Chemical Potential. Combining the First and Second Laws for a closed system, Considering (extensive properties)

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Berenice Simon
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1 Chemical Potential Combining the First and Second Laws for a closed system, Considering (extensive properties) du = TdS pdv Hence For an open system, that is, one that can gain or lose mass, U will also change from mass transfer. (Why?) Therefore U becomes a function of n, # of moles. Hence The chemical potential is defined as (more complete definition) For an open system.

2 The chemical potential is the change in internal energy (at constant S,V), when one mole of the substance is added/removed For Gibbs Free Energy, Substituting And we have CHEMICAL POTENTIAL, or partial molar Gibbs free energy For multiple constituents (at same p,t)

3 Equilibria for Complex Systems We need to extend our discussions of equilibrium to include systems which have: Multiple phases e.g. liquid/vapor Multiple constituents e.g., water and salt For an isolated system not in equilibrium, irreversible processes will occur spontaneously and entropy will increase until eventually it reaches a state where entropy is a maximum. In this state, all irreversible processes will have ceased and any remaining processes must be reversible. The system is then in equilibrium. For an open system at constant (c= # of constituents) the equilibrium condition is that where ENTROPY IS A MAXIMUM S These constraints represent macroscale variations: they apply to bulk system e.g.

4 For an open system that undergoes small transitions at constant U Internal Energy is a minimum! Need to generalize the condition of equilibrium for system with phases and c chemical constituents. Where we are heading: We would like to get the criteria for equilibrium between phases, and make it general enough to cover more than one chemical component

5 Let each chemical constituent have its chemical potential represented by Let the number of moles of each species be Heterogeneous System: a system consisting of 2 or more phases which are separated from each other by a surface of discontinuity in one or more of the intensive variables. Question: what condition on the intensive variables are necessary and sufficient to ensure equilibrium after the constraint that the phases are isolated is relaxed? Assumption: no chemical reactions occur and the bulk system remains isolated (from surroundings) After the constraint (phase partition) is relaxed (removed), each phase behaves as an independent, open system (one that is free to exchange mass with another phase). Assuming no chemical reactions occur, (so mass transfer can t take place by chemical reaction), the condition for equilibrium in each phase is Internal Energy is a minimum!

6 An important property of extensive variables, such as U, is that the total U is of course additive Summation is over all phases i.e. Additivity property applies to other variables as well: Now write the condition for equilibrium as, Expression for du for one phase is, For each constituent Chemical potential of each constituent Each chemical constituent is free to exist in any phase. (1)

7 Generalizing to the present system (sum over phases) α=1 Now from the statement of equilibrium, are subject to constraints according to, (2) (3) The total variation (change) of must be identically zero (for system as a whole.) Consider a two phase system denoting the two phases. For this system (1) becomes, (4) Invoking the constraints in (3) above, For each of the constituents

8 Therefore (4) becomes, (5) Since are independent, arbitrary variations, (5) can only be satisfied when the coefficients are identically zero. i.e. Internal Energy is a minimum Implies that the criteria for equilibria (extend to > 2 phases) are:

9 Gibbs Phase Rule Consider a multi-phase, multi-constituent system that is in equilibrium, characterized by a common T and p between phases and by the mole fractions of each component in various phases. Gibbs phase rule allows for the determination of the VARIANCE of a system, that is, the number of intensive variables that can be freely specified without causing the system to depart from its equilibrium point. Denote the mole fractions of the k th component in the j th phase as Total # of moles in j th phase For a system of φ phases and c components, there are φ c mole fractions. (Each constituent exists in each phase.) Hence the total number of intensive variables at equilibrium is, Not all intensive variables are independent. To find the number of truly independent intensive variables, we must examine their DEPENDENCY, or constraints that the intensive variables must satisfy to still maintain equilibrium.

10 Constraints to be applied to the multiphase system: 1. Mass Conservation: the sum of the mole fractions must sum to unity This equation holds for each phase, thus in general there are constraints arising from this condition 2. Constraint on chemical potentials: At equilibrium, for each component, This condition gives constraints for each constituent. Applying this condition to each of the c constituents gives c constraints So, the total variance is w = 2 +ϕ c ϕ c (ϕ 1) = 2 ϕ + c

11 w = 2 + c ϕ Gibbs Phase Rule c is the number of components φ is the number of phases w is the number of independent intensive variables ( degrees of freedom ) APPLICATIONS 1. Homogenous fluid (or gas) Homogenous; one phase only Two intensive variables may be freely specifiable.

12 2. Mixture of two gases Free variables are any two among relative concentration of the gases. The third freely specified variable is the 3. Liquid water is equilibrium with water vapor MONOVARIANT! Since vapor pressure (saturation vapor pressure in this case) is function of T only, free variable is just T (or if T is specified), which is the Clausius Clapeyron Equation 4. For liquid water in equilibrium with vapor and ice, No Freedom This is the triple point. Single choice of

13 Surface Tension Concept of surface tension has particular relevance to cloud physics. Phases in contact are separated by a thin transitional film a few molecules thick. gas transition layer drop interior (Not to scale here of course) Molecules in transition layer are subject to a net inward force due to molecules in the droplet interior. F outward deep interior F inward Center of mass Atkins, Physical Chemistry: a molecule in the bulk has a lower potential energy than one free in the gas, and it takes energy to dig out a molecule from the bulk and deposit it in the gas. so the molecules are under the influence of a force which tends to draw them into the bulk.

14 Liquid Transition layer Molecules in drop interior experience a symmetrical attractive force these forces are the Van der Waal forces that keep liquid intact at normal molecular separation distances these forces are attractive. Molecules in surface layer experience an inward directed attractive force. This net inward force gives rise to a tension on the surface of the droplet, which has units of force per unit length, or energy per unit area. The force F, when multiplied by the distance, is the amount of work that must be done to move a molecule from the drop s interior to the surface layer. F is the attractive force in the liquid.

15 {Inward pull exerted on the surface layer molecules results in a surface tension} Energy / unit area or force / unit length Surface tension represents energy stored in the surface of a drop since work must be expended to change the shape of a liquid volume. Work is essentially the work that must be done to overcome intermolecular force when moving a molecule from interior to surface. {Surface tension is the work required to change the area of a surface by one unit} Surface tension allows a liquid volume to achieve a minimum surface to volume ratio (a sphere!) - more molecules can be bulk rather than surface molecules

16 Contribution of Surface Tension to U We have seen that du for a reversible process is, Properties of surface phase allow it to be regarded as an independent phase, so then Where is the contribution from surface tension Increase in surface area Work done in system

17 Equilibrium Conditions for Two Phases Considering the Surface Phase Separating Them Consider 2 bulk phases and a surface phase separating them, Consider each phase to be isolated from one another initially, and then remove the partitions separating the phases. Seek conditions for equilibrium. Assume bulk system remains isolated from environment. The extensive variables for this system are, Surface area of interface (with negligible volume) Condition for equilibrium is included here if there are more than one curved interface. If only one curved interface, this is really not a constraint.

18 For a 2 phase system we can expand the equilibrium condition as before, now including the phase, phase phase Additivity (1) Now the constraints that must be satisfied for the equation to satisfy equilibrium conditions are,

19 Make the following assumptions: 1. Assume the system undergoes a small variation in which 2. The phase is a sphere of radius Therefore can be written in terms of and is, Use in expression With these assumptions, the constraints are,

20 As before, represent arbitrary, independent variations. Hence (2) will be satisfied only when the coefficients vanish. The conditions for equilibrium are then, generalizing the assumptions above to any set of variations, (2) Internal Energy is a minimum! The mechanical equilibrium condition is the so called Laplace formula, Surface tension Curvature term a, pressure in drop interior is greater than pressure adjacent to drop by the term. Pressure in interior must be greater to prevent drop collapse due to inward directed tension force.

21 Calculations: Water-Vapor Interface vapor a droplet What is pressure difference between the drop interior and that in the surrounding vapor? Assume vapor is saturated, and So pressure difference is about 1.5 atm.

22 Another look at the Laplace relationship a provides the difference in outside and inside pressures of a spherical drop in equilibrium with its vapor. The Laplace formula can also be derived from the definition of surface tension, using only the concept of work. Reversible work that must be done on a system in order to increase its surface area A by an amount da is a Consider this system to be in equilibrium vapor pressure for drop of radius a Frictionless piston Piston supplies a pressure P 2 to the liquid liquid is extruded through an infinitely fine syringe to form the drop.

23 The applied pressure P 2 is just enough to cause the drop to grow. (equilibrium is maintained though ) Differential work done by piston on bulk liquid Differential work done by drop on surrounding vapor, The work required to increase the surface area must balance these: For spherical drop Substituting Laplace Formula

24 Phase Rule for Systems with Curved Interfaces For systems with phases and c constituents with non-curved interfaces, Now need to generalize this result to include curved interfaces. Now we have What is variance of this system, w? w=intensive variables - constraints Constraints are dependencies among the intensive variables. Another intensive variable considered in this case is the surface excess, which arises due to adsorption of any constituent into the curved interfaces. Surface excess

The nature of the interface (Gibbs adsorption isotherm) IDEALIZED REAL There is some nonzero thickness associated with the interface And in a multicomponent system, species concentrations vary across

25 The nature of the interface (Gibbs adsorption isotherm) IDEALIZED REAL There is some nonzero thickness associated with the interface And in a multicomponent system, species concentrations vary across this distance. The thin layer near the interface may be rich or deficient in some of the species (e.g., a detergent accumulates at an oil-water interface). Please note notation change in the next slide (source: Atkins, Physical Chemistry)

26 Gibbs adsorption isotherm (continued) For phases α and β, if the bulk (homogeneous) phases contain amounts of species J n J (α ) Then we can deduce how much is in the interface : n J (σ ) and n J (β ) (α = n J - n ) (β ) [ J + n J ] This excess amount of material can be expressed as an amount per unit area of the interface by defining the surface excess: Γ J = n (σ ) J σ where σ is the area of the interface. Both the number of moles at the interface and the surface excess property can take on positive or negative values. We can apply criteria of equilibrium to deduce that (σ ) σdγ + n J dµ J = 0 (constant T) J and to derive the Gibbs surface tension equation : dγ = Γ J dµ J J

27 Intensive variables for this system VARIABLE NUMBER 1. Common T 2. Adsorption of c constituents into curved interfaces 3. Specification for mole fractions of each bulk phase 4. Pressure in each bulk phase (pressure generally not same due to curvature) 5. Mean radius of curvature for each curved interface Total number of specified variables, Now examine constraints 1. Mole fractions must sum to 1 in each bulk phase -This condition does not apply to surface phases 2. Chemical potential for each constituent must be equal between all phases 3. Mechanical equilibrium must hold at all interfaces Total constraints, Hence total variance is

28 EXAMPLE 1: System of uniform T, in which a droplet of radius a is surrounded by pure H 2 O vapor Hence Because a took the place of v or ρ These variations that describe system: Specify one variable, say T, and study dependence of EXAMPLE 2: System of uniform T in which a drop of pure H 2 O of radius a is surrounded by humid air of total pressure p. H 2 O air To study the dependence of, must hold T and p fixed. a

29 NUCLEATION THERMODYNAMICS Consider the formation of a pure water droplet by condensation from the vapor phase. This process will first be considered without any foreign particle or nucleating surface. This process is known as -HOMOGENEOUS NUCLEATION- Process is also referred to as SPONTANEOUS NUCLEATION - Formation of embryonic droplet by chance collision of several tens to several hundred H 2 O molecules Question: Under what conditions does the embryo remain intact and grow into a cloud droplet? -We will examine the relationship between vapor pressure and surface curvature KELVIN S EQUATION We will study this system (droplet and vapor molecules) from the perspective of Gibbs free energy.

30 Consider the following system: vapor molecules R T, e v T,e v temperature and vapor pressure in drop s environment constants. chemical potentials of vapor and liquid molecules number of vapor molecules prior to embryo formation number of vapor molecules after embryo formation number of molecules in embryo

31 Change in Gibbs free energy, (1) We are most interested in conditions for which or Consists of embryo plus remaining free molecules Hence System Gibbs free energy prior to embryo formation (2) This is an equilibrium process with respect to vapor so after nucleation embryo formation. Number of molecules n in the embryo is the same before and Unit volume

process with respect to vapor so after nucleation embryo formation.

32 Change in Gibbs free energy, (1) We are most interested in conditions for which or Consists of embryo plus remaining free molecules Hence System Gibbs free energy prior to embryo formation (2) This is an equilibrium process with respect to vapor so after nucleation embryo formation. Number of molecules n in the embryo is the same before and Unit volume We will use this to substitute, so that we can make the equation a function of R

33 Equation (2) can be written as, Bulk thermodynamic term Surface energy term Need to derive an expression for, difference in Gibbs free energy between liquid vapor phases or (3) Energy required to form curve surface with surface tension For constant T, and using dµ = (v l v v )dp v v dp (4) µ l µ v = v v de v (5) For vapor Use the Ideal Gas Equation and integration limits go from being the saturated liquid (same as saturated vapor) to the actual vapor: µ l µ v = ktd lne v e e s (6)

34 Note the limits of integration When When equilibrium is established and non-equilibrium exists, Integration yields, µ l µ v = kt ln e e s S Hence (3) becomes, With Consider plot of vs

water Plays dominant role in radiation All three phases emit and absorb in longwave radiation

4.,4. water Plays dominant role in radiation All three phases emit and absorb in longwave radiation Some shortwave (solar) radiation is absorbed by all phases of water Principal role in the shortwave radiation