02 Question (1 point) aSee page 376 The vaporization of 1 mole of liquid water (

Question

02 Question (1 point) aSee page 376 The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. 11,0(1) + 40.7kJ 1,0(g) Assume 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. 1st attempt Part 1 (0.5 point) See Periodic Table O See Hint Calculate the work done onor by the system when 4.25 mol of liquid H20 vaporizes. Part 2 (0.5 point) O See Hint Calculate the water's change in internal energy kJ

Explanation / Answer

PART1

W = -P*dV

dV = 30.62-18.8*10^-3 = 30.6012 L = 30.6012*10^-3 m3

P = 1 atm = 101325 Pa

W = -101325*30.6012*10^-3

W = -3100.66 J

PART 2

dU = Q-W

Q = 40.7 kJ = 40700 J

dU = 40700--3100.66

dU = 43800.66 J

dU = 43.8 kJ/mol

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