Assuming that water vapour is an ideal gas, the internal energy change `(Delta U)` when `1 mol` of water is vapourised at `1` bar pressure and `100^(@)C`, (Given: Molar enthalpy of vapourization of water at `1` bar and `373K=41 kJ mol^(-1)` and `R=8.3J mol^(-1)K^(-1)`) will be:

Options

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

Answer

C

Explanation

`H_(2)(O)_((l))rarr H_(2)O_((g))`
`Delta H=Delta H+Delta n_(g)RT`
`41=Delta E+1xx8.314xx10^(-3)x373`
`therefore Delta E=37.9 kJ mol^(-1)`