The density of vapours of a substance of molar mass 18 g at 1 atm pressure and 500 K is 0.36 kg `m^(-3)`. The value of compressibility factor Z for the vapours will be (Take R = 0.082 L atm `mol ^(-1) K^(-1)`

Options

A. 0.69
B. 0.82
C. 1.22
D. 1.45

Answer

To find the compressibility factor \( Z \) for the vapours, we will follow these steps:

### Step 1: Write down the given data
- Molar mass \( M = 18 \, \text{g/mol} \)
- Density \( \rho = 0.36 \, \text{kg/m}^3 = 0.36 \, \text{g/dm}^3 \)
- Pressure \( P = 1 \, \text{atm} = 101.3 \, \text{kPa} \)
- Temperature \( T = 500 \, \text{K} \)
- Gas constant \( R = 0.082 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1} \)

### Step 2: Convert density to appropriate units
Since we will be using the gas constant in L atm, we can convert the density from kg/m³ to g/L:
\[ \rho = 0.36 \, \text{kg/m}^3 = 0.36 \times 1000 \, \text{g/m}^3 = 0.36 \, \text{g/L} \]

### Step 3: Calculate the volume using the density
Using the formula for density:
\[ \rho = \frac{m}{V} \]
We can rearrange it to find the volume \( V \):
\[ V = \frac{m}{\rho} \]
Where \( m \) (mass) can be calculated from the molar mass:
\[ m = 18 \, \text{g} \]
Now substituting the values:
\[ V = \frac{18 \, \text{g}}{0.36 \, \text{g/L}} = 50 \, \text{L} \]

### Step 4: Calculate the compressibility factor \( Z \)
The compressibility factor is given by the formula:
\[ Z = \frac{PV}{RT} \]
Substituting the values we have:
\[ P = 1 \, \text{atm}, \quad V = 50 \, \text{L}, \quad R = 0.082 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1}, \quad T = 500 \, \text{K} \]
\[ Z = \frac{(1 \, \text{atm})(50 \, \text{L})}{(0.082 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1})(500 \, \text{K})} \]
Calculating the denominator:
\[ Z = \frac{50}{0.082 \times 500} = \frac{50}{41} \approx 1.22 \]

### Final Answer
The value of the compressibility factor \( Z \) for the vapours is approximately **1.22**.

---