At constant temperature on increasing the pressure of a gas by `5%` will decrease its volume by -

Options

A. `5%`
B. `5.26%`
C. `4.26%`
D. `4.76%`

Explanation

To solve the problem of how much the volume of a gas decreases when the pressure is increased by 5% at constant temperature, we can use Boyle's Law, which states that the product of pressure and volume for a given mass of gas is constant at a constant temperature. This can be mathematically expressed as:

\[ P_1 V_1 = P_2 V_2 \]

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
Let the initial pressure be \( P_1 \) and the initial volume be \( V_1 \).

2. **Determine Final Pressure**:
The pressure is increased by 5%, so the final pressure \( P_2 \) can be expressed as:
\[ P_2 = P_1 + 0.05 P_1 = 1.05 P_1 \]

3. **Apply Boyle's Law**:
According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
Substituting \( P_2 \):
\[ P_1 V_1 = (1.05 P_1) V_2 \]

4. **Cancel \( P_1 \)**:
Since \( P_1 \) is present on both sides of the equation, we can cancel it out (assuming \( P_1 \neq 0 \)):
\[ V_1 = 1.05 V_2 \]

5. **Solve for \( V_2 \)**:
Rearranging the equation to find \( V_2 \):
\[ V_2 = \frac{V_1}{1.05} \]

6. **Calculate the Change in Volume**:
The decrease in volume can be calculated as:
\[ \Delta V = V_1 - V_2 = V_1 - \frac{V_1}{1.05} \]

7. **Express the Decrease in Volume**:
Factor out \( V_1 \):
\[ \Delta V = V_1 \left(1 - \frac{1}{1.05}\right) = V_1 \left(\frac{1.05 - 1}{1.05}\right) = V_1 \left(\frac{0.05}{1.05}\right) \]

8. **Calculate the Percentage Decrease**:
To find the percentage decrease in volume, we use the formula:
\[ \text{Percentage Decrease} = \frac{\Delta V}{V_1} \times 100\% \]
Substituting \( \Delta V \):
\[ \text{Percentage Decrease} = \frac{V_1 \left(\frac{0.05}{1.05}\right)}{V_1} \times 100\% = \frac{0.05}{1.05} \times 100\% \]

9. **Final Calculation**:
\[ \text{Percentage Decrease} = \frac{5}{1.05} \approx 4.76\% \]

### Conclusion:
The volume of the gas decreases by approximately **4.76%** when the pressure is increased by 5% at constant temperature.