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

The process is thermally insulated, hence it is adiabatic.
For an adiabatic process, the relationship between pressure and volume is expressed as:

\[P_1 V_1^\gamma = P_2 V_2^\gamma\]

where \( \gamma = \frac{C_P}{C_V} = \frac{5}{3} \) for a monoatomic gas.

Let \( P_1 \) and \( P_2 \) represent the initial and final pressures, and \( V_1 \) and \( V_2 \) represent the initial and final volumes, respectively.
The final volume is given as \( \frac{1}{8} \) of the initial volume:

\[V_2 = \frac{1}{8} V_1\]

Applying the adiabatic relation:

\[P_1 V_1^\gamma = P_2 \left(\frac{1}{8} V_1\right)^\gamma\]

Simplification yields:

\[P_2 = P_1 \times 8^\gamma = P_1 \times 8^{\frac{5}{3}}\]

Given that \( 8^{\frac{5}{3}} = 8 \), the equation becomes:

\[P_2 = 8 P_1\]

Therefore, the ratio of the final pressure to the initial pressure is \( \frac{P_2}{P_1} = 8 \).