Calculate the water's change in internal energy

Question

The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic.

H2​O(l)+40.7kJH2​O(g)

Assume that at exactly 100.0°C and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62 L, respectively.

Answer 1

∆U = q + w

∆U = change in internal energy

q = heat gained or lost by the system

w = work done by or done on the system

q = 40.7 kJ

w = -P∆V

∆V = change in volume = 1 mol x 30.62 L/mol - 0.001880 L = 30.6 L

P = 1 atm

w = - 1 atm x 30.6 L = -30.6 Latm x 101.325 J/Latm = 3102 J = 3.102 kJ

∆U = 40.7 kJ - 3.1 kJ = 37.6 kJ