In an adiabatic expansion of air (assume it a mixture of `N_2` and `O_2`), the volume increases by `5%`. The percentage change in pressure is:

Options

A. `7%`
B. `6%`
C. `4%`
D. `3%`

Answer

To solve the problem of the percentage change in pressure during an adiabatic expansion of air (assumed to be a mixture of nitrogen and oxygen), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Adiabatic Process**: In an adiabatic process, the relationship between pressure (P), volume (V), and the adiabatic index (γ) is given by the equation:
\[ PV^\gamma = \text{constant} \]
where γ (gamma) for diatomic gases (like N₂ and O₂) is approximately \( \frac{7}{5} \) or 1.4.

2. **Initial and Final Volumes**: Let the initial volume \( V_1 = V \) and the final volume \( V_2 = 1.05V \) (since the volume increases by 5%).

3. **Using the Adiabatic Relation**: From the adiabatic relation, we can write:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Rearranging gives:
\[ \frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma \]

4. **Substituting Values**: Substitute \( V_1 = V \) and \( V_2 = 1.05V \):
\[ \frac{P_2}{P_1} = \left(\frac{V}{1.05V}\right)^\gamma = \left(\frac{1}{1.05}\right)^\gamma \]

5. **Calculating the Ratio**: Since \( \gamma = \frac{7}{5} \):
\[ \frac{P_2}{P_1} = \left(\frac{1}{1.05}\right)^{\frac{7}{5}} \]

6. **Calculating \( \left(\frac{1}{1.05}\right)^{\frac{7}{5}} \)**:
First, calculate \( \frac{1}{1.05} \):
\[ \frac{1}{1.05} \approx 0.9524 \]
Now raise this to the power of \( \frac{7}{5} \):
\[ 0.9524^{1.4} \approx 0.93 \]

7. **Finding the Percentage Change in Pressure**:
Thus, we have:
\[ P_2 \approx 0.93 P_1 \]
The change in pressure is:
\[ \Delta P = P_1 - P_2 = P_1 - 0.93 P_1 = 0.07 P_1 \]
The percentage change in pressure is:
\[ \text{Percentage Change} = \left(\frac{\Delta P}{P_1}\right) \times 100 = \left(\frac{0.07 P_1}{P_1}\right) \times 100 = 7\% \]

### Final Answer:
The percentage change in pressure is **7%**.