Question

A certain amount of H2O vapor (gas) is condensed isothermally and reversibly to liquid water at 100oC, where the liquid occupied a volume of 675 cm3. The standard enthalpy of vaporization of water at 100oC is 40.66 kJ mol-1, the density of liquid water at 100oC is 0.958 g/cm3. Find w, q, U, and H for this condensation process. (You may assume that the volume of liquid H2O is negligible compared to that of H2O vapor, and that the vapor follows the perfect gas equation of state.)

Answer 1

mass of H2O = 675 x 0.958 = 646.65 g

water vapour condensed isothermally and reversibly.

so for isothermal process temperature is constant.

U = n Cv dT = 0

H = n Cp dT = 0

so U , H are zero .

so

work q = -w

enthalpy of vaporization of water at 100oC is 40.66 kJ mol-1

q = n .H

    = (646.65 / 18) x 40.66 x 10^3

    = 1459.6 kJ

q = 1459.6 kJ

w = - 1459.6 kJ