The value of vander waal's constant a and b for two gases X and
Y are 4.6,0.04 (for X) and 5.1, 0.06 (for Y). Which can be
liquified easily an appolication of pressure :-

Options

A. X
B. Y
C. Both 1 & 2
D. None of these

Answer

To determine which gas, X or Y, can be liquefied more easily when pressure is applied, we will analyze the Van der Waals constants A and B for both gases.

### Step-by-Step Solution:

1. **Understand Van der Waals Constants:**
- The Van der Waals constant \( a \) represents the strength of intermolecular forces. A lower value of \( a \) indicates weaker intermolecular forces, making it easier to liquefy the gas.
- The Van der Waals constant \( b \) represents the effective size of gas molecules. A lower value of \( b \) indicates that the molecules occupy less volume, which also facilitates liquefaction.

2. **Given Values:**
- For gas X:
- \( a_X = 4.6 \)
- \( b_X = 0.04 \)
- For gas Y:
- \( a_Y = 5.1 \)
- \( b_Y = 0.06 \)

3. **Compare the Values of A:**
- Compare \( a_X \) and \( a_Y \):
- \( a_X (4.6) < a_Y (5.1) \)
- This indicates that gas X has weaker intermolecular forces compared to gas Y.

4. **Compare the Values of B:**
- Compare \( b_X \) and \( b_Y \):
- \( b_X (0.04) < b_Y (0.06) \)
- This indicates that gas X occupies less volume per molecule compared to gas Y.

5. **Conclusion:**
- Since both \( a_X < a_Y \) and \( b_X < b_Y \), gas X has both weaker intermolecular forces and a smaller effective size. Therefore, gas X can be liquefied more easily than gas Y when pressure is applied.

### Final Answer:
Gas X can be liquefied more easily than gas Y.