It is observed that `H_(2)` and `He` gases always show positive deviation from ideal behaviour i.e. `Zgt1`. This is because

Options

A. (a) The value of a is very large due to high attraction forces
B. (b) Weak intermolecular forces of attraction due to which a is very small and `a/V^(2)` is negligible
C. (c) the value of b is very large due to large size of the molecules
D. (d) both a and b are very small and negligible .

Answer

To understand why hydrogen (H₂) and helium (He) gases show positive deviation from ideal behavior, we need to analyze the concept of the compressibility factor (Z) and the Van der Waals equation for real gases.

### Step-by-Step Solution:

1. **Understanding Ideal Gas Behavior**:
- Ideal gases follow the ideal gas law, which is represented by the equation:
\[ PV = nRT \]
- For ideal gases, the compressibility factor \( Z \) is equal to 1, which means that the behavior of the gas perfectly follows the ideal gas law.

2. **Van der Waals Equation**:
- Real gases deviate from ideal behavior, and this deviation can be described by the Van der Waals equation:
\[ \left( P + \frac{a}{V^2} \right)(V - nb) = nRT \]
- Here, \( a \) is a measure of the attractive forces between molecules, and \( b \) is the volume occupied by the gas molecules.

3. **Positive Deviation from Ideal Behavior**:
- For hydrogen and helium, both gases are very light and have small molecular sizes. This results in weak intermolecular forces.
- Due to their small size, the repulsive forces dominate over the attractive forces.

4. **Analyzing the Constants \( A \) and \( B \)**:
- In the context of the Van der Waals equation, \( A \) (related to intermolecular forces) is very small for H₂ and He because of their weak attraction.
- The value of \( B \) (related to the volume occupied by the gas molecules) is also small since both gases have small molecular sizes.

5. **Calculating the Compressibility Factor**:
- Since \( A \) is negligible, the term \( \frac{A}{V^2} \) becomes insignificant.
- Therefore, the pressure \( P \) calculated using the Van der Waals equation will be greater than that predicted by the ideal gas law:
\[ PV > nRT \]
- This leads to the conclusion that the compressibility factor \( Z \) is greater than 1:
\[ Z = \frac{PV}{nRT} > 1 \]

6. **Conclusion**:
- The positive deviation of hydrogen and helium from ideal gas behavior is primarily due to their weak intermolecular forces of attraction, which results in a negligible \( A \) value in the Van der Waals equation. Hence, the compressibility factor \( Z \) is greater than 1.

### Final Answer:
Hydrogen (H₂) and helium (He) gases show positive deviation from ideal behavior because they have weak intermolecular forces of attraction, leading to a very small value of \( A \) in the Van der Waals equation, making \( Z > 1 \).

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