What is `DeltaU` when 2.0 mole of liquid water vaporise at `100^(@)C ` ? The heat of vaporistaion , `DeltaH` vap. of water at `100^(@)C` is `40.66 kJ mol^(-1)`.

Answer

`DeltaE = 75.11 k J`

Explanation

`DeltaH= DeltaU + Deltan_(g)RT`
`40.66 = DeltaU + 1 xx 8.314 xx 10^(-3) xx 373`
`Deltan_(g)=1`
`DeltaU = 37.555 xx 2 kJ`
For two moles = 37.555 xx 2 = 75.11 kJ