Question

A monoatomic gas is stored in a thermally insulated container. The gas is suddenly compressed to $$\frac{1}{8}$$ 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, use the relation \( P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \), where \( \gamma = \frac{5}{3} \) for a monoatomic gas.

Answer 1

The Correct Option is A

We are given a monoatomic gas in a thermally insulated container. This means the process is adiabatic, and the gas undergoes a sudden compression. For an adiabatic process, the relationship between pressure and volume for an ideal gas is given by the following equation: \[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma}, \] where \( P_1 \) and \( P_2 \) are the initial and final pressures, \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( \gamma \) is the adiabatic index, which for a monoatomic gas is \( \gamma = \frac{5}{3} \). We are given that the final volume is \( \frac{1}{8} \) of the initial volume, so: \[ V_2 = \frac{V_1}{8}. \] Now, substituting into the adiabatic relation: \[ P_1 V_1^{\frac{5}{3}} = P_2 \left( \frac{V_1}{8} \right)^{\frac{5}{3}}. \] Simplifying: \[ P_1 V_1^{\frac{5}{3}} = P_2 \cdot V_1^{\frac{5}{3}} \cdot \left( \frac{1}{8} \right)^{\frac{5}{3}}. \] Cancelling \( V_1^{\frac{5}{3}} \) on both sides: \[ P_1 = P_2 \cdot \left( \frac{1}{8} \right)^{\frac{5}{3}}. \] Now, \( \left( \frac{1}{8} \right)^{\frac{5}{3}} = \frac{1}{8^{\frac{5}{3}}} = \frac{1}{32} \), so: \[ P_2 = P_1 \times 8. \] Thus, the ratio of final pressure to initial pressure is: \[ \frac{P_2}{P_1} = 8. \] Therefore, the correct answer is (1) 8.