`1` mol of a real gas obeys `P(V_(m) - b) = RT` , where `'b'` and `'R'` are constants. If the compressibility factor of gas is `1. 11` and occupied volume of the gas is `Xb` then determine value of `X//2`.

Explanation

To solve the problem, we start with the given equation for the real gas:

1. **Given Equation**:
\[ P(V_m - b) = RT \]
where \( b \) and \( R \) are constants.

2. **Compressibility Factor**:
The compressibility factor \( Z \) is given as:
\[ Z = 1.11 \]
The formula for the compressibility factor is:
\[ Z = 1 + \frac{Pb}{RT} \]

3. **Substituting the Value of Z**:
Substitute \( Z = 1.11 \) into the compressibility factor equation:
\[ 1.11 = 1 + \frac{Pb}{RT} \]
Rearranging gives:
\[ \frac{Pb}{RT} = 1.11 - 1 = 0.11 \]

4. **Rearranging for \( \frac{RT}{P} \)**:
From the above equation, we can express \( \frac{RT}{P} \):
\[ \frac{RT}{P} = \frac{b}{0.11} \]

5. **Using the Ideal Gas Equation**:
From the original equation \( P(V_m - b) = RT \), we can express \( V_m - b \):
\[ V_m - b = \frac{RT}{P} \]

6. **Equating the Two Expressions**:
Now we have two expressions for \( \frac{RT}{P} \):
\[ V_m - b = \frac{b}{0.11} \]
Setting them equal gives:
\[ V_m - b = \frac{b}{0.11} \]

7. **Solving for \( V_m \)**:
Rearranging the equation:
\[ V_m = b + \frac{b}{0.11} \]
To combine the terms, we can factor out \( b \):
\[ V_m = b \left(1 + \frac{1}{0.11}\right) = b \left(1 + 9.09\right) = b \cdot 10.09 \]
Thus, we approximate:
\[ V_m \approx 10b \]

8. **Relating \( V_m \) to Occupied Volume**:
We know from the problem that the occupied volume of the gas is \( Xb \):
\[ V_m = Xb \]
Setting the two equations equal:
\[ 10b = Xb \]

9. **Solving for \( X \)**:
Dividing both sides by \( b \) (assuming \( b \neq 0 \)):
\[ X = 10 \]

10. **Finding \( \frac{X}{2} \)**:
Finally, we need to find \( \frac{X}{2} \):
\[ \frac{X}{2} = \frac{10}{2} = 5 \]

Thus, the final answer is:
\[ \frac{X}{2} = 5 \]