The normal boiling point of a liquid X is 400 K. `DeltaH_("vap")` at normal boiling point is 40 kJ/mol. Select correct statement(s) :

Options

A. `DeltaS_("vaporisation")lt100` J/mol.K at 400 K and 2 atm
B. `DeltaS_("vaporisation")lt10` J/mol.K at 400 K and 1 atm
C. `DeltaG_("vaporisation")lt 0` at 410 K and 1 atm
D. `DeltaU=43.32` kJ/mol.K at 400 K and 1 atm

Explanation

To solve the problem, we need to analyze the statements given in the question based on the provided data about liquid X. We have the normal boiling point (400 K) and the enthalpy of vaporization (ΔH_vap = 40 kJ/mol).

### Step-by-Step Solution:

1. **Calculate ΔS_vap (Entropy of Vaporization)**:
\[ \Delta S_{vap} = \frac{\Delta H_{vap}}{T} \]
Given:
- ΔH_vap = 40 kJ/mol = 40,000 J/mol (conversion to Joules)
- T = 400 K

Substituting the values:
\[ \Delta S_{vap} = \frac{40,000 \, \text{J/mol}}{400 \, \text{K}} = 100 \, \text{J/mol·K} \]

2. **Evaluate Statement A**:
- Statement A claims that ΔS_vap is less than 100 J/mol·K at 400 K and 2 atm.
- At higher pressure (2 atm), the boiling point increases, leading to a decrease in ΔS_vap. Thus, this statement is **correct**.

3. **Evaluate Statement B**:
- Statement B claims that ΔS_vap is less than 10 J/mol·K at 400 K and 1 atm.
- We calculated ΔS_vap to be 100 J/mol·K at 1 atm, which contradicts this statement. Thus, this statement is **incorrect**.

4. **Calculate ΔG_vap (Gibbs Free Energy of Vaporization)**:
\[ \Delta G_{vap} = \Delta H_{vap} - T \cdot \Delta S_{vap} \]
For T = 410 K:
\[ \Delta G_{vap} = 40,000 \, \text{J/mol} - 410 \, \text{K} \cdot 100 \, \text{J/mol·K} \]
\[ \Delta G_{vap} = 40,000 \, \text{J/mol} - 41,000 \, \text{J/mol} = -1,000 \, \text{J/mol} \]
Since ΔG_vap is negative, Statement C is **correct**.

5. **Calculate ΔU (Internal Energy Change)**:
Using the relation:
\[ \Delta H = \Delta U + \Delta N_g RT \]
Assuming ΔN_g = 1 (1 mole of gas produced from liquid):
\[ 40 \, \text{kJ/mol} = \Delta U + (1) \cdot (8.314 \, \text{J/mol·K}) \cdot (400 \, \text{K}) \]
Converting 40 kJ to J:
\[ 40,000 \, \text{J/mol} = \Delta U + 3,325.6 \, \text{J/mol} \]
\[ \Delta U = 40,000 \, \text{J/mol} - 3,325.6 \, \text{J/mol} = 36,674.4 \, \text{J/mol} \approx 36.67 \, \text{kJ/mol} \]
Since Statement D claims ΔU is 43.32 kJ/mol, which is incorrect, Statement D is **incorrect**.

### Summary of Correct Statements:
- Statement A: Correct
- Statement B: Incorrect
- Statement C: Correct
- Statement D: Incorrect

### Final Answer:
The correct statements are A and C.

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