Comprehension-2
A mono atomic ideal gas is filled in a non conducting container. The gas can be compressed by a movable non conducting piston. The gas is compressed slowly to `12.5%` of its initial volume.
The ratio of initial adiabatic bulk molulus of the gas to the finla value of adiabatic bulk modulus of the gas is

Options

A. `32`
B. `1`
C. `1//32`
D. `4`

Answer

To solve the problem, we need to find the ratio of the initial adiabatic bulk modulus of a monoatomic ideal gas to its final value after the gas has been compressed to 12.5% of its initial volume.

### Step-by-Step Solution:

1. **Understand the Process**:
- The gas is compressed slowly to 12.5% of its initial volume. This indicates that the process is adiabatic (no heat exchange), which means \( \Delta Q = 0 \).

2. **Initial and Final Volumes**:
- Let the initial volume be \( V_0 \).
- The final volume after compression is given by:
\[ V_f = 12.5\% \text{ of } V_0 = \frac{12.5}{100} V_0 = \frac{V_0}{8} \]

3. **Bulk Modulus Definition**:
- The bulk modulus \( \beta \) is defined as:
\[ \beta = -V \frac{dP}{dV} \]
- For an adiabatic process, the relationship between pressure and volume is given by:
\[ PV^\gamma = \text{constant} \]
- Here, \( \gamma \) for a monoatomic ideal gas is \( \frac{5}{3} \).

4. **Differentiating the Adiabatic Condition**:
- Differentiate \( PV^\gamma = \text{constant} \) with respect to volume \( V \):
\[ P \gamma V^{\gamma - 1} + V \frac{dP}{dV} \gamma V^{\gamma - 1} = 0 \]
- Rearranging gives:
\[ P \gamma + V \frac{dP}{dV} = 0 \implies \frac{dP}{dV} = -\frac{P \gamma}{V} \]

5. **Substituting into Bulk Modulus**:
- Substitute \( \frac{dP}{dV} \) into the bulk modulus formula:
\[ \beta = -V \left(-\frac{P \gamma}{V}\right) = P \gamma \]
- Thus, the bulk modulus for the initial state is:
\[ \beta_1 = P_0 \gamma \]
- And for the final state:
\[ \beta_2 = P_f \gamma \]

6. **Finding the Pressure Ratio**:
- From the adiabatic condition:
\[ P_0 V_0^\gamma = P_f V_f^\gamma \]
- Substitute \( V_f = \frac{V_0}{8} \):
\[ P_0 V_0^\gamma = P_f \left(\frac{V_0}{8}\right)^\gamma \]
- Rearranging gives:
\[ P_f = P_0 \left(\frac{V_0}{\frac{V_0}{8}}\right)^\gamma = P_0 \cdot 8^\gamma \]

7. **Calculating the Ratio of Bulk Moduli**:
- Now, the ratio of the bulk moduli is:
\[ \frac{\beta_1}{\beta_2} = \frac{P_0 \gamma}{P_f \gamma} = \frac{P_0}{P_f} \]
- Substitute \( P_f = P_0 \cdot 8^\gamma \):
\[ \frac{\beta_1}{\beta_2} = \frac{P_0}{P_0 \cdot 8^\gamma} = \frac{1}{8^\gamma} \]

8. **Substituting \( \gamma \)**:
- For a monoatomic gas, \( \gamma = \frac{5}{3} \):
\[ 8^\gamma = 8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^5 = 32 \]
- Therefore:
\[ \frac{\beta_1}{\beta_2} = \frac{1}{32} \]

### Final Answer:
The ratio of the initial adiabatic bulk modulus to the final adiabatic bulk modulus is:
\[ \frac{\beta_1}{\beta_2} = \frac{1}{32} \]