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. `+43.76`
D. `40.66`

Answer

A

Explanation

`DeltaH_(vap)^(@)=40.66 kJ mol^(-1)`
`T=273+100=373 K`, `DeltaU=?`
`DeltaH=DeltaU+Deltan_(g) RT `or `DeltaU=DeltaH-Deltan_(g)RT`
`Deltan_(1)(g)=1-0=1`
`:. DeltaU=40.66 kJ mol^(-1)-1xx8.314xx10^(-8)JK^(-1)mol^(-1)xx373 K`
`=(40.66-3.101) kJ mol^(-1)`
`=37.559 kJ mol^(-1)`