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 C P of lead may be taken as 29.3J/(mol K)
Solution:
Qabsorb Qincrease Qmelt W
1 2
mv
2
1 2
nMv
n(C p T H melting )
2
Qabsorb W n[29.3 (327 25) 4.8 10 ] 3
1 2
3
n 207.2 10 v
2
V 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
Q Burn ing in g P
2500 10 3 4.1868 10467000( J )
W / t Q Burn ing in g / t
Solution
Pincrea sin g
10467000
121( J / S )
24 60 60 20 h mg 75 9.8 245( J / S ) 60 t 3
1.3 One cubic decimeter (1 dm ) of water is broken into droplets having a diameter of one micrometer (1 um) at 20 ℃. (a)what (a) what is the total area of the droplets? (b)Calculate (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/cm 2) is also used for surface tension dyn/cm) Solution
S total nS Single
(1 101 )3 4 3
4 (0.5 10 6 ) 2 6 103 (m2 )
(0.5 10 6 )3
W S 72.75 10
5 2
(3 103 6 10 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 (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 (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? (a) Adiabatic Adiabatic
Solution:
T T 0
(
P P0
R / C P
)
1 T 298 ( ) 0.4 10 (b)
T
W nC p
118( K )
1 (500 5) 101325 10 3 10 50 101325 10 3
R 298 T T 0 T 298 118 180( K )
118( K )
2.5 R
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 T 0. The valve on the tank is o !ened 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 t èe gaS in the tank? The heat cap!city mf the gas, C p 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 C p 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.
Adiabati Adiabaticc
solution
T T 0
(
P P0
T T 0 (
R / C P
)
P0
0
P0
R / C P
)
T 0
1.6 Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 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 H 290 8[ Kcal / g mol ] 17.89 94.05 57.80
FOR
DATA:
CH 4 ( g ) CO2 ( g ) H 2 O( g )
solution
2O2 CO2 2 H 2O H 298 ( H CO 2 H H O H CH ) (94.05 2 57.80 17.89)
CH 4
2
2
H 298 191.76( Kcal / g mol )
4
191.76 103 10
3
1 3
0.3048
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% O 2 and 79% N2) (a)Assuming (a) Assuming complete combustion, what is the composition of the flue gas (the gas following combustion)? (b)What (b) What is the temperature of the gas, assuming no heat loss?
(c)The (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
3
m /h)
assuming
that
for
gas
H1600-H298=1200KJ/KG (d)A (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 DAT DA TA STP means T=298K, P=1atm for CH 4
C P (cal / mol C )
CO 2
16
H 2 O
13.7
N 2 O2
11.9 8.2 8.2 (a ) CH 4
2O2 CO2 2 H 2 O
CO2 %
Solution
3 2 1.1
1 79
8.71%
2 (1.1 1) 21 H 2 O% 2CO2 % 17.43% N 2 % 72.12% O2 % 0.87%
(b)
C p,i X i 0.01[13.7 8.71 11.9 17.43 8.2 (72.12 0.87)] 9.25(cal / mol 191.76 1000 T T 0 T 298 2104( K ) 9.25 11.48 C p , p
C )
(C ) P 1200 2000 400000 2800000( KJ / h)
2800000 10 191.76 1000 9.25 11.48(1600 298) 3
V Consumin g
(
11.48
4.1868
1 0.0224
3214(m 3 / h) )
(d ) C p ,r
C p ,i X i [16 8.2 2.2
H H 298
(
C
p , p ,i
ni
100 21
] / 11.48 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 ) 3 2800000 10 V Consumin g 1644(m 3 / h) 189570 9.25 11.48(1600 800) 1 ( ) 4.1868 11.48
0.0224
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 H T-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
a(700 298)
Solution
b
a bTdT [aT 2 T ]
C P dT b
(700 2
2 b
a(1000 298)
2
2 T 298
298 2 ) 12113
(1000 2
298 2 ) 22803
a 35.62 b 0.011
35.62 0.011T H 29T 8 35.62(T 298) 0.0055(T 2 298 2 )
C P
1.9 A fuel gas containing 40% CO, 10% CO2, and the rest N 2 (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
H 0 f , 298,CO 110458 J / mol H 0 f , 298,CO 393296 J / mol 2
28.45 3.97 103 T 0.42 105 T 2 ( J / molK ) 3 5 2 C P ,CO 44.35 9.20 10 T 8.37 10 T ( J / molK ) C P ,CO
2
C P ,O2
19.92 4.10 10 3 T 1.67 105 T 2 ( J / molK )
C P , N 2
29.03 4.184 103 T ( J / molK ) 2CO O2 2CO2 when 1mole fuel need 1mole air ( N 2 / O2 4) then reation CO 20% CO2 5% N 2 65% O2 10% production CO2 27.8% N 2 72.2%
H i , 298 H f , 298, r ni H f , 298, p ni 393296 110458 282838( J / mol ) 0
0
(a) C p , r
C
p , p , i
ni 0.2C P ,CO 0.05C P ,CO2 0.65C P , N 2 0.1C P ,O2
28.77 4.38 10 3 T 0.67 105 T 2 ( J / molK ) C p , r
C
p , p , i
ni 0.278C P , CO2 0.722C P , N 2
33.28 5.58 10 3T 0.19 105 T 2 ( J / molK ) H
H i , 298ni ( C p , p , i ni C p , r , i ni )dT
282838 0.2
T
298
(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
733
28283.8 2 (28.77T 2.19 10 3T 2 0.67 105 T 1 ) 733 298
Solution
(1.18T 0.321T 2 0.499T 1 ) T 298 0 T ?
H H i, 29 8ni T
282838 0.2 29 8
(
C
n
p , p , i i
C p , r ,i ni )dT
(33.28 5.58 10 3T 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 2.19 10 3 T 2 0.67 105 T 1 ) 1250 29 8 (1.18T 0.321T 2 0.499T 1 ) T 29 8 0 T
(b)
282838 0.2
1250 Q29 8 282838 0.2 1250
Q29 8
1250
29 8 1250
29 8
(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 2
O2
CO2 ,what is the enthalpy of
reaction ( H 0 ) at 298 K ? (b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas? (c)If (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 H f at 298 K CO 110000( J / mol) CO2
393000( J / mol)
Heat capacities [J/(mol K)] to be used for this problem N 2=33, O2=33, CO=34, CO2=57 (a) H 0
H 0 f , 298, r ni H 0 f , 298, P ni 110000 393000 283000( J / mol )
(b) fuel CO%
Solution
0.5 1 0.25 / 0.2 22.2
product CO2 %
(C )C p, P C r , P
100 11.1
22.2%, N 2 % 66.6%,
O2 % 11.1%
25%, N 2 % 75%,
C i , p, P X i 33 0.666 33 0.222 34 0.111 33( J / mol K )
C i , r , P X i 57 0.25 33 0.75 39( J / mol K )
H H 0
n pC p , P dT 0
283000 0.222 (0.889 39)(T 298) 0 T 2110( K )
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 Air : n(O2 ) 0.104 n( N 2 ) 0.416
DaTA(k J?mol)
H f (kJ / mol ) 110.523 393.513
FOR CO CO2
C P [ J / m o l K ]
??
FOR CO
33 57
CO2 H 2O ( g )
50 34
N 2 , O2
Solution (a )CO H 2
1 2
1 2
O2
O2
CO2
H 2O
Fuel :
Flue :
n(CO ) 0.12
Air :
n(CO2 ) 0.24
n(O2 ) 0.08
0.36 n ( H 2O ) 0.04 n ( N 2 ) 0.92 n (CO 2 )
n( N 2 ) 0.32
n( H 2 ) 0.04 n( N 2 ) 0.6
H H COCO H H H O 0.12 (393.51 110.523) 0.04 (241.8 0) 43.6308KJ 2
C
P ,r ,i
T
ni
2
2
0.36C CO 0.04C H O 0.92C N 0.36 57 0.04 50 0.92 34 53.8( J / K ) 2
43.6308 103
2
2
810( K )
53.8 T 1108.98( K )
(b)repeat the calculation for 30% excess0combustion air at 298K H H CO CO H H H O 0.12 (393.51 110.523) 0.04 (241.8 0) 2
2
2
43.6308KJ C P,r ,i ni 0.36C CO 0.04C H O 1.016C N 0.024C O 0.36 57 0.04 50 1.016 34 0.024 34 57.88( J / K ) 43.6308 103 T 753.8( K ) 2
57.88 T 1051 8( K )
2
2
2
Fuel :
Flue :
Air Air :
n(CO ) 0.12
n(CO2 ) 0.36
n(O2 ) 0.104
n(CO2 ) 0.24
n( H 2O) 0.04
n( N 2 ) 0.416
n( H 2 ) 0.04
Maret 8, 2013
n( N 2 ) 1.016
n( N 2 ) 0.6
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 :
Flue :
Air :
n(CO) 0.12
0.36 n ( H 2O ) 0 .04 n ( N 2 ) 0.92 n (CO 2 )
n(O2 ) 0.08
n(CO2 ) 0.24
n( N 2 ) 0.32
n( H 2 ) 0.04 n( N 2 ) 0.6
H H COCO H H H O H fuel 700298 2
2
2
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 C P,r ,ini 0.36C CO 0.04C H O 0.96C N 0.36 57 0.04 50 0.92 34 53.8( J / K ) 2
T
59.373 10
3
53.8 T 1401.6( K )
2
2
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? Fuel :
Air :
Flue :
n(CO) 0.12
n(O2 ) 0.08
n (CO 2 ) 0.36
n(CO2 ) 0.24
n( N 2 ) 0.32
n ( H 2 O ) 0.048
n( H 2 ) 0.04
n( H 2O)
n( N 2 ) 0.6
15 760 15
0.4 0.008
n ( N 2 ) 0.92
H H COCO H H H O 2
2
2
0.12 (393.51 110.523) 0.04 (241.8 0) 43.6308KJ C P,r ,i ni 0.36C CO 0.048C H O 0.92C N 0.36 57 0.048 50 0.92 34 54.2( J / K ) 43.6308 103 T 805( K ) 2
2
2
54.2 T 1103( 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 S ,1803 L
1803
C P , S dT
1798
1798
C P , L dT H L ,1798 0 S
1803
H LS ,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 Cu 2O (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 Cu 2O=-41900 H2O=-59210 solution
Cu2O H 2
Cu H 2O ,exothermic reaction H 59210 41900 17310(cal / mol)
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, C p,s=26(J/molK), C p,L=29(J/molK), Melting point=932K, Heat of fusion=10700J/mol Solution
H l ,1000
932
298
C P , S dT
1000
932
C P , L dT H LS
26 (932 298) 29 (1000 932) 10700 27184( J / mol ) P
27184 1 1000 27 3600
279.7(W ) 0.28(kW )
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, C p,s,Al=26(J/molK),
C p,s,
Al2O3=104J/mol,
heat
formation
of
Al2O3=-1676000J/mol Solution;
1676000
1
27 2 T 302( K )
1
16 1.5 27 27 102
99
104 T
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: C p=4.184J/g k, Density=1g/cm3; for copper: molecular weight=63.54g/mol C p=7cal/mol k, heat of fusion=3120 cal/mol QCopper / min min
Solution: Q
[3120 7 (1356 400)]
1000 63.54 min min
1(80 20)V V 2.573 103 (m3 / min) Water
/ min 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 C p=4.184J/g k, Density=1g/cm Solution:
W 4.184 (60 20)
3
5 1000 60
13947(W )
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/m what is the internal energy change for the evaporation of water?
3
PV 101325
Solution:
%
373 273
0.0224 3101( J / mol )
3101
7.6% 2261 18 U W Q 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, evaporator, liquid water and water vapor. 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 about 10 atm) 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 4.18 80 (1000 x), x 138( g ) irreversible
The problems of the second law 2
2.1 The solar energy flux is about 4J cm /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
1
3 0
5 6
5
7
5 6 1 2
,+* %&# * .+ %)(+ $.+
4 6
02
)*($1 #
('%&$/ % #
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:
T H T L T H
0.75P P L
Q H
T H T L T L
P L
3
273 20
4
35 20
0.25 746 643(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 converted to ice at 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 efficiencies involved have their largest possible value. P
Solution:
T H T L
P L
273
0.25 746 334m 20 M 60m 457( g ) P L
T L
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 50%
Solution:
T H T L
0.5P' P
T L
Q L
T H T L T L
P L
300 4.216 84 2 4.216 60
P' 393(W ) 0.52(hp)
2.5 if
a fossil fuel fuel power plant operating between 540 and 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. boiler. (a)assume (a) assume that the efficiencies are equal to the theoretical maximum values
(b)assume (b) assume the power plant efficiency is 70% of maximum and that coefficient coefficient of performance of the heat pump p ump is 10% of maximum (c)if (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 P2
T H ,1 T L ,1 T H ,1
T H , 2 T L , 2 T H , 2
P H ,1
P H , 2
P1 P2
540 50
P H ,1
540 273 P H , 2 8.98 P H ,1
25 5 273 25
P H , 2
(b) P H , 2 0.6286 P H ,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 C p is 77J /(mol K) from 273K to 373K
Solution: U 0( J ) S n1C P ln(
323 323 T E T ) n2C P ln( E ) 0.5C P ln( ) 0.5C P ln( ) 0.933( J / K ) 373 273 T 1 T 2
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 (a) what is the maximum electrical work that can be produced by the plant per joule of heat provided provided to the boiler? (b)How (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 efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g (c)Electricity (c) Electricity is used to heat a home at 25 ℃ when the out door temperature is 10 ℃ by passing a current through resistors. 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
T H T L T H
Q H
540 30 540 30
(c ) 1 0.89( J )
W
T H T L
(b ) 29.0m
T L T H T L
500 3600
Q H
m 69371(kg ) 69.3(ton)
T H
Q H
273 25 25 10
1 19.9( J )
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 (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 (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/cm 3 m
solution: (a) 18 40000 1000, m 0.45( g ) (b)
m
18
40000 1000
100 273 100 25
, m 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 is maintained at -20℃ by a refrigerator. refrigerator. The aluminum is being fed into 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 L
1000
28 (480 20) 27 T T L 30 20 PW H P L P L T L 273 20
Solution: P
102474 (W ) 102.5(kW )
2.10 an electric power generating plant has a rated output of 100MW. 100MW. The boiler of the plant operates at 300 ℃. The condenser operates at 40℃
(a)at (a) at what rate (joules per hour) must heat be supplied to the boiler? (b)The (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 (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/cm Solution:
(c) P H
( a) P H
T H T H
T L
P
300 273 300 40
2.2 10 (W ) 11 Q H P H t 7.9 10 ( J )
T H T H T L
8
P
540 273 540 40
108
3
(b)
4.3 1011 ( J ) V 106 10 4.184 Q L V 1.03 104 ( m3 ) Q L
108
1.626 108 (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 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:
(a)W (b) P
T H T L T H mgh
1000
Q H
20 4 20 273
1 0.055( J )
3600 1.06 106 (kW / h)
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 5
rink of 10 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. reservoir. The ice in the rink is to be maintain at a temperature of
15 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:
( a) P (b)Q H
T H T L T L
P L
20 15 273 15
273 20 5 10 273 15
105 / 3600 3.77(kW )
1.14 105 (kJ )
2.13 (a) 2 Al N 2
2 AlN
solution: (b) H 152940(cal / mol )
(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 ln10 68.81(cal / molK )
2.14 0
S m( 10
solution:
C P , ICE T
dT
H m T m
40
C P ,WATER T
0
dT )
273 336 273 40 4.184 ln )12000 263 273 273 22574( J / K )
(2.1ln
2.15 W W 2
T H T L T L
Q L
300 77 77
1000 2896 ( J )
70428( J )
2.16 W W
T H T L T L T H T L T H
Q L
Q H
300 4.2 4.2
83.3 5866.7( J )
300 4.2 300
(83.3 1.5 8.314(300 4.2)) 3719.4( J )
2.17 ( a ) T 0
U O Q W n
pdV 1 8.314 298 ln10 5704( J )
(b) S nR ln
P0 P
1 8.314 ln10 19.1( J / K )
(c)Q 0 ( d ) yes
2.18 500 60
T H T L T L
335m
20 0 273
335m
m 1222( g )
Property Relations
1. At -5C, 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)
G RT ln Solution:
P H 2O ,ice P H 2O , water
8.314 (273 5) ln
3.012
3.163 8.314 268 ln 0.9523 108.9 J / mol
H 5.85 103 J / mol G H T S H G 5850 (108.9) S 22.23 J /(mol K )
T
268
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 300C. 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 25C. 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 5050 A-B alloy (per gram of alloy)? (b) To To what temperature does it apply ? (3.5, 94) Solution:
(a)
C P ,bath
100cal / K 418 J / mol
Q C P,bathT / 2 100 0.2 / 2 10cal / g
This is the heat of mixing.
(b) The heat capacity capacity of CP, alloy :
C P ,alloy
C P,bath T
2 (300 0.4 25) 100 0.4 2 274.6
0.072cal /( g K )
Assuming that the calorimeter can be applied to the maximum of TC, the for mixing to form 1 gram of alloy: Q1
C P ,bath (300 T ' ) 10
,
Q2
C P ,alloy (T T ' ) ,
Q1
Q2
C P ,bath (300 T ' ) 10 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 0C?(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 0C, G =0,
S
H T m
6030 273
TmS = H
22.09 J /(mol.K )
(b) At 0C, G =0 ©
C P,ice
0.5cal /( g.K ) 0.5 4.18 18 J /( mol.K ) 37.62 J /( mol.K )
C P , water
1.0cal /( g.K ) 1.0 4.18 18 J /( mol.
a reversible process can be designed as follows to do the calculation:
（2）
Ice, 0C
water, 0C
（1）
（3）
ice, -5 C
water, -5 C （4）
H fu H (1) H ( 2) H (3) 273 273
268 268
C p ,ice dT H C p , water dT 268 268
273
273
268
(C p ,ice C P ,water )dT H
(37.62 75.24) 5 6030 5841.9 J / mol （d）
S ( 4 ) S (1) S ( 2 ) 3( 3) C p ,ice
27 3
26 8
dT S T 27 3(C p ,ice C P , water )
26 8
T
(37.62 75.24) ln
C p , water
26 8
T
27 3
dT
dT S
273 268
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 as the ones listed listed as as data; (c) The 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
“
S 298 27.28 J / K .mol mol . o
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 29 8
given above ？
DATA： （for iron）
-1
C p = 24 J K mol-1 Compressibility
= 6 10-7 atm – 1
Linear coefficient of thermal expansion = 15 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) V iron
(b)
mol weight density
55.85 g / mol 7.87 g / cm 3
7.10cm 3 / mol 7.10 10 6 m 3 / mol
S V P T T P S V V 3V l P T
for iron: S 3V iron l ,iron P T 3 7.10 10 6 15 10 6 3.2 10 10 (m 3 /(mol.K ))
S iron 3.2 10 10 P (c) S iron 3.2 10 10 (100 1) 1.013 10 5 3.2 10 10 99 1.013 10 5 320.9 10 5 3.21 10 3 ( J / mol.K ) error %
S iron 100% 1.12 10 2 o S 298
Equilibrium
1.
At 400C, 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)
V ice
V water
18g / mol 0.92 g / cm
3
18g / mol 1g / cm
3
19.57cm3 / mol 19.57 10 6 m 3 / mol 18cm3 / mol 18 10 6 m 3 / mol
V fus 1.57 10 6 m 3 / mol
According to the Clapeyron equation:
dP dT
dP
1
V fus 1
V fus
H fus
T
H fus
T
dT
take definite integration of the above:
501.01 3105
1.01 310
ln
T
273
5
dP
H fus V fus
1
T
27 3
T
dT
V fus 49 1.013 10 5 H fus 1.57 10 6 49 1.013 10 5 6009
0.013
T 272.8K
(b)
P
150 0.01 3
50 10 3 lb / in.2 50 10 3 6897 Pa 345 10 6 Pa
P 345 10 6 Pa
(c ) ln
T
273
V fus 345 10 6 H fus 1.57 10 6 345 10 6
T 249.46K
6009
0.09 -4
1. At 400C, liquid zinc has a vapor pressure of 10 atm. Estimate the boiling temperature of zinc, knowing that its heat of vaporation is approximately approximately 28kcal/mol. (4.3,117) Solution:
H vap 28kcal / mol 4.18 28 103 J / mol 117.04 J / 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
(
T 2
1 T 1
)
ln
ln
P1
P2
1 10
4
H vap R
(
1 T 2
117.04 103 8.314
(
1 T 1 1
T 2
)
1 673
)
1202 K T 2
The boiling point of zinc is 1202K. 2. Trouton’s rule is expressed as follows: H vap 90T b in joules per mole, where T b 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 90T b
P
1
d (ln P)
ln P
H vap R
6823 298
(
1 298
1 630
)
10.83 22.90 10.83 12.07
P=5.7310-6 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: vapor
V l
18 g / mol 1g / cm
3
18cm 3 / mol 18 10 6 m 3 / mol
pressure
pressure, G v G l
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
3167.36Pa
P = 0.16Pa
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 liquid and vapor are the same. same. (a) Wr Write ite
an equation 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, (4 .7, p117) Solution:
P
1
d (ln P)
ln P
H vap 1 R
30685 T
(
T
1 2450
)
ln P
255000 8.314
(
1 T
1 2450
)
104.08
6. Zinc may exist as a solid, a liquid, or a vapor. The equilibrium pressure-temperature 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 atm)
15755 T
755 ln( ln(T ) 19.25 and 0.755
the vapor pressure of liquid
Zn varies with T as: ln P(atm) 15246 1.255 ln(T ) 21.79 . Calculate: (a) T
The boiling point of Zn under 1 atm; (b) The triple-point temperature; 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 T
1.255 ln(T ) 21.79 0
15246 1.255T ln( ln(T ) 21.79T 0
To solve the equation, let y1 = 21.79T-15246, y2 = 1.255TlnT. Plot y-T of the functions, and the intersection is the answer. answer.
30000 28000 26000 24000 22000 20000 18000 16000 14000 12000 y 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 T
1.255 ln(T ) 21.79
15755 T
509 0.5 ln(T ) 2.54 0 T
0.755 ln(T ) 19.25
509 0.5T ln(T ) 2.54T 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 5000 y 4500
4000 3500 3000 2500 2000 1500 1000 40 0
60 0
8 00 00
1 00 000
1 20 200
14 00 00
16 00 00
18 00 00
2 00 00 0
2 20 20 0
T, K
The intersection is 695.3K, and this is the triple point temperature. (c ) d (ln P)
15246 T
2
1.255 T
21.79
(15246 1.255T )(
1
T 2 1 (15246 1.255T )d ( ) T
H vap R
d (ln P)
)dT
H vap R
1 d ( ) T
15246 1.255T
H vap R(15246 1.255T ) 126755 10.43T
At T b =1180K, H vap 126755 10.43T 114.4 103 J / mol 114.4kJ / mol (d) For solids, ln P
15755
0.755 ln(T ) 19.25 T 15755 0.755 )dT d (ln P) ( 2 T T 1 (15755 0.755)( 2 )dT T 1 (15755 0.755)d ( ) T
d (ln P)
H fus R
1 d ( ) T
H fus
H fus R(15755 0.755T )
15755 0.755T
R
At triple point, T tr = 695.4K H fus 8.314 (15755 0.755T ) 126.6 103 J / mol 126.6kJ / mol
(e)
d ( H fus ) C P dT
C P
d ( H fus ) dT
0.755
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 that of the gas.
Find the fraction of material material in
the vapor phase in the initial and final states. (4.9, P118) Solution: In the initial state, P1V 1
n1 RT ,
n1
P1V 1 RT 1
T 1
300K ,
10130 10
3
8.314 300
P1
4.06 10 3 mol
mol(vapor)% 4.06%
In the final state, P 2=2 atm, V2 = 1L According to Clayperon equation: RT ln
ln T 2
2 1
P2 P1
H vap 1 1 R T 2 T 2
5000
(
1
8.314 T 2
458.6 K
1 300
)
1atm 1.013 105 Pa, V 1 1 L
P2V 2
n2 RT 2 ,
n2
P2V 2 RT 2
2 10130 10 3 8.314 458.6
5.3 10 3 mol
mol(vapor)% 5.3%
8. The melting point of gold is 1336K, and vapor pressure of liquid gold is given by: ln P(atm) 23.716
43522 T
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 T
2
1.222 T
)dT
(43522 1.222)( (43522 1.222)d ( H vap
1 T 1
2
) dT
) T R(43522 1.222T ) 361841 10.19T
(a) At 1336K,
H vap 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? C g
r
a
p h
C i d t
e i a
(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 graphit e is 2.25g/cm 2.25g /cm3
DAT DA TA
Density of diamond is 3.51g/cm3
H f o ( 29 8) (kJ / mol) H f o ( 29 8) ( J / mol.K )
Diamond
1.897
Graphite Solution: (a)
0 C graphite
C diamond
H H f o ,diamond H f o , graphite 1897kJ / mol
2.38 5.74
S S f o ,diamond S f o , graphite 5.74 2.38 3.36 J /(mol.K ) G H T S 1897 298 (3.36) 2898.28 J / mol
(b) No, diamond is not thermodynamically stable relative to graphite at 298K. (c )
G V diamand P
12 10 6 3.51
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 graphiteP 2898.28 V diamond ( P )
(V graphite V diamond )(1 N ) 10130 2898.28 12 10 6 12 10 6 10130 2898.28 N 3.51 2.25 0.194 N 2898.28 0 N 14939 (atm)
T '
T '
C p 0, T C p dT 0, T
C p T
dT 0
H 90 0 H 29 8 1897 J / mol
S 900 S 298 3.36 J / mol.K
G H T S 1897 900 900 3.36 4921 J / Chemical Equilibrium 1. Calculate the partial pressure of monatomic hydrogen in hydrogen in hydrogen gas at 2000K and 1atm. 1
For
2
H 2 ( g ) H ( g ) o H 298 217990 J o S 298 49.35 J / K
1
For the reaction : C p C P , H ( g )
1 2
C P , H 2
2
H 2 ( g ) H ( g )
3
1
2
2
8.314 31 3.035
2000
o o H 2000 H 29o 8 298 C P dT H 298 C P (2000 298)
217990 3.035 1702 212824 J 2000
o H 2000 H 29o 8 29 8 C P dT H 29o 8 C P (2000 298)
217990 3.035 1702 212824 J 2000
o S 2000 S 29o 8 29 8 C P dT S 29o 8 C P ln
49.35 3.035 ln
2000 298
2000 298
43.57 J
0 0 0 G2000 H 2000 T S 2000 212824 2000 43.57 125684 J
G
o 2000
RT ln
RT ln K RT ln
P H 2
P H ( g )
P H 2 ( g )
ln
1/ 2
125684 ln
P H ( g ) P H 2 P H 2
1/ 2
RT ln
1/ 2
P H ( g )
P H 2
P H ( g )
125684 8.314 2000
P H 2( g ) 1, P H 2 1
1 P H ( g )
1/ 2
7.562 P H 0.0005atm
7.56
2. For the reaction :
Co ( s )
1 O2 ( g ) CoO ( S ) 2
G o 59850 19.6T , where G o is in calories and T is in Kelvin. (a) Calculate the oxygen equilibrium pressure (atm) over Co and CoO at 1000C. (b) What is the uncertainty in the value calculated in part a if the error in Ho term is estimated to be 500 cal? (5.2, p144) Solution:
(a)
At
1000 C,
=-59850+19.6T=-59850+19.6(1000+273) =-34899.2cal = -1458.79J/mol At equilibrium: G o RT ln K RT ln ln PO2 PO2
1 PO2
1/ 2
1 2
RT ln PO2
145879 J
27.6
1.07 10 12 atm o
(b) uncertainty in H = 500cal/mol = 2090J/mol o
So uncertainty in G = 500cal/mol = 2090J/mol That means: 1 2 1 2
RT ln PO' 2 RT ln
ln
PO' 2 PO 2
PO' 2 PO 2
PO' 2 PO 2
1 2
RT ln PO 2
2090
2090
0.25
1.286
P PO 2
28.6%
o Similarly, Similarly, uncertainty in H =- 500cal/mol =- 2090J/mol o
G = -2090J/mol
o
G
1 2
RT ln PO' 2
1 2
RT ln PO 2
2090
'
PO 2
ln
PO 2
0.25
'
PO 2 PO 2
0.779
P PO 2
22.1%
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 P total = 1 atm. forAg 2 O 7300 H f forAg
DATA
cal / mol
Standard Entropy at 298K
Assume that C p = 0 for the decomposition reaction. reaction. Solution: (a)
Ag2O = 1/2O2 + 2Ag
H o H f o , AgO 7300cal / mol 30514 1
S o 2S Ag , 29 8 S O2, 29 8 S Ag 2O, 29 8 2
1
2 10.2 49 29.1 66.044 J / mol.K 2
G o H o T S o 30514 T S o 30514 66.044T
when Ag2O begins to decompose, G G o RT ln J 0 ie 30514 66.044T RT ln PO 2 0
(a)in (a) in pure oxygen at 1 atm,
RTlnP O2 = 0 30514-66.044T = 0 T = 462K
(b)in (b)in air at P total = =1 atm , P O2 =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 2
O2 ( g ) Cu2 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 2
O2 ( g ) Cu2 O( s) ,
o
39850+15.06T. G (in calories ) is – 39850+15.06T.
(5.4, p145) Solution: (a) When the equilibrium is reached, G G o RT ln J G o－ ln PO 2
T=
500C
1 RT ln PO2 2
0
4.18 ( 39850 15.06T ) 1 RT 2
= 773K 4.18 ( 39850 15.06 773) 1 8.314 773 2 1.14 10 26 atm
ln PO 2
PO 2
36.69
(b)at (b) at T=300C=573K, Although the equilibrium P O2 is very low, kinetically the reaction is not favoured and reaction speed is very slow. So 300 C is not
suitable
at
At T=800C=1073K, lnPO2 =-22.2, PO2 =2.2810-10 atm. At 800C, 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. either. (c ) The equilibrium oxygen pressure remains the same when the total pressure increases, which means a higher purity level of N 2 . 5. The solubility of hydrogen(P H2 = 1 atm ) in liquid copper at 1200C 7.34cm3(STP) per 100g of copper. Hydrogen in copper exists in monatomic form. (a) Write the chemical equation for the dissolution of H 2 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 melt of copper at 1200C
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 H 2 content to 0.1 cm3(STP) per 100g ? Note: STP means standard temperature and pressure(298K and 1 atm). (5.5, p145) Solution: (a) H 2(g) = 2H (b) H K
1
a
2
P H 2 1
K a 2
1
P H 22
1atm, H 7.34cm3 / 100gCu
is a constant,
[ H ] 1/ 2 H 2
P
[ H ]' ' 1/ 2 H 2
(P )
(P
' 1/ 2 H 2
)
1/ 2 P H 2 [ H ]'
[ H ]
0.1 7.34
0.0136
( P H ' 2 ) 0.00019atm 18.56 10130 18.8Pa
(c ) The amount of H2 needed to be brought out by Ar is: PV
n
RT
10130 (0.5 0.1) 10 6 8.314 298
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. P H ' 2V bubble nRT 4.05 10 2 P H ' 2 4.05 10 2 / 18.8 0.00215m 3 2.15 L
2.15L Ar is needed to be bubbled into the melt.
6. The following equilibrium data have been determined for the reaction:
NiO NiO(s) CO( g ) Ni(s) CO2 ( g ) -3
T( C)
K 10
663
4.535
716
3.323
754
2.554
793
2.037
852
1.577
(a) (b) (c) o
o
(d)Plot (d) Plot the data using appropriate axes and find H , K and G at 1000K; (e)Will (e) Will an atmosphere of 15%CO 2, 5%CO, and 80%N2 N2 oxidize
nickel at 1000K? (5.6, p145) Solution: (a)
d ln K a
H o
1 d ( ) R T
Plot ln K a ~ 1 / T 8.6
Kduishu Linear Fit of Data1_Kduishu
8.4
8.2
8.0 a
K n l 7.8
89:; <%'+,=(++-$,>5#
7.6
7.4
7.2 0.88 0.88
0.90 0.90
0.92 0.92 0.94 0.94
0.96 0.96
0.98 0.98
1/T, 10
1.00 1.00
1.02 1.02 1.04 1.04
1.06 1.06
1.08 1.08
-3
. d ln K a dT
H o R
6003 H o R 6003 49909 J
At T=1000K, lnK a =8.01, K a = 3010 o G1000 RT ln K a 8.314 1000 8.01 66600 J 66.6kJ
(b) G G
o
RT ln J RT ln K a RT ln J RT ln
J
15% 5%
J K a
3 K a
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 1
1
1
1
N K a, N 2 PN 22 ， H K a , H 2 PH 22 , For N2 dissolving :
[ N ] 1/ 2
P N 2
[ H ]
For H2 dissolving： P1/ 2 H 2
[ N ]' ( P N ' 2 )1 / 2
[ H ]' ( P H ' 2 )1 / 2 3
（a）For dissolving N2, P N2 = 1 atm, [N]=35cm /100g melt, ‘
[ N ]
( P N ' 2 )1 / 2 [ N ] 1/ 2 N 2
P
35 (0.5)1 / 2 24.75cm 3 / 100g
melt
3
’
similarly: [H] =24.75cm /100g melt 3
total gas : [H] ＋[N] = 49.5 cm /100g melt (b)
3
[H] =24.75 cm /100g melt
(c ) [H]＋[N] =
1/2
1/2
[N](0.33) /1+[H](0.33) /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 ) G o 667000 25.0T ln T 510T free energy change
Gibbs
is
.
(J)
(a) Calculate the
equilibrium pressure of SiO gas at 1100K; (b) For the reaction above, o
o
calculate H and S at 1100K; (c) Using the Ellingham chart (Figure 5.7), estimate the pressure of oxygen (O 2) in equilibrium
with the materials materials in the furnace.
(5.8, P146)
2 (a) G o RT ln K RT ln PSiO 2 RT ln PSiO
At 1100K, o
G =667000+25.0TlnT-510T
= 667000+25.01100ln1100-510 1100 =667000+192584-561000 =298584 -2RTlnPSiO =298584 lnPSiO =-16.32 PSiO = 8.110-8 (atm) (b
o
)G =667000+25.0TlnT-510T =-RTlnK ln K
667000 RT
d ln K （
H o R
667000 RT
2
25 R
ln T 510
25 RT
667000
) dT (
25 R
667000 R
25 R
T )d (1 / T )
H o R
d (1 / T )
T
H o 667000 25T o
T = 1100K, H = 639500J S o
H o G o T
667000 298584 1100
334.9 J / K
-30
(c ) PO2 =10 atm 9. What is the pressure of uranium (gas) in equilibrium with uranium dicarbide G f o
DAT DA TA: At 2263K, Vapor (5.9,p146)
pressure
for UC2 is – 82,000 82,000 cal/mol of
pure
uranium
is:
atm, uranium ) 25.33 ln P(atm
Solution:
T
(T in K )
G f o 82000cal / mol 342760 J
U ( g )
C ( s ) UC 2( s )
G f o ,UC 2 RT ln K RT ln( RT ln Pu ( g )
ln Pu ( g )
100000
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 T
25.35
100000 2263
18.89
0.6 108 (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
the 2 Fe H 2
3 H 2 2 Fe 3 H 2 O reaction?
3 O2 Fe 2 O3 2 1 O2 H 2 O 2
Is
What is the enthalpy change, in joules, for it
exothermic
G o 810250 254.0T G o 246000 54.8T
(5.10)
or
endothermic?
Solution: 3 O 2 Fe 2 O 3 2 1 H 2 O 2 H 2 O 2 2 Fe
Fe2 O3
(1)
G1o 810250 254 .0T
( 2)
G 2o 246000 54.8T
3 H 2 2 Fe 3 H 2 O
(3)
G3o
G3o 3G2o G1o 3 24600 54.8T (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 DAT DA TA
CaCO3
CaO CO2
for the pressure press ure of carbon dioxide in equilibrium equilibriu m with CaO
and CaCO3: Temperature (K)
Pressure (atm)
1030
0.10
921
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)
H o 1 d , K PCO R T H o 1 d ln PCO 2 d R T o PCO 2, 2 H 1 1 ln PCO 2,1 R T 2 T 1 o 0.01 1 H 1 ln 0.1 R 921 1030 H o 166528 J d ln K
2
the reaction is endothermic endothermic (b)P (b) PCO2 =1atm ln
1 0.1
H o 1 R
(
T
1 1030
)
T 1168K
At 1168K, the equilibrium pressure of CO 2 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). 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
G f o 174000 48.7T
CO
G f o 28000 20.2T
(a). The reaction is: MgO(s) C (s) CO( g ) Mg( g ) (b).
G o G f o ,CO G f o , MgO 146000 68.9T G o RT ln K RT ln(PCO ( g ) P Mg ( g ) ) (146000 68.7T )
PCO
P Mg
(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 (SiO 2), 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; o
(b)Using the accompanying data, where G 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
1 O 2 ( g ) H 2 O ( g ) 2 Si O 2 SiO 2 ( s )
H 2 ( g )
G o 246000 54.8T G o 902000 174T
5.13, P147 Solution: (a)
Si( s) 2 H 2O( g ) SiO2 (s) 2 H 2 ( g )
(b) 1 O2 ( g ) H 2 O ( g ) (1) 2 Si O 2 SiO 2 ( s ) ( 2)
H 2 ( g )
G (o1) 246000 54.8T G (o2) 902000 174T
Si(s ) 2 H 2 O( g ) SiO2 ( s) 2 H 2 ( g )
(3)
G3o
G3o G2o 2G1o 902000 174T ( 246000 54.8T ) 2 410000 64.4T G3o RT ln K 410000 64.4T 410000 64.4 1273 31 At T 1273 K , ln K 8.314 1273 K 2.9 1013
(c ) 2
P H 2 ( g ) 13 K 2.9 10 P H 2O ( g ) P H 2 ( g ) P H 2O ( g ) P H 2O ( g ) P H 2 ( g )
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 (BaCO 3) in a furnace open to the atmosphere (P = 1 atm). (a)Write (a) Write the equation of the decomposition (witherite and BaO are immiscible). (b)Based (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) (endothermic) or evolved (exothermic). (exothermic). (c)How (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 f o ( 29 8)
H f o ( 29 8)
Solution: (a) BaCO BaCO3 ( s) BaO BaO ( s) CO2 ( g ) (b) CP = 0 H H f o ,CO 2( 298) H f o , BaO( 29 8) H f o , BaCO3( 29 8) 94 133 (219) 64kcal 267.52kJ
the reaction is endothermic (c ) At 298K, G29o 8 G f o ,CO 2, 29 8 G f o ,CaO, 29 8 G f o ,CaCO3, 29 8 94 126 272 52kcal 217.36kJ G29o 8 H 29o 8 T S 29o 8 H 29o 8 G29o 8 (267.52 217.36) S 29o 8 168 J / mol.K 298
T
G H T S o T
o T
o T
when PCO2=1 atm,
GT o 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: MgCl2 (l ) H 2 ( g ) Mg ( g ) 2 HCl ( g )
(a)Calculate (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 (b) Calculate the maximum vapor pressure of H2O that can be
tolerated in the hydrogen without causing the oxidation of the Mg vapor. DATA o
Reaction
G at 1200C
Mg(g)+Cl2(g)
=
MgCl (l)
－425484 J
H2 (g) + Cl2(g)
=
2HCl(g)
－207856 J
Mg(g) +1/2O2(g)
= MgO(s)
H2 (g) + 1/2O2(g)
－437185 J －165280J
= H2O(g)
（5．15， p48）
Solution: (a)
MgCl2 (l ) H 2 ( g ) Mg ( g ) 2 HCl ( g )
Mg(g)+Cl2(g)
=
(1)
G1o o 2
MgCl (l)
(2)
G
2HCl(g)
(3)
G
－425484 J
H2 (g) + Cl2(g)
=
－207856 J
G1o G3o G2o 207856 425484 217628 J
o 3
G RT ln K RT ln o 1
2 P HCl P Mg ( g ) (g)
P H 2( g )
8.314 1473 ln
2 P HCl P Mg ( g ) (g )
P H 2( g )
217628 ln
2 P HCl P Mg ( g ) (g)
P H 2( g )
2 P HCl P Mg ( g ) (g )
P H 2( g ) P H 2 ( g )
＝17.78
5.27 10 7
P Mg ( g ) P HCl 1,
P HCl
2 P Mg ( g )
let PMg(g) = x, PHCl = 2 x, PH2 = 1-3 x 1 3 x x (2 x) x
Mg(g) +
2
5.27 10 7
1.6 10 3 (atm)
H2O(g) = MgO(s)+ H2 (g)
Mg(g) +1/2O2(g)
= MgO(s)
o 4
(4)
G
o 5
－
o 6
－
(5)
G
(6)
G
437185 J H2 (g) + 1/2O2(g)
= H2O(g)
165280J
G4o G5o G6o 437185 (165280) 271905 J G4o RT ln K RT ln
P H 2( g ) P H 2O ( g ) P Mg ( g )
8.314 1473 ln
P H 2 ( g ) P H 2O ( g ) P Mg ( g )
271905 ln
P H 2 ( g ) P H 2O ( g ) P Mg ( g )
1.6 10
＝22.2
3
P H 2O ( g ) 1.6 10 3
ln P H 2O ( g ) P H 2O ( g )
22.2
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 (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 Table 5.1. (5.16, 148) Solution:
H 2 O( g ) C (s) H 2 ( g ) CO( g )
(a)
K
PCO ( g ) P H 2( g ) P HO 2( g )
G o G f ,CO G f , H 2O 111710 87.65 1100 (246740 54.81 1100) 21676 J
G o RT ln K ln K 2.3 K 9.97
(b) let P H 2 ( g ) K
x atm, x
2
1 2 x x 4.14 atm
PCO ( g )
x atm,
P H 2O ( g )
(10 2 x) atm
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
the
RT ln Zn
following
equation
for
temperature
1000-1500K:
2 38300 xCu , where xCu is the mole fraction of copper.
Calculate the partial pressure of zinc P Zn over a solution of 60 mol % copper and 40 mol % zinc at 1200K. 1200K . The vapor pressure of pure zinc
is 1.17 atm at 1200K. (7.1, p196) Solution: ln Zn
2 38300 xCu
RT
2 38300 0.6Cu 1.38 8.314 1200
0.25 Zn x Zn 0.25 0.4 0.1 P Zn p a Zn 0.1 1.17 0.117(atm)
Zn
a Zn P Zn
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 According to Gibbs – Duhem Duhem Equation: x Zn d (ln a Zn ) xCu d (ln aCu ) 0 x Zn d (ln Zn ) x Zn d (ln x Zn ) xCu d (ln a Cu ) xCu d (ln x Cu ) 0 x Zn d (ln x Zn )＋ xCu d (ln xCu ) dx Zn
dxCu dx Zn d (1 x Zn ) 0
x Zn d (ln Zn )＋ xCu d (ln Cu ) 0 d (ln Cu )
ln
0
x Zn
d (ln Zn )
xCu
d (ln Cu )
xCu
1
xCu
1 ln Cu
38300 RT
x Zn xCu
38300 RT
38300 RT
2 xCu dxCun x Z n
2 x Zn dxCu 1
38300 RT
2 x Zn dx Zn
2 x Zn
3. (a) At 900K, is Fe3C a stable compound relative to pure Fe and graphite?( 7.3, 196) (b)At (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 DATA
AT 900K 90 0K,,
3Fe C ( graphite)
Fe3C
G o 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 N H 0.94 g
9
a H 0.96 g
1
0.89 0.85 0.75 0.65 0.53 0.43 0.33 0.20 0.06 3
1
3
3
7
7
0
7
3
0.92 0.90 0.84 0.76 0.65 0.54 0.43 0.27 0.09 9
8
0
5
0
2
2
8
2
(7.4,196) Solution: N H 0.94 g
9
a H 0.96 g
1
0.89 0.85 0.75 0.65 0.53 0.43 0.33 0.20 0.06 3
1
3
3
7
7
0
7
3
0.92 0.90 0.84 0.76 0.65 0.54 0.43 0.27 0.09 9
8
0
5
0
2
2
8
2
Hg 1.01 1.04 1.06 1.12 1.17 1.21 1.24 1.31 1.34 1.46
3
Plot lnHg ~xHg
lnB Linear Fit of Data1_lnB
0.40 0.35 0.30 0.25 g H
0.20
n l
0.15
896?<+'-.&@+'-.,A6?
0.10 0.05 0.00
0.0
0.2
0.4
0.6
0.8
1.0
xHg
d (ln Bi )
ln
0
x Hg x Bi
d (ln Hg ) x Bi
d (ln Bi ) 0.391
ln Bi
1
1 ( x Bi
x Hg x Bi
(0.391)dx Hg
1)dx Bi
0.391(ln x Bi x Hg )
when xBi = 0.4, xHg =0.6 ln Bi Bi
0.391(ln 0.4 0.6) 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
solution
(cm3/mol)
is
given
by
:
80 x B 2.5 x A x B
(a)Compute (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 (b) Compute the volume of mixing as a function of composition. (7.5, 196) Solution: V 100 x A
the
calculated
80 x B 2.5 x A x A
partial
variables
are
as
follows:
V A
V 100 2.5 x B x A T , P , x B
V B
V 80 2.5 x A 82.5 2.5x B x B T , P, x B
105
100 partial volume of A partial volume of B molar volume of the solution
95
l o m / 90 3 m c 85
80
0.0
0.2
0.4
0.6
0.8
1.0
xB
(b)
100, V B 80 V M V (100(1 x B ) 80 x B V (100 x B ) 2.5 x A x B V A
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 473C, the system Pb-Sn exhibits regular solution behavior, and the activity coefficient of Pb is given by:
log Pb 0.321 x Pb
Write the corresponding equation of the variation of
2
Sn
.
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 = 43905-43.644 43905- 43.644T T,
Melting point
=987K For MgF2:
G = 58702-38.217T, 58702-38.217T,
Melting point
=
1536K The free energy of mixing (J/mol) for liquid mixture MgCl 2 and MgF2 is given by: x MgF (2556 25( x MgF xMgCl )) . G Mix 2 RT ( x MgCl ln x MgCl x MgF ln x MgF ) x MgCl 2
2
2
2
2
2
2
2
Compute the maximum solubility of MgF 2 in liquid MgCl 2 at 900C. MgCl2 does not dissolve in solid MgF 2. (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 ? DAT DA TA
Quantities Quantit ies of Mixing Liquid Alloys at 1000K
x Mg
0.1
G M (cal cal / mol)
H M (cal cal / mol)
S M (cal cal / mol)
C p [cal /(mol.K )]
-800
-300
0.5
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 ZnSO 4 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 Assume that both solid solutions are ideal? Data: for Fe
Fe( s)
1 2
O2
FeO(s)
For Mn
Mn( s)
1 2
O2
MnO(s)
G 384700 72.8T
G 259000 62.55T
Solution: (a) F=(5-2-1-1)-2+1=0 G 0
(b) P
1 2
RT ln PO2
O2 1 exp(
2 G 0
PO2 2 exp(
RT
2G 0 RT
) 3.0 10 21 ) 2.6 10 33
P 0.999 PO2 1 0.001PO2 2 3.0 10
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 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% mol% of 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 emperatur e
Phase
Composition Composi tion
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) 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 (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 answer parts A-C (a)at (a) at
what temperature does the the first first solid solid formed formed and what is the the
composition of that solid? (b)At (b) At what temperature does the last liquid solidify, and what is the composition of the last liquid (c)Which (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 BaCl 2-NaCl-KCl-CaCl2 assume for the purpose of this problem that there is no solid solubility of the compounds in one another. (a)on (a) on the diagram trace the path of the liquid composition when a material consisting of BaCl 2 60%, NaCl 20%, KCl-CaCl220% is cooled from 900 ℃. Assume that equilibrium is maintained at all temperature.
(b)Sketch (b) Sketch two blank ternary diagram and draw in the isothermal section at 750 and 650 ℃ (c)What (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
(b) 750℃ 20
Soli 750
750 Na
Li
Soli
KCll CaCl KC CaCl 20
600℃ BaCl2
20%
Solid
Liquid NaCl
KCl, CaCl2 20%
(c) 20%liquid;
30%BaCl2+ 20%NaCl+50% KCl-CaCl2
