{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2D_Course_Notes_-_Part_II

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

This preview shows pages 1–10. Sign up to view the full content.

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 ± = 0, at const P & T 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
3 Chemical Potential - The Experiment ° Const T & P. ° Add heptane drop by drop and record free energy per mole of mixture . ³ n 1 G 1 nG Moles of added heptane Raw Data If component 1 is benzene where is G 1 on this graph?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6 Does μ i depend upon composition? ° Yes! ° Questions? G Heptane Benzene G 1 G 2 x 1a μ 1a x 1b μ 1b
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 μ x x G + =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
8 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 G i = your ability when you are alone μ i = your ability when you are in a team
9 Gibbs-Duhem Equation for μ Leant d(nG) = (nV)dP+(-nS)dT+ μ 1 dn 1 + μ 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.