3.012 PS 7 Thermo solutions Issued: Fall 2003 Graded problems due:

Size: px

Start display at page:

Download "3.012 PS 7 Thermo solutions Issued: Fall 2003 Graded problems due:"

Magdalene Walton
6 years ago
Views:

1 3.012 PS 7 Thermo solutions Issued: Fall 2003 Graded problems due: Graded problems: 1. Analysis of equilibrium phases with a binary phase diagram. Shown below is the phase diagram for the binary system magnesium-lead at 1 atm pressure. Supposing a sample initially at temperature T A is slowly heated to temperature T E, prepare a list of the phase equilibria at each of the noted temperatures A-E. At each temperature, give: a. Which phases are present. b. The phase fraction of each phase. c. The composition of each phase. T E T D T C T B T A PS 7 1 of 11 12/12/03

2 Solutions can be found using the lever rule and a ruler (to measure the relative lengths of the tie lines). The phase fractions and compositions needed only to be approximate since you are reading them from the graph directly. T Phases T A Mg 2 Pb f Mg 2Pb = 0.55 Mg x 2 Pb B = 0.81 α f α = 0.45 x B 0.02 T B Mg 2 Pb f Mg 2 Pb = 0.36 Mg x 2 Pb B = 0.81 α f α = 0.64 x B 0.24 T C Eutectic L α + β f Eu = 0.10 Eu x B = 0.68 α f α = 0.90 x B 0.40 T D α f α 0.0 (infinitesimal amount remaining) x B 0.15 Liquid f L 1.0 L x B 0.44 T E Liquid f L = 1.0 L x B 0.44 Note that the phase diagram gave an x-axis (at the bottom) in weight fraction, not mole fraction. It was fine to report the compositions as weight fractions, they are given in the table above denoted by the lowercase x s. Since we specified a heating path from A to E (as opposed to a cooling path from E to A), you didn t have to break down the phases at T A and T B into primary vs. eutectic α contributions (we would need to know the original cooling behavior from the melt that was used to prepare the sample at T A to specify this information). 2. Construction of a phase diagram from free energy data. A new material comprised of an A-B binary system is being studied. The free energy of the solid phase of this system is described by the regular solution model, while the liquid phase forms an ideal solution. Thermodynamic data: L J S J µ A,o = 175 µ A,o = 5,000 mole mole L J S J µ A,o = 1,000 µ B,o = 4,000 mole mole J Ω = 20,000 mole a. Derive an expression for the molar free energy of the solid solution phase as a function of composition ( ) and temperature. Note that the free energy of mixing is not equivalent to the free energy of the solid solution. We derived the expression for the free energy of mixing of the regular solution model in class; it is given by: PS 7 2 of 11 12/12/03

3 ΔG mix,rs = Ω (1 ) + RT [(1 )ln( 1 ) + ln ] The definition of the free energy of mixing is: ΔG mix,rs = G solution,rs G heter where G solution,rs is the free energy in the homogeneous solution, and G heter is the free energy of the unmixed heterogeneous A-B system. G heter is always given by: G heter = G A,o X A + G B,o = µ A,o X A + µ B,o Using the equation for ΔG mix,rs and the equation for G heter, we can directly write that the free energy of the regular solution is: G solution,rs = ΔG mix,rs + G heter ( A,o G solution,rs = Ω (1 ) + RT [(1 )ln 1 ) + ln ] + µ (1 ) + µ B,o Note that this expression is identical to the free energy of the ideal solution, with the omega term added which accounts for the chemical interactions between the solution components. b. Prepare a plot of the free energy of the solid phase and of the liquid phase vs. composition at the following 3 temperatures: 900 K, 700 K, 500 K. c. Using your 3 plots, graph common tangents and identify the stable phases as a function of at each temperature. d. From this data, suggest what the complete phase diagram of this system must look like. The solution for parts b, c, and d are given in the following plots of the free energy curves: PS 7 3 of 11 12/12/03

4 0 T = 900K G (J/mole) solid phase liquid phase α 1 α 1 + L L α 2 + L α T = 700K G (J/mole) solid phase liquid phase common tangent PS 7 4 of 11 12/12/03

5 0 T = 500K G (J/mole) solid phase liquid phase α 1 + α α 1 α 2 These 3 temperature slices suggest that a eutectic-type phase diagram is formed in this system, with the invariant point located at T=700K and ~ PS 7 5 of 11 12/12/03

6 Optional problems: 3. Characterizing phase transitions. Shown below is a sketch of the isothermal compressibility of a new material vs. temperature. Is the discontinuity shown a characteristic of a first-order phase transition, a second-order transition, or a higher-order transition? Show why. κ Τ Recalling the definition of the isothermal compressibility: 1 V κ T = V T trans T --> We can relate κ T to the Gibbs free energy in order to characterize the order of the transition. Using the expression for the differential of G and the algebraic definition of dg, we know that the volume is a first derivative of the free energy: P T -- dg = SdT + VdP + µ i dn i G G G dg = dt + dp + dn i T P,n P T,n n i T,P G V = P T,n 1 G 1 2 G κ T = = V P V P T,n P P 2 T,n A discontinuity in a second derivative of the Gibbs free energy is characterized as a second-order transition PS 7 6 of 11 12/12/03

7 4. Thermodynamic quantities derived from free energy-composition diagrams. On the solution free energy diagram for a binary system given below, sketch or identify the following quantities: a. The standard state chemical potentials of A and B at 300 K. b. The free energy of a heterogeneous mixture of A and B with overall composition X. c. The chemical potentials of A and B at 300 K in the solution. d. The free energy change ΔG mix obtained when the heterogeneous mixture of A and B forms a homogeneous solution. T = 300K G X The free energy diagram with the requested annotation is shown below: PS 7 7 of 11 12/12/03

8 µ - A,o G heter G - µ ΔG mix B,o µ A - - µ B X Remember that the chemical potentials in the mixed solution are given by the tangent to the solution free energy curve at the desired overall composition. 5. Invariant points and congruent transitions. On the following hypothetical phase diagram, mark and label the class of the invariant points, and identify the congruent transitions. These diagrams are analyzed using the condensed phase rule - since pressure is held constant, the Gibbs phase rule reduces to: D + P = C PS 7 8 of 11 12/12/03

9 The points are marked below. Note the invariant point on the left side of the hump region containing L 1 + L 2. The invariant point is known as a monotectic, and comprises the 3 phase equilibrium L = L + R; L and L are both liquid but with two different compositions (thus they are two differentiable phases at the monotectic point). Congruent transition point invariant point 6. Free energy of ideal solutions. Prove that the enthalpy of mixing for an ideal solution is zero, and derive an expression for the entropy of mixing for an ideal solution. The chemical potential of components in an ideal solution is: µ i = µ i,o + RT ln X i Thus the molar free energy of the ideal solution is given by: G ideal = µ A X A + µ B ideal G = (µ A,o + RT ln X A )X A + (µ B,o + RT ln ) The free energy of mixing for the ideal solution is thus: ΔG mix = G ideal G heter mix ΔG = G ideal µ A,o X A µ B,o ΔG mix = RT [ X A ln X A + ln ] PS 7 9 of 11 12/12/03

10 Recalling that the derivative of the Gibbs free energy with respect to temperature is related to the entropy, we can directly calculate the entropy of mixing: Δ G mix Δ S mix = = R X A ln X A + ln ] T P, [ Finally, one can always relate enthalpy to entropy and free energy using: ΔH mix = ΔG mix + TΔS mix = ΔG mix R [ X A ln X A + ln ] = 0 7. Spinodals in binary systems. For the regular solution solid phase of the problem Construction of a phase diagram from free energy data, make the following calculations: a. Determine the critical temperature at which the solid solution would phase separate. (Note that we don t see the critical temperature on the phase diagram since the liquid phase in this system is present at temperatures above the eutectic isotherm; we would have to superheat the solid phase to see the critical temperature where the phase-separated solid becomes a homogeneous solid). In the regular solution model, the free energy is symmetric about the composition = 0.5 (thus the miscibility gap is centered at a 50/50 composition). To calculate the hypothetic critical temperature where phase separation first occurs on decreasing temperature at = 0.5, we calculate the temperature where the curvature of the free energy of mixing becomes 0- i.e. the point just before a hump appears in the free energy of mixing vs. composition diagram. We derived the expression for the curvature of the free energy of mixing in class: 2 ΔG mix,rs 1 1 X = 2Ω + RT + 2 B 1 Setting the curvature = 0 and = 0.5, we calculate the (hypothetical) critical temperature: Ω 20,000 T crit = = 2R 2(8.3144) = 1,202K b. Calculate the composition of the spinodal points at T = 500 K. Qualitatively, how would the spinodals be plotted onto the phase diagram of this system? The spinodals are located at the composition points where the free energy changes curvature. We have from the earlier problem the following expression for the free energy of the regular solution solid phase: ( A,o G solution,rs = Ω (1 ) + RT [(1 )ln 1 ) + ln ] + µ (1 ) + µ B,o PS 7 10 of 11 12/12/03

11 To find the spinodals, we calculate the partial second derivative of the molar free energy with respect to composition, and find the two points where the second derivative is zero. Note that the second derivative of the molar free energy of the solution with respect to composition is equivalent to the second derivative of the molar free energy of mixing, since the contribution from G heter in the free energy of mixing drops out of the second derivative of the free energy of mixing. 2 ΔG mix,rs 1 1 X = 2Ω + RT 2 + = 0 B 1 The two compositions that satisfy this equation at T = 500K are ~ 0.12 and PS 7 11 of 11 12/12/03

3.012 PS Issued: Fall 2003 Graded problems due:

3.012 PS Issued: Fall 2003 Graded problems due: 3.012 PS 7 3.012 Issued: 11.18.03 Fall 2003 Graded problems due: 11.26.03 Graded problems As usual, try as much as possible to explain in a narrative form your reasoning (as opposed to writing down bare