A gas occupies a volume of `250 mL` at `0.98` bar of Hg and `25^(@)C`. What additional pressure is required to reduce the volume of the gas to `200 mL` at the same temperature ?

Explanation

To solve the problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant. The formula is given as:

\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( P_2 \) = final pressure
- \( V_2 \) = final volume

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial pressure, \( P_1 = 0.98 \) bar
- Initial volume, \( V_1 = 250 \) mL
- Final volume, \( V_2 = 200 \) mL
- Temperature is constant (not needed for the calculation).

2. **Apply Boyle's Law:**
Using the formula \( P_1 V_1 = P_2 V_2 \), we need to find \( P_2 \).

3. **Rearranging the formula to solve for \( P_2 \):**
\[ P_2 = \frac{P_1 V_1}{V_2} \]

4. **Substituting the known values into the equation:**
\[ P_2 = \frac{0.98 \, \text{bar} \times 250 \, \text{mL}}{200 \, \text{mL}} \]

5. **Calculating \( P_2 \):**
- First, calculate the numerator:
\[ 0.98 \times 250 = 245 \, \text{bar mL} \]
- Now divide by the final volume:
\[ P_2 = \frac{245 \, \text{bar mL}}{200 \, \text{mL}} = 1.225 \, \text{bar} \]

6. **Calculate the additional pressure required:**
- The additional pressure \( \Delta P \) is given by:
\[ \Delta P = P_2 - P_1 \]
- Substitute the values:
\[ \Delta P = 1.225 \, \text{bar} - 0.98 \, \text{bar} = 0.245 \, \text{bar} \]

### Final Answer:
The additional pressure required to reduce the volume of the gas to 200 mL at the same temperature is **0.245 bar**.