`A` gas for which `gamma=1.5` is suddenly compressed to `1//4` th of the initial volume. Then the ratio of the final to initial pressure is

Options

A. `1:16`
B. `1:8`
C. `1:4`
D. `8:1`

Explanation

To solve the problem, we need to find the ratio of the final pressure to the initial pressure (P2/P1) after the gas is suddenly compressed to 1/4th of its initial volume (V1). We will use the adiabatic process relation for this calculation.

### Step-by-Step Solution:

1. **Identify Given Values**:
- The ratio of specific heats, \( \gamma = 1.5 \).
- The final volume \( V_2 = \frac{1}{4} V_1 \).

2. **Use the Adiabatic Relation**:
The adiabatic process is described by the equation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Rearranging this gives us:
\[ \frac{P_2}{P_1} = \frac{V_1^\gamma}{V_2^\gamma} \]

3. **Substitute the Volume Relationship**:
Since \( V_2 = \frac{1}{4} V_1 \), we can substitute this into the equation:
\[ \frac{P_2}{P_1} = \frac{V_1^\gamma}{\left(\frac{1}{4} V_1\right)^\gamma} \]

4. **Simplify the Equation**:
This simplifies to:
\[ \frac{P_2}{P_1} = \frac{V_1^\gamma}{\left(\frac{1}{4^\gamma} V_1^\gamma\right)} = \frac{V_1^\gamma}{\frac{1}{4^\gamma} V_1^\gamma} \]
The \( V_1^\gamma \) terms cancel out:
\[ \frac{P_2}{P_1} = 4^\gamma \]

5. **Calculate \( 4^\gamma \)**:
Now we substitute \( \gamma = 1.5 \):
\[ \frac{P_2}{P_1} = 4^{1.5} = (4^1) \cdot (4^{0.5}) = 4 \cdot 2 = 8 \]

6. **Final Result**:
Therefore, the ratio of the final pressure to the initial pressure is:
\[ \frac{P_2}{P_1} = 8 \]

### Final Answer:
The ratio of the final pressure to the initial pressure is \( 8 \).

---