A diatomic gas undergoes adiabatic compression and its volume reduces to half of initial volume then its final pressure would be if initial pressure of gas is P.

Options

A. `2^(1.4) P`
B. P/2
C. 2P
D. 3.07 P

Answer

To solve the problem of finding the final pressure \( P_2 \) of a diatomic gas undergoing adiabatic compression when its volume is reduced to half of its initial volume, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial pressure \( P_1 = P \)
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = \frac{V}{2} \)
- For a diatomic gas, the degrees of freedom \( f = 5 \).

2. **Calculate the Value of Gamma (\( \gamma \)):**
- The formula for \( \gamma \) is given by:
\[ \gamma = 1 + \frac{2}{f} \]
- Substituting \( f = 5 \):
\[ \gamma = 1 + \frac{2}{5} = 1 + 0.4 = \frac{7}{5} = 1.4 \]

3. **Use the Adiabatic Condition:**
- For an adiabatic process, the relationship between pressure and volume is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

4. **Substitute the Known Values into the Equation:**
- Substitute \( P_1 = P \), \( V_1 = V \), and \( V_2 = \frac{V}{2} \):
\[ P V^\gamma = P_2 \left(\frac{V}{2}\right)^\gamma \]

5. **Rearranging the Equation to Solve for \( P_2 \):**
- Rearranging gives:
\[ P_2 = P \cdot \frac{V^\gamma}{\left(\frac{V}{2}\right)^\gamma} \]
- Simplifying the right-hand side:
\[ P_2 = P \cdot \frac{V^\gamma}{\left(\frac{V^\gamma}{2^\gamma}\right)} = P \cdot 2^\gamma \]

6. **Substituting the Value of Gamma:**
- Now substitute \( \gamma = 1.4 \):
\[ P_2 = P \cdot 2^{1.4} \]

7. **Final Result:**
- Therefore, the final pressure \( P_2 \) is:
\[ P_2 = 2^{1.4} \cdot P \]

### Final Answer:
The final pressure \( P_2 \) when the volume is reduced to half during adiabatic compression is:
\[ P_2 = 2^{1.4} \cdot P \]