Chemical Potential •  Combining the First and Second Laws for a closed system, •  Considering

•  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

For an open system.

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dU = TdS − pdV

(extensive properties)

(more complete definition)

•  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

•  For multiple constituents (at same p,T)

CHEMICAL POTENTIAL, or partial molar Gibbs free energy

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

•  These constraints represent ‘macroscale’ variations: they apply to bulk system

ENTROPY IS A MAXIMUM

S

e.g.

•  For an open system that undergoes small transitions at constant

•  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

U

Internal Energy is a minimum!

•  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!

•  An important property of extensive variables, such as U, is that the total U is of course additive

•  Additivity property applies to other variables as well:

•  Now write the condition for equilibrium as,

•  Expression for dU for one phase is,

•  Each chemical constituent is free to exist in any phase.

i.e.

Summation is over all phases

For each constituent

Chemical potential of each constituent

(1)

•  Generalizing to the present system (sum over phases)

•  Now from the statement of equilibrium, are subject to constraints according to,

•  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,

•  Invoking the constraints in (3) above,

(2)

(3)

(4)

For each of the constituents

α=1

•  Therefore (4) becomes,

•  Since are independent, arbitrary variations, (5) can only be satisfied when the coefficients are identically zero.

•  i.e.

•  Implies that the criteria for equilibria (extend to > 2 phases) are:

(5)

Internal Energy is a minimum

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 kth component in the jth phase as

•  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.

Total # of moles in jth phase

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

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w = 2+ϕ ⋅c−ϕ − c ⋅ (ϕ −1)= 2−ϕ + c

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)

Two intensive variables may be freely specifiable.

Gibbs Phase Rule

Homogenous;

one phase only

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w = 2+ c−ϕ

2.  Mixture of two gases

Free variables are any two among . The third freely specified variable is the relative concentration of the gases

3.  Liquid water is equilibrium with water vapor

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,

This is the triple point. Single choice of

MONOVARIANT!

“No Freedom”

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.

•  Molecules in transition layer are subject to a net inward force due to molecules in the droplet interior.

drop interior

transition layer

gas

“deep” interior

F outward

F inward

Center of mass

(Not to scale here of course)

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.”

•  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.

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.

Transition layer

Liquid

{Inward pull exerted on the surface layer molecules results in a surface tension}

•  Surface tension represents energy stored in the surface of a drop since work must be expended to change the shape of a liquid volume.

{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

“Work” is essentially the work that must be done to overcome intermolecular force when moving a molecule from interior to surface.

Energy / unit area or force / unit length

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

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,

•  Condition for equilibrium is

•  If only one curved interface, this is really not a constraint.

′ ″

Surface area of interface (with negligible volume)

included here if there are more than one curved interface.

•  For a 2 phase system we can expand the equilibrium condition as before, now including the phase

•  Now the constraints that must be satisfied for the equation to satisfy equilibrium conditions are,

′, ″ phase

phase

Additivity (1)

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,

•  With these assumptions, the constraints are,

Use in expression

•  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,

•  The mechanical equilibrium condition is the so called Laplace formula,

•  , 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.

(2) Internal Energy is a minimum!

a

Surface tension

Curvature term

Calculations: Water-Vapor Interface

•  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.

a

vapor

droplet

•  Another look at the Laplace relationship

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

a

Frictionless piston

Consider this system to be in equilibrium

vapor pressure for drop of radius ‘a’

Piston supplies a pressure P2 to the liquid– liquid is extruded through an infinitely fine syringe to form the drop.

•  The applied pressure P2 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

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)

There is some nonzero thickness associated with “the interface”…

And in a multicomponent system, species concentrations vary across this distance.

IDEALIZED REAL

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)

Gibbs adsorption isotherm (continued)

For phases α and β, if the bulk (homogeneous) phases contain amounts of species J

Then we can deduce how much is in the “interface”:

This “excess” amount of material can be expressed as an amount per unit area of the interface by defining the surface excess:

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.

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nJ(α ) and nJ

(β )

€

nJ(σ ) = nJ - nJ

(α ) + nJ(β )[ ]

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ΓJ =nJ(σ )

σ

We can apply criteria of equilibrium to deduce that

and to derive the “Gibbs surface tension equation”:

€

σdγ + nJ(σ )

J∑ dµJ = 0 (constant T)

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dγ = − ΓJJ∑ dµJ

•  Intensive variables for this system—

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

VARIABLE NUMBER

•  EXAMPLE 1: System of uniform T, in which a droplet of radius ‘a’ is surrounded by pure H2O vapor

•  Hence

•  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 H2O of radius ‘a’ is surrounded by humid air of total pressure p.

•  To study the dependence of , must hold T and p fixed.

H2O

“air”

a

Because a took the place of v or ρ

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 H2O 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.

Consider the following system:

R

vapor molecules

T, ev

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

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 is the same before and after nucleation embryo formation.

Number of molecules ‘n’ in the embryo Unit volume

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 is the same before and after nucleation embryo formation.

Number of molecules ‘n’ in the embryo Unit volume

We will use this to substitute, so that we can make the equation a function of R

Equation (2) can be written as,

(3)

Bulk thermodynamic term

Surface energy term

Energy required to form curve surface with surface tension

Need to derive an expression for , difference in Gibbs free energy between liquid vapor phases

For constant T, and using

(4)

(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:

(6)

or

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dµ = (vl − vv )dp ≈ −vvdp

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µl −µv = − vv∫ dev

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µl −µv = − kTd lneves

e

∫

Note the limits of integration

When equilibrium is established and

When non-equilibrium exists,

Integration yields,

Hence (3) becomes,

With

Consider plot of vs

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µl −µv = −kT ln ees

S