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 \[373K=40.66kJ\,mo{{l}^{-1}}\]

Answer:

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: \[q=n{{\Delta }_{vap}}H\] \[=1\times 40.66kJ\] \[=40.66kJ\] \[{{\Delta }_{vap}}U={{\Delta }_{vap}}H-\Delta {{n}_{g}}RT\] \[{{H}_{2}}O(l)\to {{H}_{2}}O(g)\] \[\Delta {{n}_{g}}=1-0=1\] \[{{\Delta }_{vap}}U=40.66-1\times 8.314\times {{10}^{-3}}\times 373\] \[=37.56\,\,kJ\,mo{{l}^{-1}}\]