Calculate w and E when one mole of a liquid is vaporized at its boiling point (81°C) and 1.00 atm pressure?

Question

. DeltaHvap for the liquid is 23.6 kJ mol-1 at 81°C.
w=?
deltaE=?

Answer 1

Remember, the work done is P*∆V due to one mole changed from liquid to gas. While the volume of one mole liquid is negligible, the work is:
w = P*∆V = ∆n*RT = 1.00*8.314*(273+81) = 2.94 kJ
Since ∆H is 23.6 kJ/mol at 81°C (and always under constant pressure),
∆E = ∆H - w = 20.7 kJ/mol.