Assuming that water vapour is an ideal gas, the internal energy change `(DeltaU)` when 1 mole of water is vaporised at 1 bar pressure and `100^@ C`, (given : molar enthalpy of vaporization of water `"41 kJ mol"^(-1)` at 1 bar and 373 k and `R = 8.3 "J 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("vaporisation")rarrH_2O_((g))`
`Deltan_(g)=1-0=1`
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaU=DeltaH-Deltan_(g)RT`
`=41-8.3 xx10^(-3)xx373`
`= 37.9 "k mol"^(-1)`