Metallurgical Physical Chemistry

7/28/2019 Metallurgical Physical Chemistry

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Four Types of Systems1. Open systemexchange of energy and matter occurs

with its surroundings2. Closed systemexchange of energy may occur but no

transfer of matter occurs between the system and its

surroundings3. Thermally isolated systemno exchange of energy (in

the form of heat) takes place4. Mechanically isolated systemno work is done in the

system or by the systemSee ATKINS


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State of a system can be defined completely by

observable macroscopic properties of matter knownas the variable of state Pressure, Volume, Temperature, and

Composition

Used to specify the state of a thermodynamic systemand their values depend on the conditions or thestate of a thermodynamic system.i. Extensive variables are proportional to the amount of the

matter (volume, area, mass, heat capacity, free energy,entropy and enthalpy)

ii. Intensive variables are independent of the amount ofmatter (temperature, pressure, density, chemicalpotential(measure of reactivity of the system) andviscosity.


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Any two of the variable of state can besufficient to fix the state.

Equation of State

identifies the minimum number of

variables needed to define the system. Variables that can be controlled during

experiments (T and P)

Mass (both extensive)

Volume


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Boyles Law (Robert Boyle) Pressure Volume relationship at constant

temperature

P (1/V)T Charles Law

Volume-temperature relationshipV (T)P

Joseph Gay-Lussacs Law Coefficient of thermal expansion, , as the fractional

increase, with temperature at constant pressure, ofthe volume of a gas

At -273.10C the volume of gas becomes zero, allmotions stop at this temperature

= (1/Vo)(dV/dT)P = CTE ; Vo=volume of gas at 0o C


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Combination of Boyles, Charles and Gay-Lussacs Law

PV=nRTBoyles Law

PoV(Po, T) = PV(T, P)

Charles and Gay Lussacs LawV(Po, To) = V(Po, T)

To T

PoVo = PV = Constant =0.08205 L-atm/mole K

To T

From Avogadros hypothesis, volume of 1mole of all gases at STP is 22.414L.

R = 0.08205 L-atm/mole K

= 1.987 cal/mole K

= 8.3144 Joules/mole K


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Ideal-gas equations assumptions are not

valid when dealing with real gases.

Applicable assumptions are:

a. Volume of molecules may not be negligible inrelation to the volume occupied by the gas

b. Attractive forces between molecules may notbe negligible

P+ P = P + a

V2Correction term proportional to:a. Number of molecules striking

unit area of wall per second at

any instant

b. Number per unit volume of

molecules

Cohesion pressure-Measure of attractiveforce of molecules in

the bulk of the gas


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P + a (V-b) = RT

V2

pressure & volume correction

All molecules have a particular diameteror volume which is equal to four timesthe actual volume of the moleculesaccording to van der Waals becauserepulsive forces occur where theyapproach very closely

b, correction factor for ideal volume

occupied by the molecule in the container


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A. Various Definitions

B. Forms of Energy

C. Statement of the First Law of Thermodynamics

D. Reactions at Constant Volume and ConstantPressure

E. Adiabatic Process

F. Isothermal Transformation


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A. Various Definitions Principle of conservation of energy

i. Total amount of energy of an isolatedsystem remains constant but may changefrom one form to another

ii. When an amount of energy of one formdisappears, an equivalent amount of

energy of other forms appeariii. Energy cannot be created or destroyed


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a. Internal Energy Comes from atom and electron movement

Measure of energy stored in the bonds translational, vibrational, rotational and electronic effects

b. Work

Interaction between a system and its surroundings Mechanical Gravitational Surface tension Electrical Magnetic, etc

+W system has done work on thesurroundings

-W work done on the system


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Sample Problem:

Six moles of an ideal gas at 100C undergo isothermalreversible expansion against a constant external pressure of3.5 atm in a piston-cylinder apparatus. The volume of the gasis increased by a factor of 450%. Compute the workperformed as a result of the expansion.

Solution:

Initial total volume (V1)

V1 = 6*8.3144*373.16 = 52.492 x 10-3 m3

3.5* 101,325

V2 = 4.5 * V1 = 236.214 x 10-3 m3

W=V1V2 PdV = P( V2-V1) = 3.5*101,325(236.214-52.492) x 10-3 m3

W= 65,155 J


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c. Heat, Q

i. Form of energy mainly due to temperature

ii. Method of energy transfer to an assembly which are notobservable as macroscopic work

iii. Usually produces a rise in temperature when in enters asystem; flows from high to low temperature body

iv. Irreversible process

Calorie( = 4.184 J/cal)

amount of heat required to raise the temperature of 1gram of water from 14.5 to 15.5C at 1 atm pressure.

Q=CT + Q heat is added to the system orwhen it crosses the boundaryfrom the surroundings into thesystem

- Q heat flowing out of the systeminto the surroundings


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2. A copper slag whose mass mc is 75 g is heated in a laboratory ovento a temperature of 312C. The slag is then dropped into a glassbeaker containing a mass mw=220g of water. The effective heat

capacity Co of the beaker is 45 cal/K. The initial temp Ti of the waterand the beaker is 12.0C. What are the final temperatures Tf of theslag, the beaker and the water?

Heat capacity =mc

Qw

= mw

cw

(Tf

-Ti

)

Qb = mbcb(Tf-Ti)

Qc= mccc(Tf-Tc)

Qw+ Qb+ Qc= 0

mwcw(Tf-Ti) + mbcb(Tf-Ti) + mccc(Tf-Tc) = 0

Cc=0.092 cal/gK

Cw=1.00 cal/gK

Tf=

ccbww

iwwipccc

fcmccm

TcmTcTcmT


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W (joules)

1 cal =4.186 J

1 Cal = 1000 cal

Q= mcTQ= amount of heat needed to raise temperature by Tm= mass of the bodyc= heat capacity; characteristic constant of a given body

Q= nMcTn= number of moles

M= molecular massQ= nCT

C= molar heat capacity

Example:

1. How much heat is needed (a) to raise the temperature of 725g of

lead from room temperature (293 K) to its melting point (602 K)?clead=128 J/kg-K

Q = mcT= 0.725 kg (128 J/kg-K) (602-293 K)

Q = 2.87 x 104 J = 28.7 KJ


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Internal energy (U)

molar heat, work,

Q W

U= Q-WQ= molar heat absorbed by a system

W=work performed per mole on or by a system

U increased Workperformed on the system

Heat transferred into the system

U decreased Workperformed by the system

Heat transferred from the system

SystemEnergy, U


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Isochoric Process U= Q W=0; No work done on and by the system

Isobaric Process U= Q-PV Work done on the system because of volume

change (at constant pressure)

Sample problem:

When 1 g-atom of pure iron is dissolved in dilute HCl at 18C, theheat liberated is 87.03 kJ. Calculate the energy change (U)of the system.

Solution:

Fe + 2H+

H2(g)+Fe2+

U= Q-PV PV= nRT U= Q-RT ;Heat is liberated:

U= -87.03 kJ-2.415kJ=-89.44kJ


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At constant volume

QV=CVT U= CVT

if CV is not constant: U=QV= CVTdT

At constant pressure

QP=C

PT U= q

p-PV

(U2+PV2) - (U1+PV1)=qpH=U+PV for constant pressure only

H2-H1 =

H=CPT = CPTdT For 1 mole of gas

CP-CV=nR


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Constant Volume

P1V1= P2V2

=constant

Constant Temperature

1

12

2

1

1

2

exp

lnln

nRT

WPP

P

PRTV

VnRTQW


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1. Liquid Cd boils at 760C, 1 atm pressure.The heat of evaporation, Hev is 23.87kcal/mole. Calculate the incrementalchange in internal energyaccompanying volatilization of 1 moleof liquid Cd at the boiling temperature.

H = U + PV


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A. Definition

B. Thermodynamic Relations Involving

Entropy


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Entropy-measure of state of order and disorder of a system

dU=TdS-PdV*dS is independent of the path

a. Entropy in Isothermalphase transition

b. Entropy calculations when temperature changes1. Constant pressure

2. Constant Volume

T

HS TrTr

TdCSP

T

Tln2

1

TdT

CS V

T

T 21


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Entropy changes for irreversible processes

Entropies in the reaction can be calculated as follows:

),(),(

),(),(

),(),(

),(),(

),(),(

11

,

,.

,

)3(

)2(

)1(

TsTl

TsTl

TsTl

TsTl

TsTl

AA

AA

AA

AA

AA

lpm

pmpm

pml

S1S2S3

Ssyst

pm

fT

T

sPlP

syst

T

T

sP

pm

f

T

T

lP

T

HdT

T

CCS

dTT

CS

THS

dTT

CS

pm

pm

pm

,

3

,

2

1

,

1

1

,

,

1


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Sample Problem:

Calculate the entropy changes of a system andsurroundings for the case of the freezing ofsupercooled liquid silver at 800C. The melting pointof silver is 961 C and the heat of fusion is 2.69kcal/mole. (HF=2690)

Ag(l,1073K)Ag(s,1073K)1. Ag(l,1073K)Ag(l,1234K) S1= (CP(l)/T) dT2. Ag(l,1234K)Ag(s,1234K) S2=-Hf /Tm,p3. Ag(s,1234K)Ag(s,1073K) S3= (CP(s)/T) dT

molKcalS

xxS

dTTxxTS

syst

syst

syst

/20.2

18.21073

1

1234

11036.0*)2()10731234(1004.2

1073

1234ln21.2

123426901036.01004.209.530.7

22

53

1234

10733531


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To solve for the H of the surroundings,the heat of fusion at 800C has to be

calculated as follows:

molKcalS

molKcalS

molcalH

xxH

dTTxTxH

univ

surr

f

syst

f

syst

systf

/33.053.220.2

/53.21073

42.2717

/2717

26901073

1

1234

1

1036.0)10731234(1004.2*5.0)10731234(21.2

26901036.01004.209.530.7

)1073,(

5223

)1073,(

1234

1073

253)1073,(


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A. Entropy at Standard Zero

B. Third Law of Thermodynamics

C. Entropy of Reactions

D. Entropy of Reaction withVariation in Temperature


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All substance have the same entropies at absolute zerotemperature

Third law of thermodynamics:

The entropy of any homogeneous substance which is incomplete internal equilibrium may be taken as zero at 0K.

Entropy of Reaction

S298(rxn) =nS298(products) -nS298(reactants)Example:

2Al+3/2 O2Al2O3 at 298K

S298(Al) =28.33J/mol KS298(O2) = 205.02J/mol K

S298(Al2O3) =50.90 J/mol K

S298(rxn) =50.90-2*28.33-(3/2)*205.02=-313.29 J/mol K

= Sformation(298)


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For processes at temperature combinations otherthan T=298K and T=T K, the general equation is asfollows:

dTT

CSS

T

T

p

TrxnTrxn

2

112 )()(


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A. Heat Capacity and Heat Content

B. Enthalpy or Heat Content

C. Heat of Formation

D. Heat of Transformation

E. Heat of Reaction

F. Hess Law

G. Variation of Enthalpy Change With Temperature

H. Adiabatic Flame Temperature

I. Helmholtz Free Energy

J. Calculation of Free Energy Change

K. Gibbs free energy and the Equilibrium Constant

+ H exothermic, evolves heat- H endothermic, requires heat


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1. Given: T =298 k, P=1atm and

a. W(s)+O2(g)WO2(s) Hf = -560.7 kJ/mol

b. 3WO2(s)+O2(g)W3 O8(s) Hrxn = -550.2 kJ/mol

c. W3 O8(s) +1/2 O2(g)3WO3(s) Hrxn = -278.3 kJ/mol

Find enthalpy change for the reaction producing WO3(s)

2. The standard enthalpies of formation of severalminerals at 968K are as follows:

a. Al6Si2O13(mullite) Hf(968K) = 42.2kJ/mol

b. Al2O3(corundum) Hf(968K) = 31.8kJ/molc. SiO2(quartz) Hf(968K) = -15.3kJ/mol

Calculate H for the production of mullite fromcorundum and quartz at 968 K.

S C G


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PHASE CHANGEalways accompanied by release or absorption of

heatPhase change from solid to liquid

Q m, Q=Lf mLf =latent heat of fusion;

constant character of body

Lc =latent heat of combustion;

Ls =latent heat of sublimation;

Phase change from liquid to gas

Q=Lv m

Lv =latent heat of vaporization;

constant character of body

* Steam is hotter due to the temperature it absorbed in the process of vaporization

Heats of fusion and vaporization

reversible - temperature used in phasechange

Temperature

Timesolid state

liquid stategas state

Substance Melting Point(C) Lf(J/kg) Boiling Point(C) Lv(J/kg)

H -253.91 58.6 X 103 252.89 452 X 103O2 -218.79 13.8 X 103 183.0 213 X 10 3

H2O 0.00 334 X 103 100.00 225 X 103Ag 960.80 88 X 103 2193 2336 X 103


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Example:

How much heat is needed(a) to raise the temperature of 725g of lead

from room temperature (20C or 293K) to itsmelting point?

C = 1285 /kg KTm= 602 K

Q = mc (Tf-Ti)= 2.87 x 106 J

Q= 287 kJ(b) How much additional heat is required to

melt the lead at its melting point?Q= Lf mLf= 23.2 J/ kgQ=16.8 kJ


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3. Calculate the standard heat of formation of solidPbO from liquid Pb and O2 gas at 527C. Themelting point of lead is 327C and its latent heat of

fusion is 1.15kcal/mole. The molar heat capacity atconstant pressure of liquid Pb is

CP(Pb(l))=7.75-0.74x10-3Tcal/molK.

Hf(PbO,298)= -52.4 kcal/mol

CP(PbO)=10.6 + 4.0x10-3T cal/molK.CP(Pb)= 5.63 + 2.33x10

-3T cal/molK.

CP(O2)=7.16 + 1.0x10-3T0.4 x105T-2 cal/molK.


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Maximum attainable temperature ofcombustion products when reaction occurscompletely at 298K under adiabaticconditions. Reaction where heat neither enters or leaves the

system Combustion of fuel in a confined system

Fuel + oxidant (at 298)combustion products(at very high temp, Tm)

This rection can be performed in two imaginary steps:1. fuel + oxidantat298combustion productsat 2982. Combustion products at298 combustion products at Tm

Reaction in 1 is always exothermic(combustion reaction)

Available thermal energy is used to heat up combustionproducts from 298 to Tm


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Depend on the initial and final states of the system

When P and T are constant

G= dHTdS

Reaction at constant temperature and

volumeF=Q-TS=U-TS (Helmholtz Free Energy)

Independent of the path taken

At constant T and P:G = 0 reaction is at equilibrium G0 not spontaneous in the specified direction


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DCp0, thus GT= H298 -T S298


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For a system in equilibrium Rate of forward reaction=rate of backward

reaction

Keq=K1/K2 ; K1 & K2 are specific rate constants Activity, ai= i Ci i =activity coefficient characteristic of a given ionic

specie Ci =concentration of specie expressed in moles

ai pure substance =1

i ionic specie in very dilute solutions =1 ai =partial pressure; for gases at low pressures

Keq =exp(G/RT)


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Vant Hoffs Isochore

d ln Kp=H

dT T2

Kp = - H + constant (at any temp)

RT

Clausius-Clapeyron Equationd ln PA= Le

dT RT2 PA=vapor saturation pressure


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1. Calculate the vapor pressure of Mn inmolten steel (1600C) if Le=226 kJ/mole

at 2095 C.


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Daltons Law

The total pressure Ptof an ideal gas mixture isequal to the sum of the pressures exerted byeach component

G1=RT lnPi=RT ln(XiPT)

Le Chateliers Principle

When a system, which is at equilibrium, issubjected to the effects of externalinfluence, the system moves in that directionwhich tends to nullify the effects of theexternal influence.


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1. Calculate the equilibrium PO2 over Ni at 1200C. Determine the corresponding air pressure

below which NiO will begin to dissociate.

2. Calculate the equilibrium ratio PH2/PH20 for theoxidation of chromium in water vapor at 1000

C.

3. Characterize the relative thermal stability ofSi3N4 and BN in a mixture of Si3N4 and BN at 1atm pressure. Assume all components arepure. BN has been sugested as an abradablehigh tem[erature coating. Si3N4 is utilized inhigh temperature ceramic applications.


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Heterogeneous Condensed Phase

Dissociation of CO2

2CO2(g)2CO(g) +O2(g)Example:1. Consider the following reaction in the roast reduction

reaction(2nd stage converster) at 1000K, in coppersmelting

2Cu2O(l)+Cu2S(l)6Cu(l)+SO2(g)Given the following data:

2Cu(l) + O2(g)Cu2O(l) GT =-40,500-3.92TlogT +29.5T2Cu(l) + S2(g)Cu2S(l) GT =-30,610+6.80T

1/2 S2(g) +O2(g)SO2(g) GT =-86,620+ 17.31T

G1000K=


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2. Consider the Kroll processTiO2(s)+2Cl(g)TiCl4(g)+O2(g) T=1000C

Determine the partial pressures of the product gasesbefore the reaction ceases to produce thetitanium chloride if the operation is done at 1atmpressure.

Given:

Ti(s) + 2Cl2(g)TiCl4(g) GT = -180,700-1.8logT +34.65TTi(s) + O2(g)TiO2(g) GT = -2,184,600+ 41.74T

C(s) + O2(g)CO(g) GT = -26,700- 20.95T


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Raoults Law The vapor pressure exerted by a dissolved

component A, PA, in a homogeneouscondensed solution is equal to the product ofthe atomic fraction of A in the solution, XA, andthe vapor pressure of pure A, PA, at the

temperature of the solution PA= XAPA aA=XA

Sixty moles of an ideal gas mixture at 5atmosphere pressure contains 15 moles ofS2(g) in contact with microcrystalline quartz,SiO2(s). If analysis of the quartz reveals that itcontains no sulfur impurities, calculate theactivity of S2(g). (Ps2(g)=1atm)


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Metallurgical Physical Chemistry