Standard enthalpy of vaporisation`DeltaV_(vap).H^(Theta)` for water at `100^(@)C`is`40.66kJmol^(-1)`.The internal energy of Vaporization of water at` 100^(@)C("in kJ mol"^(-1))`is

Options

A. `+37.56`
B. `-43.76`
C. `+42.76`
D. `+40.66`

Explanation

To find the internal energy of vaporization of water at 100°C, we can use the relationship between enthalpy change (ΔH) and internal energy change (ΔE). The formula we will use is:

\[ \Delta H = \Delta E + \Delta N_g RT \]

Where:
- ΔH = standard enthalpy of vaporization
- ΔE = internal energy of vaporization
- ΔN_g = change in the number of moles of gas
- R = universal gas constant (8.314 J/mol·K)
- T = temperature in Kelvin

### Step-by-Step Solution:

1. **Identify Given Values:**
- Standard enthalpy of vaporization, ΔH = 40.66 kJ/mol
- Temperature, T = 100°C = 373 K (since T(K) = T(°C) + 273.15)
- R = 8.314 J/mol·K = 0.008314 kJ/mol·K (to convert J to kJ)

2. **Calculate ΔN_g:**
- The reaction for the vaporization of water is:
\[ \text{H}_2\text{O (l)} \rightleftharpoons \text{H}_2\text{O (g)} \]
- In this reaction, there is 1 mole of gaseous product (H₂O gas) and 0 moles of gaseous reactants (H₂O liquid).
- Therefore, ΔN_g = 1 - 0 = 1.

3. **Substitute Values into the Equation:**
- Now we can substitute the values into the equation:
\[ \Delta E = \Delta H - \Delta N_g RT \]
\[ \Delta E = 40.66 \text{ kJ/mol} - (1)(0.008314 \text{ kJ/mol·K})(373 \text{ K}) \]

4. **Calculate RT:**
- Calculate RT:
\[ RT = 0.008314 \text{ kJ/mol·K} \times 373 \text{ K} = 3.101 \text{ kJ/mol} \]

5. **Final Calculation of ΔE:**
- Now substitute RT back into the equation:
\[ \Delta E = 40.66 \text{ kJ/mol} - 3.101 \text{ kJ/mol} \]
\[ \Delta E = 37.559 \text{ kJ/mol} \]

6. **Round the Result:**
- Rounding to two decimal places, we get:
\[ \Delta E \approx 37.56 \text{ kJ/mol} \]

### Conclusion:
The internal energy of vaporization of water at 100°C is approximately **37.56 kJ/mol**.