Manganese oxide formation by heat treatment of MnCO3 in air.

MnCO3 + ½ O2MnO2+CO2

<500 C Reaction 1

>500 C Reaction 2

2 MnCO3 + ½ O2Mn2O3 + 2 CO2

Note that in reaction 1, there is a net increaseOf ½ mole of gas for each mole of Mn, and for reaction 2 there is a net increase of ¾ of a mole of gas for each mole of Mn. What can you say about the entropy change in each reaction? How does this help explain the temperature dependence between the two reactions?

ab

c

xy

z

a

b

c

x

y

z

0.2 um

As the manganese oxide particles form from the carbonate salt, they begin to grow together, or ‘sinter’. The figure below is a TEM micrograph of neck formation during the

sintering of Mn2O3 particles.

Why do the necks get larger and the pores get smaller as the heat treat time and or temperature increases?

Pore

Neck

Neck

Sintering of Nickel powder. The time lapse photography illustrates Neck formation and coarsening.

rdr

rddrrd

Volumed

Aread 2

)3/4(

)4(

)(

)(3

2

ΔP

r

Curved surfaceFlat surface

P=Patmσ=Patm + 2γ/rP

ΔP=2γ/r

dG S dT V dP

1

m

i

Gi dni

‘Master’ Equation of Thermodynamics

For an isothermal process with no change in composition

dP P 1 P odG G P 1 G P o

dG S dT V dP

1

m

i

Gi dni

where Gi is the chemical potential per atom of type i and dni is the change in number of atoms of

type i. We know that an increase in pressure will increase the chemical potential of the components,i. Consider an incremental increase in pressure from the ambient pressure Po to a Pressure P1. Letthe the free energy at ambient pressure Po be given by G Po . Then using the master equation at

constant volume, temperature and composition in a one component system, we get .

G P 1 G P o V P( )( )

Divide both sides by the number of atoms in the 'system'

V

N

G

NChemicalPotential Gi

where is the atomic volume. Let

dP P1 Po

Substitution and rearrangement relates the chemical potential at two pressures as

Gi P1 Gi P0 P

dG S dT V dP

1

m

i

Gi dni

where Gi is the chemical potential per atom of type i and dni is the change in number of atoms of

type i. We know that an increase in pressure will increase the chemical potential of the components,i. Consider an incremental increase in pressure from the ambient pressure Po to a Pressure P1. Letthe the free energy at ambient pressure Po be given by G Po . Then using the master equation at

constant volume, temperature and composition in a one component system, we get .

G P1 G Po V P( )( )

Divide both sides by the number of atoms in the 'system'

V

N

G

NChemicalPotential G i

where is the atomic volume. Let

dP P1 Po

Substitution and rearrangement relates the chemical potential at two pressures as

Gi P1 Gi P0 P

Divide both sides by the number of atoms in the system=N

G i P 1 G i P o P ΔP=2γ/r for inside a spherical particle.

Positive and Negative CurvatureCorrugated Surface Example (2D)

solid

vapor

ΔP=γ/rPositive Curvature

ΔP=-γ/rNegative Curvature

Atoms move from

high free energy to low.

G i P 1 G i P o P

Two sphere model

r

r1

r2

The neck has a negative curvature component (-1/ρ), acting to reduce the pressure relative to the spherical surface.

P 1

r

1

P 1

r 1

1

r 2