If water vapour is assumed to be a perfect gas, molar enthalpy change for vapouration of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)`. Calculate the internal energy change when
a) 1 mol of water is vapourised at 1 bar and `100^(@)C`
b) 1 mol of water liquid is converted into ice.

Explanation

(i) The change `H_2O (l) to H_2O(g)`
`triangleH =triangleU + trianglen_g RT`
or `triangleU =triangleH -trianglen_g RT,` substituting the values, we get
`triangleU=" 41.00 kJ mol"^(-1) : -1 xx 8.3"J mol"^(-1) K^(-1) xx "8.3 J mol"^(-1) K^(-1) xx 373 |K|`
`"= 41.00 kJ mol"^(-1) -"3.096 kJ mol"^(-1)`
`=37.904" kJ mol"^(-1)`
ii) The change `H_(2)O (l) to H_2O(s)`
There is negligible change in volume,
So, we can put `P triangleV = trianglen_g` RT= 0 in this case, `triangleH cong triangleU`
so, `triangleU="41.00 kJ mol"^(-1)`