Question

Calculate the change in entropy when one mole of metallic aluminum is heated at one bar pressure from an initial temperature of 25 °C to a final temperature of 750 °C. The molar heat capacities of solid and liquid aluminum at one bar pressure are 29.2 J mol' K and 31.75 J mol' K, respectively. The specific enthalpy of fusion of aluminum at its melting point (660.46 °C) is 396.57 J g! The molar mass of aluminum is 26.98 g mol!.

Answer 1

Data: AS? A: 1 mol T = 25°C + 273 = 298 K T, = 750° + 273 = 1023 K Ce A (s) = 29.2 J/ mol K Ce A 0 = 31.75 J/mol K 4A (Latent heat of fusion) = 396.57 J/g mm A = 26.98 g/mol To get from Al solid to liquid, Al experiences a process consisting of 3 sections: 1) Aluminum raises its temperature from 298K to 933 K. 2) Aluminum melts (at constant T = 660º = 933K) 3) Aluminum raises its temperature from 933K to 1050 K. 1050 K 41 = mx Ce A) (1050K -933K) 933 K q = mx L A 298 K 9. = mx Ce Al(s) X (933K -298K) Calculate , Calculate q, 9, = 1 mol x 29.2 J/mol Kx (635 K) 42 = 26.98 G x 396.57 J/g 9, = 18542J 9, = 10699.46 J Calculate : 9, = 1-mol x 31.75 J/mol Kx (635 K) 9, = 20161.25 J Calculate AS, Calculate AS: Calculate AS AS, 91 49, -—q 48, 20161.25 J 18542 J 298 K AS, = 62.22 J/K AS, = 10699.46 AS, - 1050 K 933 K AS, - 11.46 J/K AS3 = 19.20 J/K Entropy change is : Adding the three contributions we get the total increase in entropy: AS = (AS, + AS, + AS3) AS = (62.22 J/K + 11.46 J/K + 19.20 J/K) AS = 92.88 J/K