Answers to Assigned Problems from Chapter 2

Answers to Assigned Problems from Chapter 2 1 mol of ice has a volume of g cm 3 = cm3 1 mol of water has a volume of g cm 3 = cm3 V(ice water) = cm3 mol 1 (PV) = 63 atm dm3 = J mol 1 H = 6025 J = U + (PV) = U ( J mol 1) U = 6025 J mol 1 = kJ mol 1 (Difference between H and U is only J mol 1.) Work done on the system = J mol 1 1 mol of water at 100 C has a volume of g cm 3 = cm3 Volume of steam = 596 g cm 3 = 30 cm3 Volume increase = 30 199 cm3 mol 1 = dm3 mol 1 (PV) = atm dm3 mol 1 = J mol 1 = kJ mol 1 H = 4063 J mol 1 = U + ( J mol 1) U = kJ mol 1 Work done by the system = w = kJ Heat the water from 10 C to 0 C: (1) q1 = CP dT = CP(T2 T1) = 753 J mol 1 Freeze the water at 0 C: (2) q2 = 6025 J mol 1 Cool the ice from 0 C to 10 C: (3) q3 = 377 J mol 1 Net q = 753 6025 377 = 5649 J mol 1

Answers to Assigned Problems from Chapter 2 2.2. 1 mol of ice has a volume of 18.01 g/0.9168 g cmŒ3 = 19.64 cm3 1 mol of water has a volume of 18.01 g/0.9998 g cmŒ3

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