The compressibility factor for a real gas at high pressure is .

Options

A. `1`
B. `1 + Pb//RT`
C. `1-Pb//RT`
D. `1+RT//Pb`.

Explanation

To solve the question regarding the compressibility factor (Z) for a real gas at high pressure, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Compressibility Factor (Z)**:
The compressibility factor (Z) is defined as the ratio of the molar volume of a real gas to the molar volume of an ideal gas under the same conditions of temperature and pressure. It is given by the equation:
\[ Z = \frac{PV}{RT} \]
where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature.

2. **Behavior of Real Gases at High Pressure**:
At high pressures, the distance between gas molecules decreases significantly. This leads to an increase in the repulsive forces between the molecules, while the attractive forces become negligible. Therefore, we can assume:
- The attractive force (A) approaches 0.
- The repulsive force (B) becomes significant.

3. **Applying the Van der Waals Equation**:
For real gases, we can use the Van der Waals equation:
\[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \]
At high pressure, we can neglect the attractive term (A = 0), simplifying the equation to:
\[ P(V - b) = RT \]

4. **Rearranging the Equation**:
Rearranging the simplified equation gives:
\[ PV - Pb = RT \]
Dividing through by RT, we have:
\[ \frac{PV}{RT} - \frac{Pb}{RT} = 1 \]

5. **Expressing Compressibility Factor (Z)**:
From the previous step, we can express Z as:
\[ Z - \frac{Pb}{RT} = 1 \]
Therefore, we can solve for Z:
\[ Z = 1 + \frac{Pb}{RT} \]

6. **Conclusion**:
At high pressure, the compressibility factor (Z) for a real gas can be expressed as:
\[ Z = 1 + \frac{Pb}{RT} \]
This indicates that Z is greater than 1, which is consistent with the behavior of real gases under high pressure.

### Final Answer:
The compressibility factor for a real gas at high pressure is given by:
\[ Z = 1 + \frac{Pb}{RT} \]