What is the compressibility factor (Z) for 0.02 mole of a van der Waals's gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible.
Given : RT=20 L atm `mol^(-1)` and a=1000 atm `L^(2) mol^(-2)`

Options

A. 2
B. 1
C. 0.02
D. 0.5

Explanation

To find the compressibility factor (Z) for the given van der Waals gas, we will follow these steps:

### Step 1: Write down the formula for the compressibility factor (Z)
The compressibility factor (Z) is defined as:

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

### Step 2: Identify the known values
From the problem, we have:
- Number of moles (n) = 0.02 moles
- Pressure (P) = 0.1 atm
- \(RT\) = 20 L atm/mol
- \(a\) = 1000 atm L²/mol²
- Size of gas molecules is negligible, implying \(b\) can be ignored.

### Step 3: Use the van der Waals equation to find the volume (V)
The van der Waals equation is given by:

\[ \left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT \]

Since the size of gas molecules is negligible, we can ignore \(b\), simplifying the equation to:

\[ \left(P + \frac{a n^2}{V^2}\right)V = nRT \]

Substituting the known values into the equation:

\[ \left(0.1 + \frac{1000 \times (0.02)^2}{V^2}\right)V = 0.02 \times 20 \]

Calculating \(0.02 \times 20\):

\[ 0.02 \times 20 = 0.4 \]

Thus, we have:

\[ \left(0.1 + \frac{1000 \times 0.0004}{V^2}\right)V = 0.4 \]

### Step 4: Rearranging the equation
This can be rearranged to:

\[ 0.1V + \frac{0.4}{V} = 0.4 \]

Multiplying through by \(V\) to eliminate the fraction:

\[ 0.1V^2 + 0.4 = 0.4V \]

Rearranging gives:

\[ 0.1V^2 - 0.4V + 0.4 = 0 \]

### Step 5: Solve the quadratic equation
To simplify, multiply through by 10 to eliminate the decimal:

\[ V^2 - 4V + 4 = 0 \]

Factoring gives:

\[ (V - 2)^2 = 0 \]

Thus, we find:

\[ V = 2 \text{ L} \]

### Step 6: Substitute back to find Z
Now that we have \(V\), we can substitute back into the formula for Z:

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

Substituting the known values:

\[ Z = \frac{0.1 \times 2}{0.02 \times 20} \]

Calculating the numerator:

\[ 0.1 \times 2 = 0.2 \]

Calculating the denominator:

\[ 0.02 \times 20 = 0.4 \]

Thus:

\[ Z = \frac{0.2}{0.4} = 0.5 \]

### Final Answer
The compressibility factor (Z) is:

\[ \boxed{0.5} \]