Report

Thermodynamics of surface and interfaces Define : Consider to be a force / unit length of surface perimeter. (fluid systems) If a portion of the perimeter moves an infinitesimal of distance in the plane of the surface of area A, the area change dA is a product of that portion of perimeter and the length moved. Work term - dA; force x distance, and appears in the combined 1st and 2nd laws of thermodynamics as dU TdS pdV dN i i i dA Strictly speaking , is defined as the change in internal energy when the area is reversibly increased at constant S, V and Ni (i.e., closed system). For a system containing a plane surface this equation can be readily integrated : U TS PV N i i A i and rearranging for yields. 1 U TS PV i N i A i where U – TS + PV is the Gibbs free energy of the system, i.e., the actual energy of the system And N i i i is the Gibbs free energy of the materials comprising the system, i.e., the energy of the system as if it were uniform ignoring any variations associated with the surface Thus is an excess free energy due to the presence of the surface. def Surface Excess Quantities Macroscopic extensive properties of an interface separating bulk phases are defined as a surface excess. There is a hypothetical 2D “dividing surface” defined for which the parameters of the bulk phases change discontinuously at the dividing surface. def The excess is defined as the difference between the actual value of the extensive quantity in the system and that which would have been present in the same volume if the phases were homogeneous right up to the “ Dividing Surface ” i.e., x x s total The real value of x in the system x x The values of x in the homogeneous and phases Concept of the Gibbs Dividing Surface Extensive property Density Distance perpendicular to the surface For a 1 component system the position of the dividing surface is chosen such that the two shared areas in the figure are equal. This yields a consistent value (equal to zero ) for the surface excess. For a multicomponent system the position of the dividing surface that makes some Ni equal to zero will be unlikely to make all the other Nj ≠i = 0. By convention, N1, the surface excess of the component present in the largest amount (i.e., the solvent) is made zero by appropriate choice of dividing surface. Alternatively if we consider a large homogeneous crystalline body containing N atoms surrounded by plane surfaces then if U0 and S0 are the energy and entropy / per atom, the surface energy per unit area Us is defined by U N U AU N U A 0 s 0 where U is the total energy of the system. :U s Similarly S S N AS 0 s Consider once again the combined form of 1st and 2nd laws including the surface work term. dU TdS pdV dN i i dA i Substitution of the definition of G leads to dG SdT Vdp i dN i dA i If the surface is reversibly created in a closed system (Ni fixed) at constant T and P. G A T ,P ,N i is always the free energy change appropriate to the constraints imposed on the system. Since for the bulk phases and the surface terms vanish, the combined 1st and 2nd law take the form dU T dS pdV i dN dN i i and dU T dS pdV i i i and for the total system dU T dS pdV dN i i dA i From the definition of surface excess: dU T dS pdV s s s 0 i X s X X X i dN s i dA By Def. Integration yields, N U TS s s i A i i Forming the Gibbs-Duhem relation : 0 S dT N i d i Ad s i so d sdT d i Gibbs-Adsorption Equation i i where s S s s A ; i Ni A Solid and liquid Surfaces In a nn pair potential model of a solid, the surface free energy can be thought of as the energy/ unit -area associated with bond breaking. : work/ unit area to create new surface = n 2 A where n/A is the # of broken bonds / unit-area and the is the energy per bond i.e., the well depth in the pair-potential. Then letting A = a2 where a lattice spacing 2a 2 pair potential If the solid is sketched such that U(r) the surface area is altered a a da A A dA r the energy d The total energy of the surface U S Asurf .is changed by an amount. dU dA Ad and dU dA f A d dA Surface Stress and Surface Energy Unit Cube W1=2 1 fxx Split Stretch W2 w1 The difference in the work per unit area required for the constrained stretching (fix dimension in the y direction while stretching along the x-direction) is defined as the surface stress, fxx. This is the excess work owing to the presence of the surfaces. fxx w2=2(+d)(1+dx) 1+dx Shuttleworth cycle relating surface stress, f and surface energy, . Surface Stress and Surface Energy Relation between fij and Consider 2 paths to get to the same final state of the deformed halves. Path I - The cube is first stretched Path II - The cube is first separated and then separated. and then stretched. WI = w1 + w2 WII = W1 W2 = w1 + 2(1+dx) ( D) = 2 W2 = w1 + 2 2 D 2 dx where Dxx (= dx/1) has caused a change D in . Since WI = WII, w1 + 2 2 D 2 Dxx = 2 W2 work/unit area = (W2 - w1)/2Dxx = fxx = + D /Dxx Surface stress, surface free energy and chemical equilibrium of small crystals Recall that for finite-size liquid drops in equilibrium with the vapor. (see condensation discussion) l 0 2 r Vl Equil. cond. where Vl is the molar vol. of the liquid. For a finite-size solid of radius r the internal pressure is a function of the size owing to the surface stress {isotropic surface stress}. s 0 2f r Vs The pressure difference between the finite-size solid in equil. with the liquid is Ps Pl 2f r Consider the equilibrium between a solid sphere and a fluid containing the dissolved solid. r The total energy of the system is given by dU TdS pdV dN i dA =0 i d U T d S s ( so lid ) T d S l ( liq u id ) T d S ( su rfa ce ) p s d V s p l d V l p d V N + dN dN s 1 s 1 l 1 l 1 N dN s i i2 s i i dN i dA l l i2 Gibbs dividing surface set for component 1, other components are not allowed to cause area changes. dU T dS s ( solid ) T dS l ( liquid ) T dS ( surface ) p s dV s p l dV l + 1 dN 1 1 dN 1 dA s s l l Consider the variation dU = 0 under the indicated constraints, dU 0; dS s ( solid ) dS l ( liquid ) dS ( surface ) 0, dV 0 dV S dV l ; dV S dV l dN 1 0 dN 1 dN 1 0; S dN 1 dN 1 l S l Making the substitutions dU ( p s p l ) dV s ( 1 1 ) dN 1 dA 0 s l s and for a sphere, dA = (2/r)dVs 0 ( p s p l ) dV s ( ) dN s 1 l 1 s 1 2 r dV s since dV s / dN 1 V s / N 1 V o the m olar volum e s s 2 ( p s pl ) Vo r s 1 l 1 A lso since, p S p l s 1 l 1 2f r 2( f ) r Vo Now consider an N component solid of which components 1, ….. k are substitutional and k +1, …. N are interstitial. Note that the addition or removal of interstitial atoms leaves AL unchanged. Then N il N is U s TS s P 2 / r V s is N is U s TS s P 2 f / r V s i 1 N i 1 and N ( i 1 is il ) N is 2 ( f )V s / r For interstitial exchange : fluid --interstitial--- solid dA L 0 is il , ⓐ i k 1,....... N For substitutional exchange : fluid -- substitutional --- solid k ( is il ) N is 2 ( f )V s / r , i 1 .... k i 1 k and defining N is N 0 and V s / N 0 as V 0 the molar volume. i 1 ⓑ is il 2 ( f )V 0 / r , i 1 .... k Examples of how finite – size effects alter equilibria (1) Vapor pressure of a single – component solid l ( p e ) R T ln P / Pe s ( p e ) 2V 0 f / r V 0 ( P Pl ) 0 in com parision to f term using ⓑ s l 2V 0 f / r RT ln p / p l 2 ( f )V 0 / r RT ln p / p l 2 V 0 / r same result as earlier (2) Solubility of a sparingly soluble single component solid : il i ( C e ) R T ln C / C e is i ( C e ) 2V 0 f / r using ⓑ RT ln C / C l 2 V 0 / r (3) Melting point of a single component solid : l (Tm ) S l Tm T s (T m ) S s T m T where Sl and Ss are molar entropies. 2f r see Clausius – Clapyron Vo Equation using ⓑ Tm T 2 V0 r 1 Sl Ss 2 V 0 Tm r Lf , Sl Ss Lf Tm (4) Vapor pressure of a dilute interstitial component in a non-volatile matrix ( H in Fe….) If the interstitial vaporizes as a molecule: nx x n or if it reacts with a vapor species, A, forming a compound AmXn mA nX Am X n The chemical potential of X in the vapor is related to the partial pressure P of Xn or AmXn by xl p l RT ln n p pl and for the solid xs p l 2V x f / r when Vx is the molar volume of X in the solid. Using ⓐ RT ln p / p l 2 n V x f / r indicating that f determines the change in vapor pressure