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

(a) `H_(2)O Leftrightarrow H_(2) O_(g) " "triangleH=41J`
`triangleH=triangleU+trianglen_(g) RT" "trianglen_(g)=1`
`41=triangleU+8.314 xx 10^(-3) xx 373 " "R=8.314 xx 10^(-3)kJ`
`triangleU=41-3.101 ="37.899 KJ/mole"`
(b) `H_(2)O Leftrightarrow H_(2)O_("ice")`
`triangleH=triangleU [trianglen=0]`
`triangleH=triangleU="41kJ/mole"`