At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behavior.
Their equation of state is given as `P = (RT)/(V-b)` at T. Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs P?

Options

A. Xe
B. Ne
C. Kr
D. Ar

Explanation

To determine which gas among Ne, Ar, Xe, and Kr will exhibit the steepest increase in the plot of the compressibility factor (Z) versus pressure (P), we can follow these steps:

### Step-by-step Solution:

1. **Understand the Compressibility Factor (Z)**:
The compressibility factor Z is defined as:
\[ Z = \frac{PV}{RT} \]
For real gases, Z deviates from 1, especially under high pressures.

2. **Use the Given Equation of State**:
The equation of state provided is:
\[ P = \frac{RT}{V - b} \]
Rearranging this gives us:
\[ V = \frac{RT}{P} + b \]

3. **Substituting into the Compressibility Factor Equation**:
Substitute \(V\) from the rearranged equation into the Z equation:
\[ Z = \frac{P\left(\frac{RT}{P} + b\right)}{RT} = 1 + \frac{Pb}{RT} \]

4. **Analyze the Relationship**:
From the equation \(Z = 1 + \frac{Pb}{RT}\), we see that Z is directly proportional to the van der Waals constant \(b\):
\[ Z \propto b \]
This indicates that the larger the value of \(b\), the steeper the increase of Z with pressure P.

5. **Identify the Van der Waals Constant (b)**:
The value of \(b\) is related to the size of the gas atoms. Larger atoms have larger values of \(b\). The order of atomic size for the given gases is:
- Neon (Ne) < Argon (Ar) < Krypton (Kr) < Xenon (Xe)

6. **Conclusion**:
Since Xenon (Xe) has the largest atomic size, it will have the largest value of \(b\) and therefore will exhibit the steepest increase in the plot of Z versus P.

### Final Answer:
The gas that will exhibit the steepest increase in the plot of Z versus P is **Xenon (Xe)**.