Question

06. Question (2 points) The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. H,0(1) +40.7kJ → H2O(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 30.62L. respectively.

Answer 1

dus) we kname that. W = PAU APVE AORT where we use here. W = ART on= moles of product of gas subitance - moles of Reactant of gas substance, Step find work done for this Ron Moll) + 40.7 KJ > N2009) an= 3.65 -0 = 3.65 mol .:(3.65 moles of liquid to give 3.65 moles of gasto W = DnRT where R= 8.314J/molok W = 3.65 mol X08-314314373K w = 11319. 15 de We 11.32 KJ Step Enthlpy of Ron= 40. I kJ/mol Enthly chang An for 3.65 mol = 3.65840.7 molok P, PPPPPPPPPPPТТІЛІГІ = 148.56 kJ we know, AU=Ah-w OU= 148.56 - 11.32 = 137.24KT su = Chenye in internal Energy /37.2 kJ Thank you may it help you. dus