Calculate the values of `Delta H and DeltaU` in the vaporisation of 90g of water at `100^(@)C` and 1 atm pressure. The latent heat of vaporisation of water at the same temperature and pressure `=540cal*g^(-1)`

Explanation

90g of water`=(90)/(18)=5` mol of water,
Vaporisation of water:`H_(2)O(l)toH_(2)O(g)`
Therefore, in the vaporisation of 1 mol of water, `Deltan=+1`.
Hence, for the vaporisation of 5 mol water, `Deltan=+5`.
So, the amount of heat required to vaporise 90g (5 mol) of water`=540xx90=48600cal`.
As the vaporisation porcess occurs at constant pressure (1 atm), the heat absorbed=enthalpy change.
`therefore`The change in enthalpy in the vaporisation of 90 g of water, `DeltaH=48600cal`.
`therefore`The change in internal energy in the vaporisation of 90g of water, `DeltaU=DeltaH-DeltanRT`
`=48600-5xx1.987xx(273+100)=44894.24cal`.