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 need to apply the principles of an adiabatic process, where the relationship between pressure (P) and volume (V) is given by the equation:

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

where:
- \( P_1 \) is the initial pressure,
- \( V_1 \) is the initial volume,
- \( P_2 \) is the final pressure,
- \( V_2 \) is the final volume,
- \( \gamma \) is the adiabatic constant (given as 1.4).

### Step 1: Define Initial Conditions
Let:
- Initial pressure \( P_1 = P \)
- Initial volume \( V_1 = V \)

### Step 2: Define Final Conditions
According to the problem, the final volume \( V_2 \) is given as:
\[ V_2 = \frac{1}{32} V_1 = \frac{1}{32} V \]

### Step 3: Substitute Values into the Adiabatic Equation
Using the adiabatic equation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Substituting the known values:
\[ P V^\gamma = P_2 \left( \frac{1}{32} V \right)^\gamma \]

### Step 4: Simplify the Equation
Now, simplify the right-hand side:
\[ P V^\gamma = P_2 \left( \frac{1}{32^\gamma} V^\gamma \right) \]

### Step 5: Cancel Common Terms
Since \( V^\gamma \) appears on both sides, we can cancel it:
\[ P = P_2 \left( \frac{1}{32^\gamma} \right) \]

### Step 6: Solve for Final Pressure \( P_2 \)
Rearranging the equation gives:
\[ P_2 = P \cdot 32^\gamma \]

### Step 7: Substitute the Value of \( \gamma \)
Substituting \( \gamma = 1.4 \):
\[ P_2 = P \cdot 32^{1.4} \]

### Step 8: Calculate \( 32^{1.4} \)
We can express \( 32 \) as \( 2^5 \):
\[ 32^{1.4} = (2^5)^{1.4} = 2^{5 \cdot 1.4} = 2^7 \]

### Step 9: Final Expression for \( P_2 \)
Thus, we have:
\[ P_2 = P \cdot 2^7 \]

### Step 10: Calculate \( 2^7 \)
Calculating \( 2^7 \):
\[ 2^7 = 128 \]

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