For an adiabatic process if volume becomes `1^(nd)/(32)` of initial value then pressure become (Take y= 1.4, if P is initial pressure)

Options

A. 2 P`
B. 4 P`
C. 128 P`
D. 8 P`

Answer

To solve the problem, we will use the relationship for an adiabatic process, which states that \( P V^{\gamma} = \text{constant} \). Here, \( \gamma \) (gamma) is given as 1.4, and we need to find the final pressure \( P_2 \) when the volume \( V_2 \) becomes \( \frac{1}{32} V_1 \).

### Step-by-Step Solution:

1. **Understand the Adiabatic Condition**:
The relationship for an adiabatic process is given by:
\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]

2. **Substitute the Given Values**:
We know that \( V_2 = \frac{1}{32} V_1 \). Therefore, we can substitute \( V_2 \) into the equation:
\[ P_1 V_1^{\gamma} = P_2 \left(\frac{1}{32} V_1\right)^{\gamma} \]

3. **Simplify the Equation**:
This can be rewritten as:
\[ P_1 V_1^{\gamma} = P_2 \frac{V_1^{\gamma}}{32^{\gamma}} \]
Now, we can cancel \( V_1^{\gamma} \) from both sides (assuming \( V_1 \neq 0 \)):
\[ P_1 = P_2 \frac{1}{32^{\gamma}} \]

4. **Rearranging to Find \( P_2 \)**:
Rearranging the equation gives:
\[ P_2 = P_1 \cdot 32^{\gamma} \]

5. **Substituting \( \gamma \)**:
Now substitute \( \gamma = 1.4 \):
\[ P_2 = P_1 \cdot 32^{1.4} \]

6. **Calculating \( 32^{1.4} \)**:
We can express \( 32 \) as \( 2^5 \):
\[ 32^{1.4} = (2^5)^{1.4} = 2^{5 \cdot 1.4} = 2^{7} \]

7. **Final Calculation**:
Therefore, we have:
\[ P_2 = P_1 \cdot 2^{7} \]
Since \( 2^{7} = 128 \):
\[ P_2 = 128 P_1 \]

### Final Result:
Thus, the final pressure \( P_2 \) is:
\[ P_2 = 128 P \]