Lecture_5b

Partial molar free energy partial molar free energy

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Unformatted text preview: we get VE = V − n WVW nE æ 100 g/0.914 g mL −1 ö − 2.775 mol  17.4 mL mol − ø è = 1.085 mol −1 = 56.33 mL mol . A third example of using Eq. (5.33) is provided in Homework Assignment # 6. Partial molar Free Energy: Partial molar free energy is commonly refered to as the “chemical potential,” and 5.7. Raoult’s Law and Activities Consider a binary system consisting of a solution and a vapor phase, each containing components A and B. The equilibrium condition is: G sol = G vap (or l sol = l vap ) and G sol = G vap (or l sol = l vap ). B B B B A A A A If the vapor phase behaves ideally, we may write P P G vap = G ,vap + n A RT ln æ PA ö , and G vap = G ,vap + n B RT ln æ PB ö . B B A A ø è è ø Now, at equilibrium, we may write P P G sol = G ,vap + n A RT ln æ PA ö , and G sol = G ,vap + n B RT ln æ PB ö , or B B A A ø è è ø (5.70,5.71) PA ö PB ö l sol = l ,vap + RT ln æ P  ø and l sol = l ,vap + RT ln æ P  ø . B B A A è è For the pure liquids in equilibri...
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This document was uploaded on 02/28/2014 for the course CHEM 311 at LA Tech.

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