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)`.

Explanation

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)`