2D_Course_Notes_-_Part_II

2D_Course_Notes_-_Part_II - Partial Properties and Chemical...

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1 Partial Properties and Chemical Potential - Gaskell P 201 Partial properties ( ) [ ] M nM n i i P T n j = / , , summability M = x M i i i ± dM = M dx x dM i i i i i i ± ± + dM = ( M/ P) T,x dP + ( M/ T) P,x dT + M i i ± dx i ± i i i M d x = ( M/ P) T,x dP + ( M/ T) P,x dT Gibbs-Duhem equation: x dM i i i ± M = x M x M 1 1 2 2 + dM = x dM M dx x dM M dx 1 1 1 1 2 2 2 2 + + + = M dx M dx 1 1 2 2 + = ( ) M M dx 1 2 1 ² ( ) M M x dM dx 1 2 1 = + / ( ) M M x dM dx 2 1 1 = ² / Measure M ~ x 1 to get partial M s eg. V
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2 Partial Properties and Chemical Potential - Gaskell P 201 Objectives: develop thermodynamic framework to deal with non ideal mixtures of chemicals. Key concepts: mixture: solution, alloy, blend give daily life examples - why some mixed and others not partial properties - chemical potential - Gibbs/Duhem equation mixing gases ideal to non ideal - fugacity and fugacity coefficient mixing liquids - ideal to non ideal - ideal solution - excess properties
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3 Chemical Potential - The Experiment Add heptane drop by drop and record free energy per mole of mixture . n 1 G 1 nG Moles of added heptane Raw Data If component 1 is benzene where is G 1 on this graph?
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4 Converting Data to Mole Fractions G = (nG)/(n 1 +n 2 ) X 1 = n 1 /(n 1 +n 2 ) Fill this in Show G 1 and G 2 0 X 1 1 G Heptane Benzene Questions G 1 G 2
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5 Defining Chemical Potential, μ i Draw a tangent 0 X 1 1 Chemical potentials are the intercepts . μ 1 μ 2 Q: What s the unit G Heptane Benzene G 1 G 2
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6 Does μ i depend upon composition? Yes! Questions? G Heptane Benzene G 1 G 2 x 1a μ 1a x 1b μ 1b
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7 An important equation for μ 1 Slope of blue line is μ 1 G Mole Fraction of Chemical 1 G 1 G 2 G x 1 0 1 dG/dx 1 = μ 1 - μ 2 From geometry: μ 1 = G + x 2 (dG/dx 1 ) μ 2 = G x 1 (dG/dx 1 ) μ 2 Rearranging gives: 2 2 1 1 μ x x G + =
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Why chemical potential so important ? (nG) = f(P, T, n 1 , n 2 ) d(nG) = [d(nG)/dP] T,n1,n2 dP+ [d(nG)/dT] P,n1,n2 dT [d(nG)/dn 1 ] T,P,n2 dn 1 + [d(nG)/dn 2 ] T,P,n1 dn 2 = (nV)dP+(-nS)dT+ μ 1 dn 1 + μ 2 dn 2 (nV) = [d(nG)/dP] T,n1,n2 known -(nS) = [d(nG)/dT] P,n1,n2 known μ 1 = [d(nG)/dn 1 ] T,P,n2 μ 2 = [d(nG)/dn 2 ] T,P,n1
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This homework help was uploaded on 04/17/2008 for the course MATLS Matls 2D03 taught by Professor Zhu during the Fall '05 term at McMaster University.

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2D_Course_Notes_-_Part_II - Partial Properties and Chemical...

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