Thermodynamic parameters in a binary system

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Kerrie Thompson
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1 Thermodynamic parameters in a binary system Previously, we 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.

2 Let us consider a binary system with elements and For our analysis, we consider mol of and mol of, so that + = That means we consider total one mole of the system. That further means we consider total number of atoms equal to the vogadro number, N o (= 6.23x 23, where i = N i /N o. N i is the number of atoms of element i. 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 is G and one mole of is G. So before ing when kept separately, mole of and mole of will have the free energy of G and G, respectively. Total free energy before ing G = G + G

3 fter ing there will be change in free energy. Total free energy after ing G = G + G G is the free energy change of the alloy because of ing G = G G ( H TS = ( H H T ( S = H TS S G = H T S So, once we determine the change in free energy because of ing, we can determine the total free energy after ing. Let us first determine, the enthalpy change because of ing ( change in entropy because of ing ( S H and the Note that system always tries to decrease enthalpy and increase entropy for stability.

4 The change in enthalpy because of ing, DH We take the following assumptions: The molar volume does not change because of ing. ond energies between the pure elements do not change with the change in composition We neglect the role of other energies. fter ing, the system can have three different types of bonding, -, - and - Enthalpy of ing can be expressed as H = N Z ε N is the vogrado number, Z is the coordination number The change in internal energy ε = ε ( ε + ε ε ε ε is the bond energy between and is the bond energy between and is the bond energy between and 2 It can be written as H = ξ where ξ = Z ε N

5 There can be three situations: Situation : Enthalpy of ing is zero H = ξ = That means ε = ( ε + ε 2 There will be no preference to choose neighbouring atoms. toms can sit randomly at any lattice points.

6 Situation 2: Enthalpy of ing is less than zero H = ξ < That means ( ε ε ε < + 2 ecause of transformation internal energy will decrease. That means transformation is exothermic. toms will try to maximize - bonds. Situation 3: Enthalpy of ing is greater than zero H = ξ > That means ( ε ε ε > + 2 ecause of transformation internal energy will increase. That means transformation is to be endothermic. toms will try to maximize - and - bonds.

7 Slope/maximum/minimum of the enthalpy of ing curve H = ξ = ξ 2 ( d ( H d = ξ ( 2 d H d t maximum/minimum ( = This implies =.5. That means maximum or minimum will be at =.5 Further, d ( H d at lim = ξ That means the slope at the beginning has a finite value of ξ

8 The change in entropy because of ing, DS 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 = k ln w (k is the oltzmann constant and w is the measure of randomness S = S S = k ln w k ln = k ln w since atoms at their pure state before ing can be arranged in only one way If we consider the random solid solution, then w ( n + n! n!! = n and n are the number of atoms of and n

9 Following Stirling s approximation ln N!= N ln N N So S can be written as S = k ln w = k = k n ln {[ ( n + n ln( n + n ( n + n ] [ n ln n n ] [ n ln n n ]} n n + n + n ln n n + n Number of atoms can be related to the mole fraction, and the vogadro number, N o following n = N n = N + = n + n = N S = kn = R [ ln + ln ] [ ln + ln ] where, R is the gas constant

10 Slope/maximum of the entropy of ing curve 6 5 Entropy of ing, S ( S d d = R ln Composition, ( ( + ln ( + = R ln d ( S d = at maximum, this corresponds to =.5 Further, the slope at is infinite. That means the entropy change is very high in a dilute solution

11 s mentioned earlier the total free energy after ing can be written as G = G + G where G = G + G G = H T S H = ξ S = R [ ln + ln ] So G can be written as G = ξ + RT [ ln + ln ] Following, total free energy of the system after ing can be written as G = G + G + ξ + RT [ ln + ln ]

Recitation: 12 12/04/03

Recitation: 12 12/04/03 Recitation: 12 12/4/3 Regular Solution Solution: In an ideal solution, the only contribution to the Gibbs free energy of ing is the configurational entropy due to a random ture: ΔG id G id = x + x µ µ