Real gases exert less pressure when compared with ideal gases due to the attraction between the gas molecules. Reduction in pressure of real gases `prop` `("Concentration of gas")^(n)`. What is the value of n ?

Answer

To solve the problem, we need to determine the value of \( n \) in the relationship that describes how the reduction in pressure of real gases is proportional to the concentration of gas raised to the power of \( n \).

### Step-by-Step Solution:

1. **Understanding Real Gases vs. Ideal Gases**:
- Real gases exert less pressure compared to ideal gases due to intermolecular attractions. In ideal gases, these attractions are negligible.

2. **Using the Van der Waals Equation**:
- The Van der Waals equation for \( n \) moles of a real gas is given by:
\[ \left( p + \frac{a n^2}{V^2} \right)(V - nb) = nRT \]
- Here, \( a \) and \( b \) are Van der Waals constants specific to each gas.

3. **Identifying Pressure Correction**:
- The term \( \frac{a n^2}{V^2} \) represents the pressure correction due to intermolecular forces. This indicates that the actual pressure \( p \) is reduced by this term.

4. **Relating Concentration to Pressure Reduction**:
- The concentration \( c \) of the gas can be expressed as:
\[ c = \frac{n}{V} \]
- Therefore, we can express the pressure reduction in terms of concentration.

5. **Finding the Relationship**:
- The reduction in pressure \( \Delta p \) can be expressed as being proportional to the pressure correction term:
\[ \Delta p \propto \frac{a n^2}{V^2} \]
- Substituting \( c \) into this equation gives:
\[ \Delta p \propto \frac{a c^2}{V^2} \]
- Since \( c = \frac{n}{V} \), we can rewrite this as:
\[ \Delta p \propto c^2 \]

6. **Conclusion**:
- From the above relationship, we see that the reduction in pressure of real gases is directly proportional to the concentration squared. Therefore, the value of \( n \) is:
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).