Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)` (If water vapour is assumed to be perfect gas). Find out of the internal energy change. If 1 mole of water is vaporised at 1 bar pressure and `100^(@)C`.

Options

A. `+37.904kJmol^(-1)`
B. `-37.904kJmol^(-1)`
C. `44.096kJmol^(-1)`
D. `-44.096kJmol^(-1)`

Answer

A

Explanation

`H_(2)O(l)toH_(2)O(g)`
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaU=41.00" kJ "mol^(-1)-8.3JK^(-1)mol^(-1)xx373K`
`=41.00-3.096" kJ "mol^(-1)=37.904" kJ "mol^(-1)`