Assuming that water vapour is an ideal gas, the internal energy change `(DeltaU)` when 1 mol of water is vapourised at vapourisation of water at 1 bar and 373K=41 `kJmol^-1` and R=8.3 `Jmol^-1K^-1` will be

Options

A. 41.00 kJ `mol^(-1)`
B. 4.100 kJ `mol^-1`
C. 3.7904 kJ `mol^(-1)`
D. 37.904 kJ `mol^(-1)`

Answer

D

Explanation

(d) Given `DeltaH=41kJ" "mol^(-1)=41000Jmol^(-1)`
`T=100^(@)C=273+100=373K`
n=1
`DeltaU=DeltaH-DeltanRT=41000-(1xx8.314xx373)`
=37898.88 J `mol^(-1)cong37.9kJmol^(-1)`