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InterviewSolution
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Water is boiled under a pressure of `1.0 atm`. When an electric current of `0.50A` from a `12V` supply is passed for `300`s through a resistance in thermal contact with it, it is found that `0.789 g` of water is vaporied. Calculate the molar internal enegry and enthalpy chnages at boiling point `(373.15K)`. |
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Answer» Solution : The vaporization occurs at constant pressure therefore the enthalpy change is equal to the work done by the heater, `Delta H = 0.50 xx 12 xx 300 = 1800 J = + 1.8 kJ` `:.` Molar enthalpy of vaporization `Delta H = (Delta H)/("mole of " H_(2)O) = (Delta H)/(n_(H_(2))O)` `= (1.8)/(((0.798)/(18))) = 40.6 kJ mol^(-1)` `= Delta U + PDelta V = Delta U + Delta n_(g) RT = Delta U + RT` also. `( :. H_(2)o(l) hArr H_(2)O (g) , Delta n_(g) = 1)` `Delta U =` molar int ernal enrgy change `= Delta H - RT = 40.6 -8.314 xx 10^(-3) xx 373.15 = 37.5 kJ mol^(-1)` |
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