Question

A monoatomic gas is stored in a thermally insulated container. The gas is suddenly compressed to $ \frac{1}{8} $th of its initial volume. Find the ratio of final pressure to initial pressure.

• A. 8
• B. 16
• C. 4
• D. 32

Hint:

For an adiabatic process, the relation \( P V^\gamma = \text{constant} \) holds. Use this relation to find the pressure-volume relationship in such processes.

Answer 1

The Correct Option is A

Since the process is thermally insulated, it is an adiabatic process.
For an adiabatic process, the relation between pressure and volume is given by: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] where \( \gamma = \frac{C_P}{C_V} = \frac{5}{3} \) for a monoatomic gas. Let \( P_1 \) and \( P_2 \) be the initial and final pressures, and \( V_1 \) and \( V_2 \) be the initial and final volumes, respectively.
We are given that the final volume is \( \frac{1}{8} \) of the initial volume: \[ V_2 = \frac{1}{8} V_1 \] Using the adiabatic relation: \[ P_1 V_1^\gamma = P_2 \left(\frac{1}{8} V_1\right)^\gamma \] Simplifying: \[ P_2 = P_1 \times 8^\gamma = P_1 \times 8^{\frac{5}{3}} \] Since \( 8^{\frac{5}{3}} = 8 \), we get: \[ P_2 = 8 P_1 \] Thus, the ratio of final pressure to initial pressure is \( \frac{P_2}{P_1} = 8 \).