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_2O (l) overset("vaporization")to H_2O(g)`
`Deltan_g = 1 - 0 = 1`
`DeltaH = DeltaU + Delta n_g RT`
`DeltaU = DeltaH - Deltan_g RT`
` = 41 - 8.3 xx 10^(-3) xx 373 = 37.9 kJ mol^(-1)`