On a humid day, the mole fraction of water vapour in air at 298 K is 0.0265 bar. If total vapour pressure of air is 0.980 bar and vapour pressure of water at 298 K is 0.0315 bar, the relative humidity is

Options

A. 82.5
B. 88.6
C. 76.4
D. 28.5

Explanation

To find the relative humidity, we can follow these steps:

### Step 1: Understand the formula for relative humidity
Relative humidity (RH) is defined as the ratio of the partial pressure of water vapor in the air to the saturation vapor pressure of water at the same temperature, multiplied by 100 to express it as a percentage.

\[ \text{Relative Humidity (RH)} = \left( \frac{P_{\text{H}_2\text{O}}}{P_{\text{H}_2\text{O}}^{\text{sat}}} \right) \times 100 \]

Where:
- \( P_{\text{H}_2\text{O}} \) = partial pressure of water vapor in the air
- \( P_{\text{H}_2\text{O}}^{\text{sat}} \) = saturation vapor pressure of water at 298 K

### Step 2: Calculate the partial pressure of water vapor in the air
We are given the mole fraction of water vapor (\( \chi_{\text{H}_2\text{O}} \)) and the total vapor pressure (\( P_{\text{total}} \)). We can calculate the partial pressure of water vapor using the formula:

\[ P_{\text{H}_2\text{O}} = \chi_{\text{H}_2\text{O}} \times P_{\text{total}} \]

Substituting the values:
- \( \chi_{\text{H}_2\text{O}} = 0.0265 \)
- \( P_{\text{total}} = 0.980 \, \text{bar} \)

\[ P_{\text{H}_2\text{O}} = 0.0265 \times 0.980 = 0.0260 \, \text{bar} \]

### Step 3: Use the saturation vapor pressure
We are given the saturation vapor pressure of water at 298 K:
- \( P_{\text{H}_2\text{O}}^{\text{sat}} = 0.0315 \, \text{bar} \)

### Step 4: Calculate the relative humidity
Now we can substitute the values into the relative humidity formula:

\[ \text{Relative Humidity (RH)} = \left( \frac{0.0260}{0.0315} \right) \times 100 \]

Calculating this gives:

\[ \text{Relative Humidity (RH)} = \left( 0.826 \right) \times 100 \approx 82.6\% \]

### Final Answer:
The relative humidity is approximately **82.6%**.

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