A swimmer coming out from a pool is covered a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298K ? Calculate the internal energy of vaporisation at100^(@)C. Delta_(vap) H^(@) for water at 373K = 40.66 kJ mol^(_1).

A swimmer coming out from a pool is covered a film of water weighing a

1.	A swimmer coming out from a pool is covered a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298K ? Calculate the internal energy of vaporisation at100^(@)C. Delta_(vap) H^(@) for water at 373K = 40.66 kJ mol^(_1).
Answer» <html><body><p></p>Solution :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> is `: <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) O(l ) rarr 18 g H_(2)O(g) ` <br/> No. of moles of 18 g `H_(2) O = (18g )/(18 g mol^(-1))= 1 mol` <br/> `Delta n_(g) = 1-0 =1 mol` <br/> `:. Delta _(vap) U^(@) = V_(vap) H^(@) - Delta n_(g) RT = 40.66 k J mol^(-1) - (1 mol ) (8.314 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-3) k J K^(-1) mol^(-1) ) ( 298 K) ` <br/>`= 40.66 k J mol^(-1)- 3.10 kJ mol^(-1) = 37. 56 kJ mol^(-1)`</body></html>