Standard enthalpy of vapourisation `Delta_("vap")H^(-)` for water at `100^@C` is `40.66 kJ mol^(-1)`. The internal energy of vapourisation of water at `100^@C` (in kJ `mol^(-1)`) is (Assume water vapour to behave like an ideal gas).

Options

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

Explanation

To find the internal energy of vaporization (ΔU) of water at 100°C, we can use the relationship between enthalpy (ΔH) and internal energy (ΔU):

\[ \Delta H = \Delta U + nRT \]

Where:
- ΔH is the enthalpy of vaporization (given as 40.66 kJ/mol)
- ΔU is the internal energy of vaporization (what we need to find)
- n is the number of moles (we will take it as 1 mole)
- R is the ideal gas constant (8.314 J/(K·mol))
- T is the temperature in Kelvin

### Step 1: Convert the temperature to Kelvin
The temperature is given as 100°C. To convert this to Kelvin:
\[ T(K) = 100 + 273 = 373 \text{ K} \]

### Step 2: Convert ΔH from kJ to J
The enthalpy of vaporization is given in kJ/mol, so we need to convert it to J/mol:
\[ \Delta H = 40.66 \text{ kJ/mol} = 40.66 \times 1000 \text{ J/mol} = 40660 \text{ J/mol} \]

### Step 3: Substitute values into the equation
Now we can substitute the known values into the equation:
\[ 40660 = \Delta U + (1)(8.314)(373) \]

### Step 4: Calculate nRT
Calculate \(nRT\):
\[ nRT = 1 \times 8.314 \times 373 = 3101.122 \text{ J/mol} \]

### Step 5: Rearrange the equation to solve for ΔU
Now we can rearrange the equation to solve for ΔU:
\[ \Delta U = 40660 - 3101.122 \]

### Step 6: Perform the calculation
Now, calculate ΔU:
\[ \Delta U = 40660 - 3101.122 = 37558.878 \text{ J/mol} \]

### Step 7: Convert ΔU back to kJ
Finally, convert ΔU back to kJ:
\[ \Delta U = \frac{37558.878}{1000} = 37.558878 \text{ kJ/mol} \]

### Final Answer
The internal energy of vaporization of water at 100°C is approximately:
\[ \Delta U \approx 37.56 \text{ kJ/mol} \]