Question

Please just answer Part 2

The vaporization of 1 mole of liquid water (the system) at 100.9°C, 100 atm, s endothermic. H0HOg) 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 5th attempt d See Periodic Table See Hint (0.5 point) Part 1 Calculate the work done on or by the system when 1.85 mol of liquid H20 vaporizes. 5736.23319:J (0.5 point) 9 See Hint Part 2 Calculate the water's change in internal energy. kJ 36.8

Answer 1

Part 2 : water's change in internal energy = 65.559 kJ

Explanation

heat gained by water = (enthalpy of vaporization) * (moles of water)

heat gained by water = (40.7 kJ/mol) * (1.85 mol)

heat gained by water = 75.295 kJ

work done = -5736.233192 J = -5.736 kJ

change in internal energy = heat gained + work done

change in internal energy = 75.295 kJ + (-5.736 kJ)

change in internal energy = 75.295 kJ - 5.736 kJ

change in internal energy = 69.559 kJ