The density of vapour of a substance `(X)` at 1 atm pressure and 500 K is `0.8 kg//m^(3)`. The vapour effuse through a small hole at a rate of `4//5` times slower than oxygen under the same condition. What is the compressibility factor `(z)` of the vapour ?

Options

A. 0.974
B. 1.35
C. 1.52
D. 1.22

Explanation

To find the compressibility factor (Z) of the vapor, we can follow these steps:

### Step 1: Determine the Molar Mass of the Vapor
Given that the vapor effuses at a rate of \( \frac{4}{5} \) times slower than oxygen, we can use Graham's law of effusion, which states that the rate of effusion is inversely proportional to the square root of the molar mass.

Let:
- \( R_X \) = Rate of effusion of vapor
- \( R_{O_2} \) = Rate of effusion of oxygen

From the problem, we have:
\[ \frac{R_X}{R_{O_2}} = \frac{4}{5} \]

According to Graham's law:
\[ \frac{R_X}{R_{O_2}} = \sqrt{\frac{M_{O_2}}{M_X}} \]

Where \( M_{O_2} \) is the molar mass of oxygen (32 g/mol) and \( M_X \) is the molar mass of the vapor.

Setting the two equations equal gives:
\[ \frac{4}{5} = \sqrt{\frac{32}{M_X}} \]

### Step 2: Solve for Molar Mass of the Vapor
Squaring both sides:
\[ \left(\frac{4}{5}\right)^2 = \frac{32}{M_X} \]
\[ \frac{16}{25} = \frac{32}{M_X} \]

Cross-multiplying gives:
\[ 16M_X = 32 \times 25 \]
\[ M_X = \frac{800}{16} = 50 \text{ g/mol} \]

### Step 3: Calculate the Compressibility Factor (Z)
The compressibility factor \( Z \) is given by the formula:
\[ Z = \frac{PV_m}{nRT} \]

Where:
- \( P \) = Pressure = 1 atm
- \( V_m \) = Molar volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = Temperature = 500 K

First, we need to find \( n \):
\[ n = \frac{m}{M_X} \]
Where \( m \) is the mass of the vapor. We can relate density (D) to mass and volume:
\[ D = \frac{m}{V} \implies m = D \cdot V \]

Using the ideal gas law, \( PV = nRT \), we can express \( V_m \):
\[ V_m = \frac{RT}{P} \]

Substituting \( n \) into the equation for \( Z \):
\[ Z = \frac{P \cdot \frac{m}{D}}{RT} = \frac{P \cdot D \cdot V}{D \cdot R \cdot T} = \frac{P \cdot M_X}{D \cdot R \cdot T} \]

### Step 4: Substitute Values
Substituting the known values:
- \( P = 1 \text{ atm} \)
- \( M_X = 50 \text{ g/mol} = 0.05 \text{ kg/mol} \)
- \( D = 0.8 \text{ kg/m}^3 \)
- \( R = 0.0821 \text{ L·atm/(K·mol)} \) (convert to consistent units if necessary)
- \( T = 500 \text{ K} \)

Now substituting into the equation:
\[ Z = \frac{1 \cdot 0.05}{0.8 \cdot 0.0821 \cdot 500} \]

Calculating:
\[ Z = \frac{0.05}{0.8 \cdot 0.0821 \cdot 500} \]
\[ Z = \frac{0.05}{32.84} \approx 1.52 \]

### Final Answer
The compressibility factor \( Z \) of the vapor is approximately **1.52**.

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