Assuming that water vapour is an ideal gas, the internal energy (`DeltaU`) when 1 mol of water is vapourized at 1 bar pressure and `100^@C`. (Molar enthalpy of vapourization of water at 1 bar and 373 K =41 kJ `mol^(-1)` and `R=8.3Jmol^(-1)K^(-1)`) will be

Options

A. `4.1 kJ mol ^(-1)`
B. `3.7904 kJ mol ^(-1)`
C. `37.904 kJ mol ^(-1)`
D. `41.0 kJ mol ^(-1)`

Explanation

The reaction involved is `H _(2) O _(1) to H _(2) O (g) , ` where `Delta n _(g) = 1 mol , Delta _(vap) H = 41kJ mol ^(-1).` Now `Delta U = Delta H - Delta n _(g) RT`
`=41 kJ mol ^(-1) -1 xx 8.3 xx 10 ^(-3) kJ K ^(-1) mol ^(-1) xx 373 K = 41 - 3 .096 kJ mol ^(-1) = 37.904 kJ mol ^(-1)`