Question

2.7 Calculate the entropy changes of the system and surroundings for the case of isothermal freezing of one mole of supercooled liquid silver at 850°c, when the surroundings are also at the same temperature 961 C (1234 K). Given: Melting point of silver 2,690 cal/mole (11,255 J/mol) Heat of fusion of silver at 961°c 5.092.04 x 103 T ca1/deg/mole (21.30+8.54x10 J/K/mol) Cp,<Ag> 7.30 cal/deg/mole (30.54 J/K/mol) CP(Ag)

in thermodynamics, how do i answer 2.7?

Answer 1

Sime enthalpy is an odditive peroperty so the standard state entropy at any temperature will be due to the conteributions of entropies of the phases, transition of phases. Pos the given condition of freezing of silver, 1. Agco (1234K) Agis (1234K) 9610 = 1234k 2. - 850'C - 1123K Ag (3) (1234K) Ag 13 (1123k) Total entropy = As, & DS2 AS = - Oljus . There is a transition in phase at constant Temperature , As the transition of phase is from liquid to solid , energy is lost his the system and so phius is negative. Ds, = - 11,255 Indo 1234 K value taken from question. os, = -9.12 Jkt mott.

In second is no change temperature change of entropy, there of phase by but the of system decreases, DS₂ = 30.54 Julmot & dT 1123 AS = cpat on integrating Ale = cpfen T. I DS2 = 30.54 x [du T] 11 Jul mare & 30.54 x (en 1123) Jurmat = 30.54 x 60.094) Jki matt Dla = - 2.87 Himore 1234 1123 1234 Jut not Therefore, Total entropy ASF AS, & DS₂ =(-9.12 - 2.87) T Kimon DS 2 - 11.99 Julmon This is the end scopy change for system. As surroundings are also in same temperature = +11.99 Jur molt Asmorronding