`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. `1:6`
B. `1:8`
C. `1:4`
D. `8:1`

Answer

To solve the problem, we need to find the ratio of the final pressure \( P_2 \) to the initial pressure \( P_1 \) when a gas is compressed to \( \frac{1}{4} \) of its initial volume. Given that the value of \( \gamma \) (gamma) is \( 1.5 \), we can use the relationship between pressure and volume for an adiabatic process.

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process:**
For an adiabatic process, the relationship between pressure and volume is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume.

2. **Setting Up the Volumes:**
We know that the gas is compressed to \( \frac{1}{4} \) of its initial volume, so:
\[ V_2 = \frac{1}{4} V_1 \]

3. **Substituting Values:**
Substitute \( V_2 \) into the adiabatic equation:
\[ P_1 V_1^\gamma = P_2 \left(\frac{1}{4} V_1\right)^\gamma \]

4. **Simplifying the Equation:**
This can be rewritten as:
\[ P_1 V_1^\gamma = P_2 \left(\frac{V_1^\gamma}{4^\gamma}\right) \]
Rearranging gives:
\[ P_2 = P_1 \cdot 4^\gamma \]

5. **Calculating the Pressure Ratio:**
Now, we need to find the ratio \( \frac{P_2}{P_1} \):
\[ \frac{P_2}{P_1} = 4^\gamma \]

6. **Substituting the Value of Gamma:**
Since \( \gamma = 1.5 \):
\[ \frac{P_2}{P_1} = 4^{1.5} \]

7. **Calculating \( 4^{1.5} \):**
We can express \( 4^{1.5} \) as:
\[ 4^{1.5} = (2^2)^{1.5} = 2^{3} = 8 \]

8. **Final Result:**
Therefore, the ratio of the final pressure to the initial pressure is:
\[ \frac{P_2}{P_1} = 8 \]
This can be expressed as:
\[ P_2 : P_1 = 8 : 1 \]

### Conclusion:
The ratio of the final pressure to the initial pressure is \( 8 : 1 \).