If the volume of the gas containing n number of molecules is V, then the pressure will decrease due to force of intermolecular attraction in the proportion

Options

A. `n//V`
B. `n//V^2`
C. `(n//V)^2`
D. `1//V^2`

Explanation

To solve the problem of how the pressure of a gas decreases due to intermolecular attractions, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to determine how the pressure of a gas containing 'n' molecules in volume 'V' decreases due to intermolecular attractions.

2. **Identify the Relevant Equation**: The Van der Waals equation is applicable here because it accounts for intermolecular forces. The equation is given by:
\[ \left( P + \frac{aN^2}{V^2} \right) (V - Nb) = NRT \]
where:
- \( P \) = pressure of the gas
- \( N \) = number of moles (or molecules)
- \( R \) = universal gas constant
- \( T \) = temperature
- \( a \) and \( b \) = constants specific to the gas

3. **Rearranging the Equation**: We can rearrange the Van der Waals equation to express pressure \( P \):
\[ P = \frac{NRT}{V - Nb} - \frac{aN^2}{V^2} \]
Here, the second term \(-\frac{aN^2}{V^2}\) represents the decrease in pressure due to intermolecular attractions.

4. **Analyzing the Decrease in Pressure**: The term \(-\frac{aN^2}{V^2}\) indicates that the pressure decreases in proportion to \( \frac{N^2}{V^2} \). Since we are interested in how the pressure decreases due to intermolecular attraction, we can express this as:
\[ \text{Decrease in Pressure} \propto \frac{N^2}{V^2} \]

5. **Relating to Given Options**: The problem provides options in terms of \( n \) and \( b \). If we consider \( V \) to be proportional to \( b \) (the volume occupied by one mole of gas), we can express the decrease in pressure as:
\[ \text{Decrease in Pressure} \propto \frac{N}{b^2} \]

6. **Final Conclusion**: Therefore, the pressure will decrease due to the force of intermolecular attraction in the proportion:
\[ \frac{N}{b^2} \]
This corresponds to option **C**.