Unformatted text preview: Mat E 510 Thermodynamics of Solids Mat Mat E 510 Thermodynamics of Solids Lecture #6 Yury Gogotsi
A.J. Drexel Nanotechnology Institute and gy Department of Materials Science & Engineering, Drexel University, Phil Philadelphia, Pennsylvania, USA USA Explanation Explanation of Activity
What is happening physically  what does the vapor pressure tell us? Solid> vapor: Atoms/molecules in the two phases have same G. The partial molar free energy of B, GB, is the same in both phases. For pure solid B, XB =1 _ In gas: GB = GBo + RTln PB/PBo = GBo + RT ln aB with XB = 1, PB = PBo aB = 1 Rate of evaporation = rate of condensation, this depends on PBo. In ideal solution, activity falls with XB. Fewer B atoms, but rate of evaporation per site unchanged. So PB falls with XB. In nonideal, dilute, solution, each B atom in solution has the same environment ( all A) but different binding than to B atoms. Activity coefficient, γB, is constant. In more concentrated solution, environment is changing, becoming more B rich, so γB begins to approach 1. Raoult's law regime, most B atoms are surrounded by B so γB >1. Alpha‐Sn Beta‐Sn Real solution 1 atm 1 M Cu + 1M Sn Solid phases only Real solution 1 atm 1 M Cu + 1M Sn Solid and liquid phases Determining Determining Activity
In non ideal solutions, ΔHmix and ΔVmix ≠ 0. Solutions can be exothermic ΔHmix < 0. B atoms more strongly bound to A than to B. Or they can be endothermic ΔHmix > 0. B atoms less strongly bound to A than to B. We now need a means of measuring either (i) the activ ity, aI of components or the partial molar free energies, GI, µI, of both components. What are the means of doing this? i) Vapor pressure. But PIo must be measurable. Metals or ceramics do not smell. PIo < 106 torr. Sometimes possible: Zn, Hg, in metallic alloys. With mass spectrometry, lower PIo possible. ii) From reactions involving gases: (FeNi) + H2O > FeO + H2 (FeO pure) Fe greater (more negative) ΔG to form oxide. Ni is more "noble". FeO is free of NiO, aFeO > 1 Activity from the Electrochemical Cell
iii) Electrochemical cells (cathode) Al Ag vs Al (anode) with molten salt electrolyte. Al0 > Al3+ + 3 e. Valency (h) =3 Cell reaction for 1 mole w' = ΔGR = Eo h F Eo is the open circuit ( zero current) cell voltage. F is the "Faraday", NAv e = 6 x1023 x 1.6 x 1019C F = 96,000C _ ΔGR = GAl  GAlo = RT ln aAl So from cell voltage we can measure both the activity and the partial molar free energy of the the less noble element. As with oxidation, it requires that the more noble element does not react. _ That is we can measure GA of one, more active component but not the other more noble one. _ _ From d(GA/T)/ d(1/ T) we can get HA at a range of compositions. _ Then by use of Gibbs Duhem for Enthalpy we get HB. ( B being the noble metal) _ _ Also, from d(GA)/ d(T) we can get SA at a range of compositions. But we cannot solve the GibbsDuhem equation for S or G since as XB > 1, SA > +∞ and GA > ∞. So graphical solution not possible. Use activity coefficient of A, γA. This is not only finite as XB > 1, but it tends to a constant _ _ XA dGA + XB dGB = 0 _ GB = GBo + RT ln aB = GBo + RT ln XB+ RT ln γB _ So: dGB = RT dln XB+ RTdln γB value (Henry's law.) Gibbs GibbsDuhem Equation
So: XART dlnXA+ XART dlnγA+ XBRT dlnXB + XBRTdlnγB = 0 XA dlnXA+ XA dlnγA+ XB dlnXB + XB dlnγB = 0 But XAdlnxA + XBdlnxB = XAdXA/XA + XBdXB/XB = dXA+ dXB = 0 Since XA+ XB = 1 So the new form of the GibbsDuhem eq. becomes: becomes: XAdlnγA + XBdlnγB = 0 CH4 = C + H2 Ideal vs real solution in gas phase at 1000 atm Regular Solutions
These have ideal entropy of mixing (true random solid solutions) but non zero enthalpy of mixing. Few solutions are ideal, but many approach being regular. Also, we can model but Also we regular solutions and used the result to understand a range of material behaviors for example unmixing and ordering as T > 0K. First some TD results for regular solutions: ΔSMIX =  R(xA ln xA + xB ln xB) ΔGMIX = RT(xAln aA + xB ln aB) ΔGMIX = RT(xA ln γAxA + xB ln γBxB) ΔGMIX = RT(xAlnγA + xAlnxA + xB lnγB + xB ln xB) ΔGMIX = RT(xAlnγA + xAlnxA + xB lnγB + xB ln xB) = ΔHMIX TΔSMIX = ΔHMIX + RT(xAln xA + xB ln xB) So: ΔHMIX = RT(xAlnγA + xB ln γB) Also excess quantities: ΔGEX = ΔGMIX  ΔGID = ΔHMIX Also ΔHEX = ΔHMIX  ΔHID = ΔHMIX (If regular solution). (and, if regular) = RT(XAlnγA + XB ln γB) Quasi Quasi Chemical Model of Regular Solutions
We model the Enthalpy by assuming that the bond energies, hAA, between two A atoms/molecules, and the equivalent, hBB and hAB, are constant and independent of composition , xB. This This assumption is reasonably good for liquid solutions but it omits a critical factor for solid (crystalline) solutions. Usually the atomic size varies (DA≠ DB) so as the composition varies the lattice parameter varies. Thus the bond energies cannot be constant due to elastic misfit. So the model is only perfect for liquids, glasses and crystals where the atoms have the same size. The elastic misfit caused by differences of size limits solid solubility in systems where complete mixing easily occurs in the liquid. So be aware of this limit of the QC model. Now we define: PAA = the NUMBER of AA bonds PBB = the NUMBER of BB bonds PAB = the NUMBER of AB bonds Then H = PAA hAA + PBB hBB + PAB hAB Z is the Coordination Number = # of bonds (nearest neighbours) per atom /molecule ( Z = 8 in bcc 12 in fcc 4 in Si) Then Number of Atoms: NA = PAB/Z + 2PAA/Z. 2 A atoms in each AA bond NB = PAB/Z + 2PBB/Z. So: PAA = 0.5 (NA Z  PAB) and PBB = 0.5 (NB Z  PAB) H = 0.5NAZhAA+0.5NBZhBB+ PAB{hAB0.5(hAA+hBB)} But But ΔHMIX = H  xAHAo  xBHBo and xAHAo = 0.5NAZhAA and xBHBo = 0.5NBZhBB Since we have, in pure A, xA moles of A, with 0.5NAZ AA bonds. So ΔHMIX = PAB{hAB0.5(hAA+hBB)} Now to determine PAB in a random solid solution: Probability of a site hav ing an A atom = xA. it Probability of an adjacent site being B = xB. So, the probability of the bond being AB or BA is then: Total # of bonds in one mole = 0.5 Z NAV. So: PAB = 2 xA xB 0.5 Z NAV = xA xB Z NAV ΔHMIX = xA xB Z NAV {hAB0.5(hAA+hBB)} ΔHMIX = xA xB Ω , where Ω = Z NAV {hAB0.5(hAA+hBB)} Ω is Interaction Parameter 2 xA xB When hAB = 0.5(hAA+hBB) (average) Ω = 0 and ΔHMIX = xA xB Ω = 0 Ideal solution. Exact ideality is very rare. When hAB < 0.5(hAA+hBB), Ω and ΔHMIX < 0. Stronger bonding in solution. When hAB > 0.5(hAA+hBB), Ω and ΔHMIX > 0. Weaker bonding in solution. Swalin ( p146) gives the useful result that: HAMIX = HA  HAo = (1  xA)2 Ω HBMIX = HB  HBo = (1  xB)2 Ω This can be found easily to be true: ΔHMIX = xA (HA  HAo) + xB(HB  HBo) ΔHMIX = xA (1  xA)2 Ω + xB (1  xB)2 Ω But xA = 1  xB ΔHMIX = Ω xA xB { (1  xA) + (1  xB) } ΔHMIX = Ω xA xB {2  (xA + xB)} = Ω xA xB as shown above (top of the page) So using these steps we can readily go in reverse back to Swalin's equation Next more Next, more useful equations: ln γA = (1  xA)2 Ω /RT and ln γB = (1  xB)2 Ω /RT From: _ _ ΔGMIX = xA(GA GAo) + xB (GB  GBo) ΔGMIX = xA(ΔGA) + xB (ΔGB ) _ _ ΔHMIX = xA(HA HAo) + xB (HB  HBo) ΔHMIX = xA(ΔHA) + xB (ΔHB ) ΔSMIX = xA(SA SAo) + xB (SB  SBo) ΔSMIX = xA(ΔSA) + xB (ΔSB ) Comparing terms: _ _ ΔGA = GA GAo = RT ln a A = ΔHA  TΔSA ΔGB = GB GBo = RT ln a B = ΔHB  TΔSB Application to unmixing (Spinodal decomposition)
Ω > 0. So: γB > 1 Examples: NaClKCl, Al Zn… Occurs at TC = Ω /2R Ω = Z NAV{hAB0.5(hAA+hBB)} hAB > 0.5(hAA+hBB) To show TC = Ω /2R ln γB = (1  xB)2 Ω /RT aB = xBγB = xB exp{(1  xB)2 Ω /RT} daB/dxB = exp{(1xB)2Ω/RT} + xB (2xB  2) Ω /RT exp{(1  xB)2 Ω /RT} but at TC, (daB/dxB)X(B)=0.5 = 0 Div ide thru by exp{(1xB)2 Ω /RT} 0 = 1 + xB(2xB  2) Ω /RT = 1 + (0.5 1) Ω /RTC 0 = 1  Ω /2RTC So So Ω = 2RTC or TC = Ω /2R Remember: Remember: Ω > 1 , hAB > 0.5(hAA+hBB) – Bonding between unlike atoms/molecules weaker than that between like atoms. Examples: Oil and water. Metals and oxides Illustration Illustration of Omega Effect on ΔGm ΔHMIX = xA xB Ω Ω = Z NAV {hAB‐0.5(hAA+hBB)} Ω negative – γi > 1, giving positive deviation from Raoult’s law Free energy From Swalin Omega Omega effect on activity Ω negative – attraction between A‐B stronger than between like neighbors. For positive Ω – the reverse is true From Swalin ...
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This note was uploaded on 02/12/2010 for the course MAT E 510 taught by Professor Yury during the Summer '09 term at Drexel.
 Summer '09
 Yury

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