer 1

The Correct Option is B

For a thermally insulated (adiabatic) process, the relation between pressure and volume is given by: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] Where: - \( P_1 \) and \( V_1 \) are the initial pressure and volume, - \( P_2 \) and \( V_2 \) are the final pressure and volume, - \( \gamma = \frac{5}{3} \) is the adiabatic index (ratio of specific heats). Given that the gas is compressed to \( \frac{1}{8} \) of its initial volume, we have: \[ V_2 = \frac{1}{8} V_1 \] Substituting this into the adiabatic equation: \[ P_1 V_1^\gamma = P_2 \left( \frac{1}{8} V_1 \right)^\gamma \] Simplifying: \[ P_1 V_1^\gamma = P_2 \times \left( \frac{1}{8} \right)^\gamma V_1^\gamma \] Canceling \( V_1^\gamma \) from both sides: \[ P_1 = P_2 \times \left( \frac{1}{8} \right)^\gamma \] Since \( \gamma = \frac{5}{3} \), we have: \[ P_1 = P_2 \times \left( \frac{1}{8} \right)^{\frac{5}{3}} \] Now, calculate \( \left( \frac{1}{8} \right)^{\frac{5}{3}} \): \[ \left( \frac{1}{8} \right)^{\frac{5}{3}} = \frac{1}{8^{\frac{5}{3}}} = \frac{1}{32} \] Therefore: \[ P_2 = P_1 \times 32 \] Thus, the ratio of final pressure to initial pressure is: \[ \frac{P_2}{P_1} = 32 \] Therefore, the correct answer is Option (2) — \( 32 \).