Standard enthalpy of vaporisation`DeltaV_(vap).H^(Theta)` for water at `100^(@)C` is `40.66kJmol^(-1)`.The internal energy of Vaporization of water at` 100^(@)C("in kJ mol"^(-1))`is

Options

A. `+37.56`
B. `-43.76`
C. `+42.76`
D. `+40.66`

Answer

A

Explanation

`H_(2)O(l) overset(100^(@)C)to H_(2)O(g)`
`Delta_("vap")H^(@)=Delta_("vap")E^(@)+ Deltan_(g)RT`
`Delta_("vap")H^(@)=` enthalpy of vaporisation
`=40.66 KJ "mol"^(-1)`
For the above reaction, `Deltan_(g)=n_(p)-n_(r)=1-0=1`
R=8.314
`T=100^(@)C = 273 +100=373 K`
`therefore ` 40.66 KJ `"mol"^(-1)=Delta_("vap")E^(@)+ 1 xx 8.314 xx 10^(-3)xx373`
`Delta_("vap")E^(@)=40.66 KJ "mol"^(-1) -3.1 KJ "mol"^(-1)`
`=+37.56 KJ "mol"^(-1)`