The compressibility factor `Z = ((PV)/(nRT))` of a gas above `T = (a)/(Rb)` will be :

Options

A. less than unity
B. grater than unity
C. equal to unity
D. none of these

Explanation

To solve the problem regarding the compressibility factor \( Z \) of a gas above the temperature \( T = \frac{A}{Rb} \), we can follow these steps:

### Step 1: Understand the Compressibility Factor
The compressibility factor \( Z \) is defined as:
\[ Z = \frac{PV}{nRT} \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of the gas, \( R \) is the universal gas constant, and \( T \) is the temperature.

### Step 2: Identify the Temperature Condition
The problem states that we are considering the behavior of the gas above the temperature \( T = \frac{A}{Rb} \). This temperature is known as the Boyle's temperature, where the gas behaves ideally.

### Step 3: Analyze the Behavior Above Boyle's Temperature
At the Boyle's temperature, gases exhibit ideal behavior, meaning that \( Z = 1 \). However, when the temperature increases beyond this point, the gas begins to deviate from ideal behavior.

### Step 4: Determine the Effect of Increased Temperature
As the temperature increases above the Boyle's temperature, the kinetic energy of the gas molecules increases, leading to more significant interactions between them. This results in the gas becoming less compressible, which means that the compressibility factor \( Z \) will increase.

### Step 5: Conclude the Value of \( Z \)
Since \( Z = 1 \) at the Boyle's temperature and increases with temperature above this point, we can conclude that:
\[ Z > 1 \]

### Final Answer
The compressibility factor \( Z \) of a gas above \( T = \frac{A}{Rb} \) will be greater than unity:
\[ Z > 1 \]

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