A sample of air is saturated with benezed (vapour pressure = 100 mm Hg at 298 K,750 mmHg pressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is

Options

A. `2250` torr
B. `2150` torr
C. `2050` torr
D. `1950` torr

Explanation

To solve the problem, we need to determine the final pressure of a system that is initially saturated with benzene vapor and then isothermally compressed to one third of its initial volume.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of benzene at 298 K, \( P_{vapor} = 100 \, \text{mm Hg} \)
- Total pressure of the system, \( P_{total} = 750 \, \text{mm Hg} \)

2. **Calculate the Pressure of Air:**
- The pressure of air, \( P_{air} \), can be calculated using the formula:
\[ P_{total} = P_{air} + P_{vapor} \]
- Rearranging gives:
\[ P_{air} = P_{total} - P_{vapor} = 750 \, \text{mm Hg} - 100 \, \text{mm Hg} = 650 \, \text{mm Hg} \]

3. **Initial and Final Volume:**
- Let the initial volume be \( V_1 \).
- The final volume after compression is:
\[ V_2 = \frac{V_1}{3} \]

4. **Apply Boyle's Law:**
- Boyle's Law states that for a given mass of gas at constant temperature, the product of pressure and volume is constant:
\[ P_1 V_1 = P_2 V_2 \]
- Rearranging to find \( P_2 \) gives:
\[ P_2 = \frac{P_1 V_1}{V_2} \]

5. **Substituting Values:**
- Substitute \( P_1 = P_{air} = 650 \, \text{mm Hg} \) and \( V_2 = \frac{V_1}{3} \):
\[ P_2 = \frac{650 \, \text{mm Hg} \cdot V_1}{\frac{V_1}{3}} = 650 \, \text{mm Hg} \cdot 3 = 1950 \, \text{mm Hg} \]

6. **Calculate Final Pressure of the System:**
- The final pressure of the system, \( P_{final} \), includes the vapor pressure of benzene:
\[ P_{final} = P_2 + P_{vapor} = 1950 \, \text{mm Hg} + 100 \, \text{mm Hg} = 2050 \, \text{mm Hg} \]

### Final Answer:
The final pressure of the system is \( 2050 \, \text{mm Hg} \).

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