Post on 18-Nov-2015
description
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:
)/(363
102.20721]108.4)25327(3.29[
21
21
)(
233
22
smV
vnn
WQ
nMvmvW
HTCnQQQ
absorb
meltingpmeltincreaseabsorb
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
Solution
)/(24560208.975
)/(121606024
10467000//
)(104670001868.4102500
sin
3
SJthmgP
SJtQtWP
JQ
gincrea
Burning
Burning
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
)(6.436)106103(1075.72
)(106)105.0(4)105.0(
34
)101(
2325
2326
36
31
JSW
mnSS Singletotal
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:
)(180118298
)(1185.2
298101013255010
10101325)5500(1)(
)(118)101(298
)(
)(
0
3
3
4.0
/
00
KTTT
KR
RnCWT
b
KT
PP
TTAdiabatica
p
CR P
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 oened 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 te gaS in the tank? The heat cap!city
mf the gas, Cp and Cv each ay be(assumed to be c/nsuant over th
temperature rang!spanNed by the dperiment. 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.
solution
0/
0
00
/
00
)0
(
)(
TP
PTT
PP
TTAdiabatic
P
P
CR
CR
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
DATA:
)()()(
2
2
4
gOHgCOgCH
FOR
80.5705.9489.17
]/[0298 molgKcalH
solution
)1000/(9.2610252103048.0
110
1076.191)/(76.191
)89.1780.57205.94()2(22
3333
3
298
298
2224
422
SCFBtumolgKcalH
HHHHOHCOOCH
CHOHCO
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
2
2
2
2
4
ON
OHCOCHfor
2.82.89.117.13
16)/( CmolcalCP
Solution
)(210448.1125.9
100076.191298
)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(
%87.0%%12.72%
%43.17%2%
%71.8)11.1(2
21791.123
1%
22)(
0
,,
2
2
22
2
2224
KTTT
CmolcalXCCb
ON
COOH
CO
OHCOOCHa
iippp
)/(1644)
0224.011868.4
48.11)8001600(48.1125.9189570(
102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191
)(
)/(87.848.11/]21
1002.22.816[
)(
)/(3214)
0224.011868.4
48.11)2981600(48.1125.9100076.191(
102800000)/(280000040000020001200
)(
33
min
,,,,298
,,
33
min
hmV
molgcal
dTnCnCHH
CmolcalXCC
d
hmV
hKJPC
gConsu
iirpiipp
iiprp
gConsu
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.
Solution
)298(0055.0)298(62.35
011.062.35011.062.35
22803)2981000(2
)2981000(
12113)298700(2
)298700(
]2
[
22298
22
22
2982
TTHTC
ba
ba
ba
TbaTbTdTadTCH
TP
TP
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
molJH
molJH
COf
COf
/393296
/1104580
,298,
0,298,
2
)/(10184.403.29
)/(1067.11010.492.19
)/(1037.81020.935.44
)/(1042.01097.345.28
3,
253,
253,
253,
2
2
2
molKJTC
molKJTTC
molKJTTC
molKJTTC
NP
OP
COP
COP
Solution ?
0)499.0321.018.1(
)1067.01019.277.28(28.28283
1067.01038.477.28
9.0)1019.01058.528.33(2.0282838
)(
)/(1019.01058.528.33
722.0278.0)/(1067.01038.477.28
1.065.005.02.0)(
)/(282838110458393296
%2.72%8.27
%10%65%5%20
)4/(11
22
29812
733298
1523
733
253
253
298
,,,,298,
253
,,,,,
253
,,,,,,,
0,298,
0,298,298,
2
2
2
2
2
22
22
22
222
T
TTT
TTT
dTTT
dTTT
dTnCnCnHH
molKJTT
CCnCCmolKJTT
CCCCnCCa
molJnHnHH
NCOproductionONCOCOreationthen
ONairmoleneedfuelmolewhen
COOCO
T
T
T
iirpiippii
NPCOPiipprp
OPNPCOPCOPiipprp
ipfirfi
dTTTQ
dTTTQ
bT
TTT
TTT
dTTT
dTTT
dTnCnCnHH
T
T
T
iirpiippii
9.0)1019.01058.528.33(2.0282838
9.0)1019.01058.528.33(2.0282838
)(
0)499.0321.018.1(
)1067.01019.277.28(28.28283
1067.01038.477.28
9.0)1019.01058.528.33(2.0282838
)(
2531250
298
1250298
2531250
298
1250298
29812
1250298
1523
1250
253
253
298
,,,,298,
1.10 (a) for the reaction 2221 COOCO ,what is the enthalpy of
reaction ( 0H ) 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 fH at 298 K
)/(393000)/(110000
2 molJCOmolJCO
Heat capacities [J/(mol K)] to be used for this problem N2=33,
O2=33, CO=34, CO2=57
Solution
)(21100)298)(39889.0(222.0283000
0
)/(3975.03325.057
)/(33111.034222.033666.033)(
%,75%%,251.11100
2.22%
%1.11%%,6.66%%,2.222.0/25.01
5.0%)(
)/(283000393000110000)(
,0
,,,
,,,
22
22
0,298,
0,298,
0
KTT
dTCnHH
KmolJXCC
KmolJXCCC
NCOproduct
ONCOfuelb
molJnHnHHa
Ppp
iPriPr
iPpiPp
iPfirf
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?
DaTA(k J?mol)
2COCOFOR
513.393523.110
)/( molkJH f
22
2
2
,)(
ONgOH
COCOFOR ??
3 45 05 73 3
]/[ Km o lJCP
Solution
OHOH
COOCOa
222
22
21
21)(
6.0)(04.0)(
24.0)(12.0)(
:
2
2
2
NnHnCOnCOn
Fuel
32.0)(08.0)(
:
2
2
NnOn
Air
92.0)(04.0)(36.0)(
:
2
2
2
NnOHn
COnFlue
)(98.1108
)(8108.53
106308.43
)/(8.533492.05004.05736.092.004.036.0
6308.43)08.241(04.0)523.11051.393(12.0
3
,, 222
222
KT
KT
KJCCCnC
KJHHH
NOHCOiirP
OHHCOCO
(b)repeat the calculation for 30% excess0combustion air at 298K
416.0)(104.0)(
:
2
2
NnOn
Air
)(8.1051
)(8.75388.57
106308.43)/(88.57
34024.034016.15004.05736.0
024.0016.104.036.06308.43
)08.241(04.0)523.11051.393(12.0
3
,, 2222
222
KT
KT
KJ
CCCCnCKJ
HHH
ONOHCOiirP
OHHCOCO
6.0)(04.0)(
24.0)(12.0)(
:
2
2
2
NnHnCOnCOn
Fuel
416.0)(104.0)(
:
2
2
NnOn
Air
024.0)(016.1)(
04.0)(36.0)(
:
2
2
2
2
OnNn
OHnCOn
Flue
Maret 8, 2013
(C)what is the adiabatic flame temperature when the blasp furnace
gas is preheated to 700K (the dry air is at 298K)
6.0)(04.0)(
24.0)(12.0)(
:
2
2
2
NnHnCOnCOn
Fuel
32.0)(08.0)(
:
2
2
NnOn
Air
92.0)(04.0)(36.0)(
:
2
2
2
NnOHn
COnFlue
)(6.1401
)(6.11038.53
10373.59
)/(8.533492.05004.05736.096.004.036.0373.59
)346.02804.05724.03312.0()298700()08.241(04.0)523.11051.393(12.0
3
,,
298700
222
222
KT
KT
KJCCCnCKJ
HHHH
NOHCOiirP
fuelOHHCOCO
(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?
6.0)(04.0)(
24.0)(12.0)(
:
2
2
2
NnHnCOnCOn
Fuel
008.04.015760
15)(
32.0)(08.0)(
:
2
2
2
OHn
NnOn
Air
92.0)(048.0)(36.0)(
:
2
2
2
NnOHn
COnFlue
)(1103
)(8052.54
106308.43)/(2.543492.050048.05736.0
92.0048.036.06308.43)08.241(04.0)523.11051.393(12.0
3
,, 222
222
KT
KT
KJCCCnC
KJ
HHH
NOHCOiirP
OHHCOCO
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
)/(310355.75.0)17981803(105.1541.53100
0223
1798,
1798,
1798
1803 ,
1803
1798 ,1803,
molcalH
HdTCdTCHLS
SLLPSP
LS
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)/(173104190059210
222
molcalHOHCuHOCu ,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
)(28.0)(7.2793600
1100027
27184)/(2718410700)9321000(29)298932(26
1000
932 ,
932
298 ,1000,
kWWP
molJ
HdTCdTCH SLLPSPl
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;)(600
)(302
104102
9927
275.1161
22711676000
KTKT
T
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
Solution:min)/(10573.2
)2080(1min/min54.63
1000)]4001356(73120[min/
33 mVVQ
Q
Water
Copper
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: )(139476010005)2060(184.4 WW
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?
Solution:
)/(375971822613101
%6.7182261
3101%
)/(31010224.0273373101325
molJQWU
molJVP
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 )(125,3341000)10018.42261( gmm
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), HV=2261J/g, Hm=334 J/g
Solution:leirreversib
gxxx )(138),1000(8018.4)8018.48022261(
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
)(25.6
)(746
60
104
27390
)2590(
2
4
mS
Wt
WP
StQT
TTW
H
H
LH
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.
Solution:
)(64374625.0203520273
43
75.0
WP
PT
TTP
QT
TTW
L
LL
LH
HH
LH
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.
Solution:
)(45760
33474625.020273
gmM
mP
PT
TTP
L
LL
LH
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.
Solution:
)(52.0)(393'60
284216.4
216.4300'5.0
%50
hpWP
PT
TTPP
QT
TTW
LL
LH
LL
LH
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:
1,2,
2,1,
21
2,2,
2,2,2
1,1,
1,1,1
98.825273525
27354050540
)(
HH
HH
HH
LH
HH
LH
PP
PP
PP
PT
TTP
PT
TTPa
.,)(
6286.0)(
1,2,
notisbokisac
PPb
HH
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:
)/(933.0)273323ln(5.0)
373323ln(5.0)ln()ln(
)(0
22
11 KJCCT
TCnTTCnS
JU
PPE
PE
P
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?
Solution:
)(3.69)(69371
36005000.29
)(
)(89.013054030540
)(
tonkgmTT
Tm
b
JQT
TTW
a
LH
L
HH
LH
)(9.191102525273
)(
JQ
QT
TTW
c
H
HH
LH
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
solution:)(23.2,
25100273100100040000
18)(
)(45.0,10004000018
)(
gmmb
gmma
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:)(5.102)(102474
202732030
)20480(2827
1000
kWWPPT
TTP
P
LLL
LHW
L
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: )(109.7
)(102.2
1040300273300)(
11
8
8
JtPQW
PTT
TPa
HH
LH
HH
)(1003.1
184.41010
)(103.4)(
34
6
11
mVQV
JQb
L
L
noW
PTT
TPcLH
HH
)(10626.1
1040540273540)(
8
8
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: )/(1006.13600
1000)(
)(055.0127320420)(
6 hkWhmgPb
JQT
TTWa HH
LH
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: )(1014.110
1527320273)(
)(77.33600/10152731520)(
55
5
kJQb
kWPT
TTPa
H
LL
LH
2.13
solution:
)/(81.6810ln314.877.45277.6282.4)/(152940)(
)/(67.4977.45277.6282.4)()/(152940)(
22)( 2
molKcalSmolcalHd
molKcalScmolcalHb
AlNNAla
2.14
solution:
)/(22574
12000)273
40273ln184.4273336
263273ln1.2(
)(40
0
,0
10
,
KJ
dTT
CTHdT
TC
mS WATERPm
mICEP
2.15
)(70428
)(2896100077
77300
2 JW
JQT
TTW LL
LH
2.16
)(4.3719))2.4300(314.85.13.83(300
2.4300
)(7.58663.832.4
2.4300
JQT
TTW
JQT
TTW
HH
LH
LL
LH
2.17
yesdQc
KJPPnRSb
JpdVnWQ
OUTa
)(0)(
)/(1.1910ln314.81ln)(
)(570410ln298314.81
0)(
0
2.18
)(1222
335273
02033560500
gm
mmT
TTL
LH
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)
Solution: molJ
PP
RTGwaterOH
iceOH
/9.1089523.0ln268314.8163.3012.3ln)5273(314.8
ln,
,
2
2
molJH /1085.5 3
)/(23.22268
)9.108(5850 KmolJT
GHS
STHG
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: molJKcalC bathP /418/100,
(a) gcalTCQ bathP /102/2.01002/,
This is the heat of mixing.
(b) The heat capacity of CP, alloy : )/(072.0
6.27424.0100
)254.0300(2,
,
Kgcal
TCC bathPalloyP
Assuming that the calorimeter can be applied to the maximum of
T C, the for mixing to form 1 gram of alloy:
10)'300(,1 TCQ bathP , )'(,2 TTCQ alloyP , 21 QQ
)'(10)'300( ,, TTCTC alloyPbathP
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, Tm S = H
)./(09.22273
6030 KmolJTHSm
(b) At 0 C, G =0
)./(62.37)./(1818.45.0)./(5.0, KmolJKmolJKgcalC iceP
)./(24.75)./(1818.40.1)./(0.1, KmolJKmolJKgcalC waterP
a reversible process can be designed as follows to do the
calculation:
molJ
HdTCC
dTCHdTC
HHHH
waterPicep
waterpicep
fu
/9.584160305)24.7562.37(
)(273
268 ,,
268
273 ,
273
268 ,
)3()2()1(
d
Ice, 0 C water, 0 C
water, -5 C ice, -5 C
1
2
3
4
)./(39.21
09.22268273ln)24.7562.37(
)(
3
273
268
,,
268
273
,273
268
,
)3()2()1()4(
KmolJ
SdTTCC
dTT
CSdT
TC
SSS
waterPicep
waterpicep
38.10939.212689.5841)4()4()4( STHG
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
molKJSo ./28.27298 . What percentage error would result if one assumed
that the specific entropy at 298K and 100 atm were equal to the
value of oS298 given above
DATAfor iron Cp = 24 J K-1mol-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) molmmolcm
cmgmolg
densityweightmolV iron /1010.7/10.7/87.7
/85.55 3633
(b) PT T
VPS
lV
T
VVPS 3
for iron:
))./((102.3
10151010.73
3
310
66
,
Kmolm
VPS
ironlironT
PS iron10102.3
( c )
)./(1021.3109.32010013.199102.3
10013.1)1100(102.3
35
510
510
KmolJ
S iron
2
298
1012.1%100% oiron
SSerror
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) molmmolcmcmg
molgV ice /1057.19/57.19/92.0/18 363
3
molmmolcmcmgmolgV water /1018/18/1
/18 3633
molmVfus /1057.136
According to the Clapeyron equation:
dTTH
VdP
TH
VdTdP
fus
fus
fus
fus
1
1
take definite integration of the above:
dTTV
HdP
T
fus
fus
273
10013.150
10013.1
155
013.06009
10013.1491057.1
10013.149273
ln
56
5
fus
fus
HVT
KT 8.272 (b) PaPainlbP 6323 1034568971050./1050
301.0150
PaP 610345
(c )
09.06009
103451057.1
10345273
ln
66
6
fus
fus
HVT
KT 46.249
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: molJmolJmolkcalH vap /04.117/102818.4/28 3
According to Claperon equation in vapor equilibrium:
)1()(lnT
dR
HPd vap
)11()(ln12
2
1 TTRH
Pd vapP
P
)11(ln122
1
TTRH
PP vap
)67311(
314.81004.117
101ln
2
3
4 T
KT 12022
The boiling point of zinc is 1202K.
2. Troutons rule is expressed as follows: bvap TH 90 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 Troutons rule to estimate the heat of vaporization of
mercury.
Solution: bvap TH 90
)6301
2981()(ln
1 RH
Pd vapP
07.1283.1090.2283.102986823ln P
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: molmmolcmcmgmolgV l /1018/18/1
/18 3633
vapor pressure changes with the total external
pressure, lv GG
)(ln 1,1,
2,eTl
e
e PPVPP
RT
)2.316710130(1018ln 61,
2,
e
e
PP
RT
000051.11,
2,
e
e
PP PaPe 36.31672, 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 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: )2450
11()(ln1 TR
HPd vap
P )2450
11(314.8
255000lnT
P
08.10430685lnT
P
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:
25.19)ln(755.015755)(ln TT
atmP and the vapor pressure of liquid
Zn varies with T as: 79.21)ln(255.115246)(ln TT
atmP . 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
079.21)ln(255.115246 TT
079.21)ln(255.115246 TTT
To solve the equation, let y1 = 21.79T-15246, y2 = 1.255TlnT. Plot
y-T of the functions, and the intersection is the answer.
400 600 800 1000 1200 1400 1600 1800 2000 2200
-6000-4000-2000
02000400060008000
1000012000140001600018000200002200024000260002800030000
y
T, K
y1 y2
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;
25.19)ln(755.01575579.21)ln(255.115246 TT
TT
054.2)ln(5.0509 TT
054.2)ln(5.0509 TTT
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.
400 600 800 1000 1200 1400 1600 1800 2000 22001000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
7500
8000
y
T, K
y11 y22
The intersection is 695.3K, and this is the triple point temperature.
(c )
)1()255.115246(
)1)(255.115246(
79.21255.115246)(ln
2
2
TdT
dTT
T
TTPd )1()(ln
Td
RH
Pd vap
TR
H vap 255.115246
TTRH vap 43.10126755)255.115246(
At Tb =1180K, molkJmolJTH vap /4.114/104.11443.10126755 3
(d) For solids,
)1()755.015755(
)1)(755.015755(
)755.015755()(ln
25.19)ln(755.015755ln
2
2
Td
dTT
dTTT
Pd
TT
P
)1()(lnT
dR
HPd fus
TR
H fus 755.015755 )755.015755( TRH fus
At triple point, Ttr = 695.4K
molkJmolJTH fus /6.126/106.126)755.015755(314.83
(e) dTCHd Pfus )(
755.0)(
dTHd
C fusP
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, LVPaatmPKT 1,10013.11,300 1511
molRT
VPnRTnVP 33
1
111111 1006.4300314.8
1010130,
%06.4)%(vapormol
In the final state, P2=2 atm, V2 = 1L
According to Clayperon equation:
KTT
TTRH
PPRT vap
6.458
)30011(
314.85000
12ln
11ln
2
2
221
2
molRT
VPnRTnVP 33
2
2222222 103.56.458314.8
10101302,
%3.5)%(vapormol
8. The melting point of gold is 1336K, and vapor pressure of liquid
gold is given by:
)(ln222.143522716.23)(ln KTT
atmP . (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)
TTRH vap 19.10361841)222.143522(
(a) At 1336K, kJJH vap 34810348133619.10361841 3
(b) Solution not possible;
(c) Solution not possible.
9. (a) At 298K, what is the Gibbs free energy change for
the following reaction? d i a m o n dg r a p h i t eCC
)1()222.143522(
)1)(222.143522(
)222.143522()(ln
2
2
Td
dTT
dTTT
Pd
(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.
DATA Density of graphite is 2.25g/cm3
Density of diamond is 3.51g/cm3
)/()298( molkJH
of
)./()298( KmolJHof
Diamond 1.897 2.38
Graphite 0 5.74
Solution: (a) diamondgraphite CC
molkJHHH o graphitef
odiamondf /1897,,
)./(36.338.274.5,, KmolJSSSo
graphitefo
diamondf molJSTHG /28.2898)36.3(2981897
(b) No, diamond is not thermodynamically stable
relative to graphite at 298K. (c ) molJPVG diamand /29.34101309951.3
1012 6
(c ) Assuming N atm , G = 0, reversible processes as
following can be designed to realize this,
)(14939028.2898194.0
28.28981013051.31012
25.21012
28.289810130)1)(VV(
)(V28.2898V
66
)3()2()1()4(
atmNN
N
NPP
GGGG
diamondgraphite
diamondgraphite
0,0,0
''dT
TC
dTCCT
T
pT
T pp
molJHH /1897298900
graphite, 298K,1atm
1
4
2
graphite, 298K, N atm diamond, 298K, N atm
diamond, 298K,1atm
(3)
KmolJSS ./36.3298900
molJSTHG /492136.39001897
Chemical Equilibrium
1. Calculate the partial pressure of monatomic hydrogen in hydrogen
in hydrogen
gas at 2000K and 1atm.
For )()(21
2 gHgH
KJSJH
o
o
/35.49
217990
298
298
For the reaction : )()(21
2 gHgH
035.33121314.8
23
21
2,)(, HPgHPpCCC
J
CHdTCHH Po
Poo
2128241702035.3217990
)2982000(2982000
2982982000
J
CHdTCHH Po
Poo
2128241702035.3217990
)2982000(2982000
2982982000
J
CSdTCSS Po
Poo
57.43298
2000ln035.335.49
2982000ln298
2000
2982982000
JSTHG 12568457.432000212824020000200002000
56.72000314.8
125684ln125684ln
lnlnln
)(
2/1
)(
2/1
)(
2/1
2/1)(
2000
22
2
2
gH
H
gH
H
gH
H
H
gHo
PP
PP
RT
PP
RTPP
RTKRTG
atmPP
PPP
HgH
HgHgH
0005.0562.71ln
1,1
)(
2)(2)(2
2. For the reaction : )()(21)( 2 SCoOgOsCo
TGo 6.1959850 , where oG 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
part a if the error in Ho term is estimated to be 500 cal? (5.2,
p144)
Solution: (a) At 1000 C, Go
=-59850+19.6T=-59850+19.6 (1000+273)
=-34899.2cal = -1458.79J/mol
At equilibrium:
atmP
P
JPRTP
RTKRTG
O
O
OO
o
12
2/1
1007.1
6.27ln
145879ln211lnln
2
2
2
2
(b) uncertainty in Ho = 500cal/mol = 2090J/mol
So uncertainty in Go = 500cal/mol = 2090J/mol
That means:
%6.28P286.1
25.0ln
2090ln21
2090ln21ln
21
22
'2
2
'2
2
'2
2'2
OO
O
O
O
O
O
OO
PPP
PP
PPRT
PRTPRT
Similarly, uncertainty in Ho =- 500cal/mol =- 2090J/mol
Go = -2090J/mol
%1.22P779.0
25.0ln
2090ln21ln
21
22
'2
2
'2
2'2
OO
O
O
O
OO
PPP
PP
PRTPRT
3. Calculate the temperature at which silver oxide (Ag2O) 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 molcalOforAgH f /73002
Assume that Cp = 0 for the decomposition reaction.
Solution: (a) Ag2O = 1/2O2 + 2Ag
30514/7300, molcalHH o AgOfo
KmolJ
SSSS OAgOAgo
./044.661.2949212.102
212 298,2298,2298,
TSTSTHG oooo 044.663051430514
when Ag2O begins to decompose,
0ln044.6630514
0ln
2O
o
PRTTieJRTGG
(a) in pure oxygen at 1 atm, RTlnPO2 = 0
30514-66.044T = 0
T = 462K
(b) in air at Ptotal = =1 atm , PO2 =0.21
Standard Entropy at 298K
[cal/(mol.K)] Ag2O 29.1 O2 49.0 Ag 10.2
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: )()(21)(2 22 sOCugOsCu
(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 )()(21)(2 22 sOCugOsCu , G
o (in calories ) is 39850+15.06T.
(5.4, p145)
Solution: (a) When the equilibrium is reached,
0Pln21ln
2Ooo RTGJRTGG
RT
TPO
21
)06.1539850(18.4ln 2
T = 500 C = 773K
atmP
P
O
O
262
2
1014.1
69.36773314.8
21
)77306.1539850(18.4ln
(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) 21221 Ha PKH gCucmHatmPH 100/34.7,1 32
21
aK is a constant,
PaatmPH
HPPP
HPH
H
HH
HH
8.181013056.1800019.0)(
0136.034.71.0
][]'[)(
)(][][
'2
2/122/1'
22/1'2
'
2/12
(c ) The amount of H2 needed to be brought out by Ar is:
molRT
VPn 66
106.1298314.8
10)1.05.0(10130
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.
LmPnRTVP
H
bubbleH
15.200215.08.18/1005.4
1005.432'
2
2'2
2.15L Ar is needed to be bubbled into the melt.
6. The following equilibrium data have been determined for the
reaction:
(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
T( C) K 10-3 663 4.535 716 3.323 754 2.554 793 2.037 852 1.577
)()()()( 2 gCOsNigCOsNiO
nickel at 1000K? (5.6, p145)
Solution: (a) )1(lnT
dRHKd
o
a
Plot TKa /1~ln
0.88 0.90 0.92 0.94 0.96 0.98 1.00 1.02 1.04 1.06 1.087.2
7.4
7.6
7.8
8.0
8.2
8.4
8.6
lnKa =2.01+6003(1/T)
lnK a
1/T, 10-3
Kduishu Linear Fit of Data1_Kduishu
.
JRHRH
dTKd ooa 4990960036003ln
At T=1000K, lnKa =8.01, Ka = 3010
kJJKRTG ao 6.666660001.81000314.8ln1000
(b)a
aa
o
KJ
KJRTJRTKRTJRTGG
3%5%15
lnlnlnln
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
21221
, NNa PKN ,21
22
1
, HHa PKH For N2 dissolving : 2/1'
2
'
2/12 )(
][][
NN PN
PN
For H2 dissolving 2/1'2
'
2/12 )(
][][
HH PH
PH
aFor dissolving N2, PN2 = 1 atm, [N]=35cm3/100g melt,
meltgcmP
NPNN
N 100/75.24)5.0(35][)(
][ 32/12/12
2/1'2
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:
the Gibbs free energy change (J)
is . (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
)(2)( 2 gSiOSiOsSi
TTTGo 510ln0.25667000
with the materials in the furnace. (5.8, P146)
(a) SiOSiOo PRTPRTKRTG ln2lnln 2
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
TH
TRR
H
TdRHTdT
RRdT
RRTKd
TRRT
K
o
o
o
25667000
25667000
)/1()/1()25667000()T
25667000ln
510ln25667000ln
2
T = 1100K, Ho = 639500J
KJT
GHSoo
o /9.3341100
298584667000
(c ) PO2 =10-30 atm
9. What is the pressure of uranium (gas) in equilibrium with uranium
dicarbide
DATA: At 2263K, for UC2 is 82,000 cal/mol
Vapor pressure of pure uranium is:
(5.9,p146)
ofG
Solution:
)(2)()( ssg UCCU
vapor pressure of uranium:
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:
OHFeHOFe 2232 323 What is the enthalpy change, in joules, for
the reaction? Is it exothermic or endothermic?
TGOHOH
TGOFeOFe
o
o
8.5424600021
0.254810250232
222
322 (5.10)
)(10000033.25),(ln KinTT
uraniumatmP
JmolcalGof 342760/82000
)(102.1ln
342760ln
ln)1ln(ln
8)(
)(
)()(
2,
atmP
PRT
PRTP
RTKRTG
gu
gu
gugu
oUCf
)(106.0
89.182263
10000035.2510000033.25),(ln
8)(
)(
atmPT
uraniumatmP
gu
gu
Solution:
TGOHOH
TGOFeOFe
o
o
8.54246000)2(21
0.254810250)1(232
2222
1322
oGOHFeHOFe 32232 )3(323
JHT
TTGGG
o
ooo
722506.8972250
)0.254810250(8.542460033
3
123
The reaction is an endothermic one.
11.Calcium carbonate decomposes into calcium oxide and carbon
dioxide according to the reaction 23 COCaOCaCO
DATA for the pressure of carbon dioxide in equilibrium with CaO
and CaCO3:
(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)
Temperature (K) Pressure (atm) 1030 0.10 921 0.01
JHRH
TTRH
PP
Td
RHPd
PKT
dRHKd
o
o
o
CO
CO
o
CO
CO
o
1665281030
19211
1.001.0ln
11ln
1ln
,1ln
121,2
2,2
2
2
the reaction is endothermic
(b) PCO2 =1atm
KT
TRH o
1168
)1030
11(1.0
1ln
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:
TGCO
TGMgOof
of
2.2028000
7.48174000
(a). The reaction is: )()()()( gMggCOsCsMgO
(b). )7.68146000()ln(ln
9.68146000
)()(
,,
TPPRTKRTG
TGGG
gMggCOo
oMgOf
oCOf
o
MgCO PP
(c ) Ptotal = 1 atm, PCO = 0.5 atm, PMg =0.5 atm 18.4)7.68146000()5.05.0ln( TRT
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 TGsSiOOSi
TGgOHgOgH
o
o
174902000)(
8.54246000)()(21)(
22
222
5.13, P147
Solution: (a) )(2)()(2)( 222 gHsSiOgOHsSi
(b)
TGsSiOOSi
TGgOHgOgH
o
o
174902000)2()(
8.54246000)1()()(21)(
)2(22
)1(222
oGgHsSiOgOHsSi 3222 )3()(2)()(2)(
13
3
123
109.2
311273314.8
12734.64410000ln,1273
4.64410000ln4.64410000
2)8.54246000(1749020002
K
KKTAt
TKRTGT
TTGGG
o
ooo
(c )
ppmPPPP
PP
K
gH
gOH
gOH
gH
gOH
gH
186.010186.01038.5
1
1038.5
109.2
66
)(2
)(2
6
)(2
)(2
13
2
)(2
)(2
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)]
ofG )298(
ofH )298(
CO2 -94 -94 BaO -126 -133 BaCO3 -272 -291 (Assuming that CP,CO2+ CP, BaO = CP,BaCO3)
Solution: (a) )()()( 23 gCOsBaOsBaCO
(b) CP = 0
kJkcal
HHHH o BaCOfo
BaOfo
COf
52.26764)219(13394)298(3,)298(,)298(2,
the reaction is endothermic
(c ) At 298K,
oT
oT
oT
ooo
ooo
oCaCOf
oCaOf
oCOf
o
STHG
KmolJT
GHS
STHGkJkcal
GGGG
./168298
)36.21752.267(
36.2175227212694
298298298
298298298
298,3,298,,298,2,298
when PCO2=1 atm,
KTTieGoT
15921682675201,0
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 H2 gas
over liquid MgCl2 at 1200 C. The reaction in this last step is:
)(2)()()( 22 gHClgMggHlMgCl
(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
515 p48
Solution: (a) oGgHClgMggHlMgCl 122 )1()(2)()()(
Mg(g)+Cl2(g) = MgCl (l) (2) Go2
425484 J
H2 (g) + Cl2(g) = 2HCl(g) (3) Go3
207856 J
JGGG ooo 217628425484207856231
Reaction Go at 1200 C Mg(g)+Cl2(g) = MgCl (l) 425484 J H2 (g) + Cl2(g) = 2HCl(g) 207856 J Mg(g) +1/2O2(g) = MgO(s) 437185 J H2 (g) + 1/2O2(g) = H2O(g) 165280J
)()()(2
7
)(2
)(2
)(2
)(2
)(2
)(2
)(2
)(2
1
2,1
1027.5
78.17ln
217628
ln1473314.8lnln
)(
)(
)()(
gMgHClHClgMggH
gH
gMg
gH
gMg
gH
gMg
gH
gMgo
PPPPPP
PP
P
PP
P
PP
P
PPRTKRTG
gHCl
gHCl
gHClgHCl
let PMg(g) =x, PHCl = 2x, PH2 = 1-3x
)(106.1
1027.5)2(
31
3
72
atmxxx
x
Mg(g) + H2O(g) = MgO(s)+ H2 (g) (4) Go4
Mg(g) +1/2O2(g) = MgO(s) (5) Go5
437185 J
H2 (g) + 1/2O2(g) = H2O(g) (6) Go6
165280J
JGGG ooo 271905)165280(437185654
)(1028.2
2.22ln
2.22106.1
106.1
2.22ln
271905
ln1473314.8lnln
10)(2
)(2
3)(2
3
)()(2
)(2
)()(2
)(2
)()(2
)(24
atmP
PP
PPP
PPP
PPP
RTKRTG
gOH
gOH
gOH
gMggOH
gH
gMggOH
gH
gMggOH
gHo
16. A common reaction for the gasification of coal is:
)()()()( 22 gCOgHsCgOH
(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: )()()()( 22 gCOgHsCgOH
(a) )(2
)(2)(
gHO
gHgCO
PPP
K
JGGG OHfCOf
o
21676)110081.54246740(110065.871117102,,
97.93.2lnln KKKRTGo
(b)
atmxx
xK
atmxPatmxPatmxPlet gOHgCOgH
14.4
97.921
)210(,,2
)(2)()(2
(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 following equation for temperature 1000-1500K: 238300ln CuZn xRT , 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:
)(117.017.11.0
1.04.025.025.0
38.11200314.8
6.03830038300ln
22
atmaPPxa
RTx
Znp
ZnZn
ZnZnZn
Zn
CuCuZn
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:
2
11
1
ln
0
Cu
Cu
38300ln
238300238300
238300)(ln
)(ln)(ln
0)(ln)(ln0)1()(ln)(ln
0)(ln)(ln)(ln)(ln0)(ln)(ln
ZnCu
Zn
x
ZnCu
x
Zn
Cun
x
CuCu
ZnCu
ZnCu
ZnCu
CuZnZn
ZnZnCuZnCuZnZn
CuCuCuCuZnZnZnZn
CuCuZnZn
xRT
dxxRT
dxxRT
dxxRTx
xd
dxxd
dxdxxddxdxdxxdxxdx
xdxadxxdxdxadxadx
nZCu
Cu
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, JGCFeCFe ographite 34633 3)(
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
g
0.94
9
0.89
3
0.85
1
0.75
3
0.65
3
0.53
7
0.43
7
0.33
0
0.20
7
0.06
3
aH
g
0.96
1
0.92
9
0.90
8
0.84
0
0.76
5
0.65
0
0.54
2
0.43
2
0.27
8
0.09
2
(7.4,196)
Solution:
NH
g
0.94
9
0.89
3
0.85
1
0.75
3
0.65
3
0.53
7
0.43
7
0.33
0
0.20
7
0.06
3
aH
g
0.96
1
0.92
9
0.90
8
0.84
0
0.76
5
0.65
0
0.54
2
0.43
2
0.27
8
0.09
2
Hg 1.01
3
1.04 1.06 1.12 1.17 1.21 1.24 1.31 1.34 1.46
Plot ln Hg ~xHg
0.0 0.2 0.4 0.6 0.8 1.0
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
lnHg=0.395-0.391x
Hgln
Hg
xHg
lnB Linear Fit of Data1_lnB
)(ln391.0ln
)11(391.0)(ln
)391.0()(ln)(ln
1
ln
0
HgBiBi
Bi
x
BiBi
HgBi
HgHg
Bi
HgBi
xx
dxx
d
dxxx
dxx
d
Bi
when xBi = 0.4, xHg =0.6
107.11.0)6.04.0(ln391.0ln
Bi
Bi
7.5 For a given binary system at constant T and P, the liquid molar
volume of the solution (cm3/mol) is given by :
BABA xxxxV 5.280100
(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 calculated partial variables are as follows:
AABA xxxxV 5.280100
BxPTA
A xxVV
B
5.2100,,
BAxPTB
B xxxVV
B
5.25.825.280,,
0.0 0.2 0.4 0.6 0.8 1.0
80
85
90
95
100
105
cm3 /m
ol
xB
partial volume of A partial volume of B molar volume of the solution
(b)
BA
BBBM
BA
xxxVxxVV
VV
5.2)100(80)1(100(
80,100
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: 2132.0log PbPb x .
Write the corresponding equation of the variation of 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 = 43905-43.644T, Melting point
=987K
For MgF2: G = 58702-38.217T, Melting point =
1536K
The free energy of mixing (J/mol) for liquid mixture MgCl2 and
MgF2 is given by:
))(252556()lnln(222222222 MgClMgFMgFMgClMgFMgFMgClMgClMix
xxxxxxxxRTG .
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
Mgx
)/( molcalGM )/( molcalHM )/( molcalSM )]./([ KmolcalCp
0.1 -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 ZnSO4 exist in equilibrium?
Solution:
(a) 422 ZnSOOZnS
(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 TG
sFeOOsFe
55.62259000
)(21)( 2 For Mn
TG
sMnOOsMn
8.72384700
)(21)( 2
Solution: (a) F=(5-2-1-1)-2+1=0
(b)
21
330
210
0
100.32001.01999.0
106.2)2exp(2
100.3)2exp(1
ln21
22
2
2
2
OO
O
O
O
PPPRT
GP
RTGP
PRTG
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) CuAg solution (b)F=C-P=(2-1)-1=0 (c) Ag phase in the
solution increases and Cu 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
8 38.5
99 0
1000+ Liquid 70 50.8
9 49.2
98 0
1000- Liquid _ 0
7 63.7
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 , (12%) (b) 2300 , liq(95%) (c) 8.2%
(composition is 24% )+91.8%(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-CaCl2BaCl2+ NaCl+ KCl-CaCl2
20%
20%
BaCl2
NaCl KCl, CaCl2
(b) 750
600
20%
20%
BaCl2
NaCl KCl, CaCl2
Liquid
Solid
20%
20%
BaCl2
NaCl
KCl, CaCl2
Liquid
Solid 750
Solid
750
(c) 20%liquid; 30%BaCl2+ 20%NaCl+50% KCl-CaCl2