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

Options

A. `+4376`
B. `+40.66`
C. `+37.56`
D. `-43.76`

Explanation

To find the internal energy of vaporization (\( \Delta U \)) of water at \( 100^\circ C \), we can use the relationship between enthalpy (\( \Delta H \)) and internal energy:

\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]

### Step 1: Identify the given values
- Standard enthalpy of vaporization (\( \Delta H \)) = \( 40.66 \, \text{kJ/mol} \)
- The ideal gas constant (\( R \)) = \( 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \)
- Temperature (\( T \)) = \( 100^\circ C = 373 \, \text{K} \)

### Step 2: Calculate \( \Delta N \)
- The change in the number of moles of gas (\( \Delta N \)) during the vaporization of water:
- For the reaction \( \text{H}_2\text{O (l)} \rightarrow \text{H}_2\text{O (g)} \), we have:
- Moles of gaseous products = 1 (water vapor)
- Moles of gaseous reactants = 0 (liquid water)
- Therefore, \( \Delta N = 1 - 0 = 1 \)

### Step 3: Substitute values into the equation
Now we can substitute the known values into the equation:

\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]

Substituting the values:

\[ 40.66 = \Delta U + 1 \cdot 0.008314 \cdot 373 \]

### Step 4: Calculate \( \Delta N \cdot R \cdot T \)
Calculating \( 1 \cdot 0.008314 \cdot 373 \):

\[ \Delta N \cdot R \cdot T = 0.008314 \cdot 373 = 3.101 \, \text{kJ} \]

### Step 5: Rearranging the equation to find \( \Delta U \)
Now we can rearrange the equation to solve for \( \Delta U \):

\[ \Delta U = \Delta H - \Delta N \cdot R \cdot T \]

Substituting the values:

\[ \Delta U = 40.66 - 3.101 \]

### Step 6: Perform the final calculation
Calculating \( \Delta U \):

\[ \Delta U = 40.66 - 3.101 = 37.559 \, \text{kJ/mol} \]

Thus, the internal energy of vaporization of water at \( 100^\circ C \) is approximately:

\[ \Delta U \approx 37.56 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta U \approx 37.56 \, \text{kJ/mol} \]

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