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If water vapour is assumed to be a perfe...

If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mol 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 vaporised at 1 bar pressure and `100^(@)`C.
(b) 1 mol of water is converted into ice.

Text Solution

Verified by Experts

(a)`H_(2)O(l) to H_(2)O(g)`
`DeltaH = DeltaU + Deltan_(g)RT`
or `DeltaU = DeltaH - Deltan_(g)RT`
`DeltaH = 41.0 kJ mol^(-1), Deltan_(g) = 1 -0 = 1, T = 273 + 100 = 373 K`
`R = 8.314 J mol^(-1)K^(-1) = 8.314 xx 10^(-3) kJ mol^(-1) K^(-1)`
`therefore DeltaU = 41.0 -1 xx 8.314 xx 10^(-3) xx 373`
`= 41.0 -3.10 = 37.90 kJ mol^(-1)`
(b) `H_(2)O(l) to H_(2)O(s)`
There is negligible change in volume so that
`p DeltaV = Deltan_(g) RT = 0`
`therefore DeltaU = DeltaH = 41.0 kJ mol^(-1)`
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