Assuming that water vapour is an ideal gas, the internal energy change `(Delta U)` when `1 mol` of water is vapourised at `1` bar pressure and `100^(@)C`, (Given: Molar enthalpy of vapourization of water at `1` bar and `373K=41 kJ mol^(-1)` and `R=8.3J mol^(-1)K^(-1)`) will be:

Options

A. `41.00 kJ mol^(-1)`
B. `4.100 kJmol^(-1)`
C. `3.7904 kJ mol^(-1)`
D. `37.904 kJmol^(-1)`

Explanation

To find the internal energy change (ΔU) when 1 mole of water is vaporized at 1 bar pressure and 100°C, we can use the relationship between the change in internal energy (ΔU) and the change in enthalpy (ΔH):

\[ \Delta U = \Delta H - \Delta N_g RT \]

### Step 1: Identify the given values
- Molar enthalpy of vaporization (ΔH) = 41 kJ/mol = 41,000 J/mol
- R (universal gas constant) = 8.3 J/(mol·K)
- Temperature (T) = 100°C = 373 K (since 100°C = 373 K)
- Number of moles of gaseous products (ΔN_g) = 1 (for 1 mole of water vapor) - 0 (for liquid water) = 1

### Step 2: Calculate ΔN_g
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 1 - 0 = 1 \]

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

\[ \Delta U = \Delta H - \Delta N_g RT \]

Substituting the values:

\[ \Delta U = 41000 \, \text{J/mol} - (1)(8.3 \, \text{J/(mol·K)})(373 \, \text{K}) \]

### Step 4: Calculate the second term (ΔN_g RT)
Calculating the second term:

\[ \Delta N_g RT = 1 \cdot 8.3 \cdot 373 = 3093.9 \, \text{J/mol} \]

### Step 5: Substitute back to find ΔU
Now substituting this back into the equation for ΔU:

\[ \Delta U = 41000 \, \text{J/mol} - 3093.9 \, \text{J/mol} \]

\[ \Delta U = 37906.1 \, \text{J/mol} \]

### Step 6: Convert ΔU to kJ/mol
To convert ΔU from J to kJ:

\[ \Delta U = \frac{37906.1 \, \text{J/mol}}{1000} = 37.9061 \, \text{kJ/mol} \]

### Step 7: Final answer
Thus, the internal energy change (ΔU) when 1 mole of water is vaporized at 1 bar pressure and 100°C is approximately:

\[ \Delta U \approx 37.91 \, \text{kJ/mol} \]