Thermodynamics

N. Zabaras 1

INTRODUCTION TO THERMODYNAMICS OF MATERIALS

Thermodynamic systems, extensive and intensive variables Energy, work and heat Important state functions: H, F , G, etc. Reversible and spontaneous processes, entropy The first and second laws of thermodynamics Conditions of equilibrium for single- or multi-component multi-phase al-

loys

Introduction to thermodynamics of materials Materials Process Design and Control Laboratory

N. Zabaras 2

INTRODUCTION TO THERMODYNAMICS OF MATERIALS

Thermodynamics deals with energy and its transformations Structure of materials determines properties. Thermodynamics is used to:

understand how the properties of materials are affected by thermody-namic processes (processes that involve changes in the material en-ergy)

to control properties through its determination of structure Thermodynamics includes two subjects:

Classical thermodynamics: Phenomenological in nature focuses on the macroscopic properties

of materials and their relationships Statistical thermodynamics: Based on how the material behaves at the microscopic level allows

the description of the macroscopic properties


N. Zabaras 3

THERMODYNAMIC VARIABLES

The state of a material system is described with the value of certainvariables of the system: T,P,V,S,c, etc.

A thermodynamic process is a real or virtual change in the system stateaffected by varying the variables of the system such that the system is nolonger at equilibrium. A process is completed when the system reachesits new equilibrium state.

State variables are independent of the system history and depend only onthe system state and not on the process by which the system was taken tothat state. For any state property,

R(A B) = RBRA Process variables are defined only when a specific path is specified i.e.

by how the process was carried out. Examples include the heat q and workw.


N. Zabaras 4

EXTENSIVE AND INTENSIVE VARIABLES Intensive variables do not depend on the system size and can be defined

at any point in the system: Examples include T , P, etc. Not additive as they do not scale with the system size

Extensive variables depend on the system size and cannot be defined ata point in the system, e.g. V , m, N (number of atoms), etc. Additive as they scale with the system size, e.g. V+ = V+V

Derived intensive quantities can be defined by scaling with anotherextensive quantity

Densities can be introduced by scaling with total volume: mV ,NV , etc.

The molar species number is defined as Nj =Nj

Ntotal, also denoted as

Xj ( j Xj = 1). We will later introduce the partial molar value of anyextensive property E as Ei = ENi|T,p,Nj such that E = i NiEi

Partial quantities are used to associate chemical species with intensivequantities, e.g. the partial pressure of species j: Pj = XjP ( j Pj = P)


N. Zabaras 5

THERMODYNAMIC SYSTEM AND ITS BOUNDARY

Thermodynamic system separated from its surroundings from its systemboundary

System boundaries can be: rigid, impermeable, adiabatic or any combi-nations

System types include: Isolated No energy or matter may be passed through its boundary

(e.g. the universe) Closed or open allowing or not exhange of matter Adiabatic No heat transfer can pass through its boundary Single- or multi-component Reacting or non-reacting Homogeneous (single phase) or heterogeneous (multiple phases)


N. Zabaras 6

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Energy is the potential to do work it can only be measured indirectly interms of other system variables such as m,v (e.g. E = 12mv2), etc.

First law of thermodynamics The energy of the universe is constant Any change in the change of the energy of a system is balanced by an

opposite change in the energy of the surroundings. This change in theenergy of the surroundings is transferred to the system as heat or workthrough the system boundary

When we talk on the energy of a system we usually mean the internalenergy U : The energy associated with the atomic motions in the material(the kinetic and gravitational energy sources are typically ignored)

The internal energy U is intrinsic to the body


N. Zabaras 7

THE FIRST LAW OF THERMODYNAMICS

We define the energy of the system as a state variable U , e.g.U = U(P,V ) and U = U(P2,V2)U(P1,V1)

For all processes, the change of U for the system is balanced by an equaland opposite change of U of the surroundings

The 1st law of thermodynamics is stated as: dU = q+w q is the heat transfered to the system and w is the work done on the

system w represents energy transfers due to forces other than temperature

gradients w = i Fi dxi in terms of generalized forces and displacements, e.g.

the compression work for a pure fluid: w = PdV (note dV < 0 forcompression)

q and w are not exact differentials, e.g. H dw = H PdV = 0


N. Zabaras 8

FORMS OF WORK dw = V E dD+V H dB+Vi ji j + . . .POLARIZATION WORK: dw = V E dD

Let E be the applied electric field and D the electric displacement P the polarization of the material due to local charge displacement

The following relations are applied:D = 0E+P, 0 = the permittivity of free space

For isotropic materials: P = 0E, is the dielectric susceptibility andthus D = E, = 0(1+)=permittivity of the material

dw = V E dD w = 12VE2

P = 0E = dielectric susceptibility tensorThe response of the material de-pends on the orientation of E withrespect to the crystal orientation


N. Zabaras 9

FORMS OF WORK dw = V E dD+V H dB+Vi ji j + . . .MAGNETIC WORK: dw = V H dB

Let H be the applied magnetic field and B the magnetic induction of thelocal net field

I the magnetization of the material (magnetic moment per unit volume)The following relations are applied:

B = 0H+ I, 0 = the permeability of vacuum For isotropic materials: I = 0H, is the magnetic susceptibility

> 0 the induced magnetic moments align with H paramagneticmaterial

< 0 the induced magnetic moments aligns antiparallel with H dia-magnetic material

Thus B = H, = 0(1+)=magnetic permeability of the materialdw = V H dB w = 12V H2


N. Zabaras 10

HEAT AND WORK

Heat transfer occurs when there is a temperature gradient Heat flows from regions of high T to regions of low T

Heat is the transfer of energy accomplished through random and chaoticatomic motions Heat transfer is thus a disordered process

Work is an ordered process of organized atomic motions


N. Zabaras 11HEAT CAPACITIES

Constant volume heat capacity CV : CV = qT |V = UT |V Constant pressure heat capacity CP: CP = qT |P One can show that generally CP > CV

Consider two ideal (fluid) systems that receive the same q under con-stant V and P conditions, respectivelly:

q =

R TVT0 CV dT =

R TpT0 CPdT (using mean value theorem) =

TVT0TPT0

After the end of the constant V process, with a subsequent processone should be able to expand the fluid and bring it to the same condi-tions as those at the end of the constant P pressure. At this additionalprocess, work is done on the surroundings. We can store this workand pass it back as heat to the system (under constant V ) thus leadingto an increase in T : Thus TV > TP > 1

larger difference between CP and CV for gases rather then liquids for solids the difference between CP and CV is negligible


N. Zabaras 12

THE FIRST LAW FOR IDEAL PROCESSES

General form with compression work: dU = qPdV Adiabatic processes q = 0: dU =PdV Isothermal processes dT = 0: dUT = qT PT dVT Isobaric processes dP = 0: dUP = qPPPdVP Isometric processes dV = 0: dUV = qV

For all processes, to computer dU , one needs to know an equation of state P(T,V ) comes from the specific properties of a material (material property)

and the heat flow dq


N. Zabaras 13

THE FIRST LAW FOR IDEAL GAS PROCESSES: U = U(T ) andPV = nRT (or P V = RT ), R = gas constant = 8.3144 joules/(degree mole)For an ideal gas, the particles do not interact and have no volume: U = U(T ) CV = UT |V = dUdT CP = qT |P = UT |P +PVT |P = CV +nR CP = CV +nR, or CP = CV +R

Isothermal processes dT = 0: dUT = 0 = qT PT dVT = qT nRTVT dVT qT =wT = nRT lnV2V1 =nRTln

P2P1

Adiabatic processes q = 0: dU CV dT =PdV =nRTV dV CV dTT =nRdVV CV lnT2T1 =nRln

V2V1

T2T1

= (V1V2)nR/CV = (P2P1)

nR/CP or PV = constant,= CP/CV

van der Waals extension to the ideal gas law:(P+ a

V 2)( V b) = RT , a,b constants


N. Zabaras 14

A NEW STATE FUNCTION: ENTHALPY H = U +PV

Defined by subtracting from the internal energy the compression energy H is the heat associated with constant pressure processes:

dH = dU +PdV +dPV (using the 1st law) dH = q+V dPdH|P = q|P

Thus H can be interpreted as the available thermal energy at constant P For constant P and small PdV (small dV and small P):

dH = U +PdV UThis approximation is applicable to condensed phases

Can now redefine CP as follows:CP = qT |P = HT |P


N. Zabaras 15

NATURAL DIRECTION OF PROCESSESREVERSIBLE PROCESSES

The natural direction of processes is one which increases the entropy(randomness) of the universe transform organized motion (e.g. work) to disorganized motion (e.g.

heat)

A process is reversible when it can reverse itself with an infinitesimaldriving force: they involve no production of entropy and occur infinitelyslowly e.g. solidification of a pure material at the melting temperature


N. Zabaras 16

SPONTANEOUS PROCESSES

A process is spontaneous if it proceeds without an external driving forceand produces entropy equalization of T when two bodies of initially unequal T are brought in

contact conversion of work into heat e.g. heat flows down a temperature gradient without an external force spontaneous processes are irreversible: cannot create order from disor-

der As a spontaneous process continues, the system is brought near equilib-

rium and the ability of the system for a spontaneous change is decreased


N. Zabaras 17

THE SECOND LAW OF THERMODYNAMICS

Entropy is a state function that can be expressed in terms of measurablecoordinates of the system and is an intrinsic property of the system related tothe degree of disorder in the system

The change of the entropy of the universe (system and surroundings) is posi-tive and zero only for reversible processes:

dStotal 0, dStotal = 0 only for reversible processes

As one of the many alternative statements of the 2nd law, we can also statethat heat is not observed to spontaneously pass from a cold to a hot body

The entropy of a system can be computed by a reversible path from an arbi-trary chosen reference state by integrating the heat absorbed by the systemdivided by the system temperature


N. Zabaras 18

REVERSIBLE PROCESSES: dSrev = dqrevT

For an ideal gas undergoing a reversible process: dqrevT = CV dTT + nRdVV ,which implies that dqrevT is integrable over segments of dT and dV andthus at least for ideal gases a state function

We call this perfect differential dS, the new state function S calledentropy

Any process can be considered as anumber of segments of constant Tand V

For a reversible process, the only heat involved is the transfer of energybetween the system and the surroundings via heat

No additional random motion is generated in the system or in the sur-roundings


N. Zabaras 19

ORDER AND DISORDER

The disorder (randomness) of the system is measured in terms of quantaof energy, atom arrangements and motion, etc. A material is more ordered as a crystalline solid than liquid

The system disorder is related to the detail required to specify the micro-scopic state of the system

Example: Isothermal expansion (dU = 0) of an ideal gasdU = 0 = dqrev pdV dqrev = pdV = nRT dVV qrev = nRTlnV2V1

S = nRlnV2V1 For V2 > V1, S > 0: the expanded gas has higher entropy as its random-

ness has increased (more places for an atom to be!)


N. Zabaras 20

EQUIVALENT FORMS OF THE SECOND LAW

No process is possible in which heat (disordered motion) is absorbed froma resorvoir and is completely converted to work (ordered motion)

The entropy change dS of a system is associated with a transfer of entropyfrom the surrondings dStrans (which can be > 0 or < 0) and from entropydSprod produced within the system (always dSprod > 0):

dS = dStrans +dSprod, dSprod > 0


N. Zabaras 21

COMBINED FIRST AND SECOND LAWSU is a state function, i.e. U = U(S,V ): dU = US |V dS+ UV |SdV

For reversible processesdqrev = T dS and considering only mechanical work, dw =pdV . The com-bined first and second laws take the form:

dU = T dS pdV

T = US |V and P =UV |SFor irreversible processes

Since U is a state function, one can find a reversible process that connects thestates (S +dS,V +dV ) and (S,V ). This will result in the same dU as for theirreversible process and all equations given above for reversible processes arestill applicable!

S(any state) =

R any stateT=0

dqrevT

T = 0 taken as the reference state:(3nd Law of thermodynamics)


N. Zabaras 22

THE WORK DONE BY THE SYSTEM IS MAXIMUM FORREVERSIBLE OR QUASISTATIC PROCESSES

Consider a change of the material state from A to B. Using the 1st law:UBUA = q+w

The entropy change of the system that absorbs heat dq from a resorvoir attemperature Tr is: dSs = dqTr +d, d 0 from the 2nd law

From the first law: dq = dUsdw dSs = dUsdwTr +ddw = TrdSsdUsTrd dw TrdSsdUs For a constant Tr (large reservoir) by integration of the equation above:

w Tr(SBSA) (UBUA)wmax Conversely the work which must be done on a system to bring it to a given

state is minimum for reversible processes. For irreversible processes agreater amount of work is needed & energy is dissipated as heat


N. Zabaras 23

HEAT STORED DURING PHASE CHANGES OF PURE MATERIALSH(Tm0o)H(Tm +0o) = Hsolidification =Hmelting

T (temperature)

H (e

ntha

lpy)

warm water

cold ice

hot ice

freezing water

P=1 atm

c P (water)

T=0 C

P (ice) c

Hot ice (T > 0) melts, under-cooled water (T < 0) solidifiesand Suniverse > 0

Cold ice (T < 0) does not melt,warm water (T > 0) does not so-lidify if they did Suniverse < 0

Suniv = 0 for phase transforma-tion only at Tm = 0 for H2O at 1atm H lT Sl = HsT Ss at Tm

Variation of enthalpy with tempera-ture at constant pressure for a purematerial (here water)

For P=constant, H = q = CPT The change H at the transforma-

tion temperature Tm is the heat ab-sorbed during the transformation:H(Tm0o)H(Tm +0o) =Hsolidification =Hmelting

Phase transformations can beexothermic (heat released, e.g.solidification) or endothermic (heatabsorbed, e.g. melting)


N. Zabaras 24

EXAMPLE OF ENTROPY CREATION IN THE SOLIDIFICATION OFAN UNDERCOOLED MELT AT CONSTANT P

Consider 1 mole of undercooled lead at 590K (Tm = 600 K)cP(l) = 32.43.1 103T joules/degree,cP(s) = 23.6+9.75 103T joules/degree andlatent heat of lead qrev = 4810 joules/mole

To find Ssys we follow the reversible patha b c d: Ssys =

R ba

cP(l)dTT +

qrevTm +

R dc

cP(s)dTT =7.997 joules/degree

The heat entering the constant-T reservoir at 590 K is given as: H =Hab+Hbc+Hcd =

R ba cP(l)dTqrev

R dc cP(s)dT = 4799 joules

Thus the change of the entropy Sreservoir = H590 = 8.134 joules/degreeand the entropy created is: S = Ssys +Sreservoir = 7.994 + 8.134 =0.137 joules/degree


N. Zabaras 25

TWO NEW STATE FUNCTIONS: THE HELMHOLTZ FREEENERGY F AND GIBBS FREE ENERGY G

F = U T S F is defined by subtracting from the internal energy the thermal energy dF = dU TdSSdT dF =SdT pdV dF|T =pdV |T F represents the compression work that can be done by the system for

constant T

G = U +PV TS = HTS Represents the available internal energy after the thermal energy and the

compressive energy are removed Represents the available energy that can be extracted from the system at

constant T and P Related to the energy associated with the internal degrees of freedom of

the system at constant T and P


N. Zabaras 26

PHASE TRANSFORMATIONS AT FIXED V and T or FIXED P and TTransformation with fixed V in a reservoir of fixed T

Fsys = UsysTSsysSsysT Fsys = q pV TSsysSsysT From the 2nd law: Ssys = qT +Suniv Finally: Fsys = TSsysTSuniv pV TSsysSsysT Fsys =TSuniv pV SsysT

Fsys|V,T =TSunivTransformation with fixed P in a reservoir of fixed T

Gsys = Usys +PV +VPTSsysT Ssys Gsys = q+VPTSsysTSsys

From the 2nd law: Ssys = qT +Suniv Finally: Gsys = TSsysTSuniv +VPTSsysTSsys Gsys =TSuniv +VPT Ssys

Gsys|T,P =TSunivIntroduction to thermodynamics of materials Materials Process Design and Control Laboratory

N. Zabaras 27

SUMMARY OF EQUILIBRIUM CONDITIONS FOR PHASETRANSITIONS

Suniv = 0

Applicable to all cases All subsystems affected by the process need to be included

Fsystem = 0

Applicable for isothermal constant volume phase transitions You need to consider only the system under phase change

Gsystem = 0

Applicable for isothermal phase transitions with constant P You need to consider only the system under phase change


N. Zabaras 28

THE PHASE WITH THE LOWEST VALUE OF G INDICATESWHICH PHASE IS MOST STABLE

Tm T (temperature)

G

Liquid

Solid

P = constant

Molar free energies of solid and liquidphases versus temperature

The phase with the smallest G at agiven T is the most stable phase at thatparticular T

At equilibrium Teq Tm: Gsolid =Gliquid

Plot of phase fractions f s, f l versus Hduring a phase transformation

Plot of the molar Gibbs free energyversus H [2] note that during thetransformation, the molar Gibbs freeenergy of each phase is equal regard-less of how much material is present


N. Zabaras 29

VARIOUS COEFFICIENT RELATIONS USING LEGENDRETRANSFORMATIONS

Legendre transformations are used to replace the differential CidXi with thedifferential XidCi in the combined 1st and 2nd laws: dU = T dSPdV

Legendre transformationsH = U (PV ) = U +PV, F = U T S & G = U +PV TS

U(S,V ) :H(S,P) :F(T,V ) :G(T,P) :

dU = TdSPdVdH = TdS+V dP

dF =SdT PdVdG =SdT +V dP

dH|P = qrev dF |T = wrev

Coefficient relationsT = US |V = HS |P

P =UV |S =FV |TS =FT |V =GT |P

V = HP |S = GP |TIntroduction to thermodynamics of materials Materials Process Design and Control Laboratory

N. Zabaras 30

MAXWELL RELATIONS

These four relations are derived using the property relations given earlierand applying the 2nd derivative property of continuous functions Y (X1,X2, . . . ,Xn)(here for U, H, F and G):

dY = C1dX1 + +CndXn 2YXiXj =2Y

XjXi CiXj =

CjXi

Coefficient relationsT = US |V = HS |PP =UV |S =FV |TS =FT |V =GT |PV = HP |S = GP |T

Maxwell relationsTV |S =PS |VTP |S = VS |PSV |T = PT |VSP|T =VT |P


N. Zabaras 32

THE STATE FUNCTIONS S,V,U,H,F,GIN TERMS OF T AND P

dS = CPT dT VdPdV = VdT VdP

dU = TdS pdV = (CPPV)dT +V (PT)dPdH = TdS+V dP = CPdT +V (1T)dP

dF =SdT PdV =(S+PV)dT +PVdPdG =SdT +V dP


N. Zabaras 33

OPEN AND CLOSED SYSTEMS

All earlier developments were for closed systems, e.g. dU = dq + dw,dS dqT , etc. These developments are not applicable to open systems or closed sys-

tems that undergo irreversible changes in composition All earlier developments are of course still applicable for the universe

Let us see why dU = T dS pdV becomes ambiguous For an isolated system dU = dV = 0 dS = 0 However dS can increase because of a chemical reaction of because of

mixing of various substances that were initially separated

Let us similarly see why dG =SdT +V dP becomes ambiguous Consider dT = dP = 0 dG = 0 However dG can increase just by doubling the amount of the system

at constant T and P (G is extensive)


N. Zabaras 34

OPEN AND CLOSED SYSTEMS

Two state variables (e.g. T and P) are not sufficient to define the stateof an open system

We need to introduce additional state variables related to the compositionand size of the system


N. Zabaras 35

THE CHEMICAL POTENTIAL

Let ni be the number of moles of substance i. For variable ni, U = U(S,V,n1,n2, . . . ,nk) and:

dU = US |V,nidS+UV |S,nidV +

k

i=1

Uni|S,V,n jdni

For constant n1,n2, . . . ,nk, dU = T dS pdV T = US |V,ni andP =UV |S,ni. We can thus write:

dU = T dSPdV +k

i=1

Uni|S,V,n jdni = TdSPdV +

k

i=1

idni

where the chemical potential i is defined as follows:

i =Uni|S,V,n j (an intensive property)

i represents the tendency of a substance to diffuse from one phase toanother


N. Zabaras 36

INTRODUCING THE CHEMICAL POTENTIAL THROUGH G

G = G(T,P,n1, . . . ,nk) dG = GT |P,nidT + GP |T,nidP+ki=1 Gni|T,P,n jdni For constant n1,n2, . . . ,nk (closed system), dG =SdT +V dP S = GT |P,ni and V = GP |T,ni. We can thus write:

dG =SdT +V dP+k

i=1

Gni|T,P,n jdni =SdT +V dP+

k

i=1

idni

The definition of i = Gni|T,P,n j is the same as i =Uni|S,V,nj

Indeed, recall that: dU = T dSPdV +ki=1 idni Add on both sides: d(PV TS) to derive:

dG = T dSPdV +k

i=1

idni+d(PVT S)=SdT +V dP+k

i=1

idni

i is thus the amount by which the capacity of the phase for doing workother than work of expansion is increased per unit amount of substancei added for an infinitesimal addition at constant T and P


N. Zabaras 37

COMBINED FIRST AND SECOND LAWS FOR AN OPEN SYSTEM

U = U(S,V,n1,n2, . . . ,nk): dU = T dSPdV +ki=1 idni G = G(T,P,n1,n2, . . . ,nk): dG =SdT +V dP+ki=1 idni H = H(S,P,n1,n2, . . . ,nk): dH = T dS+V dP+ki=1 idni F = F(T,V,n1,n2, . . . ,nk): dF =SdT PdV +ki=1 idni

i =Uni|S,V,nj = Gni |T,P,n j =

Hni|S,P,nj = Fni|T,V,n j


N. Zabaras 38

ki=1 idni AS A WORK TERM

dU = T dSPdV +k

i=1

idni

For a reversible change in composition of a closed system,ki=1 idni can be interpreted as the chemical work (workother than compression work)

For open systems, we cannot interpret ki=1 idni as work:When there is simultaneous transfer of energy andmass as in open systems, the term heat is ambiguousand we cannot interpret T dS as heat so the remain-ing terms in the equation above cannot be interpretedas work! (see Denbigh, section 2.7)


N. Zabaras 41

INTEGRATION OF THE BASIC EQUATIONS

Let us consider dU = T dSPdV +ki=1 idni Let the phase under examination be enlarged from V to kV T , P & i remain unchanged (intensive variables). Thus we can write:

U = TSPV +k

i=1

ini () Because U , S and ni are extensive variables, U = (k 1)U , S =

(k1)S and ni = (k1)ni and Equation () becomes:(k1)U = T (k1)SP(k1)V +ki=1 i(k1)ni. Finally:

U = TSPV +k

i=1

ini

dH = TdS+V dP+ki=1 idni H = T S+ki=1 ini U (PV ) dF =SdT PdV +ki=1 idni F =PV +ki=1 ini U TS dG =SdT +V dP+ki=1 idni G = ki=1 ini U (PV )TS


N. Zabaras 42

DERIVING G =ki=1 ini USING THE CONCEPT OFHOMOGENEOUS FUNCTIONS

With T and P being intensive variables, we note that:G(T,P,n1, . . . ,nk) = G(T,P,n1, . . . ,nk)

e.g. G is homogeneous in ni of degree 1 Differentiation with respect to leads to the following:

dG(T,P,n1, . . . ,nk)d = G(T,P,n1, . . . ,nk)

or

k

i=1

G(T,P,ni)(ni)

(ni) = G(T,P,ni)

Finally, we obtain: ki=1 ini = G(T,P,n1, . . . ,nk)


N. Zabaras 43

THE GIBBS-DUHEM EQUATION Consider our initial fundamental equation:

dU = T dSPdV +k

i=1

idni

Using U = T SPV +ki=1 ini dU = SdT +T dSV dP pdV +

k

i=1

nidi +k

i=1

idni

Subtracting the two equations leads to the Gibbs-Duhem equation:SdT +V dP

k

i=1

nidi = 0

If there are k substances in the phase, the number of independentintensive variables is: k +1

For a multiphase alloy, there are Gibbs-Duhem equations:SdT +V dP

k

i=1

ni di = 0, for each phase = 1, . . . ,


N. Zabaras 44

MOLAR QUANTITIES FOR A ONE COMPONENT ALLOY Consider a phase with one component: dU = TdSPdV +dn () We have already defined the molar variables: V = V

n, S = S

n, etc. From

these equations, we can write:

dU = nd U + Udn, dS = nd S+ Sdn, dV = nd V + V dn

Equation () can thus be written as:nd U + Udn = T (nd S+ Sdn)P(nd V + V dn)+dn

or as follows:

nd U = Tnd SPnd V +( U + T SP V )dn () We have shown that G =ki=1 ini (for k = 1 here)

= Gn= U+PVT S

n= U +P V T S

Equation () is then simplified as follows: d U = T d SPd V


N. Zabaras 45

FOR BOTH OPEN AND CLOSED SYSTEMSWITH ONE COMPONENT

d U = T d SPd Vd H = Td S + V dP

d F = SdT Pd Vd G d = SdT + V dP

Note that e.g. G = G(T,P), etc.


N. Zabaras 46

PARTIAL MOLAR QUANTITIES FOR A MULTI-COMPONENTALLOY

We will like to define a quantity Ei of a property E obeying an equationsimilar to E = En that the molar quantity E of one-component alloyssatisfies, such as:

E =k

i=1

Eini

Define the partial molar value of E as:

Ei =Eni|T,P,n j

From E(T,P,n1, . . . ,nk) dE = ET |p,nidT + EP|T,nidP+ki=1 Eidni Integration of the above equation leads to: E =ki=1 Eini


N. Zabaras 47

PARTIAL MOLAR QUANTITIES FOR A MULTI-COMPONENTALLOY

In summary: U = ki=1Uini, S = ki=1 SiniH = ki=1 Hini, F = ki=1 FiniV = ki=1Vini, G = ki=1 Gini

Absolute values of Ei are never known they must be computed wrt areference state (same as that of E)

Note that i Gi


N. Zabaras 48

PARTIAL MOLAR QUANTITIES FOR A MULTI-COMPONENT ALLOY

Ei are intensive variables (independent of the size of the system) but inaddition to T and P depend on the relative proportions of the variouscomponents, e.g.

Ei = Ei(T,P,x1,x2, . . . ,xk1), xi = ni/k

i=1

ni

The relations between partial molar quantities are similar to those be-tween the parent quantities:

H = U +PV Hi = Ui +PViG = HTS i = HiT Si

Using dG =SdT +V dP+ki=1 idni, we can derive the following:iP |T,ni,n j = Vni|T,P,n j = Vi, and

iT |P,ni,n j = Sni|T,P,n j =Si

Combining the eqs. above gives: i = HiT Si = Hi +T iP |P,ni,n j, iTT |P,ni,n j =HiT 2


N. Zabaras 49

PARTIAL MOLAR QUANTITIES IN TERMS OF MOLE FRACTIONS Let us start with E =ki=1 Eini, dE = ki=1 Eidni +ki=1 dEini () But dE = ET |P,nidT + EP |T,nidP+ki=1 Eidni () Subtracting Equs. (*) and (**) leads to:ET |P,nidT + EP |T,nidPki=1 dEini = 0 (for E = G, the Gibbs-Duhem Equ.) For constant T and P, the above equations lead to the following:

ki=1 nidEi = 0 or ki=1 xidEi = 0 (with xi = ni/ki=1 ni) As a result Ei = Ei(T,P,x1, . . . ,xk1) k1 independent mole fractions

di =SidT +VidP+k1i=1 ixi |T,P,x j dxi where we used the results shownearlier: iP |T,ni,n j = Vi, and iT |P,ni,n j =Si

Similarly using Hi = TSi +i, dHi = TdSi +VidP+k1i=1 ixi |T,P,x j dxi Finally, dUi = T dSiPdVi +k1i=1 ixi |T,P,x j dxi


N. Zabaras 50

REVISITING THE EQUILIBRIUM CONDITIONS

From the 2nd law: Suniv 0 For an isolated system, we also can write: Ssys 0

For an isolated system there are no interactions with the surroundings For a given process that changes the system state, all changes in the

entropy of the universe are localized within the system Since there is no heat transfer between the system and its surroundings,

all changes in Suniv represent entropy generated within the system

For an isolatedsystem:

Suniv 0Suniv = Ssurr +Ssys 0Ssys 0

Since Ssys can only increase, is there a maximum in the system entropy? In an isolated system at equilibrium Ssys is maximum


N. Zabaras 51

EQUILIBRIUM CONDITIONS Spontaneous processes increase the entropy of the system The driving forces in a spontaneous process are the potential for entropy

increases When the driving forces for spontaneous change in the system become

exhausted, the system reaches equilibrium In an isolated system at equilibrium Ssys is maximum For any possible variation S with U = V = 0, the following is

true: S|U=0,V=0 0 S is maximum at constant U and V Equivalently: For any possible variation U with S = V = 0:U |S=0,V=0 0U is minimum at constant S and V

U = US |VS+UV |SV U = TS 0

At equilibrium the state is maintained without external driving forces


N. Zabaras 52

HOW DO YOU CHECK IF A SYSTEM IS AT EQUILIBRIUMWe maximize the entropy of an isolated system The state conditions are the same regardless if the system is isolated

or not

To check if the system is in equilibrium, we perform a virtual ex-periment: We isolate the system and check if its state remainsunchanged.

Note that systems that appear to be in steady-state conditions are notnecessarily at equilibrium

The internal conditions of the system need to be responsible for thesteady-state conditions and not an external driving force For example, a system corresponding to steady-state temperature

conditions within an applied temperature gradient is not at equi-librium (isolating the system will lead to temperature changes!)


N. Zabaras 53

THERMODYNAMIC EQUILIBRIUM OF TWO SEPARATEDREGIONS (T,P) AND (T ,P) [2]

Consider the state of the two regions. The followingholds: V = V+V , U = U+U, S = S+SConsider an arbitrary virtual change with U = V = 0U = TSPV S = UT + P

TV Similarly, S = UT +

PTV

For the whole system: S = UT + P

TV+ U

T +PTV

For an isolated system: V =V , U =U

Finally, we can write: S = ( 1T 1T)U+(P

T P

T)V

For an isolated heterogeneous system, the entropy is maximum (sys-tem at equilibrium) when S = 0, i.e.:T = T (thermal equilibrium) no heat flow between and

P = P (mechanical equilibrium) no volume changes of or


N. Zabaras 54

THERMODYNAMIC EQUILIBRIUM OF AONE-COMPONENT TWO-PHASE SYSTEM

Consider the state of the two phases:(T,P,V,S,n, . . .) and (T ,P,V ,S,n, . . .)

The following holds for extensive variables:V = V+V U = U+US = S+Sn = n+n

Consider an arbitrary process:dU = TdSPdV+dn dS = dUT + P

TdV Tdn

where the chemical potential is defined as: = Un |S,V Similarly, dS = dUT + P

TdV

Tdn Finally, for the whole system:

dS = dUT +PTdV

Tdn+ dU

T +PTdV

Tdn


N. Zabaras 55

THERMODYNAMIC EQUILIBRIUM OF AONE-COMPONENT TWO-PHASE SYSTEM

For the whole system:dS = dUT +

PTdV

Tdn+ dU

T +PTdV

Tdn For an isolated system the following hold:

dV =dV dU =dUdn =dn

Finally we can write:dS = ( 1T 1T)dU+(P

T P

T)dV ( T

T)dn

For an isolated unary two phase system, the entropy is maximum(system at equilibrium) when dS = 0, i.e.:

T = T (thermal equilibrium) no heat flow between and P = P (mechanical equilibrium) no volume changes of or

= (chemical equilibrium) no changes in n and n


N. Zabaras 56

EQUILIBRIUM MINIMUM INTERNAL ENERGY We showed earlier that:

dU = TdSPdV+dndU = T dSPdV +dn

Since the system is isolated:dV =dV dS =dSdn =dn

Finally we can write for the whole system:dU = (TT )dS (PP)dV+()dn

Minimization of the internal energy for an isolated system leads to:T = T (thermal equilibrium)

P = P (mechanical equilibrium) = (chemical equilibrium)

e.g. the same conditions as for equilibrium in an isolated system


N. Zabaras 57

EQUILIBRIUM MINIMUM ENTHALPYFOR P=CONSTANT

dH = TdS+V dP+dndH = T dS+V dP+dn

If we isolate the system, we obtain:dS =dSdn =dn

For the whole system:dH = (TT )dS+V dP+V dP+()dn

Let us minimize H with constant P: dP = dP = dP = 0:T = T (thermal equilibrium)

dP = dP = dP = 0 (constant pressure constraint) = (chemical equilibrium)

Equilibrium is thus obtained when H is minimized under constantpressure for an isolated system (constant material and entropy)


N. Zabaras 58

EQUILIBRIUM MINIMUM F FOR T =CONSTdF =SdTPdV +dndF =SdT PdV +dn

For a rigid and impermeable boundary:dV =dV dn =dn

For the whole system:dF =SdTSdT (PP)dV+()dn

Let us minimize F with constant T : dT = dT = dT = 0dT = dT = dT = 0 (constant temperature constraint)

P = P (mechanical equilibrium) = (chemical equilibrium)

Equilibrium is thus obtained when F is minimized under constanttemperature and the boundary is rigid and impermeable


N. Zabaras 59

EQUILIBRIUM MINIMUM G FOR T & P CONSTANTdG =SdT+VdP+dndG =SdT +V dP+dn

For a system with impermeable boundary:dn =dn

For the whole system:dG =SdTSdT +VdP+V dP+()dn

Let us minimize G with constant T and P:dT = dT = dT = 0 (constant temperature constraint)

dP = dP = dP = 0 (constant pressure constraint) = (chemical equilibrium)

Equilibrium is thus obtained when G is minimized under constanttemperature and pressure for a system with impermeable walls


N. Zabaras 60

THE CHEMICAL POTENTIAL k FOR AMULTI-COMPONENT MULTI-PHASE ALLOY

For each phase: dU = TdSPdV+Ck=1 k dnk The chemical potential k is defined as: k = U

nk|S,V,n j

A chemical potential gradient means that there is a driving forcefor diffusion

As species move from regions of high potential to regions oflow potential, the potential is lowered in the regions from wherespecies came and increases in the regions where they arrived

Equilibrium conditions imply elimination of all driving forces fordiffusion, i.e. of all chemical potential gradients


N. Zabaras 61

EQUILIBRIUM OF MULTI-COMPONENT MULTI-PHASE ALLOYS

For each phase: dU = TdSPdV+Ck=1 k dnk Rearranging gives the follows:

dS = dU

T+

P

TdV

C

k=1

kT

dnk

Finally for the entire system: dS ==1 dS

dS =

=1{dU

T+

P

TdV

C

k=1

kT

dnk }

For an isolated system the following holds:

=2

dV =dV 1,

=2

dU =dU1,

=2

dnk =dn1k, for each k


N. Zabaras 62

EQUILIBRIUM OF MULTI-COMPONENT MULTI-PHASE ALLOYS Can easily write:

=1

dUT

=

=2

dUT

+dU1T 1

=

=2

dUT

=2 dU

T 1=

=2

(1

T 1

T 1)dU

=1

PdVT

= . . . =

=2

(P

T P

1

T 1)dV

Similarly using =2 nk =dn1k, we can write:

=1

C

k=1

kT

dnk =

=2

C

k=1

(kT

1k

T 1)dnk

dS = =2{( 1T 1T 1)dU+(P

T P1

T 1)dVCk=1(

k

T1kT 1)dn

k }

At the maximum entropy dS = 0 (equilibrium), we conclude that for allphases : T = T 1, P = P1 and k = 1k for all k


N. Zabaras 63

THE GIBBS PHASE RULE

Consider phases with C components in equilibrium The following number of equations is available (the T and P eqs not ac-

counted here but are explictitly introduced in the number of variables):

(GibbsDuhem Equations)SdT +VdPk

i=1

ni di = 0, and

C(1) ( chemical potential conditions) k = 1k TOTAL : (CC +)

The number of unknowns is: 2 (temperature and pressure) + C (con-centrations); a total of C +2

The number of degress of freedom f (the number of variables that I canchange and still remain in equilibrium) is then equal to:

f = C+2


N. Zabaras 64

REFERENCESMaterial presented in this lecture has been compiled from the following refer-ences

1. Thermodynamics, C. B. Musgravehttp://chemeng.stanford.edu/ charles/mse202/

2. Thermodynamics of materials, W. Craig Carterhttp://pruffle.mit.edu/3.00/

3. Introduction to Metallurgical Thermodynamics, D. R. Gaskell4. The Principles of Chemical Equilibrium, K. Denbigh


Documents

Thermodynamics