question_answer 1)

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.

Answer:

(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\[\Delta H\approx \Delta U\].