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 1: Understand the Compressibility Factor (Z)
The compressibility factor (Z) is defined as:
\[ Z = \frac{PV}{RT} \]
Where:
- \( P \) = pressure
- \( V \) = molar volume
- \( R \) = universal gas constant
- \( T \) = temperature

### Step 2: Analyze the Given Equation of State
The equation of state provided is:
\[ P = \frac{RT}{V - b} \]
Where \( b \) is the van der Waals constant, which represents the excluded volume per mole of gas.

### Step 3: Rearranging the Equation for Z
From the equation of state, we can express \( V \) in terms of \( P \):
\[ V = \frac{RT}{P} + b \]
Substituting this back into the Z equation gives:
\[ Z = \frac{P\left(\frac{RT}{P} + b\right)}{RT} \]
This simplifies to:
\[ Z = 1 + \frac{Pb}{RT} \]

### Step 4: Identify the Relationship Between Z and P
From the equation \( Z = 1 + \frac{Pb}{RT} \), we can see that:
- The slope of the graph of \( Z \) versus \( P \) is given by \( \frac{b}{RT} \).
- Since \( R \) and \( T \) are constants for the gases at a given temperature, the steepness of the increase in Z with respect to P depends solely on the value of \( b \).

### Step 5: Compare the Values of b for Each Gas
The van der Waals constant \( b \) generally increases with the size of the gas molecules. The order of size for the gases is:
- Ne (Neon) < Ar (Argon) < Kr (Krypton) < Xe (Xenon)

Thus, we can conclude:
- Ne has the smallest \( b \)
- Ar has a larger \( b \) than Ne
- Kr has a larger \( b \) than Ar
- Xe has the largest \( b \)

### Step 6: Determine Which Gas Exhibits the Steepest Increase
Since the slope of the Z vs P graph is proportional to \( b \), the gas with the largest \( b \) will exhibit the steepest increase in the plot of Z versus P. Therefore, among Ne, Ar, Kr, and Xe, **Xenon (Xe)** will exhibit the steepest increase in the plot of Z versus P.

### Final Answer:
**Xenon (Xe)** will exhibit the steepest increase in the plot of Z vs P.

---