If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and `100^(@)C` is 41kJ `"mol"^(-1)`. Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and `100^(@)C.`

Explanation

To calculate 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 enthalpy change (ΔH) and internal energy change (ΔU). The formula we will use is:

\[ \Delta H = \Delta U + \Delta N_{gas} \cdot R \cdot T \]

Where:
- ΔH is the molar enthalpy change (given as 41 kJ/mol)
...