• # question_answer 5) If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mole of water at 1 bar and$100{}^\circ C$is$41kJ\text{ }mo{{l}^{-1}}$. Calculate the internal energy change, when : (i) 1 mole of water is vaporised at 1 bar pressure and$100{}^\circ C$ (ii) 1 mole of water is converted into ice.

(i) ${{H}_{2}}O(l)\to {{H}_{2}}O(g)$ $\Delta H=41KJmo{{l}^{-1}}$ $\Delta U=?$ $\Delta n=1-0=1$ $\Delta H=\Delta U+\Delta nRT$ $41=\Delta U+1\times 8314\times {{10}^{-3}}\times 373$ $\Delta U=37.9kJ$ (ii) ${{H}_{2}}O(l)\to {{H}_{2}}O(s)$ $P\Delta V\approx 0$. There is no change in volume in the process of freezing, hence$\Delta H\approx \Delta U$.