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A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporization at 100°C. Delta_("vap") H^( Theta ) for water at 373K = 40.66 "kJ mol"^(-1)

Answer» <html><body><p></p>Solution :We can represent the process of <a href="https://interviewquestions.tuteehub.com/tag/evaporation-16312" style="font-weight:bold;" target="_blank" title="Click to know more about EVAPORATION">EVAPORATION</a> as `18g H_(2) O_((l)) overset("vaprisation")(to) 18g H_(2) O_((g))` <br/> No. of moles in `18 g, H_(2) O_((l))` is `= (18g)/( 18 "g mol"^(-1) )=1` mol <br/> `Delta_("vap") U= Delta_("vap") <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^( Theta ) - <a href="https://interviewquestions.tuteehub.com/tag/pdeltav-2211139" style="font-weight:bold;" target="_blank" title="Click to know more about PDELTAV">PDELTAV</a> = Delta_("vap") H^( Theta ) - Deltan_(g) RT` <br/> (assuming steam behaving as an ideal gas) <br/> `Delta_("vap") H^( Theta ) - Delta n_(g) RT = 40.66 "kJ mol"^(-1) -(1)` <br/> `(8.314 "<a href="https://interviewquestions.tuteehub.com/tag/jk-521675" style="font-weight:bold;" target="_blank" title="Click to know more about JK">JK</a>"^(-1) "mol"^(-1) ) (373 K) (10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) "kJ J"^(-1))` <br/> `Delta_("vap") U^( Theta ) = 40.66 "kJ mol"^(-1) -3.10 "kJ mol"^(-1)` <br/> `=37.56 "kJ mol"^(-1)`</body></html>


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