Standard enthalpy of vapourisation `DeltaH^(@)` forwater is `40.66 KJ mol^(-1)`.The internal energy of vapourisation of water for its 2 mol will be :-

Options

A. `+43.76`
B. `+40.66`
C. `+37.56`
D. None of these

Explanation

To find the internal energy of vaporization of water for 2 moles, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Standard enthalpy of vaporization of water, \( \Delta H^\circ = 40.66 \, \text{kJ/mol} \)
- We need to find the internal energy of vaporization for 2 moles of water.

2. **Understand the relationship between enthalpy and internal energy:**
The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)) can be expressed as:
\[ \Delta H = \Delta E + \Delta N_g RT \]
where:
- \( \Delta N_g \) = change in the number of moles of gas
- \( R \) = universal gas constant \( = 8.314 \, \text{J/(mol K)} \)
- \( T \) = temperature in Kelvin

3. **Calculate \( \Delta N_g \):**
- For the vaporization of water (liquid to gas), we have:
- 1 mole of water vapor (product)
- 0 moles of water (reactant)
- Thus, \( \Delta N_g = 1 - 0 = 1 \).

4. **Convert the gas constant \( R \) to kJ:**
\[ R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \]

5. **Convert the boiling point of water to Kelvin:**
- The boiling point of water is \( 100^\circ C \).
- In Kelvin, \( T = 100 + 273 = 373 \, \text{K} \).

6. **Substitute the values into the equation:**
\[ 40.66 = \Delta E + (1)(0.008314)(373) \]

7. **Calculate \( \Delta N_g RT \):**
\[ \Delta N_g RT = 0.008314 \times 373 \approx 3.10 \, \text{kJ} \]

8. **Rearranging the equation to find \( \Delta E \):**
\[ \Delta E = 40.66 - 3.10 = 37.56 \, \text{kJ/mol} \]

9. **Calculate for 2 moles:**
\[ \Delta E \text{ for 2 moles} = 2 \times 37.56 = 75.12 \, \text{kJ} \]

### Final Answer:
The internal energy of vaporization of water for 2 moles is approximately **75.12 kJ**.