What is the density of wet air with 75% relative humidity at 1 atm and 300 K? Given : vapour pressure of `H_(2)O` is 30 torr and average molar mass of air is 29 g `mol^(_1)` .

Options

A. 1.614 g/L
B. 0.96 g/L
C. 1.06 g/L
D. 1.164 g/L

Explanation

To find the density of wet air with 75% relative humidity at 1 atm and 300 K, we can follow these steps:

### Step 1: Calculate the Partial Pressure of Water Vapor
The formula for relative humidity (RH) is given by:

\[ \text{RH} = \left( \frac{P_{H_2O}}{P_{H_2O}^{\text{sat}}} \right) \times 100 \]

Where:
- \( P_{H_2O} \) = partial pressure of water vapor
- \( P_{H_2O}^{\text{sat}} \) = saturation vapor pressure of water at the given temperature

Given:
- \( \text{RH} = 75\% \)
- \( P_{H_2O}^{\text{sat}} = 30 \, \text{torr} \)

Rearranging the formula to find \( P_{H_2O} \):

\[ P_{H_2O} = \frac{\text{RH} \times P_{H_2O}^{\text{sat}}}{100} \]

Substituting the values:

\[ P_{H_2O} = \frac{75 \times 30}{100} = 22.5 \, \text{torr} \]

### Step 2: Calculate the Mole Fraction of Water Vapor
To find the mole fraction of water vapor in the air, we need to convert the partial pressure of water vapor to atm (1 atm = 760 torr):

\[ P_{H_2O} = \frac{22.5 \, \text{torr}}{760 \, \text{torr/atm}} = 0.0296 \, \text{atm} \]

### Step 3: Calculate the Mole Fraction of Dry Air
The total pressure of the air is 1 atm, so the partial pressure of dry air \( P_{dry \, air} \) is:

\[ P_{dry \, air} = P_{total} - P_{H_2O} = 1 - 0.0296 = 0.9704 \, \text{atm} \]

The mole fraction of water vapor \( X_{H_2O} \) can be calculated as:

\[ X_{H_2O} = \frac{P_{H_2O}}{P_{total}} = \frac{0.0296}{1} = 0.0296 \]

### Step 4: Calculate the Molar Mass of Wet Air
The average molar mass of air is given as 29 g/mol, and the molar mass of water \( H_2O \) is 18 g/mol. The molar mass of wet air can be calculated as:

\[ M_{wet \, air} = X_{H_2O} \times M_{H_2O} + (1 - X_{H_2O}) \times M_{air} \]

Substituting the values:

\[ M_{wet \, air} = (0.0296 \times 18) + (0.9704 \times 29) \]

Calculating:

\[ M_{wet \, air} = 0.5328 + 28.188 = 28.7208 \, \text{g/mol} \]

### Step 5: Calculate the Density of Wet Air
Using the ideal gas law, the density \( \rho \) can be calculated using the formula:

\[ \rho = \frac{PM}{RT} \]

Where:
- \( P = 1 \, \text{atm} \)
- \( M = 28.7208 \, \text{g/mol} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)

Substituting the values:

\[ \rho = \frac{1 \times 28.7208}{0.0821 \times 300} \]

Calculating:

\[ \rho = \frac{28.7208}{24.63} \approx 1.165 \, \text{g/L} \]

### Final Answer
The density of wet air with 75% relative humidity at 1 atm and 300 K is approximately **1.165 g/L**.

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