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

A monoatomic gas in a thermally insulated container undergoes a sudden adiabatic compression. The adiabatic process for an ideal gas follows the equation \( P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \), where \( P \) is pressure, \( V \) is volume, and \( \gamma \) is the adiabatic index. For a monoatomic gas, \( \gamma = \frac{5}{3} \). The final volume is \( V_2 = \frac{V_1}{8} \). Substituting these values into the adiabatic relation gives \( P_1 V_1^{\frac{5}{3}} = P_2 \left( \frac{V_1}{8} \right)^{\frac{5}{3}} \). Simplifying yields \( P_1 V_1^{\frac{5}{3}} = P_2 V_1^{\frac{5}{3}} \left( \frac{1}{8} \right)^{\frac{5}{3}} \). Cancelling \( V_1^{\frac{5}{3}} \) from both sides results in \( P_1 = P_2 \left( \frac{1}{8} \right)^{\frac{5}{3}} \). Since \( \left( \frac{1}{8} \right)^{\frac{5}{3}} = \frac{1}{32} \), the equation becomes \( P_1 = \frac{P_2}{32} \), or \( P_2 = 32 P_1 \). This means the ratio of final pressure to initial pressure is \( \frac{P_2}{P_1} = 32 \). Therefore, the correct answer is (1) 32.