Phase Transformation Binary Solution PDF

Entropy (S) Thermal Two views and not two types Configurational + other* The entropy change of a system at temperature T absorbing an infinitesimal amount of heat S config = k ln  Q in a reversible way, is:...

Entropy (S) Thermal Two views and not two types Configurational + other* The entropy change of a system at temperature T absorbing an infinitesimal amount of heat S config = k ln  Q in a reversible way, is: Zero or +ve Boltzmann constant (1.38  10−23 J/K) Q Sthermal = T No. of different configurations the system * For a system at Constant Energy, many microscopic states can give a macroscopic state of identical energy. These microscopic states could originate from various sources like Configurational, Electronic, Vibrational and Rotational states. ▪ In many cases the configurational term may be the predominant one considered. S = SConfigurational + SVibrational + SRotational + SElectronic + SNuclear  Entropy (S) ❑ One way of simply stating the ‘concept behind entropy’is → A system will, more often than not, be found in ‘states; with higher probability. (This is nothing but the statement of the obvious!) ❑ However, the implications of the above are ‘profound’. This can be best understood by considering the mixing of two ideal gases (or in the ‘toy model’ below as the mixing of 6 circles- 3 red and 3 blue, on 6 fixed lattice sites). ❑ Assuming that red and blue circles can move about randomly on the fixed sites and also assuming that the probability of the occurrence of each state is identical (i.e. no state is preferred over any other state); there are 20 possible configurations as shown in the next slide. ❑ As seen (from the figure in the next slide) the majority of the states (18/20) are ‘mixed states’ and only two are the ‘totally’unmixed ones. ❑ Hence, purely from a probabilistic point of view, mixed states occur more often than the unmixed ones. ❑ This implies, if we start with a unmixed configuration as in the figure below and the system can access all possible states with equal probability → the system will go from a unmixed state (of low entropy) to a mixed state (of higher entropy). A B 6 6.5.4 C3 = = = 20 6 Unmixed state 3 3 3.2.1 Two boxes separated by a barrier initially Unmixed state ▪ In the case of two gases initially separated by a barrier, which is allowed to mix on the removal of the barrier: the number of mixed states are very large compared to the unmixed states. Hence, if all configurational states are accessible, the system will more likely found in the mixed state. I.e. the system will go from a unmixed state to a mixed state (worded differently the system will go from ‘order’to ‘disorder). ▪ On the other hand it is unlikely (improbable) that the system will go from mixed state to a unmixed state. (Though this is not impossible → i.e. a mixed system can spontaneously get ‘unmix’itself!!) Mixed states with ‘various’ degrees of mixing Note: the profoundness of the concept of entropy comes from its connection to energy (via T) 18 mixed states 2 unmixed states * We assume that all states have equal probability of Unmixed state 39 occurring and are all accessible  Entropy change due to mixing of two pure elements ❑ Let us consider the entropy change due to mixing of two pure crystalline elements A & B (a simple case for illustration of the concept of entropy). ❑ The unmixed state is two pure elements held separately. The mixed state (for now assuming that the enthalpy of mixing is negative- i.e. the elements want to mix) represents an atomic level mixing of the two elements. ❑ Let the total number of lattice sites (all equivalent) be N. ❑ The Entropy of the unmixed state is zero (as in pure crystalline elements atoms are indistinguishable and hence represent one state). Spure A = Spure B = k ln(1) =0 ❑ In the mixed state the entropy of the system increases (Smixed state) N!  = NC n = ❑ The number of permutations possible in the mixed system is  A (N − nA )! (nA )! N! S = k ln  = k ln S = S mixed state − S pure elements ( A& B) (n A )! (nB )! N! Zero S = k ln  = k ln (N − n A )! nA ! n n An useful formula for evaluating ln(factorials) is the Stirling’s approximation: n! ~ 2 n   e Ln(r !) = r ln(r) − r r! = r ~  asymptotically equal, e = 2.718…  Entropy change during melting ❑ At the melting point of a material when heat is supplied (Q) to the material it does not lead to an increase in the temperature. Instead, the absorbed heat leads to melting- i.e. the energy goes into breaking of bonds in the solid and consequently a transformation in the state of the material (solid → liquid). The entire process of melting takes place at a constant temperature (Tm). The heat absorbed is called the Latent Heat of Fusion (Hfusion). ❑ Suppose we take a mole of Al atoms melt then the change in entropy can be calculated as below. ❑ In the solid state the atoms are fixed on a lattice (of course with vibrations!) and this represents a ‘low entropy’ state. On melting the entropy of the system increases as the atoms are free to move around and may configurations are possible. From this point of view often Entropy is considered as a measure of disorder (however, it must be clear that the phrase ‘measure of disorder’ is used with the understanding of the context) Data: Enthalpy of fusion (Hf) = 10.67 kJ/mole, Melting Point (Tm) = 933.4 K (660.25C) Q H f 10.67 103 Smelting Al = = = = 11.43 J / K / mole T Tm 933.4 Single component system This line slopes upward as at constant T if we increase the P the gas will liquefy as liquid has lower volume (similarly the reader should draw horizontal lines to understand the effect of pressure on the stability of various phases- and rationalize the same). Phase fields of non-close packed structures shrink under higher pressure Phase fields of close packed structures expand under higher pressure These lines slope downward as: Under higher pressure the phase with higher packing fraction (lower volume) is preferred  Gibb’s free energy change with temperature in a single component system ❑ An isolated system always tries to maximize the entropy. That means the system is stable when it has maximum possible entropy. ❑ Instead of considering isolated system, we need to consider the system which interacts with the surroundings because heat transfer is always with respect to the surroundings only. ❑ Any transformation is possible only when dS+dSSurrou ≥ 0 where dS is the change in entropy of the system. ❑ In a reversible process, when system can transform back to its previous state it is equal to zero and in an irreversible process when the system cannot come back to its previous state is greater than zero. ❑ We need to find the stability with respect to another term, for the sake of convenience, which can be used without referring to the surroundings. We shall show now that free energy is rather a suitable property to define stability of the phases. ❑ Let us consider that the system absorbs some amount of heat δQ from the system. Since the surrounding is giving away the heat, we can write Q dS surrou = − T  Gibb’s free energy change with temperature in a single component system ❑ We have seen before that in an isobaric system δQ = dH. So we can write dH dS − 0 T dH − TdS  0 ❑ We are considering isobaric system. If we consider the transformation at a particular temperature ( T constant, dT = 0) then dH − TdS − SdT  0 d (H −TS)  0 dG  0 ❑ So we have derived a more reasonable relation, which can be used without referring to the surroundings. ❑ In an reversible process, such as allotropic transformation or the transformation from solid to liquid or liquid to solid the free energy change is zero. There will be many irreversible transformations (later we shall see these kinds of diffusion controlled transformations), where dG is less than zero. ❑ This also indicates that a system will be stable when it has minimum free energy, so that it is not possible to get anymore dG less than zero by any further transformation.  Stability of the phases in a single component system ❑ One component can have different stable phases at different temperature and pressure ranges, for example, solid, liquid and the gas phase. ❑ One phase is said to be stable, when it has lower free energy compared to other phases in a particular range of temperature and pressure. ❑ Let us consider constant pressure. ❑ To compare the stability of different phases in a particular range of temperatures, we need to calculate the free energy of the phases with respect to temperature. ❑ To determine Gibb’s free energy at a particular temperature, we need to determine H and S. Similarly it can be calculated at every temperature to gain knowledge on the change in free energy with temperature. ❑ As mentioned previously the data on specific heat are available in literature and these parameters can be calculated.  Stability of the phases in a single component system ❑ From the definition, we know that the slope of the enthalpy at any temperature is equal to the specific heat at that temperature. G = H − TS dG = dH − TdS − SdT (H = E + PV ) dG = dE + PdV +Vdp − TdS − SdT (dE = Q − PdV ) dG = Q − PdV + PdV +VdP − TdS − SdT dG = TdS +VdP − TdS − SdT Q dG = VdP − SdT = dS T ❑ So at a constant pressure the slope of the free energy curve  G    = −S Typical variation of thermodynamic  T P parameters are shown in the figure. ❑ If we want to know whether a solid or liquid phase will be stable at a particular temperature or in a temperature range, we need to find free energy for both the phases.  Stability of the phases in a single component system ❑ For that we need to know the variation of specific heat with temperature for both the phases as shown in the figure. ❑ Specific heat for a liquid phase is always higher than the solid phase. That means energy required to increase the temperature by one degree for liquid phase is higher than the solid phase. ❑ Superscript, S and L denote the solid and liquid phases. ❑ Free energy for liquid phase changes more drastically compared to the free energy of the liquid phase because entropy of the liquid phase is always higher, which is the slope of the free energy. ❑ At melting point, free energy for both the phases are the same and the difference between the enthalpy of these two phases is equal to the latent heat of fusion L. 47  Stability of the phases in a single component system ❑ It must be apparent that one particular phase at certain temperature range will be stable, if the free energy, G is lower of that phase than the other phase. ❑ At low temperature range, one particular phase will be stable, which has low enthalpy, since “TS” term will not dominate. That is why solid phase is stable at low temperature range. ❑ On the other hand, at higher temperature range, phase having higher entropy will be stable since in this range “TS” term will dominate. That is why liquid phase is stable at high temperature range. ❑ This is the reason that α-Ti with close packed HCP structure is stable at low temperature range, whereas, at high temperature range β-Ti with relatively more open BCC structure is stable. ❑ If we fix a temperature, we have seen that one material/element will be stable as one type of phase with a particular state compared to other. ❑ It can stay in equilibrium if it has minimum free energy and dG=0, it indicates that for small fluctuation it does not move to another state. ❑ Only dG = 0 is not sufficient condition, since we have seen that at the melting point both the liquid any solid phases can stay together.  Stability of the phases in a single component system ❑ Let us consider two different states (different atomic arrangements) of a material, as shown in the figure. ❑ State 1 fulfills the condition dG=0, but G is not minimum. ❑ State 2 is in the most stable condition, since G is minimum and also dG= 0 ❑ Material at state 1 is called metastable state, for example, diamond, metallic glass. ❑ Material at state 2 is the equilibrium state or most stable state, for example, graphite. Given chance, material will transform from the state 1 to state 2. ❑ Transformation is possible only when driving force for transformation, ∆GTrans is negative G = Gd = G final − Ginitial = G2 − G1 ❑ However, straightforward transformation is not possible. It should cross a barrier. ❑ The barrier is ∆GHump = ∆Ga = activation energy that is the barrier for the phase transformation. The barrier comes from the need for nucleation and diffusion.  Clausius - Clapeyron Equation From previous studies dG = VdP − SdT G l = G s → G = 0 Tb ∆G consider : S → L Ts dG S = V S dT − S S dT dG L = V L dT − S L dT T Tm (V −V )dP = (S − S )dT S L S L dT V V S −V L = = S dP S S − S L dT The rate of change of transformation = dP temperature as a function of pressure dT V = dP S Is known as Clausius-Clapeyron equation  Thermodynamic parameters in a binary system ❑ Previously we are considered one element only. Now we consider interaction between two elements. This is not straightforward since elements can interact differently and thermodynamic parameters may change accordingly. ❑ Let us consider a Binary system with elements A and B. For our analysis, we consider XA mole of A and X B mole of B so that XA + X B = 1 ❑ That means we consider total one mole of the system That further means we consider total number of atoms equal to the Avogadro number, N0 (= 6.023  1023), Ni Where Xi = is the number of atoms of element i. N0 ❑ Unlike single component system, where we determine the change in free energy with temperature, in the binary case we shall find the change in free energy with the change in composition at different constant temperature at a time. ❑ Let us consider the free energy for one mole of element A is GA and one mole of B is GB.  Thermodynamic parameters in a binary system ❑ So before mixing when kept separately, XA mole of A and X B mole of B will have the free energy of XAGA and XBGB respectively Total free energy before mixing G0 = X AGA + X BGB ❑ After mixing there will be change in free energy ❑ Total free energy after mixing G = G0 + Gmix ❑ ∆Gmix is the free energy change of the alloy because of mixing Gmix = G − G0 = H − TS − (H 0 − TS0 ) = (H − H 0 ) − T (S − S0 ) Gmix = Hmix − TS mix ❑ So, once we determine the change in free energy because of mixing, we can determine the total free energy after mixing. ❑ Let us first determine, the enthalpy change because of mixing (∆Hmix) and the change in entropy because of mixing (∆Smix ) ❑ Note that system always tries to decrease enthalpy and increase entropy for stability.  The change in enthalpy because of mixing, ∆Hmix ❑ We take the following assumptions: ✓ The molar volume does not change because of mixing ✓ Bond energies between the pure elements do not change with the change in composition ✓ We neglect the role of other energies. ❑ After mixing, the system can have three different types of bonding, A-A, B-B and A-B ❑ Enthalpy of mixing can be expressed as H mix = N 0 ZX A X B  N0 - Avogrado number, Z = coordination number ❑ The change in internal energy 1  =  AB − ( AA +  BB ) 2 εAB is the bond energy between A and B εAA is the bond energy between A and A εBB is the bond energy between B and B ❑ It can be written as  H mix =  X A X B Where  = N 0 Z  The change in enthalpy because of mixing, ∆Hmix Situation 1: Enthalpy of mixing is zero Hmix =XAXB =0 1 That means  AB = ( AA +  BB ) 2 Because of transformation internal energy will increase. That means transformation is to be endothermic. Atoms will try to maximize A-A and B- B bonds. Situation 2: Enthalpy of mixing is less than zero Hmix =XAXB 0 1 That means  AB  ( AA +  BB ) 2 Because of transformation internal energy will decrease. That means transformation is exothermic. Atoms will try to maximize A-B bonds.  The change in enthalpy because of mixing, ∆Hmix Situation 3: Enthalpy of mixing is greater than zero Hmix =XAXB 0 1 That means  AB  ( AA +  BB ) 2 Because of transformation internal energy will increase. That means transformation is to be endothermic. Atoms will try to maximize A-A and B-B bonds. 55  Slope/maximum/minimum of the enthalpy of mixing curve H mix = X A X B = ( X B − X B2 ) d (H mix ) = (1− 2X B ) dX B d (H mix ) At maximum/minimum =0 This implies XB = 0.5. That means maximum dX B or minimum will be at XB =0.5 d (H mix ) Further atX B ⎯⎯ lim →0 =  That means the slope at the beginning dX B has a finite value of x  The change in entropy because of mixing, ∆Smix ❑ Since we are considering transformation at a particular temperature, the change in entropy because of the change in temperature can be neglected. ❑ We need to consider only the configurational entropy change. Configurational entropy change comes from the possibilities of different ways of arrangement of atoms. ❑ Following statistical thermodynamics the configurational entropy can be expressed as S = l ln  k is the Boltzmann constant w is the measure of randomness Smix = S − S 0 = k ln  − k ln 1 = k ln  since atoms at their pure state before mixing can be arranged in only one way ❑ If we consider the random solid solution, then (n A + nB )! = nA and nB are the number of atoms of A and B n A!nB ! ❑ Following Stirling’s approximation ln N!= N ln N − N  The change in entropy because of mixing, ∆Smix ❑ So, ∆Smix can be written as Smix = kln = k(nA +nB )ln(nA +nB )−(nA +nB )−[nA lnnA −nA ]−[nB lnnB −nB ]  nA nA  S mix = −k n A ln + nB ln  n A + nB n A + nB  Number of atoms can be related to the mole fraction, X and the Avogadro number N0 following nA = X A N 0 nB = X B N 0 X A + XB =1 n A + nB = N 0 S mix = −kN 0 [X A ln X A + X B ln X B ] = R[ X A ln X A + X B ln X B ] where, R is the gas constant  Slope/maximum of the entropy of mixing curve d (S mix )  1 1  XB = −R − ln(1− X B ) − (1− X B ) + ln X B + X B  = −R ln dX B  (1− X B ) X B 1− X B d (S mix ) = 0 at maximum, this corresponds to XB = 0.5 dX B Further, the slope at XB →0 is infinite. That means the entropy change is very high in a dilute solution As mentioned earlier the total free energy after mixing can be written as G = G0 + Gmix where G0 = X AG A + X B GB Gmix = Hmix − TS mix H mix = X A X B S mix = −R[ X A ln X A + X B ln X B ] So ∆Gmix can be written as Gmix = X A X B + RT[ X A ln X A + X B ln X B ] Following, total free energy of the system after mixing can be written as G = X AG A + X B GB + X A X B + RT[ X A ln X A + X B ln X B ]  Free energy of mixing We need to consider three situations for different kinds of enthalpy of mixing Situation 1: Enthalpy of mixing is zero G = G0 + Gmix = G0 −TS mix = X AGA + X BGB + RT[X A ln X A + X B ln X B ] ❑ With the increase in temperature, -T∆Smix will become even more negative. ❑ The values of GA and GB also will decrease. ❑ Following the slope Go might change since GA and GB will change differently with temperature.  Free energy of mixing Situation 2: Enthalpy of mixing is negative G=XAGA +XBGB +XAXB +RT[XA lnXA +XB lnXB] G0 Negative -T∆Smix ∆Hmix ❑ Here both ∆Hmix and -T∆Smix are negative ❑ With increasing temperature G will become even more negative. ❑ Note that here also G0 may change the slope because of change of GA and GB differently with temperature.  Free energy of mixing Situation 3: Enthalpy of mixing is positive G = X AG A + X B GB + X A X B + RT[ X A ln X A + X B ln X B ] G0 Positive ∆Hmix -T∆Smix ❑ ∆Hmix is positive but -T∆Smix is negative. At lower temperature, in a certain mole fraction range the absolute value of ∆Hmix could be higher than T∆Smix so that G goes above the G0 line. ❑ However to GA and GB, G will always be lower than G0 , since ∆Hmix has a finite slope, where as ∆Smix has infinite slope. The composition range, where G is higher than G0 will depend on the temperature i.e., -T∆Smix ❑ Because of this shape of the G, we see miscibility gap in certain phase diagrams. ❑ At higher temperature, when the absolute value of - T∆Smix will be higher than ∆Hmix at all compositions, G will be always lower than G0  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, nA, nB. G = G (T, P, nA, nB ……) ❑ At particular temperature and pressure, partial derivative gives G G dG = dn A + dn B n A n B = A dn A + BdnB G = A is the chemical potential of element A. It measures the change in free n A energy because of very minute change of element A. G = B is the chemical potential of element B. It measures the change in free n B energy because of very minute change of element B. ❑ It should be noted here that the change should be so minute that there should not be any change in concentration because the chemical potential is a concentration dependent parameter.  Concept of the chemical potential and the activity of elements ❑ Let us consider an alloy of total X moles where it has XA mole of A and XB mole of B. Note that x is much higher than 1 mole. ❑ Now suppose we add small amounts of A and B in the system, keeping the ratio of XA : XB the same, so that there is no change in overall composition. ❑ So if we add four atoms of A, then we need to add six atoms of B to keep the overall composition fixed. Following this manner, we can keep on adding A and B and will reach to the situation when XA mole of A and XB mole of B are added and total added amount is XA + XB = 1 ❑ Since previously we have considered that the total free energy of 1 mole of alloy after mixing is G, then we can write G = A X A + A X B ❑ Previously, we derived G = X AG A + X BGB + X A X B + RT[X A ln X A + X B ln X B ] ❑ Further, we can write X X = X 2X + X X 2 A B A B A B G = X A (G A + X 2 + RT ln X ) + X (G + X 2 + RT ln X ) B B B B A B  Concept of the chemical potential and the activity of elements Further, comparing the expressions for free energy, we can write  = G + X 2 + RT ln X A B A  A = GA + (1− X A ) 2 + RT ln X B = GB + (1− X B ) 2 + RT ln X A B In terms of activity A = G A + RT ln a A B = GB + RT ln aB So the relations between the chemical potential and activity are RT ln a A = (1− X A ) 2 + RT ln X A RT ln a B = (1− X B ) 2 + RT ln X B  Concept of the chemical potential and the activity of elements ❑ Activities and chemical potentials are determined from a free energy curve after taking a slope, as explained in the figure. ❑ If we are interested to determine the activities or chemical potentials in a binary system A-B, let say at X B* , we need to take a slope on the free energy curve at the free energy, G * and extended it to pure element, A (XB = 0) and pure element B (XB = 1). ❑ The point at which it hits NB = 0, the value corresponds to the chemical potential of element A (  *A ). ❑ From previous slide, we can write − RT ln a A = G A −  * A ❑ So, once the chemical potential is known, the activity of the element can be calculated using the above equation, as explained in the graph. ❑ It can be proved that, by taking slope and then extending to XB = 0 and XB = 1, we can find the chemical potentials. We can write G * =  *A + ab ❑ Further we can write ab = cd  ab = X * cd = X * (  * −  * )  G * =  * X * +  * X * X B* 1 B B B A A A B B Concept of the chemical potential and the activity of elements  Concept of the chemical potential and the activity of elements    A = exp  (1− X A ) 2   RT     B = exp  (1− X B ) 2   RT  In the case of positive enthalpy of mixing, activity deviates positively and in the case of negative enthalpy of mixing activity deviates negatively from the ideal mixing line.    X A → 0,  A → exp  RT  X A → 1,  A → 1   X B → 1,  B → 1 X B → 0,  B → exp   RT  Henry’s law: activity of elements is more Rault’s law: activity is equal to the mole or less the constant in a very dilute fraction near the mole fraction of 1. solution.  Equilibrium conditions between different phases ❑ As explained previously, while discussing the phase diagrams, we have shown that at certain conditions two phases, such as solid and liquid or two solid phases α and β can stay together. ❑ This can be explained with respect to chemical potential of elements. For the sake of explanation let us consider that there are two phases α and β which are staying together. ❑ If you remove dnB of element B from the β phase and add them to α phase, then the change in free energy of the α and β phases will be dG =  B dnB   dG  = −  B dnB ❑ So the total free energy change of the system will be dG =   dn −   dn = (   −   )dn B B B B B B B ❑ However we know that in equilibrium condition dG = 0. That means the chemical potential of element B in both the phases should be the same. ❑ That further means that even if a small amount of material is transferred from the β to the α phase, there will be no difference in equilibrium as long as the chemical potential of elements are the same in both the phases.  Equilibrium conditions between different phases ❑ Previously, we have shown that the system will not be in equilibrium that is it will go through irreversible transformation if dG < 0. That means ( −   )dn  0 B B B ❑ This indicates that the chemical potential of B in the α phase is less then the chemical potential of the same element in the β phase. ❑ So to reach to the equilibrium system will transfer B from the β phase to the α phase. ❑ Now we understand, why both solid and liquid phases can stay together in certain composition range. It can be understood from the common tangent between X SB and X BL chemical potential of any of the elements are the same in both the phases.