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

### Step-by-step Solution:

1. **Identify the Given Values**:
- Molar enthalpy change for vaporization (ΔH) = 41 kJ/mol
- Temperature (T) = 100°C = 373 K (since T(K) = T(°C) + 273)
- Gas constant (R) = 8.314 J/(mol·K)
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