A vessel containing `0.1 m^(3)` of air at `76 cm` of Hg is connected to an evacuated vessel of capacity `0.09 m^(3)`. The resultant air pressure is:

Options

A. `20 cm of Hg`
B. `30 cm of Hg`
C. `40 cm of Hg`
D. `60 cm of Hg`

Explanation

To solve the problem, we will use the ideal gas law and the principle of conservation of mass. The key idea is that when the air from the first vessel expands into the second evacuated vessel, the total volume increases, and we can find the new pressure using the initial conditions.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Volume of the first vessel (V1) = 0.1 m³
- Volume of the second vessel (V2) = 0.09 m³
- Initial pressure (P1) = 76 cm of Hg

2. **Convert the initial pressure to SI units:**
- We need to convert 76 cm of Hg to Pascals (Pa).
- 1 cm of Hg = 1333.22 Pa, so:
\[ P1 = 76 \, \text{cm of Hg} \times 1333.22 \, \text{Pa/cm of Hg} = 10132.8 \, \text{Pa} \]

3. **Calculate the total volume after connecting the vessels:**
- The total volume (V_total) when both vessels are connected is:
\[ V_{\text{total}} = V1 + V2 = 0.1 \, \text{m}^3 + 0.09 \, \text{m}^3 = 0.19 \, \text{m}^3 \]

4. **Apply the ideal gas law:**
- Since the temperature remains constant, we can use the formula:
\[ P1 \cdot V1 = P2 \cdot V_{\text{total}} \]
- Rearranging gives us:
\[ P2 = \frac{P1 \cdot V1}{V_{\text{total}}} \]

5. **Substitute the values:**
- Now substitute the known values into the equation:
\[ P2 = \frac{10132.8 \, \text{Pa} \cdot 0.1 \, \text{m}^3}{0.19 \, \text{m}^3} \]

6. **Calculate P2:**
- Performing the calculation:
\[ P2 = \frac{1013.28}{0.19} \approx 532.5 \, \text{Pa} \]

7. **Convert P2 back to cm of Hg:**
- To convert back to cm of Hg:
\[ P2 = \frac{532.5 \, \text{Pa}}{1333.22 \, \text{Pa/cm of Hg}} \approx 0.4 \, \text{cm of Hg} \]
- This value seems incorrect based on the initial pressure. Let's check the calculations again.

8. **Re-evaluate the pressure calculation:**
- Since we are looking for the pressure in cm of Hg directly, we can use the ratio of volumes:
\[ P2 = P1 \cdot \frac{V1}{V_{\text{total}}} \]
- Substituting the initial pressure in cm of Hg:
\[ P2 = 76 \, \text{cm of Hg} \cdot \frac{0.1}{0.19} \approx 40 \, \text{cm of Hg} \]

### Final Result:
The resultant air pressure in the connected vessels is approximately **40 cm of Hg**.