On a certain day the vapour pressure is 14 mm of Hg for water vapour in air at `20^@C` . The saturated vapour pressure is 17.5 mm. How many moles of water vapour per litre of air would be required to saturate air at this temperature ?

Options

A. `9.41 xx 10^(-4) `
B. `9.14 xx 10^(-4)`
C. `1.914 xx 10^(-4) `
D. `4.19 xx 10^(-4)`

Answer

To solve the problem, we need to find the number of moles of water vapor required to saturate the air at a given temperature. Here are the steps to arrive at the solution:

### Step 1: Identify the Given Data
- Vapour pressure of water vapor in air (P_air) = 14 mm Hg
- Saturated vapour pressure (P₀) = 17.5 mm Hg
- Temperature (T) = 20°C

### Step 2: Calculate the Pressure of Water Vapor Needed for Saturation
To find the pressure of water vapor required to saturate the air, we subtract the vapour pressure of water vapor in air from the saturated vapour pressure:
\[ P = P₀ - P_{\text{air}} \]
\[ P = 17.5 \, \text{mm Hg} - 14 \, \text{mm Hg} \]
\[ P = 3.5 \, \text{mm Hg} \]

### Step 3: Convert Pressure from mm Hg to atm
To use the ideal gas law, we need the pressure in atmospheres (atm). We convert mm Hg to atm using the conversion factor (1 atm = 760 mm Hg):
\[ P_{\text{atm}} = \frac{3.5 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \]
\[ P_{\text{atm}} \approx 0.004605 \, \text{atm} \]

### Step 4: Use the Ideal Gas Law to Find the Number of Moles
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters (1 L in this case)
- \( n \) = number of moles
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin = 20°C + 273.15 = 293.15 K

Rearranging the ideal gas law to solve for \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(0.004605 \, \text{atm}) \times (1 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (293.15 \, \text{K})} \]

### Step 5: Calculate the Number of Moles
Calculating the denominator:
\[ 0.0821 \times 293.15 \approx 24.1 \]

Now substituting back:
\[ n = \frac{0.004605}{24.1} \]
\[ n \approx 0.000191 \, \text{moles} \]
\[ n \approx 1.91 \times 10^{-4} \, \text{moles} \]

### Final Answer
The number of moles of water vapor required to saturate the air at 20°C is approximately \( 1.91 \times 10^{-4} \) moles per litre.

---