The problems of the first law 1.1 a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity CP of lead may be taken as 29.3J/(mol K)
Solution:
Qabsorb Qincrease Qmelt 1 2 mv 2 Qabsorb W
W
n(C p T
1 nMv 2 2
n[29.3 (327 25) 4.8 103 ] V
H melting )
1 n 207.2 10 3 v 2 2
363(m / s )
1.2 what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at the rate of 20 m/min
QBurning
2500 10 3 4.1868 10467000( J ) 10467000 121( J / S ) 24 60 60 20 75 9.8 245( J / S ) 60
P W /t
QBurning / t
Pincreasin g
mg
Solution
h t
1.3 One cubic decimeter (1 dm3) of water is broken into droplets having a diameter of one micrometer (1 um) at 20℃. (a) what is the total area of the droplets? (b)Calculate the minimum work required to produce the droplets. Assume that the droplets are rest (have zero velocity) Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm2) is also used for surface tension dyn/cm) Solution
Stotal
nS Single
W
S
(1 10 1 )3
4
(0.5 10 6 ) 2
4 (0.5 10 6 )3 3 72.75 10 5 2 (3 103 6 10 2 )
6 103 (m 2 )
436.6( J )
1.4 Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in an insulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.
(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm), what will be the temperature of the first gas to hit the specimen? (b)As the helium flows, the pressure in the tank drops. What will be the temperature of the helium entering the quench chamber when the pressure in the tank has fallen to 1 atm? Solution:
(a) Adiabatic T P ( ) R / CP T0 P0 298 (
T
1 0.4 ) 10
118( K )
(b) T T
W nC p T0
1 (500 5) 101325 10 3 10 50 101325 10 3 2.5R R 298 T 298 118 180( K )
118( K )
1.5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T0. The valve on the tank is oðened and the surrounding gas is allowed to flow suickly into t(e tank until the pressure insi`e the tank is equals the pressure outside. Assume that no heat flow takes place. What is the0final tempeture kf tèe gaS in the tank? The heat cap!city mf the gas, Cp and Cv each íay be(assumed to be c/nsuant over thå temperature rang!spanNed by the døperiment. You answer may be meft in terms of Cp and SvMhint: one way to approach the xroblem is to define the system as the gas ends up in the tank. hint: one way to approach the xroblem is to define the system as the gas ends up in the tank.
Adiabatic T P ( ) R / CP T0 P0
solution
T0 (
T
P0 0 R / CP ) P0
T0
1.6 Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO2 and CH4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atm NOTE: this value is a good approximation for the low calorific powder of natural gas 0 H 298 [ Kcal / g mol ]
FOR
DATA:
CO2 ( g )
17.89 94.05
H 2 O( g )
57.80
CH 4 ( g )
solution
CH 4 H 298
2O2
CO2
( H CO2
2 H 2O 2 H H 2O
H 298 191.76( Kcal / g mol)
H CH 4 )
( 94.05 2 57.80 17.89)
191.76 103 10 3
1 0.30483 103 252 103
26.9( Btu / 1000SCF )
1.7 Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O2 and 79% N2) (a) Assuming complete combustion, what is the composition of the flue gas (the gas following combustion)? (b)What is the temperature of the gas, assuming no heat loss?
(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundings average 400000 kJ/h. calculate the fuel consumption at STP (in m3/h) assuming that for gas H1600-H298=1200KJ/KG (d)A heat exchanger is installed to transfer some of the sensible heat of the flue gas to the combustion air. Calculate the decrease in fuel consumption if the combustion air is heated to 800K DATA STP means T=298K, P=1atm for CH 4 CO 2 H 2O N2 O2
C P (cal / mol
C)
16 13.7 11.9 8.2 8.2 (a) CH 4
2O2
CO2
2H 2O
1 79 3 2 1.1 2 (1.1 1) 21 H 2 O% 2CO2 % 17.43%
CO2 %
Solution
N2 %
72.12%
O2 %
0.87%
8.71%
(b) C p, p T
C p ,i X i T0
T
298
0.01[13.7 8.71 11.9 17.43 8.2 (72.12 0.87)] 9.25(cal / mol 191.76 1000 9.25 11.48
2104( K )
C)
(C ) P 1200 2000 400000
2800000( KJ / h)
2800000 10 3 191.76 1000 9.25 11.48(1600 298) 1 ( 4.1868 ) 11.48 0.0224
VConsumin g
3214(m 3 / h)
(d ) C p ,r H
(
H 298
100 ] / 11.48 21
[16 8.2 2.2
C p ,i X i
C p , p ,i ni
8.87(cal / mol
C)
C p ,r ,i ni )dT
191.76 1000 [(9.25 11.48) 8.87 11.48)](800 298) 189570(cal / g mol ) VConsumin g
2800000 10 3 189570 9.25 11.48(1600 800) 1 ( 4.1868 ) 11.48 0.0224
1644(m 3 / h)
1.8 In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined: H700-H298=12113 J/(g atom)
H1000-H298=22803 J/(g atom)
Find a suitable equation for HT-H298 and also for CP as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang. H
C P dT
a bTdT
[aT
b 2 T T ] 298 2
b (700 2 298 2 ) 12113 2 b a(1000 298) (1000 2 298 2 ) 22803 2 a 35.62 a(700 298)
Solution
b CP
0.011 35.62 0.011T
T H 298
35.62(T
298) 0.0055(T 2
298 2 )
1.9 A fuel gas containing 40% CO, 10% CO2, and the rest N2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gas
0
H f , 298,CO H
0 f , 298, CO2
110458 J / mol 393296 J / mol
C P ,CO
28.45 3.97 10 3 T 0.42 105 T 2 ( J / molK )
C P ,CO2
44.35 9.20 10 3 T 8.37 105 T 2 ( J / molK )
C P ,O2
19.92 4.10 10 3 T 1.67 105 T 2 ( J / molK )
CP, N 2
29.03 4.184 10 3 T ( J / molK ) 2CO
2CO2
O2
when 1mole fuel need 1mole air ( N 2 / O2
4)
then reation 20%
CO
5%
CO2 N2
65%
O2
10%
production CO2
27.8% 72.2%
N2
H i , 298
H
0 f , 298, r
ni
H
0 f , 298, p
393296
ni
110458
282838( J / mol )
(a) C p,r
0.2C P , CO
C p , p , i ni
3
28.77
5
4.38 10 T
C p,r
0.67 10 T
0.278C P , CO2
C p , p , i ni
5.58 10 3 T
33.28 H
T 298
2
0.65C P , N 2
2
( J / molK )
( J / molK ) C p , r , i ni ) dT
C p , p , i ni
5.58 10 3 T
(33.28
0.1C P , O2
0.722C P , N 2
0.19 105 T (
H i , 298ni
282838 0.2
0.05C P , CO2
0.19 105 T
2
)
0.9dT
T
28.77
4.38 10 3 T
0.67 105 T
2
2
2.19 10 3 T 2
dT
733
28283.8
Solution
(1.18T T
H
(
H i , 298ni
282838 0.2
T 298
( 28.77T
0.321T 2
0.499T
1
) T298
0.67 105 T
1
) 733 298
0
?
C p , r ,i ni )dT
C p , p ,i ni
(33.28 5.58 10 3 T
0.19 105 T 2 ) 0.9dT
T
28.77 4.38 10 3 T
0.67 105 T 2 dT
1250
28283.8 2 (28.77T (1.18T
0.321T 2
2.19 10 3 T 2
0.499T 1 ) T298
0.67 105 T 1 ) 1250 298
0
T (b) 1250 Q298
282838 0.2
1250 Q298
282838 0.2
1250 298 1250 298
(33.28 5.58 10 3 T
0.19 105 T 2 ) 0.9dT
(33.28 5.58 10 3 T
0.19 105 T 2 ) 0.9dT
1.10 (a) for the reaction CO
1 O2 2
CO2 ,what
is the enthalpy of
reaction ( H 0 ) at 298 K ? (b) a fuel gas, with composition 50% CO, 50% N2 is burned using the stoichiometric amount of air. What is the composition of the flue gas? (c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperature the flame may attain (adiabatic flame temperature)? DATA :standard heats of formation CO CO2
Hf
at 298 K
110000( J / mol) 393000( J / mol)
Heat capacities [J/(mol K)] to be used for this problem N2=33, O2=33, CO=34, CO2=57 0
(a) H 0
0
H f , 298, r ni
110000 393000
H f , 298, P ni
283000( J / mol )
0.5 22.2%, N 2 % 66.6%, O2 % 11.1% 1 0.25 / 0.2 22.2 product CO2 % 25%, N 2 % 75%, 100 11.1 (C )C p , P Ci , p , P X i 33 0.666 33 0.222 34 0.111 33( J / mol K ) (b) fuel CO %
Solution
Cr , P H
Ci , r , P X i H0
57 0.25 33 0.75 39( J / mol K )
n pC p , P dT
0
283000 0.222 (0.889 39)(T T 2110( K )
298)
0
1.11 a particular blast furnace gas has the following composition by (volume): N2=60%, H2=4, CO=12%, CO2=24% (a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature? (b) repeat the calculation for 30% excess combustion air at 298K
(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K) (d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will thE dlaMe temperature be affected? Air : n(O2 )
0.104
n( N 2 )
0.416
DaTA(k J?mol) FOR
H f (kJ / mol )
CO CO2
110.523
??
FOR CO
393.513
CO2
CP [ J / m o l K ] 33 57 50
H 2O ( g )
34
N 2 , O2
Solution 1 O2 2
(a)CO
1 O2 2
H2
CO2 H 2O
Fuel : n(CO )
0.12
n(CO2 )
0.24
n( H 2 )
0.04
n( N 2 )
0.6
H
H CO
CP ,r ,i ni T T
H H2
CO2
0.36CCO2
43.6308 10 53.8 1108.98( K )
Flue :
Air : n(O2 )
0.08
n( N 2 )
0.32
n(CO2 )
0.36
n ( H 2O )
0.04
n( N 2 )
0.92
0.12 (393.51 110.523) 0.04 (241.8 0)
H 2O
0.04CH 2O
0.92C N2
0.36 57 0.04 50 0.92 34
43.6308KJ 53.8( J / K )
3
810( K )
(b)repeat the calculation for 30% excess0combustion air at 298K H
H CO
CO2
HH2
H 2O
0.12 (393.51 110.523) 0.04 (241.8 0)
43.6308KJ CP , r ,i ni
0.36CCO2
0.04CH 2 O 1.016C N 2
0.024CO2
0.36 57 0.04 50 1.016 34 0.024 34 57.88( J / K ) T T
43.6308 103 57.88 1051.8( K )
753.8( K )
Fuel : n(CO )
0.12
n(CO2 )
0.24
n( H 2 )
0.04
n( N 2 )
0.6
Flue :
Air : n(O2 )
0.104
n( N 2 )
0.416
n(CO2 ) 0.36 n( H 2O) 0.04
Maret 8, 2013
n( N 2 ) 1.016 n(O2 ) 0.024
(C)what is the adiabatic flame temperature when the blasp furnace gas is preheated to 700K (the dry air is at 298K) Fuel : n(CO )
0.12
n(CO2 )
0.24
n( H 2 )
0.04
n( N 2 )
0.6
H
H CO
H H2
CO2
Flue :
Air : n(O2 )
0.08
n( N 2 )
0.32
H 2O
n(CO2 )
0.36
n ( H 2O )
0.04
n( N 2 )
0.92
H fuel 700 298
0.12 (393.51 110.523) 0.04 (241.8 0) (700 298) (0.12 33 0.24 57 0.04 28 0.6 34) 59.373KJ 0.36CCO2
CP ,r ,i ni
59.373 103 53.8 1401.6( K )
T T
0.04CH 2O
0.96C N2
0.36 57 0.04 50 0.92 34
53.8( J / K )
1103.6( K )
(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected? n(CO )
0.12
n(CO2 )
0.24
n( H 2 )
0.04
n( N 2 )
0.6
H
Flue :
Air :
Fuel :
H CO
n(O2 )
0.08
n(CO2 )
0.36
n( N 2 )
0.32
n( H 2 O )
0.048
n ( H 2O )
H H2
CO2
15 0.4 760 15
0.36CCO2
0.048CH 2O
0.36 57 0.048 50 0.92 34 T T
43.6308 10 54.2 1103( K )
n( N 2 )
0.92
H 2O
0.12 (393.51 110.523) 0.04 (241.8 0) CP ,r ,i ni
0.008
43.6308KJ
0.92C N2 54.2( J / K )
3
805( K )
1.12 A bath of molten copper is super cooled to 5℃ below its true
melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies? DATA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) CP,L=7.5(cal/mol℃), CP,S=5.41+(1.5*10-3T )(cal/mol℃) Solution H SL,1803
1803
1798
1798
1803
CP , S dT
CP , L dT
H LS,1798 0
H SL,1798 3100 5.41 5 1.5 10 3 (18032 17982 ) 0.5 7.5 5 3103(cal / mol)
1.13 Cuprous oxide (Cu2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu2O (b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction) DATA: heat of formation of 1000K in cal/mol Cu2O=-41900 H2O=-59210 solution
Cu2O H 2 H
Cu
H 2O
59210 41900 17310(cal / mol)
,exothermic reaction
1.14(a) what is the enthalpy of pure, liquid aluminum at 1000K? (b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the
minimum electric powder rating (kW) of furnace. DATA : For aluminum : atomic weight=27g/mol, Cp,s=26(J/molK), Cp,L=29(J/molK), Melting point=932K, Heat of fusion=10700J/mol Solution
H l ,1000
932
1000
298
932
CP , S dT
CP , L dT
H LS
26 (932 298) 29 (1000 932) 10700 27184( J / mol ) 27184 1 P 1000 279.7(W ) 0.28(kW ) 27 3600
1.15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected. If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, Cp,s,Al=26(J/molK),
Cp,s,
Al2O3=104J/mol,
heat
formation
of
Al2O3=-1676000J/mol Solution;
1 1676000 27 2 T 302( K ) T
1
16 1.5 27 27 102
99 104
T
600( K )
1.16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface,
where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute? DATA; for water: Cp=4.184J/g k, Density=1g/cm3; for copper: molecular weight=63.54g/mol Cp=7cal/mol k, heat of fusion=3120 cal/mol QCopper / min
Solution: Q
Water
V
/ min
[3120 7 (1356 400)]
1000 63.54 min
1(80 20)V
2.573 10 3 (m3 / min)
1.17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DATA; For water Cp=4.184J/g k, Density=1g/cm3 Solution: W 4.184 (60 20) 5 1000 13947(W ) 60
1.18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor? (b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m3 what is the internal energy change for the evaporation of water?
P V
Solution:
101325
3101 2261 18 U W Q
%
373 0.0224 3101( J / mol ) 273 7.6% 3101 2261 18 37597( J / mol )
1.19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 Solution m(2261 4.18 100) 1000 334, m 125( g ) 1.20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃ is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃ is 1.0021 Mpa( about 10 atm) Data: CP,L=4.18J/(g k),
CP,v=2.00J/(g k),
Solution:
△HV=2261J/g, △Hm=334 J/g
(2261 2 80 4.18 80) x irreversible
4.18 80 (1000 x), x 138( g )
The problems of the second law 2.1 The solar energy flux is about 4J cm2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and
a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency. Solution
W
TH
TL TH
(90 90
QH
P
W t
S
6.25(m 2 )
25) 273
4
104 St 60
746(W )
2.2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine. W
Solution:
TH TL QH TH
0.75P PL
3 4
TH
TL TL
PL
273 20 0.25 746 35 20
643(W )
2.3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is
taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value. P
Solution:
PL M
TH
TL TL
PL
273 0.25 746 334m 20 60m 457( g )
2.4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what size motor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible. W
Solution:
50%
0.5P' P
TH
TL TL
TH
P' 393(W )
QL
TL TL
PL
300 4.216 84 2 4.216 60
0.52(hp)
2.5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical power to run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler. (a) assume that the efficiencies are equal to the theoretical maximum values
(b)assume the power plant efficiency is 70% of maximum and that coefficient of performance of the heat pump is 10% of maximum (c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be more economical in terms of overall fissile fuel consumption to use a heat pump or
a furnace ? do the
calculations for cases a and b solution:
(a) P1
TH ,1 TL ,1 PH ,1 TH ,1
P2
TH , 2 TL , 2 PH , 2 TH , 2
P1
P2
540 50 25 5 PH ,1 PH , 2 540 273 273 25 PH , 2 8.98 PH ,1
(b) PH , 2
0.6286 PH ,1
(c)a is ok , b is not. 2.6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabatic enclosure. Assume that Cp is 77J /(mol K) from 273K to 373K
Solution: U S
0( J ) n1CP ln(
TE ) T1
n2CP ln(
TE ) T2
0.5CP ln(
323 ) 373
0.5CP ln(
323 ) 273
0.933( J / K )
2.7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃ and a condensate temperature of 30℃. (a) what is the maximum electrical work that can be produced by the plant per joule of heat provided to the boiler? (b)How many metric tons (1000kg) of coal per hour is required if the plant out put is to be 500MW (megawatts). Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g (c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ by passing a current through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied? (a)
Solution: W
TH TL QH TH
540 30 1 0.89( J ) 540 30
(c ) W
(b) 29.0m m
TL TH
TL
69371(kg)
500 3600
QH
TH TL QH TH 273 25 1 19.9( J ) 25 10
69.3(ton )
2.8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt (a) calculate the rate of evaporation of the water in grams per second if the water container is insulated that is no heat is allowed to
flow to or from the water except for that provided by the resistor (b)at what rate could water could be evaporated if electrical energy were supplied at the rate of 1 kw to a heat pump operating between 25 and 100℃ data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm3 m
solution: (a) 18 40000 (b)
1000, m
0.45( g )
m 100 273 40000 1000 ,m 18 100 25
2.23( g )
2.9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ by immersing them in a brine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minus power rating in kilowatts, of motor required to operate the refrigerator? Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol Solution: P
L
PW
1000 28 (480 20) 27 TH TL 30 20 PL PL TL 273 20
102474(W ) 102.5(kW )
2.10 an electric power generating plant has a rated output of 100MW. The boiler of the plant operates at 300℃. The condenser operates at 40℃
(a) at what rate (joules per hour) must heat be supplied to the boiler? (b)The condenser is cooled by water, which may under go a temperature rise of no more than 10℃. What volume of cooling water in cubic meters per hour, is require to operate the plant? (c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electric output will remain the same. Will the cooling water requirement be increased, decreased, or remain the same? Data heat capacity 4.184, density 1g/cm3 Solution:
(a) P H
TH TH
300 273 8 10 300 40
2.2 108 (W ) QH
(c ) P H
TH P TH TL
TL
P
PH t
7.9 1011( J )
(b) QL
4.3 1011( J )
V 106 10 4.184 QL V
1.03 104 (m3 )
540 273 8 10 540 40
8
1.626 10 (W ) no
2.11 (a) Heat engines convert heat that is available at different temperature to work. They have been several proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular
region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? Solution:
20 4 1 0.055( J ) 20 273
(a)W
TH TL QH TH
(b) P
mg h 3600 1.06 106 (kW / h) 1000
2.12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20 ℃ , (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine? Solution:
TH
(a) P
TL TL
(b)QH
PL
20 15 5 10 / 3600 3.77(kW ) 273 15
273 20 5 10 273 15
1.14 105 (kJ )
2.13 solution:
(a)2 Al
N2
(b) H
152940(cal / mol)
2 AlN
(c ) S
4.82 2 6.77 2 45.77
49.67(cal / molK )
(d ) H 152940(cal / mol ) S 4.82 2 6.77 2 45.77 8.314 ln 10
68.81(cal / molK )
2.14 m(
S
solution:
0 10
CP , ICE dT T
273 336 263 273 22574( J / K )
(2.1ln
40
Hm Tm 4.184 ln
0
CP ,WATER dT ) T
273 40 )12000 273
2.15 W
TH
TL
W2
70428( J )
TL
300 77 1000 77
QL
2896( J )
2.16 W W
TH
TL
300 4.2 83.3 5866.7( J ) 4.2
QL
TL
300 4.2 (83.3 1.5 8.314(300 4.2)) 3719.4( J ) 300
TH TL QH TH
2.17 (a) T 0 U O Q
W
(b) S (c)Q
n nR ln
pdV P0 P
1 8.314 298 ln 10 5704( J )
1 8.314 ln 10 19.1( J / K )
0
(d ) yes
2.18 500 60
TH
TL TL
335m
20 0 335m 273
m 1222( g )
Property Relations 1. At -5 C, the vapor pressure of ice is 3.012mmHg and that of supercooled liquid water is 3.163mmHg. The latent heat of fusion of ice is 5.85kJ/mol at -5 C. Calculate G and S per mole for the transition of from water to ice at -5 C. (3.2, 94)
RT ln
G
PH 2O ,ice PH 2O , water
3.012 3.163 8.314 268 ln 0.9523 108.9 J / mol 8.314 (273 5) ln
Solution:
5.85 103 J / mol
H G S
H
T S
H
G T
5850 ( 108.9) 268
22.23J /( mol K )
2. (1) A container of liquid lead is to be used as a calorimeter to determine the heat of mixing of two metals, A and B. It has been determined by experiment that the “heat capacity” of the bath is 100cal/ C at 300 C. With the bath originally at 300 C, the following experiments are performed;(2) A mechanical mixture of 1g of A and 1g of B is dropped into the calorimeter. A and B were originally at 25 C. When the two have dissolved, the temperature of the bath is found to have increased 0.20 C. 2. Two grams of a 50:50(wt.%) A-B alloy at 25 C is dropped similarly into the calorimeter. The temperature decreases 0.40 C. (a) What is the heat of mixing of the 50 50 A-B alloy (per gram of alloy)? (b) To what temperature does it apply ? (3.5, 94) Solution: (a)
Q
C P,bath
100cal / K
418J / mol
C P,bath T / 2 100 0.2 / 2 10cal / g
This is the heat of mixing.
(b) The heat capacity of CP, alloy :
C P ,alloy
C P ,bath
T
2 (300 0.4 25) 100 0.4 0.072cal /( g K ) 2 274.6
Assuming that the calorimeter can be applied to the maximum of T C, the for mixing to form 1 gram of alloy: Q1
,
C P,bath (300 T ' ) 10
C P,bath (300 T ' ) 10
Q2
C P,alloy (T T ' ) ,
Q1
Q2
C P,alloy (T T ' )
3. The equilibrium freezing point of water is 0 C. At that temperature the latent heat of fusion of ice (the heat required to melt the ice) is 606 3J/mol. (a) What is the entropy of fusion of ice at 0 C ? (b) What is the change of Gibbs free energy for ice
water at
0 C?(c) What is the heat of fusion of ice at -5 C ? CP(ice) = 0.5 cal/(g. C); CP(water) = 1.0 cal/(g. C). (d) Repeat parts a and b at -5 C. (3.6, p94) Solution: (a) At 0 C, G =0, S
H Tm
6030 273
Tm S = H
22.09 J /( mol.K )
(b) At 0 C, G =0 ©
CP,ice 0.5cal /( g.K ) 0.5 4.18 18J /( mol.K ) 37.62 J /( mol.K )
C P,water
1.0cal /( g.K ) 1.0 4.18 18J /( mol.K
a reversible process can be designed as follows to do the calculation:
（2） Ice, 0 C
water, 0 C
（1）
（3）
ice, -5 C
water, -5 C （4）
H
fu
H (1)
H ( 2)
273
C p ,ice dT
268 273 268
(C p ,ice
(37.62
H
268 273
C p , water dT
C P , water )dT
75.24) 5
5841 .9 J / mol
（d）
H ( 3)
6030
H
S ( 4)
S (1)
3 ( 3)
S ( 2)
C p ,ice
273 268
dT
T 273(C p ,ice
268
C p , water
268
S
273
C P , water ) T
(37.62 75.24) ln
dT
273 268
T
dT
S 22.09
21.39 J /( mol.K )
G ( 4)
H ( 4)
T S ( 4)
5841.9 268 21.39 109.38
4. (a) What is the specific volume of iron at 298K, in cubic peter per mole? (b) Derive an equation for the change of entropy with pressure at constant temperature for a solid, expressed in terms of physical quantities usually available, such as the ones listed as data; (c) The specific entropy of iron (entropy per mole )at 298K and a pressure of 100 atm is needed for a thermodynamic calculation. The tabulated “standard entropy”(at 298 K and a pressure of 1 atm) is o S298 27.28J / K .mol .
What percentage error would result if one assumed
that the specific entropy at 298K and 100 atm were equal to the value of
o S 298
given above ？
DATA： （for iron）
Cp = 24 J K-1mol-1 Compressibility = 6
Linear coefficient of thermal expansion = 15
10-7 atm –1 10-6 C-1
Density = 7.87 g/cm3 Molecular weight = 55.85g/mol Note: It may be possible to solve this problem with out using all the
data given. (3.7, 95) Solution: (a)
(b)
55.85 g / mol 7.87 g / cm 3
mol weight density
V iron
S P S P
V T
T
7.10cm 3 / mol
7.10 10 6 m 3 / mol
P
V
3V
V
l
T
for iron: S P
3V iron
l ,iron
T
3 7.10 10 3.2 10
S iron
10
6
15 10
6
(m 3 /( mol.K ))
3.2 10
10
P
(c) S iron
3.2 10
10
3.2 10
10
320.9 10
error%
S iron 100% o S 298
(100 1) 1.013 10 99 1.013 10 5
5
5
3.21 10 3 ( J / mol.K )
1.12 10
2
Equilibrium 1. At 400 C, liquid zinc has a vapor pressure of 10-4 atm. Estimate the boiling temperature zinc, knowing that its heat of evaporation is approximately 28 kcal/mol. (4.2, P116) Solution: (a)
18 g / mol 0.92 g / cm 3
V ice
V water
V fus
18 g / mol 1g / cm 3
19.57cm 3 / mol 19.57 10 6 m 3 / mol
18cm 3 / mol 18 10 6 m 3 / mol
1.57 10 6 m 3 / mol
According to the Clapeyron equation:
dP dT
1 V fus
dP
1 V fus
H
fus
T H T
fus
dT
take definite integration of the above:
50 1.013 105 5
1.013 10
ln
T 273
H fus
dP
V
fus
H
fus
V
fus
1 dT 273 T T
49 1.013 10 5
1.57 10
6
49 1.013 10 5 6009
0.013
272.8K
T
(b)
P
150 0.01 3
P
50 10 3 lb / in.2
50 10 3 6897 Pa
345 10 6 Pa
345 10 6 Pa
(c ) ln
T 273
V
fus
H
fus
345 10 6
1.57 10 6 345 10 6 6009 0.09
T
249.46K
1. At 400 C, liquid zinc has a vapor pressure of 10-4 atm. Estimate the boiling temperature of zinc, knowing that its heat of vaporation is approximately 28kcal/mol. (4.3,117) Solution:
H vap
4.18 28 103 J / mol 117.04 J / mol
28kcal / mol
According to Claperon equation in vapor equilibrium: d (ln P ) P2 P1
d (ln P )
H vap R
1 d( ) T
H vap 1 ( R T2
1 ) T1
H vap 1 ( R T2
ln
P1 P2
ln
1 10 4
T2
1 ) T1
117.04 10 3 1 ( 8.314 T2
1 ) 673
1202K
The boiling point of zinc is 1202K. 2. Trouton’s rule is expressed as follows:
H vap
90Tb
in joules per
mole, where Tb is the boiling point (K). The boiling temperature of mercury is 630K. Estimate the partial pressure of liquid Hg at 298K. Use Trouton’s rule to estimate the heat of vaporization of mercury. Solution:
H vap P
1
90Tb H vap
d (ln P)
R
(
1 298
6823 10.83 298
ln P
1 ) 630
22.90 10.83
12.07
P=5.73 10-6 atm 3. Liquid water under an air pressure of 1 atm at 25 C has a large vapor pressure that it would have in the absence of air pressure. Calculate the increase in vapor pressure produced by the pressure of the atmosphere on the water. Water has a density of 1g/cm3; the vapor pressure ( in the absence of the air pressure) is 3167.2Pa. (4.5, p116) Solution:
Vl
vapor pressure,
Gv
18 g / mol 1g / cm 3
18cm 3 / mol 18 10 6 m 3 / mol
pressure Gl
changes
with
the
total
external
RT ln
RT ln
Pe, 2 Pe,1 Pe, 2 Pe,1
Pe, 2 Pe,1
V l ( PT
Pe,1 )
18 10 6 (10130 3167.2)
1.000051
Pe, 2
P = 0.16Pa
3167.36Pa
the vapor pressure increase is 0.16Pa. 4. The boiling point of silver (P=1 atm) is 2450K. The enthalpy of evaporation of liquid silver is 255,000 J/mol at its boiling point. Assume, for the purpose of this problem, that the heat capacities of liquid and vapor are the same. (a) Write an equation for the vapor pressure of silver, in atmospheres, as a function of kelvin temperature. (b). The equation should be suitable for use in a tabulation, NOT in differential form. Put numerical values in the equation based on the data given. (4.7, p117) Solution:
P 1
d (ln P)
ln P
H vap 1 ( R T
1 ) 2450
ln P
255000 1 ( 8.314 T
1 ) 2450
30685 104.08 T
6. Zinc may exist as a solid, a liquid, or a vapor. The equilibrium pressure-temperature relationship between solid zinc and zinc vapor is giben by the vapor pressure equation for the solid. A similar relation exists for liquid zinc. At the triple point all three phases, solid, liquid, and vapor exist in equilibrium. That means that the vapor pressure of the liquid and the solid are the same. The vapor pressure
of
solid
Zn
varies
with
T
as:
ln P(atm)
15755 T
0.755 ln(T ) 19.25
and the vapor pressure of liquid
15246 T
Zn varies with T as: ln P(atm)
1.255 ln(T )
21.79 .
Calculate: (a)
The boiling point of Zn under 1 atm; (b) The triple-point temperature; (c) the heat of evaporation of Zn at the normal (1 atm) boiling point; (d) The heat of fusion of Zn at the triple-point temperature; (e)The differences between the heat capacities of solid and liquid Zn. (4.8, p118) Solution :(a) At boiling point, P=1 atm, that is lnP=0 15246 1.255 ln(T ) 21.79 T
0
15246 1.255T ln(T ) 21.79T
0
To solve the equation, let y1 = 21.79T-15246, y2 = 1.255TlnT. Plot
y
y-T of the functions, and the intersection is the answer.
30000 28000 26000 24000 22000 20000 18000 16000 14000 12000 10000 8000 6000 4000 2000 0 -2000 -4000 -6000
y1 y2
400
600
800
1000
1200
1400
1600
1800
2000
2200
T, K
From the plot, the intersection is 1180 K. So at 1180K, zinc boils. (b) At triple point, vapor pressure of solid Zn equals that of liquid Zn;
15246 1.255 ln(T ) 21.79 T
509 0.5T ln(T ) 2.54T
15755 0.755 ln(T ) 19.25 T
509 0.5 ln(T ) 2.54 T
0
0
To solve the equation, assume two functions, y1=2.54T+509; y2 = 0.5TlnT. Plot y1-T and y2-T. The intersection is the answer. y11 y22
8000 7500 7000 6500 6000 5500
y
5000 4500 4000 3500 3000 2500 2000 1500 1000 400
600
800
1000
1200
1400
1600
1800
2000
2200
T, K
The intersection is 695.3K, and this is the triple point temperature. (c ) d (ln P)
15246 1.255 T T2
21.79
1 ( 15246 1.255T )( 2 )dT T 1 ( 15246 1.255T )d ( ) T
H vap R
H vap
d (ln P)
R
1 d( ) T
15246 1.255T
R(15246 1.255T ) 126755 10.43T
At Tb =1180K,
H vap
126755 10.43T
114.4 103 J / mol 114.4kJ / mol
(d) For solids, 15755 0.755 ln(T ) 19.25 T 15755 0.755 d (ln P) ( 2 )dT T T 1 ( 15755 0.755)( 2 )dT T 1 ( 15755 0.755)d ( ) T
ln P
d (ln P)
H vap
H fus R
1 d( ) T
H fus
H fus
15755 0.755T
R
R(15755 0.755T )
At triple point, Ttr = 695.4K H fus
(e)
d ( H fus ) CP
8.314 (15755 0.755T ) 126.6 103 J / mol 126.6kJ / mol C P dT
d ( H fus )
0.755
dT
7. A particular material has a latent heat of vaporization of 5000J/mol. This heat of vaporization does not change with temperature or pressure. One mole of the material exists in a two-phase equilibrium (liquid-vapor) in a container of volume V=1L, a temperature of 300K, and pressure of 1 atm. The container (constant volume ) is heated until the pressure reaches 2 atm. (Note that this is not a small
P.) The vapor phase can be treated as an
ideal monatomic gas and the molar volume of the liquid can be neglected relative to that of the gas. Find the fraction of material in the vapor phase in the initial and final states. (4.9, P118) Solution: In the initial state, P1V1
n1 RT ,
n1
mol(vapor)%
P1V1 RT1
T1
300K , P1
10130 10 3 8.314 300
4.06 10 3 mol
4.06%
In the final state, P2=2 atm, V2 = 1L According to Clayperon equation: RT ln 2 ln 1 T2
H vap
P2 P1
R 5000 1 ( 8.314 T2
458.6 K
1 T2 1 ) 300
1 T2
1atm 1.013 105 Pa, V1
1L
P2V2
n2 RT2 ,
2 10130 10 3 8.314 458.6
P2V2 RT2
n2
mol(vapor)%
5.3 10 3 mol
5.3%
8. The melting point of gold is 1336K, and vapor pressure of liquid gold is given by: ln P(atm)
43522 T
23.716
1.222 ln T ( K )
. (a) Calculate the heat of
vaporization of gold at its melting point; Answer parts b, c, and d only if the data given in this problem statement are sufficient to support the calculation. If there are not enough data, write “solution not possible.” (b) What is the vapor pressure of solid gold at its melting point? (c) What is the vapor pressure of solid gold at 1200K ? (d) What is the v Solution: (a) d (ln P)
(
43522 T2
1.222 )dT T
1 )dT T2 1 ( 43522 1.222)d ( ) T ( 43522 1.222)(
H vap
R( 43522 1.222T )
(a) At 1336K,
H vap
361841 10.19T
361841 10.19 1336
348 103 J
348kJ
(b)“Solution not possible”; (c) “Solution not possible”.
9. (a) At 298K, what is the Gibbs free energy change for the following reaction? Cg
r
a
p
h
C i d t
e i a
m o
n
d
(b) Is the diamond thermodynamically stable relative to graphite at 298K? (c ) What is the change of Gibbs free energy of diamond when it is compressed isothermally from 1atm to 1000 atm? (d) Assuming
that
graphite
and
diamond
are
incompressible, calculate the pressure at which the two exist at equilibrium at 298K. (e) What is the Gibbs free energy of diamond relative to graphite at 900K? to simplify the calculation, assume that the heat capacities of the two materials are equivalent. Density of graphite is 2.25g/cm3
DATA
Density of diamond is 3.51g/cm3 H of ( 298) (kJ / mol) H of ( 298) ( J / mol.K )
Diamond
1.897
2.38
Graphite
0
5.74
Solution: (a)
H
C graphite
H of ,diamond
C diamond
H of , graphite 1897kJ / mol
S
S of ,diamond
G
H T S
5.74 2.38
S of , graphite
1897 298 ( 3.36)
3.36 J /( mol.K )
2898.28J / mol
(b) No, diamond is not thermodynamically stable relative to graphite at 298K. (c )
12 10 3.51
G V diamand P
6
99 10130
34.29 J / mol
(c ) Assuming N atm , G = 0, reversible processes as following can be designed to realize this, （4） graphite, 298K, N atm
diamond, 298K, N atm
(3)
（1） （2）
diamond, 298K,1atm
graphite, 298K,1atm
G( 4 )＝ G(1)＋ G( 2 )＋ G(3) ＝V graphite P 2898.28 V diamond (
P)
(V graphite V diamond )(1 N ) 10130 2898.28 N
12 10 2.25
0.194 N N
Cp
6
6
12 10 3.51
2898.28
10130 2898.28
0
14939 (atm )
0,
H 900
T' T
C p dT
0,
T'
Cp
T
T
dT
0
H 298 1897 J / mol
S900
3.36J / mol.K
S 298
H T S 1897 900 3.36
G
Chemical Equilibrium 1. Calculate the partial pressure of monatomic hydrogen in hydrogen in hydrogen gas at 2000K and 1atm. 1 H 2 (g) 2
For
1 C P,H 2 2
C P,H ( g )
o H 2000
3 8.314 2
2000
o H 298
217990 J
o S 298
49.35 J / K
1 31 2
H (g) 3.035
CP (2000 298)
o H 298
CP dT
298
o H 298
1 H 2 (g) 2
For the reaction : Cp
H (g)
217990 3.035 1702 212824 J o H 2000
2000
o H 298
217990 3.035 1702 o S 2000
2000
o S 298
298
49.35 3.035 ln 0 G2000
RT ln
PH 2
PH ( g )
1 PH ( g )
125684
PH 2( g ) 7.562
PH ( g )
RT ln
1/ 2
PH 2 ( g ) ln
0 0 H 2000 T S 2000
RT ln K
o G2000
2000 298
ln
PH 2 PH 2
1, PH 2 PH
1/ 2
1/ 2
PH ( g )
212824 J C P ln
o S 298
C P dT
C P (2000 298)
o H 298
C P dT
298
2000 298
43.57 J
212824 2000 43.57 125684 J
RT ln
PH 2
1/ 2
PH ( g )
125684 8.314 2000
1 0.0005atm
7.56
4921J / m
2. For the reaction : Go
Co( s)
59850 19.6T ,
1 O2 ( g ) 2
where
CoO( S )
G o is
in calories and T is in Kelvin.
(a) Calculate the oxygen equilibrium pressure (atm) over Co and CoO at 1000 C. (b) What is the uncertainty in the value calculated in Ho term is estimated to be 500 cal? (5.2,
part a if the error in p144) Solution:
(a)
At
1000 C,
=-59850+19.6T=-59850+19.6 (1000+273) =-34899.2cal = -1458.79J/mol At equilibrium: Go ln PO2 PO2
RT ln K
RT ln
1 PO2
1/ 2
1 RT ln PO2 2
145879 J
27.6 1.07 10
12
atm
(b) uncertainty in Ho = 500cal/mol = 2090J/mol So uncertainty in Go = 500cal/mol = 2090J/mol That means: 1 RT ln PO' 2 2 P' 1 RT ln O 2 2 PO 2 ln
PO' 2 PO 2
PO' 2 PO 2
1 RT ln PO 2 2
2090
2090
0.25 1.286
P PO 2
28.6%
Similarly, uncertainty in Ho =- 500cal/mol =- 2090J/mol Go = -2090J/mol
Go
1 1 RT ln PO' 2 RT ln PO 2 2 2 P' ln O 2 0.25 PO 2 PO' 2 PO 2
P
0.779
2090
22.1%
PO 2
3. Calculate the temperature at which silver oxide (Ag 2O) begins to decompose into silver and oxygen upon heating: (a) in pure oxygen at P = 1 atm; (b) in air at Ptotal = 1 atm. DATA
7300 cal / mol
H f forAg 2 O
Standard Entropy at 298K
Assume
[cal/(mol.K)] Ag2O 29.1 49.0 that O2Cp = 0 for the Ag 10.2
Solution: (a)
Ag2O = 1/2O2 + 2Ag
Ho So
7300cal / mol
H of , AgO 2 S Ag , 298
Ho
1 S O 2, 298 2
30514
S Ag 2O , 298
1 49 29.1 66.044 J / mol.K 2
2 10.2
Go
decomposition reaction.
T So
30514 T S o
30514 66.044T
when Ag2O begins to decompose, G
Go
RT ln J
ie 30514 66.044T
(a) in pure oxygen at 1 atm,
0 RT ln PO 2
0
RTlnPO2 = 0 30514-66.044T = 0 T = 462K
(b)in air at Ptotal = =1 atm , PO2 =0.21
ie.
30514- 66.044T + RTln0.21 = 0 T = 386K
4. One step in the manufacture of specially purified nitrogen is the removal of small amounts of residual oxygen by passing the gas over copper gauze at approximately 500 C. The following reaction takes place:
2Cu ( s)
1 O2 ( g ) 2
Cu 2 O( s)
(a) Assuming that equilibrium is reached in this process, calculate the amount of oxygen present in the purified nitrogen; (b) What would be the effect of raising the temperature to 800 C? Or lowering it to 300 C? What is the reason for using 500 C? (c) What would be the effect of increasing the gas pressure? For
2Cu ( s)
1 O2 ( g ) 2
Cu 2 O( s) ,
Go (in calories ) is –39850+15.06T.
(5.4, p145) Solution: (a) When the equilibrium is reached, Go
G
RT ln J
1 G o－ RT ln PO2 2
0
4.18 ( 39850 15.06T ) 1 RT 2
ln PO 2
T = 500 C = 773K ln PO 2 PO 2
4.18 ( 39850 15.06 773) 1 8.314 773 2 1.14 10 26 atm
36.69
(b) at T=300 C=573K, Although the equilibrium PO2 is very low, kinetically the reaction is not favoured and reaction speed is very slow. So 300 C is not
suitable at At T=800 C=1073K, lnPO2 =-22.2, PO2 =2.28 10-10 atm. At 800 C, if the equilibrium is reached, nitrogen can be of high purity level. However, at this high temperature , particles of Cu will weld together to reduce effective work surface. So it is not suitable to use this high temperature in purification either. (c ) The equilibrium oxygen pressure remains the same when the total pressure increases, which means a higher purity level of N2 . 5. The solubility of hydrogen(PH2 = 1 atm ) in liquid copper at 1200 C 7.34cm3(STP) per 100g of copper. Hydrogen in copper exists in monatomic form. (a) Write the chemical equation for the dissolution of H2 in copper; (b) What level of vacuum(atm) must be drown over a copper melt at 1200 C to reduce its hydrogen content to 0.1 cm3 (STP) per 100g?
(c) A 100g melt of copper at 1200 C
contains 0.5 cm3(STP) of H2. Argon is bubbled through the melt slowly so that each bubble equilibrates with the melt. How much argon must be bubbled through the melt to reduce the H2 content to 0.1 cm3(STP) per 100g ? Note: STP means standard temperature and pressure(298K and 1 atm). (5.5, p145) Solution: (a) H2(g) = 2H (b)
H
1
PH 2 1
Ka 2
1
K a 2 PH 22 1atm, H
7.34cm 3 / 100 gCu
is a constant,
PH1 /22 [ H ]' [H ]
[H ] PH1 /22
[ H ]' ( PH' 2 )1 / 2
( PH' 2 )
0.00019atm 18.56 10130 18.8Pa
( PH' 2 )1 / 2
0.1 7.34
0.0136
(c ) The amount of H2 needed to be brought out by Ar is: P V RT
n
10130 (0.5 0.1) 10 8.314 298
6
1.6 10 6 mol
This amount of H2 is in equilibrium with the melt in the bubble, ie. The partial pressure of H2 in the bubbles is 18.8Pa. PH' 2Vbubble PH' 2
nRT
4.05 10
4.05 10 2 / 18.8
2
0.00215m 3
2.15L
2.15L Ar is needed to be bubbled into the melt.
6. The following equilibrium data have been determined for the reaction:
NiO(s) CO( g ) Ni(s) CO2 ( g ) T( C) 663 716 754 793 852
K 10-3 4.535 3.323 2.554 2.037 1.577
(a) (b) (c) (d)Plot the data using appropriate axes and find Ho, K and Go at 1000K; (e) Will an atmosphere of 15%CO2, 5%CO, and 80%N2 N2 oxidize
nickel at 1000K? (5.6, p145) Solution: (a)
Ho 1 d( ) R T
d ln K a
Plot ln K a ~ 1 / T 8.6
Kduishu Linear Fit of Data1_Kduishu
8.4 8.2
lnKa
8.0 7.8
lnKa =2.01+6003(1/T)
7.6 7.4 7.2 0.88
0.90
0.92
0.94
0.96
0.98
1/T, 10
1.00
1.02
1.04
1.06
1.08
-3
. d ln K a dT
Ho R
6003
Ho
R 6003
49909 J
At T=1000K, lnKa =8.01, Ka = 3010 o G1000
(b)
G
RT ln K a Go J
8.314 1000 8.01 66600 J
RT ln J 15% 5%
3
RT ln K a
RT ln J
RT ln
66.6kJ
J Ka
Ka
So the atmosphere will oxidize Ni. 7. At 1 atm pressure and 1750 C, 100 g of iron dissolve 35cm3 (STP) of nitrogen. Under the same conditions, 100 g of iron dissolves 35 cm3 of hydrogen. Argon is insoluble in molten iron. How much gas will 100 g of iron dissolve at 1750 C and 760 mm pressure under an atmosphere that consists of: (a) 50% nitrogen and 50% hydrogen? (b) 50% argon and 50% hydrogen? (c) 33% nitrogen, 33 hydrogen, and 34 argon? (5.7, p145)
Solution:
N2 =2N, H2 = 2H N
1
1
K a ,2N PN 22 [N ] PN1 /22
For N2 dissolving :
[H ]
For H2 dissolving： PH1 /22
， H
1
1
K a ,2H PH 22 ,
[ N ]' ( PN' 2 )1 / 2
[ H ]' ( PH' 2 )1 / 2
（a）For dissolving N2, PN2 = 1 atm, [N]=35cm3/100g melt, [N‘ ]
( PN' 2 )1 / 2 [ N ] PN1 /22
35 (0.5)1 / 2
24.75cm 3 / 100 g melt
similarly: [H]’ =24.75cm3/100g melt total gas : [H] ＋[N] = 49.5 cm3/100g melt (b)
[H] =24.75 cm3/100g melt
(c ) [H] ＋[N] = [N](0.33)1/2 /1+[H](0.33)1/2 /1=20.10+20.10 = 40.2cm3/100g melt 8. Solid silicon in contact with solid silicon dioxide is to be heated to a temperature of 1100 K in a vaccum furnace. The two solid phases are not soluble in each other, but is known that silicon and silicon dioxide can react to form gaseous silicon monoxide. For the reaction: Si(s) SiO2
the
2SiO( g )
Gibbs
is
free
Go
667000 25.0T ln T 510T
energy
change
.
(J)
(a) Calculate the
equilibrium pressure of SiO gas at 1100K; (b) For the reaction above, calculate
Ho and
So at 1100K; (c) Using the Ellingham chart
(Figure 5.7), estimate the pressure of oxygen (O2) in equilibrium
with the materials in the furnace. (5.8, P146) (a)
2 RT ln PSiO
RT ln K
Go
2RT ln PSiO
At 1100K, Go =667000+25.0TlnT-510T = 667000+25.0 1100ln1100-510 1100 =667000+192584-561000 =298584 -2RTlnPSiO =298584 lnPSiO =-16.32 PSiO = 8.1 10-8 (atm) (b ) Go =667000+25.0TlnT-510T =-RTlnK 667000 25 ln T 510 RT R 667000 25 667000 d ln K （ )dT ( RT R RT 2 Ho 25 667000 T R R H o 667000 25T
ln K
25 T )d (1 / T ) R
Ho d (1 / T ) R
T = 1100K, Ho = 639500J So
Ho
Go T
667000 298584 1100
334.9 J / K
(c ) PO2 =10-30 atm 9. What is the pressure of uranium (gas) in equilibrium with uranium dicarbide Go
DATA: At 2263K,f Vapor (5.9,p146)
pressure
for UC2 is –82,000 cal/mol of
pure
uranium
is:
ln P(atm, uranium ) 25.33
Solution:
82000cal / mol
G of
U(g) RT ln K
G of ,UC 2 RT ln Pu ( g ) ln Pu ( g )
C( s ) RT ln(
100000 (T in K ) T
342760 J
UC2( s ) 1 ) Pu ( g )
RT ln Pu ( g )
342760
1.2 10 8 (atm )
vapor pressure of uranium: ln Pu ( g ) (atm, uranium ) 25.33 Pu ( g )
100000 100000 25.35 18.89 T 2263
0.6 10 8 (atm)
the vapor pressure is lower than the one determined by chemical reaction. It is the one in equilibrium with dicarbide.
10. The direct reduction of iron oxide by hydrogen maybe represented by the following equation: Fe2 O3
3H 2
the
reaction?
2 Fe H2
3 O2 2 1 O2 2
2Fe 3H 2 O
Fe 2 O3 H 2O
Is Go Go
What is the enthalpy change, in joules, for it 810250
exothermic 254.0T
246000 54.8T
(5.10)
or
endothermic?
Solution: 2 Fe H2
3 O2 2 1 O2 2
Fe2 O3 G3o
3 G2o
G1o
Fe 2 O3
(1) (2)
H 2O
3H 2
G1o G 2o
810250 254.0T 246000 54.8T
2Fe 3H 2 O
3 24600 54.8T
(3)
G3o
( 810250 254.0T )
72250 89.6T H 3o
72250 J
The reaction is an endothermic one. 11.Calcium carbonate decomposes into calcium oxide and carbon dioxide according to the reaction
CaCO3
CaO CO2
DATA for the pressure of carbon dioxide in equilibrium with CaO and CaCO3: Temperature (K) 1030 921
Pressure (atm) 0.10 0.01
(a) What is the heat effect ( H) of the decomposition of one mole of CaCO3 ? Is the reaction endothermic or exothermic? (b) At what temperature will the equilibrium pressure of CO2 equal one atmosphere? (5.11, P146) Solution: (a)
Ho 1 d , K R T
d ln K
Ho 1 d R T
d ln PCO 2 ln ln
PCO 2, 2 PCO 2,1 0.01 0 .1
Ho
PCO2
Ho 1 R T2
1 T1
Ho 1 1 R 921 1030 166528 J
the reaction is endothermic (b)PCO2 =1atm ln
T
1 0.1
Ho 1 ( R T
1 ) 1030
1168K
At 1168K, the equilibrium pressure of CO2 equals one atmosphere. 12. In the carbothermic reduction of magnesium oxide, briquettes of MgO and and carbon are heated at high temperature in a vacuum furnace to form magnesium (gas) and carbon monoxide(gas). (a) write the chemical reaction for the process; (b) What can you say abou the relationship between the pressure of magnesium gas and the pressure of carbon monoxide? (c) Calculate the temperature at which the sum of the pressures of Mg(gas) and CO reaches on atmosphere. With T in Kelvin, the free energies of formation, in calories, of the relevant compounds are:
MgO CO
174000 48.7T
G of G
28000 20.2T
o f
(a). The reaction is: MgO(s) C(s) (b).
Go G
o
G of ,CO RT ln K
G of , MgO
CO( g ) Mg ( g )
146000 68.9T
RT ln( PCO( g ) PMg ( g ) )
(146000 68.7T )
PCO
PMg
(c ) Ptotal = 1 atm, PCO = 0.5 atm, PMg =0.5 atm RT ln(0.5 0.5)
(146000 68.7T ) 4.18
T = 2037 K 13. Metallic silicon is to be heated to 1000 C. To prevent the formation of silicon dioxide (SiO2), it is proposed that a hydrogen atmosphere be used. Water vapor, which is present as an impurity in the hydrogen, can oxidize the silicon. (a) Write the chemical equation for the oxidation of silicon to dioxide by water vapor; (b)Using the accompanying data, where Go is in joules, determine the equilibrium constant fro the reaction at 1000 C (1273K); (c) What is the maximum content of water in the hydrogen (ppm) that is permitted if the oxidation at 1000 C is to be prevented ? (d) Check the answer to part c on the Ellingham diagram (Figure 5.7) DATA
H 2 (g) Si
O2
1 O2 ( g ) 2 SiO 2 ( s)
H 2 O( g )
Go Go
246000 54.8T 902000 174T
5.13, P147 Solution: (a)
Si(s) 2H 2 O( g )
SiO2 (s) 2H 2 ( g )
(b) H 2 (g) Si
O2
1 O2 ( g ) 2 SiO 2 ( s)
Si(s) 2H 2 O( g )
H 2 O( g )
(1) (2)
G(o1) G
o ( 2)
SiO2 (s) 2H 2 ( g ) (3)
246000 54.8T 902000 174T
G3o
G3o
2 G1o
G2o
902000 174T
( 246000 54.8T ) 2
410000 64.4T G
RT ln K
o 3
At T K
410000 64.4T 410000 64.4 1273 8.314 1273
1273K , ln K
31
2.9 1013
(c ) K
PH 2( g )
2
PH 2O ( g )
PH 2 ( g ) PH 2O ( g ) PH 2O ( g ) PH 2 ( g )
2.9 1013
5.38 10 6 1 5.38 10 6
0.186 10
6
0.186 ppm
14. Solid barium oxide(BaO) is to be prepared by the decomposition of the mineral witherite (BaCO3) in a furnace open to the atmosphere (P = 1 atm). (a) Write the equation of the decomposition (witherite and BaO are immiscible). (b)Based on the accompanying data, what is the heat effect of the decomposition of the witherite(J/mol). Specify whether heat is to be added (endothermic) or evolved (exothermic). (c) How high must the temperature be raised to raise the carbon dioxide pressure above the mineral to one atmosphere? ( 5.14, P147 ) DATA
Thermodynamic
Properties
[KCAL/(g.mol)] G of ( 298)
H of ( 298)
CO2 -94 -94 BaO -126 -133 BaCO3 -272 -291 (Assuming that CP,CO2+ CP, BaO = CP,BaCO3)
Solution: (a)
BaCO3 (s)
BaO(s) CO2 ( g )
(b) CP = 0 H of ,CO2( 298)
H
H of , BaO( 298)
94 133 ( 219)
H of , BaCO3( 298)
64kcal
267.52kJ
the reaction is endothermic (c ) At 298K, o G298
G of ,CO2, 298
G of ,CaO, 298
94 126 272 o G298 o S 298
GTo
52kcal
G of ,CaCO3, 298 217.36kJ
o o H 298 T S 298 o H 298
o G298
T H To
(267.52 217.36) 298
168 J / mol.K
T S To
when PCO2=1 atm, GTo
0, ie 2675201 168T T
1592 K
15. As the Elligham diagram indicated, Mg has a very stable oxide. Therefore Mg metal can be obtained from the oxide ore by a two-step process. First the oxide is converted to a chloride. In the second step the chloride is converted to metal Mg by passing H 2 gas over liquid MgCl2 at 1200 C. The reaction in this last step is: MgCl 2 (l )
H 2 (g)
Mg ( g ) 2HCl( g )
(a) Calculate the equilibrium pressure of H2(g), Mg(g) and HCl(g) if the total pressure is maintained constant at 1 atm. (b)Calculate the maximum vapor pressure of H2O that can be
tolerated in the hydrogen without causing the oxidation of the Mg vapor. DATA Go at 1200 C
Reaction Mg(g)+Cl2(g) H2 (g) + Cl2(g) Mg(g) +1/2O2(g) H2 (g) + 1/2O2(g)
= MgCl (l) = 2HCl(g) = MgO(s) = H2O(g)
－425484 J －207856 J －437185 J －165280J
（5．15， p48） Solution: (a)
MgCl 2 (l )
H 2 (g)
Mg(g)+Cl2(g)
Mg ( g ) 2HCl( g )
=
(1)
G1o
MgCl (l)
(2)
Go2
2HCl(g)
(3)
Go3
－425484 J H2 (g) + Cl2(g)
=
－207856 J G1o
G3o
G2o
207856 425484
217628J
RT ln K
G1o
RT ln
2 PHCl PMg ( g ) (g)
8.314 1473 ln
PH 2( g )
2 PHCl PMg ( g ) (g)
PH 2 ( g )
217628 ln
2 PHCl PMg ( g ) (g)
PH 2( g )
2 PHCl PMg ( g ) (g)
5.27 10 7
PH 2( g ) PH 2 ( g )
＝17.78
PMg ( g )
PHCl
1, PHCl
2 PMg ( g )
let PMg(g) =x, PHCl = 2x, PH2 = 1-3x 1 3x x(2 x) 2
5.27 10 7
x 1.6 10 3 (atm )
Mg(g) + H2O(g) = MgO(s)+ H2 (g) Mg(g) +1/2O2(g)
Go4
(4)
= MgO(s)
(5)
Go5
－
(6)
Go6
－
437185 J H2 (g) + 1/2O2(g)
= H2O(g)
165280J G4o
271905 PH 2 ( g ) PH 2O ( g ) PMg ( g )
PH 2O ( g )
RT ln
PH 2( g ) PH 2O ( g ) PMg ( g )
271905J
8.314 1473 ln
PH 2 ( g ) PH 2O ( g ) PMg ( g )
＝22.2
1.6 10 3 PH 2O ( g ) 1.6 10 ln PH 2O ( g )
437185 ( 165280)
G6o
RT ln K
G4o
ln
G5o
22.2
3
22.2 2.28 10
10
(atm )
16. A common reaction for the gasification of coal is: H 2 O( g ) C(s)
H 2 ( g ) CO( g )
(a) Write the equilibrium constant for this reaction and compute its value at 1100K;
(b)If the total gas pressure is kept constant at 10 atm, calculate the fraction of H2O that reacts; (c) If the reaction temperature is increased, will the fraction of water reacted increase or decrease? Explain your answer. Use the data in Table 5.1. (5.16, 148) Solution:
H 2 O( g ) C(s)
(a)
H 2 ( g ) CO( g )
PCO( g ) PH 2( g )
K
PHO 2( g )
Go
G f ,CO
G f , H 2O
111710 87.65 1100 ( 246740 54.81 1100)
RT ln K
Go
ln K
2.3
K
21676 J
9.97
(b) let
PH 2( g )
x atm, PCO( g )
x atm, PH 2O ( g )
(10 2 x) atm
2
x 1 2x 4.14 atm K
x
9.97
(c ) if the temperature is increased, the fraction of water reacted will increase since the equilibria constant increases with increasing temperature.
Solutions 1. The activity coefficient of zinc in liquid brass is given (in joules ) by RT ln
the Zn
following 2 38300 xCu
equation
for
temperature
1000-1500K:
, where xCu is the mole fraction of copper.
Calculate the partial pressure of zinc PZn over a solution of 60 mol % copper and 40 mol % zinc at 1200K. The vapor pressure of pure zinc
is 1.17 atm at 1200K. (7.1, p196) Solution: ln
Zn
Zn
2 38300 xCu RT 0.25
2 38300 0.6 Cu 8.314 1200
x Zn
0.25 0.4
0.1
P a Zn
0.1 1.17
0.117(atm )
a Zn
Zn
PZn
p Zn
1.38
2. Using the equation give in Problem 7.1, for the activity coefficient of zinc in liquid brass, derive an equation for the activity coefficient of copper using the Gibbs-Duhem equation. (7.2, 196) Solution: According to Gibbs –Duhem Equation: x Zn d (ln a Zn )
xCu d (ln aCu )
0
x Zn d (ln
x Zn d (ln x Zn )
xCu d (ln aCu )
x Zn d (ln x Zn )＋xCu d (ln xCu )
dx Zn
x Zn d (ln
0
d (ln ln 0
Zn )
Cu
d (ln
Zn
)＋xCu d (ln x Zn d (ln xCu
)
Cu
xCu
)
1 xCu 1
ln
Cu
Zn
Cu
)
dxCu
xCu d (ln xCu ) dx Zn
0
d (1 x Zn )
0
)
x Zn xCu
38300 2 xCu dxCun RT
38300 2 x Zn dxCu RT
xZ n 1
38300 2 x Zn dx Zn RT
38300 2 x Zn RT
3. (a) At 900K, is Fe3C a stable compound relative to pure Fe and graphite?( 7.3, 196) (b)At 900K, what is the thermodynamic activity of carbon in equilibrium with Fe and Fe3C ? Carbon as graphite is taken as the standard state. (c) In the Fe-C phase diagram, the carbon content of
-iron in
equilibrium with Fe3C is 0.0113 wt. %. What is the solubility of graphite in -iron at 900K? DATA
AT 900K,
3Fe C( graphite)
Go
Fe3C
3463 J
4. From vapor pressure measurements, the following values have been
determined
for
the
activity
of
mercury
in
liquid
mercury-bismuth alloys at 593K. Calculate the activity of bismuth in a 40 atom % alloy at this temperature NH 0.94 0.89 0.85 0.75 0.65 0.53 0.43 0.33 0.20 0.06 9
g
3
1
3
3
7
7
0
7
3
aH 0.96 0.92 0.90 0.84 0.76 0.65 0.54 0.43 0.27 0.09 1
g
9
8
0
5
0
2
2
8
2
(7.4,196) Solution: NH 0.94 0.89 0.85 0.75 0.65 0.53 0.43 0.33 0.20 0.06 9
g
3
1
3
3
7
7
0
7
3
aH 0.96 0.92 0.90 0.84 0.76 0.65 0.54 0.43 0.27 0.09 1
g Hg
9
8
0
5
0
2
2
8
2
1.01 1.04 1.06 1.12 1.17 1.21 1.24 1.31 1.34 1.46 3
Plot ln
Hg
~xHg
lnB Linear Fit of Data1_lnB
0.40 0.35 0.30 0.25
ln
Hg
0.20 0.15 0.10
ln Hg=0.395-0.391xHg
0.05 0.00 0.0
0.2
0.4
0.6
0.8
1.0
xHg
d (ln ln 0
ln
Bi
x Hg
)
x Bi )
d (ln
d (ln
Bi
0.391
Bi
0.391(ln x Bi
Hg x Bi 1
) (
1 x Bi
x Hg x Bi
( 0.391)dx Hg
1)dx Bi
x Hg )
when xBi = 0.4, xHg =0.6 ln
0.391(ln 0.4 0.6)
Bi
Bi
0.1
1.107
7.5 For a given binary system at constant T and P, the liquid molar volume V
100 x A
of
the
80 xB
2.5x A x B
solution
(cm3/mol)
is
given
by
:
(a) Compute the partial molar volumes of A and B and plot them, together with the molar volume of the solution, as a function of the composition of the solution; (b)Compute the volume of mixing as a function of composition. (7.5, 196) Solution:
the
100 x A
80 x B
V
calculated 2.5x A x A
partial
variables
are
as
follows:
V xA
VA
V xB
VB
100 2.5 x B T , P , xB
80 2.5 x A
82.5 2.5 x B
T , P , xB
105
100 partial volume of A partial volume of B molar volume of the solution
3
cm /mol
95
90
85
80 0.0
0.2
0.4
0.6
0.8
1.0
xB
(b)
VA
100,V B
VM
V
80
(100(1 x B ) 80 x B
V
(100 x B )
2.5 x A x B
4. For an ideal binary solution of A and B atoms, plot schematically the chemical potential of both species as a function of the composition of the solution. Indicate on the plot the molar Gibbs free energy of pure A and B. (7.6,196)
5. At 473 C, the system Pb-Sn exhibits regular solution behavior, and the activity coefficient of Pb is given by:
log
Pb
0.32 1 x Pb
Write the corresponding equation of the variation of
Sn
2
.
with
composition at 473 C. (7.7, p196)
6. MgCl2 and MgF2 are two salts that can form solutions. The Gibbs free energy of fusion(J/mol) for both compounds is given by:
For MgCl2 :
G = 43905-43.644T,
Melting point
=987K For MgF2:
G = 58702-38.217T,
Melting point
=
1536K The free energy of mixing (J/mol) for liquid mixture MgCl 2 and MgF2 is given by: GMix
2RT ( xMgCl2 ln xMgCl2
xMgF2 ln xMgF2 ) xMgCl2 xMgF2 ( 2556 25( xMgF2
xMgCl2 )) .
Compute the maximum solubility of MgF2 in liquid MgCl2 at 900 C. MgCl2 does not dissolve in solid MgF2. (7.8, 197)
7. The thermodynamic properties of Al-Mg solution at 1000K are given in accompanying table. (a) If one mole of pure liquid aluminum and one mole of pure liquid magnesium, each at 1000K, are mixed adiabatically, what will be the final temperature of the solution that is formed ? (b) What is the total change in entropy for the process ? DATA
Quantities of Mixing Liquid Alloys at 1000K
x Mg
GM (cal / mol)
0.1
-800
H M (cal / mol)
-300
S M (cal / mol)
0.5
C p [cal /( mol.K )]
7.1
0.2
-1250
-600
0.65
7.18
0.3
-1550
-750
0.8
7.26
0.4
-1700
-850
0.85
7.34
0.5
-1800
-900
0.9
7.42
0.6
-1700
-850
0.85
7.5
0.7
-1550
-750
0.8
7.58
0.8
-1250
-600
0.65
7.66
0.9
-800
gf-300
0.5
7.74
The problems of the phase rule 8.1 Zinc sulfide (ZnS) is reacted in pure oxygen to form zinc sulfate (ZnSO4) (a) write the chemical reaction representing the process (b) how many solid phases may exist in equilibrium if pressure and temperature are arbitrarily fixed? (c) if the temperature is fixed, will the pressure be determined if ZnS and ZnSO4 exist in equilibrium?
Solution: (a) ZnS 2O2
ZnSO4
(b)two (c) because F=(3-1)-3+1=0 so yes
8.2 an Fe-Mn solid solution containing 0.001 mole fraction Mn is in equilibrium with an FeO-MnO solid solution and a gaseous atmosphere containing oxygen at 1000K. How many degree of freedom does the equilibrium have? What is the composition of the equilibrium oxide solution, and what is the oxygen pressure in the gas phase? Assume that both solid solutions are ideal? Data: for Fe
Fe ( s) G
1 O2 FeO ( s) 2 259000 62.55T
For Mn
Mn( s) G
1 O2 MnO( s) 2 384700 72.8T
Solution: (a) F=(5-2-1-1)-2+1=0 1 RT ln PO2 2 (b) PO 1 exp( 2 G 0 ) 3.0 10 21 2 RT 2 G0 PO2 2 exp( ) 2.6 10 33 RT P 0.999 PO2 1 0.001PO2 2 3.0 10 G0
21
The problems of the phase diagram 9.1 (a) if an alloy of 50 atom % copper and 50 atom % silver is brought to equilibrium at 600℃ at one atmosphere pressure, what phase or phase in the accompanying Ag-Cu phase diagram are present? (b) apply the phase rule to the situation in part A, how many degrees of freedom does the system have? (c) Assume that the system described in part a is brought a new equilibrium at 700℃. Describe the physical changes you expect to
occur in the system Fig
Solution: (a)
Ag
Cu
solution
solution increases and
(b)F=C-P=(2-1)-1=0 (c) Cu
Ag
phase in the
decreases
9.6 in the accompanying eutectic equilibrium phase diagram of temperature versus mole fraction of B for the A-B system shown, note that the pressure for the diagram is constant at 1 atm. Consider an alloy containing 40 mol% of B. Fig
In the table indicate which of the phase are present in the 40% alloy and give the composition of each and the fraction present of each for the temperature shown Temperature
Phase
Composition
Fraction
1300
Liquid
60
61.5
α
8
38.5
β
99
0
Liquid
70
50.8
α
9
49.2
1000+
1000-
β
98
0
Liquid
_
0
α
7
63.7
β
98
36.3
9.8 The phase behavior of material A and B can be described using the accompanying phase diagram. Assume that A and B form ideal solutions in the liquid state and in the solid state. fig (a) if a solution containing 50 mole% B is cooled from 1300K, what is the composition of the first solid to form? What is the composition of the last liquid to solidify? (b)For this 50 mole% B solution ,estimate the fraction solid and liquid in equilibrium at 1000K.
Solution: (a) 90% is the composition of the first solid to form;10% is the composition of the last liquid drop. (b) solid (60% is the composition) is about 77% ; liquid (15% is the composition) is 23%
9.9an alloy composed of 80 atom% rhodium and 20 atom% rhenium
is being slowly cooled from 3500℃ during processing. Equilibrium is maintained at each temperature. Use the accompanying Rh-Re phase diagram to answer parts A-C (a) at
what temperature does the first solid formed and what is the
composition of that solid? (b)At what temperature does the last liquid solidify, and what is the composition of the last liquid (c) Which phase exist at 2000℃ and what is their composition? Given the fraction of each phase. How many degrees of freedom are therr in this equilibrium?
(a) 2900 ℃ , α (12%) (b) 2300 ℃ , liq(95%) (c) 8.2% α (composition is 24% )+91.8%β(85%)
9.10 the accompanying diagram represents the liquidus surface in the ternary phase diagram for BaCl2-NaCl-KCl-CaCl2 assume for the purpose of this problem that there is no solid solubility of the compounds in one another. (a) on the diagram trace the path of the liquid composition when a material consisting of BaCl2 60%, NaCl 20%, KCl-CaCl220% is cooled from 900℃. Assume that equilibrium is maintained at all temperature.
(b)Sketch two blank ternary diagram and draw in the isothermal section at 750 and 650℃ (c) What will be the fraction liquid when the liquid reaches the ternary eutectic temperature but none has solidified as a ternary eutectic? That is what will be the fraction of ternary eutectic in the material when it is all solidified? Solution: (a)liq → liq+ BaCl2 → liq+ BaCl2+ NaCl → liq+ BaCl2+ NaCl+ KCl-CaCl2→BaCl2+ NaCl+ KCl-CaCl2 BaCl2
20%
NaCl
KCl, CaCl2 20%
BaC l2
(b) 750℃ 20 %
750 ℃ Soli Na d Cl
Soli d
750 ℃
Liq uid 20 %
KCl, CaCl2
600℃ BaCl2
20%
Solid
Liquid NaCl
KCl, CaCl2 20%
(c) 20%liquid; 30%BaCl2+ 20%NaCl+50% KCl-CaCl2
