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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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