Question

Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporation of `90.0g` of water at `100^(@)C`. Assume that water behaves as an ideal gas and heat of evaporation of water is `540 cal g^(-1) (R = 2.0 cal mol^(-1)K^(-1))`.

Answer 1

Total heat change
`DeltaH = 90.0 xx 540 = 48600cal`
Now, `DeltaH = DeltaU +P DeltaV`
Here, `DeltaV =(V_(vapour) -V_(liquid)) = V_(vapour)`
(Volume of liquid is negligible as compared to volume of vapor)
`DeltaH = DeltaU +PV_(vapour) = DeltaU +nRT`
`n = (90)/(18) = 5mol, R = 2 cal K^(-1) mol^(-1), T = 373 K`
`:. DeltaH = DeltaU +5 xx2 xx 373`
`DeltaH = DeltaU +3730`
or `DeltaU = DeltaH - 3730 =48600 - 3730 = 44870cal`