The pressure P and volume V of a gas are related as `PV^(3//2)=K` where K is a constant. The percentage increase in the pressure for a diminution of `0.5%` of the volume is

Options

A. `0.25%`
B. `0.75%`
C. `1.50%`
D. `0.50%`

Answer

To solve the problem, we need to find the percentage increase in pressure \( P \) when the volume \( V \) of a gas decreases by \( 0.5\% \). The relationship between pressure and volume is given by the equation:

\[ PV^{\frac{3}{2}} = K \]

where \( K \) is a constant.

### Step-by-Step Solution:

1. **Understanding the Change in Volume**:
- A decrease of \( 0.5\% \) in volume means:
\[ \Delta V = -0.005V \]
- Therefore, the new volume \( V' \) can be expressed as:
\[ V' = V + \Delta V = V - 0.005V = 0.995V \]

2. **Using the Given Relationship**:
- From the relationship \( PV^{\frac{3}{2}} = K \), we can express pressure \( P \) as:
\[ P = \frac{K}{V^{\frac{3}{2}}} \]

3. **Finding the New Pressure**:
- Substitute \( V' \) into the equation for pressure:
\[ P' = \frac{K}{(0.995V)^{\frac{3}{2}}} \]
- Simplifying this gives:
\[ P' = \frac{K}{(0.995^{\frac{3}{2}})(V^{\frac{3}{2}})} = \frac{P}{0.995^{\frac{3}{2}}} \]

4. **Calculating the Percentage Increase in Pressure**:
- The change in pressure \( \Delta P \) can be expressed as:
\[ \Delta P = P' - P = P \left( \frac{1}{0.995^{\frac{3}{2}}} - 1 \right) \]
- To find the percentage increase:
\[ \text{Percentage Increase} = \frac{\Delta P}{P} \times 100 = \left( \frac{1}{0.995^{\frac{3}{2}}} - 1 \right) \times 100 \]

5. **Calculating \( 0.995^{\frac{3}{2}} \)**:
- Calculate \( 0.995^{\frac{3}{2}} \):
\[ 0.995^{\frac{3}{2}} \approx 0.995^{1.5} \approx 0.9925 \quad (\text{using a calculator}) \]

6. **Substituting Back**:
- Now substitute this value back into the percentage increase formula:
\[ \text{Percentage Increase} \approx \left( \frac{1}{0.9925} - 1 \right) \times 100 \]
- Calculate \( \frac{1}{0.9925} \):
\[ \frac{1}{0.9925} \approx 1.0075 \]
- Therefore:
\[ \text{Percentage Increase} \approx (1.0075 - 1) \times 100 \approx 0.75\% \]

### Final Answer:
The percentage increase in pressure for a diminution of \( 0.5\% \) of the volume is approximately **0.75%**.