The behavior of the gas becomes more ideal at
I. Very low pressure
II. Value of Z is unity
III. Very high pressure
IV. Value of Z is greater than one
Choose the correct option.

Options

A. I and II are correct
B. I and IV are correct
C. I and III are correct
D. III and IV are correct

Explanation

To determine when the behavior of a gas becomes more ideal, we can analyze the given statements based on the compressibility factor (Z) and the conditions of pressure.

### Step-by-Step Solution:

1. **Understanding the Compressibility Factor (Z)**:
- The compressibility factor (Z) is defined as \( Z = \frac{PV}{nRT} \).
- For an ideal gas, \( Z \) is equal to 1. This means that the gas follows the ideal gas law perfectly.

2. **Analyzing the Statements**:
- **Statement I: Very low pressure**:
- At very low pressures, gases tend to behave more ideally. This is because the intermolecular forces become negligible, and the volume occupied by the gas molecules becomes insignificant compared to the volume of the container. Therefore, this statement is true.

- **Statement II: Value of Z is unity**:
- If \( Z = 1 \), it indicates that the gas behaves ideally. This is the definition of an ideal gas. Therefore, this statement is also true.

- **Statement III: Very high pressure**:
- At very high pressures, gases deviate from ideal behavior due to significant intermolecular forces and the volume occupied by gas molecules. Therefore, this statement is false.

- **Statement IV: Value of Z is greater than one**:
- When \( Z > 1 \), it indicates that the gas is behaving less ideally and is deviating from the ideal gas law. This typically occurs at high pressures. Therefore, this statement is false.

3. **Conclusion**:
- The correct statements that indicate the behavior of the gas becomes more ideal are:
- I. Very low pressure
- II. Value of Z is unity

### Final Answer:
The correct options are I and II.