Introduction Definition of Terms:

Introduction

Definition of Term s:

Hom ogeneous Heterogeneous Ergotic and Non-Ergotic

(Odyssey way path)

(Ergotic try all paths => crystal)

(Non-Ergotic stuck on one path => glass) M eta-stable

Equilibrium

Extent, extensive: V, M ass

Intensive: Density, Tem perature State Function T, P, r, G, H, S,…

First Law, Energy is Conserved SdU = Sdq + Sdw = 0

Internal Energy, Heat, Work

Adiabatic, Exotherm ic, Endotherm ic

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W hat happens to the energy when I heat a m aterial?

Or How m uch heat, dq, is required to change the tem perature dT? (Heat Capacity, C) dq = C dT

C = dq/dT Constant Volum e, C^V

Constant Pressure, C^p dU = dq + dw

W ith only pV work (expansion/contraction), dw^ec = -pdV dU = dq – pdV

For constant volum e (dU)^V = dq, so

C^V = (dU/dT)^V, or the energy change with T: (dU)^V = C^V dT

dU = dq + dw = dq – pdV (only e/c work, i.e. no shaft work) Invent Entropy H = U + PV so dH = dU + pdV + Vdp

(dH)^p= dU + pdV for constant pressure

W ith only pV work (expansion/contraction), dw^ec= -pdV

Constant Volum e Com puter Sim ulation Helm holtz Free Energy, A A = U – TS = G - pV

Constant Pressure

Atm ospheric Experim ents Gibbs Free Energy, G

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Therm odynam ic Square

-S U V

H A

-p G T

H = U + PV A = U – TS = G - pV G = H – TS = A + pV

C^p - C^v= a²VT/k^T

a = (1/V) (dV/dT)^p kT = (1/V) (dV/dP)^T

We will obtain this later

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Slope is C^p, this is not defined at first order transitions

(m elting and vaporization, crystalline phase change, order/disorder transition)

Size dependent enthalpy of m elting (Gibbs-Thom pson Equation)

For bulk m aterials, r = ∞ , at the m elting point DG = DH – T^∞DS = 0 So T^∞ = DH/DS Larger bonding enthalpy leads to higher T^∞ , Greater random ness gain on m elting leads to lower T^∞.

For nanoparticles there is also a surface term ,

(DG) V = (DH – T^rDS)V + sA = 0, where Tr is the melting point for size r nanoparticle

If V = r³ and A = r² and using DS = DH/T^∞ this becom es, r = s/(DH(1– T^r/T^∞)) or T^r = T^∞ (1 - s/(rDH)

Sm aller particles have a lower m elting point and the dependence suggests a plot of T^r/T^∞ against 1/r

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Hess’ Law

Second Law: Reversibility

For an adiabatic reversible system DS = 0

In a process this is often used by engineers in calculations 1) Assum e DS = 0, calculate DH for the process 2) Use and efficiency to m odify DH to a larger value 3) Calculate the actual DS > 0 for the process

The change in entropy is tied to the concept of efficiency 100% efficient process has DS = 0

Calusius Theorem is that entropy increases or stays the sam e but can not spontaneously decrease

For a reversible process dS^rev= (dq/T)^rev For any process dS ≥ (dq/T)

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Equilibrium

For any process dS ≥ (dq/T) (Calusius Theorem )

Or dq – TdS ≥ 0

dq – TdS we can call the “Change in Free Energy”

(This is the free energy available to do work.)

For a reversible process at therm odynam ic equilibrium it is equal to 0 This is a quantitative definition for equilibrium

At constant volum e dq = dU (Sim ulations)

The Helm holtz Free Enegy is defined: A = U – TS so dA = dU –TdS - SdT dA = dU –TdS (at constant tem perature)

dA = 0 at equilibrium for constant V and T At constant pressure dq = dH (Experim ents)

The Gibbs Free Energy is defined: G = H – TS so dG = dH – TdS – SdT dG = dH – TdS (at constant tem perature)

dG = 0 at equilibrium for constant p and T

Third Law

S = k^BlnW

W is the number of states

For an infinite perfect crystal there is only one state W = 1 and lnW = 0 so S = 0

This could only occur at T = 0 for an ergodic system where there is no therm al m otion. (without therm al m otion the system can’t be ergodic so it is not possible to reach this hypothetical state)

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dU = dq + dw For only ec work dU = dq – pdV

dq = TdS for a reversible process dU = TdS – pdV

So U is naturally broken into functions of S and V

(dU/dS)^V = T (dU/dV)^S= -p

dU = (dU/dS)^V dS + (dU/dV)^S dV

dU = (dU/dS)^V dS + (dU/dV)^S dV Take the second derivative

d²U/(dSdV) =(d/dS(dU/dV)^S)^V = (d/dV(dU/dS)^V)^S= d²U/(dVdS) Using the above expressions and the m iddle two term s

-(dp/dS)^V = (dT/dV)^S

This is a M axwell Relationship and the process is called a Legendre transform ation This can be done for all four fundam ental equations, U; H; G; A

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Therm odynam ic Square

-S U V

H A

-p G T

Therm odynam ic Square

-S U V

H A

-p G T

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Derive the expression for C^p – C^V C^p- C^v = a²VT/k^T a = (1/V) (dV/dT)^p kT = (1/V) (dV/dP)^T

C^V = (dU/dT)^V

From the Therm odynam ic Square

dU = TdS – pdV so C^V = (dU/dT)^V = T (dS/dT)^V - p (dV/dT)^V Second term is 0 dV at constant V is 0

(dS/dT)^V = C^V /T Sim ilarly

C^p= (dH/dT)^p

From the Therm odynam ic Square

dH = TdS + Vdp so C^p= (dH/dT)^p = T (dS/dT)^p - V (dp/dT)^p Second term is 0 dp at constant p is 0

(dS/dT)^p= C^p/T

Write a differential expression for dS as a function of T and V

dS = (dS/dT) dT + (dS/dV) dV using expression for C above and M axwell for (dS/dV)

-S U V

H A

-p G T

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Lowest Gibbs Free Energy is the stable phase

S generally increases W ith tem perature -S U V

H A -p G T

dG = VdP - SdT

Lowest Gibbs Free Energy is the stable phase

Volum e is positive

So G increases with pressure -S U V

H A -p G T

dG = VdP - SdT

0.1 M Pa is atm ospheric presure

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Phases at Equilibrium

Consider water and water vapor at equilibrium . A m olecule of water can leave liquid water and join the water vapor due to therm al energy. At equilibrium it can just as likely leave water vapor to join liquid water. We are not considering the interfacial energy. The total num ber of m oles in the container (system ) , n^total, is fixed. But the num ber in liquid water can change by dn^liquid= -dn^vapor. This change would change the Gibbs free energy (we are doing an experim ent at constant

atm ospheric pressure). µ^liq = (dG^liq/dn^liq) = (dG^vap/dn^vap) = µ^vap at equilibrium so that the change for liquid = -the change for vapor; dn^liq = -dn^vap for conservation of m ass.

dn^liq µliq = dn^vap µvap. µ^vapis called the chem ical potential of water in the vapor phase.

Chem ical potential always has two qualifiers, of what com ponent in what phase.

µvapis the partial m olar Gibbs Free Energy of water in the vapor phase. H, S, V can also have partial m olar values in the sam e way usually signified by a bar.

Gibbs-Duhem Equation

Consider a binary system A + B m akes a solution

At constant T and p:

Fundam ental equation with chem ical potential:

So, at constant T and p:

Reintroducing the T and p dependences:

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Gibbs-Duhem Equation

Consider a binary solution in a nanoparticle.

There is significant stress on the surface, s,

com pared to the core, c. Break particle into core and surface regions with different stress, s^c, s^s, m olar com position of the solute in core, x^c, and surface, x^s. Assum e equilibrium , µ^c = µ^s. Ln term is gaseous m ixing entropy, 0 term s are for infinite dilution.

s = the stress (DP) so the contribution to chemical potential from stress is proportional to s V depending on geom etry

Corrosion in nanoparticles can be higher or lower

For the two com ponents with an electric potential the equilibrium corrosion potential can be

calculated for each com ponent which changes with particle size.

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