Calculate w and E when one mole of a liquid is vaporized at its boiling point (71°C) and 1.00 atm pressure. Hv

Question

Calculate w and E when one mole of a liquid is vaporized at its boiling point (71°C) and 1.00 atm pressure. Hvap for the liquid is 37.1 kJ mol-1 at 71°C.

Answer 1

Since the liquid volume is negligable compared to the gas volume once vaporized, we ignore the work influenced by the liquid volume. Specified pressure of 1.00 atm is abundent.
Remember that 1 atm•L = 101.3 J.
W = PV = nRT = 1*0.08206*(273+71) = 28.2 atm•L = 2.86 kJ
Hence Evap = Hvap - W = 37.1 kJ - 2.86 kJ = 34.2 kJ.