Lecture_05

Partial molar free energy partial molar free energy

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Unformatted text preview: l molar volume of ethanol? No. of moles of ethanol in 100 g of solution: 50 g/46.07 g mol–1 = 1.085 mol. No. of moles of water in 100 g of solution: 50 g/18.02 g mol–1 = 2.775 mol. Now, since V = n E V E + n W V W ,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 denoted by the letter µ: l A = G A = æ ¹G ö ç ¹n ÷ è A ø P, T , n B l B = G B = æ ¹G ö ç ¹n ÷ è B ø P, T , n A . The figure shows that the partial molar free energy changes with composition, just as partial molar volume. If we were to plot µ as a function of composition, we would get a plot that is the exact analog of the plot of partial molar volume we discussed earlier. If A and B are present in two phases α and β, the phase equilibrium condit...
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