• # question_answer 7) A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at$100{}^\circ C$. ${{\Delta }_{vap}}{{H}^{{}^\circ }}$for water at $373K=40.66kJ\,mo{{l}^{-1}}$

Information shadow: Amount of water to evaporate = 18 g = 1 mole water ${{\Delta }_{vap}}{{H}^{{}^\circ }}=40.66kJmo{{l}^{-1}}$ Problem solving strategy: Required heat for evaporation $(q)=n{{\Delta }_{vap}}H$ Internal energy of vaporisation can be calculated as, ${{\Delta }_{vap}}H={{\Delta }_{vap}}U+\Delta {{n}_{g}}RT$ Working it out: $q=n{{\Delta }_{vap}}H$ $=1\times 40.66kJ$ $=40.66kJ$ ${{\Delta }_{vap}}U={{\Delta }_{vap}}H-\Delta {{n}_{g}}RT$ ${{H}_{2}}O(l)\to {{H}_{2}}O(g)$ $\Delta {{n}_{g}}=1-0=1$ ${{\Delta }_{vap}}U=40.66-1\times 8.314\times {{10}^{-3}}\times 373$ $=37.56\,\,kJ\,mo{{l}^{-1}}$