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

The vaporisation occurs at constant pressure, therefore the enthalpy change is equal to the work done by the heater:
`DeltaH' = 0.50 xx 12 xx 300 (DeltaH = I xx V xxt)`
`= 1800 J = 1.8 kJ`
Therefore, molar enthalpy of vaporisation `DeltaH`
`= (DeltaH')/("Mole of" H_(2)O) = (DeltaH)/(nH_(2)O) = (1.8)/(((0.798)/(18))) = 40.6 kJ mol^(-1)`
Also, `DeltaH = DeltaU +P DeltaV = DeltaU +Deltan_(g)RT = DeltaU +RT`

`:. DeltaU = `Molar internal enegry change ltbr `= DeltaH - RT`
`= 40.6 - 8.314 xx 10^(-3) xx 373.15 = 37.5kJmol^(-1)`