One mole of H20(g) at 1.00 atm and 100 degrees C occupies a volume of 30.6L. When one mole of H20(g) is condensed to 1 mole of H20(l) at 1.00 atm and 100C, 40.66 kJ of heat is released. If the density of H20(l) at this temperature and pressure is .996 g/cm^3, calculate ? E for the condensation of 1 mole of water at 1.00 atm and 100 degrees C. [ H20(g) -> H20(l) ]