Technical Notes - Role of Strain Energy in Solid Solution Thermodynamics

THE function of this paper is to present certain results based on the fact that the strain energy arising from the solution of out-of-size solute atoms into the solid matrix is free energy and not internal energy, and that this strain energy is almost wholly shear strain energy as Zener1 pointed out. One natural consequence of this fact is an expression for the excess entropy of mixing due to the strain energy. Also, a derivation of the integral molar free energy of mixing partitioned to strain energy will be made using the model of Lawson2 and thermodynamics. This derivation has the advantage of showing in a simple way the assumptions involved in Lawson's treatment. The excess entropy of mixing partitioned to strain energy can be derived in two ways. The first way makes use of the fact that the strain energy is almost wholly in the form of shear strain energy. In this case, as Zener1 pointed out, the change in entropy due to the addition of shear strain energy is to a first approximation given by ?S/?? p,T = - ?lnµ/?T p,? Hence, the excess entropy of mixing partitioned to strain energy will be given by Sx ? = ?S/?? p,T Gx ? = - ?lnµ/?T p,? Gx ? where, Gx ? is the relative integral molar free en- ergy partitioned to strain energy; S is the entropy; is the shear strain energy; µ is the shear modulus; T is the absolute temperature; p is the pressure; and ? is the shear strain. Zener,1 as well as others, has pointed out that the shear strain energy stored in a compressible isotropic solid due to the insertion of an off-size incompres-sible sphere into a spherical hole in the solid, is approximately given by 2/3 µ/O (O°I - O)2 where a is the volume of spherical hole and no , is the volume of sphere. If we now consider the introduction of one mol of such spheres into an infinite volume of solid, then the relative partial molar free energy excess due to strain energy alone is GM ? = 2/3 µ/v (v ° 1 - v)2 where v is the molar volume of solvent and v°1, is the molar volume of pure solute type 1. Let us consider that these spheres are the two types of atoms in a binary solution, then there will be a similar relation for solute 2. The total relative integral molar free energy excess due to strain is then, to the approximations stated above GM ? = GM ? x1 + GM2 ? x2 = 2/3 µ/v [(v ° 1 - v)2 x1 + (v ° 1 - v)2 x2]. Lawson's relation for the strain energy can be de-