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AG(s) = Ag(l) T = 1234K H = 11300 J mol-1...

This question belongs to Metallurgical Engineering Metallurgical Thermodynamics And Kinetics
Metallurgical Thermodynamics and Kinetics
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AG(s) = Ag(l), T = 1234K; ΔH = 11300 J.mol-1 ΔCP = Cp(l) - Cp(s) = 0
When one mole of super cooled liquid silver freezes at an ambient temperature of 1000 K, the total entropy change of the system (Ag) and the surroundings is . . . . . . . . J.K-1.

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Correct Answer: Option D
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