The relative humidity of air is `80%` at `27^(@)C`. If the aqueous tension at the same temperature is 27 mm of Hg. The partial pressure of water vapour in the air will be (in mm of Hg).

Answer

To find the partial pressure of water vapor in the air given the relative humidity and aqueous tension, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Relative Humidity (RH) = 80%
- Aqueous Tension (Vapor Pressure of Pure Water, \( P^0_{H2O} \)) at 27°C = 27 mm of Hg

2. **Use the Formula for Relative Humidity:**
The formula for relative humidity is given by:
\[ \text{Relative Humidity} = \left( \frac{P_{H2O}}{P^0_{H2O}} \right) \times 100 \]
where \( P_{H2O} \) is the partial pressure of water vapor in the air and \( P^0_{H2O} \) is the vapor pressure of pure water.

3. **Substitute the Known Values:**
We can rearrange the formula to solve for \( P_{H2O} \):
\[ P_{H2O} = \left( \frac{\text{Relative Humidity}}{100} \right) \times P^0_{H2O} \]
Substituting the values:
\[ P_{H2O} = \left( \frac{80}{100} \right) \times 27 \text{ mm of Hg} \]

4. **Calculate \( P_{H2O} \):**
\[ P_{H2O} = 0.8 \times 27 = 21.6 \text{ mm of Hg} \]

5. **Final Answer:**
The partial pressure of water vapor in the air is **21.6 mm of Hg**.