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

First let's get the mass of water, cause we already have it's volume:

mass of H2O = 675 * 0.958 = 646.65 g

As water vapour condensed isothermally and reversibly, means that for an isothermal process temperature is constant, therefore:

U = nCvT = 0

H = nCpT = 0

If U and H = 0 means that

U = q + w

q = -w

We know that the enthalpy of vaporization of water at 100 °C is 40.66 kJ/mol

q = n H

Let's get the moles of water first:

moles H₂O = 646.65 / 18 = 35.925 moles

Now, calculating q:

q = 35.925 * 40.66

q = 1460.71 kJ

then w = -1460.71 kJ

Hope this helps