A gas occupies a volume of `250 cm^(3)` at `745` torr and `25^(@)C` . What additional pressure is required to reduce the volume of the gas to `200 cm^(3)` 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 a constant. This can be expressed 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 volume, \( V_1 = 250 \, \text{cm}^3 \)
- Initial pressure, \( P_1 = 745 \, \text{torr} \)
- Final volume, \( V_2 = 200 \, \text{cm}^3 \)

2. **Apply Boyle's Law:**
Using Boyle's Law, we can rearrange the formula to find \( P_2 \):
\[ P_2 = \frac{P_1 V_1}{V_2} \]

3. **Substitute the Known Values:**
\[ P_2 = \frac{745 \, \text{torr} \times 250 \, \text{cm}^3}{200 \, \text{cm}^3} \]

4. **Calculate \( P_2 \):**
\[ P_2 = \frac{186250}{200} = 931.25 \, \text{torr} \]

5. **Calculate the Additional Pressure Required:**
The additional pressure required can be found by subtracting the initial pressure from the final pressure:
\[ \text{Additional Pressure} = P_2 - P_1 \]
\[ \text{Additional Pressure} = 931.25 \, \text{torr} - 745 \, \text{torr} = 186.25 \, \text{torr} \]

### Final Answer:
The additional pressure required to reduce the volume of the gas to \( 200 \, \text{cm}^3 \) at the same temperature is \( 186.25 \, \text{torr} \).

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