In the corrections made to the ideal gas equation for real gases, the reduction in pressure due to attractive forces is directly proportional to :

Options

A. `(n)/(V)`
B. `(n^(2))/(V^(2))`
C. V - nb
D. nb

Explanation

To solve the question regarding the reduction in pressure due to attractive forces in real gases, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**: The ideal gas equation is given by \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

2. **Identify Corrections for Real Gases**: Real gases deviate from ideal behavior due to intermolecular forces. To account for these deviations, we introduce corrections to the ideal gas law.

3. **Reduction in Pressure**: The reduction in pressure in real gases is primarily due to attractive forces between gas molecules. This reduction can be expressed in terms of a corrective factor.

4. **Corrective Factor**: The correction for pressure can be expressed as:
\[ P_{\text{real}} = P_{\text{ideal}} - \text{correction} \]
The correction for pressure due to attractive forces can be represented as:
\[ \text{correction} \propto \frac{n^2}{V^2} \]
where \( n \) is the number of moles and \( V \) is the volume.

5. **Relate to the Given Variables**: In the context of the question, we need to relate the reduction in pressure to the number of moles and the volume. The reduction in pressure due to attractive forces is directly proportional to:
\[ \frac{n^2}{b^2} \]
where \( b \) is a constant related to the volume occupied by the gas molecules.

6. **Conclusion**: Therefore, the reduction in pressure due to attractive forces is directly proportional to \( \frac{n^2}{b^2} \).

### Final Answer:
The reduction in pressure due to attractive forces is directly proportional to \( \frac{n^2}{b^2} \).