MATSCI 204 Thermodynamics and Phase Equilibria Winter Chapter #4 Practice problems

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1 MATSCI 204 Thermodynamics and Phase Equilibria Winter 2013 Chapter #4 Practice problems Problem: 1-Show that for any extensive property Ω of a binary system A-B: d ( "# ) "# B, = "# + 1$ x B 2- If "# has the form shown in the figure, use a geometric interpretation of question 1 to indicate in the figure "# B, and "# A, of the solution at point P 3-The excess Gibbs free-energy of ing of Al-Zn alloys is well-represented by the following formula: $ "G XS = x Al x Zn ( 9600x Zn x Al ) 1# T ' & ) (J/mol) % 4000( a-find "S XS and "S as a function of x Al b-find "H as a function of x Al 1

2 c-find "H Al, d-calculate γ Al and a Al as a function of x Al at 500K e-plot a Al as a function of x Al and indicate in your plot the regime where Raoult s and Henry s Law apply Assuming that the solution is Henrian for x Al <1%, find Henry s coefficient for Al 2

3 Answer 1-By definition, we know that: "# = ( 1$ x B )"# A, + x B "# B, and d "# = "# A, dx A + "# B, = "# B, $ "# A, = ("# B, $ "# A, ) therefore, d "# By substituting "# A, obtained from the third equation in the first equation, we have: % "# = ( 1$ x B ) "# B, $ d ( "# ) ( ' * + x B "# B, & ) which leads directly to "# B, = "# + 1$ x B d ( "# ) 2-The geometric interpretation is shown in the following figure: The slope of the tangent at point P is d ( "# ) therefore, "# (P) + d ( "# ) is the intercept of the tangent to the curve at $ 1% x B P point P with the x B =1 axis and according to the formula it is equal to "# B, "# A, can be found by using the same formula as in the question (substituting A for B) and remembering that (1-x A )=x B and that d "# dx A "# A, = "# + 1$ x A d ( "# ) dx A = $ d "# d "# = "# $ x B, which leads to: 3

4 Therefore, this is a general procedure to obtain graphically any partial molar property of the components from the plot of the ΔΩ (x) Of course, we have applied this procedure in the notes to derive the chemical potentials in a solution from the plot of Δg (x) 3-!g XS = x Al x Zn # ( 9600x Zn x Al ) 1" T & % ( $ 4000 ' a-!s XS = " d!g XS dt If we substitute x Zn =1-x Al in the equation of "G XS, we have #!g XS = x Al ( 1" x Al )( x Al ) 1" T & % ( therefore, $ 4000 '!s XS = x Al ( 1" x Al ) x Al = x Al ( 1" x Al )( x Al ) (Jmol -1 K -1 ) 4000!s =!s XS +!s ID and!s ID = "R#$ x Al ln( x Al ) + (1" x Al )ln(1" x Al )% & $% & ' (Jmol -1 K -1 )!s = x Al ( 1" x Al ) x Al "8314 # x Al ln( x Al ) + (1" x Al )ln(1" x Al ) b-!h ID = 0 "!h =!h XS =!g XS XS +T!s By plugging in the relevant equations, we have:!h = x Al 1" x Al ( x Al ) (Jmol -1 ) c- in order to calculate "H Al, we use the formula of question 1: d (!h )!H Al, =!h + 1" x Al dx Al Since we have the ing enthalpy as function of x Al as the only independent variable, we can proceed simply with calculating the derivative We have d!h dx Al = 9600 "12000x Al "10800( x Al ) 2 After some algebra, we obtain: "H Al, = 9600 #12000x Al # 4800( x Al ) ( x Al ) 3 (Jmol -1 ) =!µ Al, XS XS d-in order to calculate γ Al we need!µ Al, since RT ln! Al Again, we use the formula derived in question 1 applied to the XS quantities: XS!G Al, XS =!µ Al, =!g XS + 1" x Al d!g XS Al, dx Al After some algebra and plugging in the temperature, we obtain: XS 2 "µ Al, = ( 1# x Al ) 8400 # 2100x Al # 6300x Al therefore, " Al = exp 1# x Al a Al =γ Al x Al = x Al exp 1" x Al (Jmol -1 ) [ $ ( 202 # 0505x Al #1516x Al )] and # ( 202 " 0505x Al "1516x Al ) [ ] 4

5 e- Al shows positive deviations from ideality Henry s activity coefficient for x Al <1% is ~734 5

6 Problem Famous scientists publish a landmark paper on Nature where they conclude that a solution of A-B that is found to have an enthalpy of ing that obeys the following law: $ "H = #14500x A x B 1# 350 ' & ) % T ( is a regular solution Will you write to the editor of Nature to voice your concerns that the famous scientists are wrong? 6

7 Answer You should call the Editor of Nature (and face the consequences of incurring the wrath of the famous scientists) because the ΔH they report cannot be appropriate for a regular solution Indeed, ΔH is a function of T, which implies that "G XS is a function of T because "H ID = 0 # "H $ "H XS It follows that "S XS the definition of a regular solution where "S = "S ID = # d "G XS dt $ 0, which is contrary to 7

8 Problem: Component A in a gas ture that is in equilibrium with a solid A-B alloy at 500 K has the following partial pressures for varying alloying compositions (in units 10-5 Torr) P A : X A : a) Determine the activity coefficient of component A in the dilute-a composition range of this alloy at 500 K b) Determine the composition range over which the Henrian behavior applies to component A c) Plot the activity of A vs Xa and show the ideal Raoultian and Henrian behavior d) Does component A in this system exhibit positive or negative deviation from ideality? 8

9 Assume that the pure component A has activity equal to one a A = x A = 1 then if a A = P A /x A we set P T = 31*10-5 Torr ( we are choosing reference states) and scale everything accordingly x P a a ideal Henrian ideal Raoultian x a) γ is the intercept at x A = 1, extrapolating dilute A behavior γ = 016/1=16 b) Henrian behavior applies over the range 0<x<04 c) Shown above d) The real behavior is a positive deviation from ideality 9

10 Problem 3 The activity coefficient of Zn in liquid Zn-Cd alloys at 435 C can be represented as: 2 ln (" Zn ) = 0875x Cd # 03x 3 Cd A (20 points) Derive the corresponding expression for the dependence of ln(γ Cd ) B (20 points) Plot Δg as a function of composition Deduce graphically the values of "µ Cd, and "µ Zn, at x Cd =07 10

11 Answer We use the integration of the Gibbs-Duhem equation as shown in class and in the notes: ln (" Cd ) = # x Zn x Zn xzn=0 x Cd $ dln " Zn = 0875x Cd 2 Since ln " Zn # 03x 3 Cd, we have dln " Zn x Cd =1-x Zn and dx Cd =-dx Zn in the integral, we obtain: = # x Zn x Zn ln " Cd x Zn $ x Zn=0 dx Zn = 0425x Zn dx Cd Replacing 2 = 175x Cd # 09x Cd which integrates to ln " 2 Cd + 03x 3 Zn As a result, "g = RT[ x Cd ln (# Cd x Cd ) + x Zn ln (# Zn x Zn )] "µ Cd, and "µ Zn, are obtained from the intercepts of the tangent to the Δg curve with the axes: "µ Zn, ~ 51 kj/mol and "µ Cd, ~ 18 kj/mol 11

12 Problem: In this problem you will see how with a few assumptions and a few measurements you can obtain a lot of information about the thermodynamics of solutions I converted in a table of numbers the activity coefficient curve of Cu as a function of composition in a Fe-Cu solution as shown in the notes You will find it on C2G either as a text file with comma and space separated columns (Cu-Fetxt) or as an Excel spreadsheet (Cu-Fexls) The first column is the mole fraction of Cu and the second column is the activity coefficient of Cu in the solution at 1550 C You will have to use your program of choice to perform some simple numerical operations a) Using the Gibbs-Duhem integration method, calculate the activity coefficient of Fe as a function of composition Plot both γ Fe and γ Cu as a function of composition on a single graph b) Plot Δh and Δg of the Fe-Cu solution as a function of composition at 1550 C, assuming that the solution behaves like a regular solution c) Calculate and plot γ Fe and γ Cu at 1800 C 12

13 Answer The Gibbs-Duhem integration method provides a method to calculate γ Fe : x Cu = X Cu [ ( X Cu )] = # xcu = 0 ln " Fe x Cu $ dln" Cu 1# x Cu x By plotting Cu as a function of ln(γ Cu ) [see figure], you can see that the integral is 1" x Cu equal to the shaded area After integration, we get: b) "h = x Cu "H Cu, + (1# x Cu )"H Fe, XS Because we have a regular solution, "µ Fe, for Cu "g = "h # T"s = "h # RT x Cu ln(x Cu ) + (1# x Cu )ln(1# x Cu ) [ ] = RT ln (# Fe ) = "H Fe, and the same is true 13

14 c) " ln # Fe "( 1/T) = $H,Fe In regular solutions, the ing enthalpies are independent of R temperature therefore the integration is straightforward:, " Fe,1800 C = " Fe,1550 C # exp $H & Fe, 1 R 2073 % 1 )/ ( ' 1823* 0 The dashed lines are the activity coefficients previously calculated at 1550 C 14

Concept of the chemical potential and the activity of elements

Concept of the chemical potential and the activity of elements Gibb s free energy, G is function of temperature, T, pressure, P and amount of elements, n, n dg G G (T, P, n, n ) t particular temperature