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 `e ^(-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 can use the formula:

\[ Z = \frac{PV}{RT} \]

where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( R \) = ideal gas constant in L atm K\(^{-1}\) mol\(^{-1}\)
- \( T \) = temperature in Kelvin

### Step 1: Calculate the volume of the vapours

Given:
- Molar mass of the substance = 18 g/mol
- Density of vapours = 0.36 kg/m³

First, we need to convert the density into g/L for easier calculations:
\[ 0.36 \text{ kg/m}^3 = 0.36 \times 1000 \text{ g/m}^3 = 360 \text{ g/m}^3 \]

Now, we can find the volume using the formula:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]

Substituting the values:
\[ \text{Volume} = \frac{18 \text{ g}}{360 \text{ g/m}^3} = 0.05 \text{ m}^3 \]

To convert this volume into liters:
\[ 0.05 \text{ m}^3 = 0.05 \times 1000 \text{ L} = 50 \text{ L} \]

### Step 2: Substitute the values into the compressibility factor formula

Now we have:
- \( P = 1 \) atm
- \( V = 50 \) L
- \( R = 0.082 \) L atm K\(^{-1}\) mol\(^{-1}\)
- \( T = 500 \) K

Substituting these values into the formula for \( Z \):
\[ Z = \frac{PV}{RT} = \frac{(1 \text{ atm})(50 \text{ L})}{(0.082 \text{ L atm K}^{-1} \text{ mol}^{-1})(500 \text{ K})} \]

### Step 3: Calculate \( Z \)

Calculating the denominator:
\[ RT = 0.082 \times 500 = 41 \text{ L atm} \]

Now substituting back into the equation for \( Z \):
\[ Z = \frac{50}{41} \approx 1.22 \]

### Conclusion

Thus, the value of the compressibility factor \( Z \) for the vapours is approximately \( 1.22 \).