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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)

Qabsorb?Qincrease?Qmelt?n(Cp?T??Hmelting)121mv?nMv2Solution: 22Qabsorb?WW?n[29.3?(327?25)?4.8?103]?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

1n207.2?10?3v22QBurning?2500?103?4.1868?10467000(J)SolutionP?W/t?QBurning/t?Pincreasing10467000?121(J/S)

24?60?60?h20?mg?75?9.8??245(J/S)t60 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

4???(0.5?10?6)33W???S?72.75?10?5?2?(3?103?6?10?2)?436.6(J)Stotal?nSSingle?(1?10?1)3?4???(0.5?10?6)2?6?103(m2)

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?

(a)AdiabaticSolution: T?(P)R/CPT0P01T?298?()0.4?118(K)10(b)W1?(500?5)?101325?10?3?T???118(K)nCp10?50?101325?10?3?2.5RR?298T?T0??T?298?118?180(K)

1.5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T0. The valve on the tank is oeened and the surrounding gas is allowed to flow suickly into t(e tank until the pressure insi`e the tank is equals the pressure outside. Assume that no heat flow takes place. What is the0final tempeture kf tèe gaS in the tank? The heat cap!city mf the gas, Cp and Cv each íay be(assumed to be c/nsuant over th? temperature rang!spanNed by the d?periment. 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.

Adiabaticsolution TP?()R/CPT0P0T?T0(P0?0R/CP)?T0P0

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

FORDATA: CH4(g)

0?H298[Kcal/g?mol]?17.89?94.05?57.80

CO2(g)H2O(g)solution

CH4?2O2?CO2?2H2O?H298??(?HCO2?2?HH2O??HCH4)??(?94.05?2?57.80?17.89)?H298191.76?1031?191.76(Kcal/g?mol)???26.9(Btu/1000SCF)?33100.3048?103?252?1031.7

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?

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

forCH4CO2H2ON2O2

CP(cal/mol??C)1613.711.98.28.2(a)CH4?2O2?CO2?2H2OCO2%?1?8.71y3?2?1.1??2?(1.1?1)21H2O%?2CO2%?17.43%N2%?72.12%O2%?0.87%(b)Cp,p??Cp,iXi?0.01[13.7?8.71?11.9?17.43?8.2?(72.12?0.87)]?9.25(cal/mol??C)T?T0??T?298?191.76?1000?2104(K)9.25?11.48

Solution

(C)P?1200?2000?400000?2800000(KJ/h)VConsuming(d)2800000?103??3214(m3/h)191.76?1000?9.25?11.48(1600?298)1(?4.1868?)11.480.0224

100]/11.48?8.87(cal/mol??C)21Cp,r??Cp,iXi?[16?8.2?2.2??H??H298??(?Cp,p,ini??Cp,r,ini)dT?191.76?1000?[(9.25?11.48)?8.87?11.48)](800?298)?189570(cal/g?mol)VConsuming2800000?103??1644(m3/h)189570?9.25?11.48(1600?800)1(?4.1868?)11.480.02241.8 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.

b?H??CPdT??a?bTdT?[aT?T2]T2982ba(700?298)?(7002?2982)?121132ba(1000?298)?(10002?2982)?228032Solution

a?35.62b??0.011CP?35.62?0.011TT?H298?35.62(T?298)?0.0055(T2?2982)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

?Hf,298,CO??110458J/mol?H0f,298,CO20

??393296J/molCP,CO?28.45?3.97?10?3T?0.42?105T?2(J/molK)CP,CO2?44.35?9.20?10?3T?8.37?105T?2(J/molK) CP,O2?19.92?4.10?10?3T?1.67?105T?2(J/molK)CP,N2?29.03?4.184?10?3T(J/molK)2CO?O2?2CO2when1molefuelneed1moleair(N2/O2?4)thenreationCO?20%CO2?5%N2?65%O2?10%productionCO2?27.8%N2?72.2%?Hi,298???H(a)Cp,r??Cp,p,ini?0.2CP,CO?0.05CP,CO2?0.65CP,N2?0.1CP,O2?28.77?4.38?10?3T?0.67?105T?2(J/molK)Cp,r??Cp,p,ini?0.278CP,CO2?0.722CP,N2?33.28?5.58?10?3T?0.19?105T?2(J/molK)?H?0f,298,rni???H0f,298,pni?393296?110458?282838(J/mol)??Hi,298ni??(?Cp,p,ini??Cp,r,ini)dTT298?282838?0.2??T(33.28?5.58?10?3T?0.19?105T?2)?0.9dT?3?733?28.77?4.38?10T?0.67?105T?2dT?28283.8?2?(28.77T?2.19?10?3T2?0.67?105T?1)733298Solution

?(1.18T?0.321T2?0.499T?1)T298?0T??

?H???Hi,298ni??(?Cp,p,ini??Cp,r,ini)dT?282838?0.2??TT298(33.28?5.58?10?3T?0.19?105T?2)?0.9dT?3?1250?28.77?4.38?10T?0.67?105T?2dT

?28283.8?2?(28.77T?2.19?10?3T2?0.67?105T?1)1250298?(1.18T?0.321T2?0.499T?1)T298?0T?(b)1250Q298??282838?0.2??1250Q298??282838?0.2??12502981250(33.28?5.58?10?3T?0.19?105T?2)?0.9dT(33.28?5.58?10?3T?0.19?105T?2)?0.9dT2981.10 (a) for the reaction

1CO?O2?CO2,what is the enthalpy of reaction (?H0) at 298 K ?

2(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?

(adiabatic flame temperature)? DATA :standard heats of formation

?Hf at 298 K

CO??110000(J/mol)CO2??393000(J/mol)0

Heat capacities [J/(mol K)] to be used for this problem N2=33, O2=33, CO=34, CO2=57

(a)?H0???Hf,298,rni???Hf,298,Pni??110000?393000?283000(J/mol)00.5?22.2%,N2%?66.6%,O2%?11.1%1?0.25/0.2Solution productCO2%?22.2?25%,N2%?75%,100?11.1(C)Cp,P??Ci,p,PXi?33?0.666?33?0.222?34?0.111?33(J/mol?K)(b)fuelCO%?Cr,P??Ci,r,PXi?57?0.25?33?0.75?39(J/mol?K)?H??H0??npCp,PdT?0283000?0.222?(0.889?39)(T?298)?0T?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:n(O2)?0.104n(N2)?0.416DaTA(k J?mol)

FOR?Hf(kJ/mol)FOR CO ?110.523COCO?393.513CO2Solution

CP[J/mol?K]?? 3357250H2O(g)34N2,O2

1(a)CO?O2?CO2 21H2?O2?H2O2Flue:Air:n(CO)?0.12n(CO2)?0.36

n(O2)?0.08 n(CO2)?0.24n(H2O)?0.04n(N2)?0.32n(H2)?0.04Fuel:n(N2)?0.6n(N2)?0.92?H??HCO?CO2??HH2?H2O?0.12?(393.51?110.523)?0.04?(241.8?0)?43.6308KJ?CP,r,iin?0.36CCO2?0.04CH2O?0.92CN2?0.36?57?0.04?50?0.92?34?53.8(J/K)3

43.6308?10?810(K)53.8T?1108.98(K)?T?(b)repeat the calculation for 30% excess0combustion air at 298K

?H??HCO?CO2??HH2?H2O?0.12?(393.51?110.523)?0.04?(241.8?0)?43.6308KJ?CP,r,iin?0.36CCO2?0.04CH2O?1.016CN2?0.024CO2?0.36?57?0.04?50?1.016?34?0.024?34?57.88(J/K)43.6308?103?T??753.8(K)57.88T?1051.8(K)Fuel:n(CO)?0.12

n(CO2)?0.24n(H2)?0.04n(N2)?0.6

Air:Flue:n(CO2)?0.36n(O2)?0.104 n(HO)?0.04January 7, 2015

2n(N2)?0.416n(N2)?1.016n(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)

Flue:Air:n(CO)?0.12 n(O)?0.08 n(CO2)?0.36

2n(CO2)?0.24n(H2O)?0.04n(N2)?0.32n(H2)?0.04n(N2)?0.92n(N2)?0.6?H??HCO?CO2??HH2?H2O??Hfuel700?298?0.12?(393.51?110.523)?0.04?(241.8?0)?(700?298)?(0.12?33?0.24?57?0.04?28?0.6?34)?59.373KJ

Fuel:?CP,r,iin?0.36CCO2?0.04CH2O?0.96CN2?0.36?57?0.04?50?0.92?34?53.8(J/K)59.373?103?1103.6(K)53.8T?1401.6(K)?T?(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:n(CO)?0.12 n(O2)?0.08n(N2)?0.32n(CO2)?0.24n(H2)?0.04n(N2)?0.6Flue: n(CO2)?0.36

n(H2O)?0.048n(N2)?0.92n(H2O)?150.4?0.008760?15?H??HCO?CO2??HH2?H2O?0.12?(393.51?110.523)?0.04?(241.8?0)?43.6308KJ?CP,r,iin?0.36CCO2?0.048CH2O?0.92CN2

?0.36?57?0.048?50?0.92?34?54.2(J/K)43.6308?103?T??805(K)54.2T?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

LHS,1803??180317981798CP,SdT??1803SCP,LdT?HL,1798?0?322

HLS,1798?3100?5.41?5?1.5?10(1803?1798)?0.5?7.5?5?3103(cal/mol)1.13 Cuprous oxide (Cu2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu2O

(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction)

DATA: heat of formation of 1000K in cal/mol Cu2O=-41900 H2O=-59210 solution

Cu2O?H2?Cu?H2O?H?59210?41900?17310(cal/mol),exothermic reaction

1.14(a) what is the enthalpy of pure, liquid aluminum at 1000K?

(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.

DATA : For aluminum : atomic weight=27g/mol, Cp,s=26(J/molK), Cp,L=29(J/molK), Melting point=932K, Heat of fusion=10700J/mol Solution

9321000Hl,1000??CP,SdT??298932SCP,LdT?HL?26?(932?298)?29?(1000?932)?10700?27184(J/mol)271841P??1000??279.7(W)?0.28(kW)273600

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

1?27?2Solution;

?T?302(K)T?600(K)1676000?1?16?1.5?27?9927?104??T102

1.16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute? DATA; for water: Cp=4.184J/g k, Density=1g/cm3; for copper: molecular weight=63.54g/mol Cp=7cal/mol k, heat of fusion=3120 cal/mol

QCopper/min?[3120?7?(1356?400)]?Solution:QWater100063.54?min/min?1(80?20)V

V?2.573?10?3(m3/min)1.17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DATA; For water Cp=4.184J/g k, Density=1g/cm3 Solution:

W?4.184?(60?20)?5?1000?13947(W)

601.18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?

(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m3 what is the internal energy change for the evaporation of water?

P?V?101325?Solution:

373?0.0224?3101(J/mol)273%?3101

?7.6"61?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, 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:

(2261?2?80?4.18?80)x?4.18?80?(1000?x),x?138(g)irreversible

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.

TH?TL(90?25)104W?QH??4?StTH90?27360Solution

W?746(W)tS?6.25(m2)P?

2.2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine.

W?Solution:

TH?TLQHTHT?TL0.75P?HPLTLPL?3273?20??0.25?746?643(W)435?20

2.3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value.

P?TH?TLPLTLSolution:

273?0.25?746?334m20M?60m?457(g)PL?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:

TH?TLQLTLTH?TL300?4.21684?2

PL??TL4.216600.5P'?P?P'?393(W)?0.52(hp)2.5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical power to run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler.

(a) assume that the efficiencies are equal to the theoretical maximum values

(b) assume the power plant efficiency is 70% of maximum and that coefficient of performance of the heat pump

is 10% of maximum

overall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b solution:

(a)P1?P2?TH,1?TL,1PH,1TH,1

TH,2?TL,2PH,2TH,2(b) P?0.6286PH,2H,1

P1?P2540?5025?5PH,1?PH,2540?273273?25PH,2?8.98PH,1(c)aisok,bisnot.2.6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabatic enclosure. Assume that Cp is 77J /(mol K) from 273K to 373K Solution:

?U?0(J)TT323323?S?n1CPln(E)?n2CPln(E)?0.5CPln()?0.5CPln()?0.933(J/K)T1T2373273temperature of 30℃.

2.7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃ and a condensate (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

through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied?

(a)(c)TH?TL540?30QH?1?0.89(J) TH540?30TL?500?3600TH?TLSolution:W?(b)W?TH?TLQHTH273?251?19.9(J)25?10

29.0mQH?m?69371(kg)?69.3(ton)2.8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt

(a) calculate the rate of evaporation of the water in grams per second if the water container is insulated that is no

heat is allowed to flow to or from the water except for that provided by the resistor

(b) at what rate could water could be evaporated if electrical energy were supplied at the rate of 1 kw to a heat

pump operating between 25 and 100℃

data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm3 m40000?1000,m?0.45(g)solution:18

m100?273(b)40000?1000,m?2.23(g)18100?25(a)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 1000Solution:PL?2728?(480??20)

T?TL30??20PW?HPL?PL?102474(W)?102.5(kW)TL273?202.10 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?

the same. Will the cooling water requirement be increased, decreased, or remain the same? Data heat capacity 4.184, density 1g/cm3

(a)PH?TH300?2738P?10TH?TL300?40811(b) QL?4.3?1011(J)

Solution:

?2.2?10(W)QH?PHt?7.9?10(J)V?106?10?4.184?QLV?1.03?104(m3)(c)PH?TH540?2738P?10TH?TL540?40

?1.626?108(W)no2.11 (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:

(a)W?(b)P?TH?TL20?4QH?1?0.055(J) TH20?273mg?h?3600?1.06?106(kW/h)10002.12 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:

(a)P?TH?TL20?155PL?10/3600?3.77(kW) TL273?15273?20510?1.14?105(kJ)273?15(b)QH?2.13

(a)2Al?N2?2AlNsolution:

(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?8.314ln10??68.81(cal/molK)

2.14

40CP,ICEC?Hm?S?m(?dT???P,WATERdT)?10TTmT00solution:

273336273?40??4.184ln)12000263273273?22574(J/K)?(2.1ln2.15 W?TH?TL300?77QL?1000?2896(J) TL77W2?70428(J)2.16

W?W?2.17

TH?TL300?4.2QL?83.3?5866.7(J)TL4.2TH?TL300?4.2QH?(83.3?1.5?8.314(300?4.2))?3719.4(J)TH300

(a)?T?0?U?OQ??W?n??pdV?1?8.314?298ln10?5704(J)(b)?S?nRln(c)Q?0(d)yesP0?1?8.314?ln10?19.1(J/K)P

2.18

500?60?TH?TL20?0335m?335m TL273Property Relations

m?1222(g)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)

?G?RTlnPH2O,icePH2O,waterSolution:

3.0123.163?8.314?268?ln0.9523??108.9J/mol?8.314?(273?5)ln

?H?5.85?103J/mol

??G??H?T?S?H??G5850?(?108.9)??S???22.23J/(mol?K)T268

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: CP,bath?100cal/K?418J/mol (a) Q?CP,bath?T/2?100?0.2/2?10cal/g

This is the heat of mixing. (b) The heat capacity of CP, alloy : CP,alloy?CP,bath??T2?(300?0.4?25)100?0.4??0.072cal/(g?K)2?274.6

Assuming that the calorimeter can be applied to the maximum of T?C, the for mixing to form 1 gram of alloy:

Q1?CP,bath(300?T')?10 ,

Q2?CP,alloy?(T?T'),

Q1?Q2

CP,bath(300?T')?10?CP,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 6063J/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 ?S??H6030??22.09J/(mol.K) Tm273 (b) At 0?C, ?G =0 ?

CP,ice?0.5cal/(g.K)?0.5?4.18?18J/(mol.K)?37.62J/(mol.K)

CP,water?1.0cal/(g.K)?1.0?4.18?18J/(mol.K)?75.24J/(mol.K) a reversible process can be designed as follows to do the calculation:

（1）（3）（2） Ice, 0?C water, 0?C ice, -5?C （4） water, -5?C ?Hfu??H(1)??H(2)??H(3)????273268273Cp,icedT??H??Cp,waterdT273268268(Cp,ice?CP,water)dT??H

?(37.62?75.24)?5?6030?5841.9J/mol （d）

?S(4)??S(1)??S(2)??3(3)????273Cp,ice268T273(Cp,ice?CP,water)TdT??S??268Cp,waterT273dT

268dT??S?(37.62?75.24)?ln?21.39J/(mol.K)273?22.09268 ?G(4)??H(4)?T?S(4)?5841.9?268?21.39?109.38

4. (a) What is the specific volume of iron at 298K, in cubic peter per mole? (b) Derive an equation for the change of entropy with pressure at constant temperature for a solid, expressed in terms of physical quantities usually available, such as the ones listed as data; (c) The specific entropy of iron (entropy per mole )at 298K and a pressure of 100 atm is needed for a thermodynamic calculation. The tabulated “standard entropy”(at 298 K and a pressure of 1 atm) is

oS298?27.28J/K.mol. What percentage error would result if one assumed that the specific entropy at 298K and

o100 atm were equal to the value of S298 given above ？

DATA：（for 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)

molweight55.85g/mol V??7.10cm3/mol?7.10?10?6m3/mol iron?3density7.87g/cm?S???V? (b) ?????????P?T??T?P?S? ???????V?V??3V?l ??P?Tfor iron:

??S??????3Viron?l,iron??P?T??3?7.10?10?6?15?10?6??3.2?10?10(m3/(mol.K))

?Siron??3.2?10?10?P

( c )

?Siron??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%??Siron?100%??1.12?10?2 o?S298Equilibrium

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) V?18g/mol?19.57cm3/mol?19.57?10?6m3/mol

ice30.92g/cmVwater?18g/mol?18cm3/mol?18?10?6m3/mol 31g/cm?Vfus??1.57?10?6m3/mol

According to the Clapeyron equation:

?HfusdP?1??dT??VfusT??Hfus1?dT???VfusT

dP??take definite integration of the above:

?50?1.013?1051.013?105dP???Hfus?Vfus1??273T?dT

Tln?VfusT??49?1.013?105273?Hfus?1.57?10?6?49?1.013?105 ?6009??0.013T?272.8K

(b) P?1500.01?3?50?103lb/in.2?50?103?6897Pa?345?106Pa

(c )

?P?345?106Pa

ln?VfusT??345?106273?HfusT?249.46K?1.57?10?6?345?1066009??0.09

?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:

?Hvap?28kcal/mol?4.18?28?103J/mol?117.04J/mol

According to Claperon equation in vapor equilibrium:

P2? d(lnP?)???Hvap1d() RT

???d(lnP)??P1?Hvap11

(?)RT2T1?Hvap11 P1?ln???(?)RT2T1P21117.04?10311

ln?4??(?)8.314T267310T2?1202K

The boiling point of zinc is 1202K. 2. Trouton’s rule is expressed as follows:

?Hvap?90Tb in joules per mole, where Tb is the boiling point

(K). The boiling temperature of mercury is 630K. Estimate the partial pressure of liquid Hg at 298K. Use Trouton’s rule to estimate the heat of vaporization of mercury. Solution:

?Hvap?90Tb

?P1d(lnP)???HvapR(11

?)298630lnP??6823?10.83??22.90?10.83??12.07 298 P=5.73?10-6 atm

3. Liquid water under an air pressure of 1 atm at 25?C has a large vapor pressure that it would have in the

absence of air pressure. Calculate the increase in vapor pressure produced by the pressure of the atmosphere on the water. Water has a density of 1g/cm3; the vapor pressure ( in the absence of the air pressure) is 3167.2Pa. (4.5, p116) Solution:

Vl?18g/mol?18cm3/mol?18?10?6m3/mol 31g/cmvapor pressure changes with the total external pressure,?Gv??Gl

RTlnPe,2Pe,1Pe,1?Vl(PT?Pe,1)

RTlnPe,2?18?10?6(10130?3167.2)

Pe,2Pe,1 P.36Pa ?P = 0.16Pa ?1.000051e,2?3167the 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:

?P1d(lnP)???Hvap11 ?25500011

(?)lnP??(?)8.314T2450RT2450lnP??30685?104.08 T6. 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:

lnP(atm)??15755?0.755ln(T)?19.25 and the vapor pressure of liquid Zn varies with T T?15246?1.255ln(T)?21.79. Calculate: (a) The boiling point of Zn under 1 atm; (b) Tas:lnP(atm)?The triple-point temperature; (c) the heat of evaporation of Zn at the normal (1 atm) boiling point; (d) The heat of fusion of Zn at the triple-point temperature; (e)The differences between the heat capacities of solid and liquid Zn. (4.8, p118)

Solution :(a) At boiling point, P=1 atm, that is lnP=0

?15246?1.255ln(T)?21.79?0 T?15246?1.255Tln(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.

300002800026000240002200020000180001600014000120001000080006000400020000-2000-4000-60004006008001000 y1 y2y120014001600180020002200T, 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?15755509?1.255ln(T)?21.79??0.755ln(T)?19.25 ?0.5ln(T)?2.54?0 TTT509?0.5Tln(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.

80007500700065006000550050004500400035003000250020001500100040060080010001200140016001800 y11 y22y20002200T,K

The intersection is 695.3K, and this is the triple point temperature. (c )

d(lnP)??152461.255??21.79TT2 d(lnP)???Hvapd(1) 1RT?(?15246?1.255T)(?2)dTT1?(?15246?1.255T)d()T

?At Tb =1180K, (d) For solids,

?HvapRT)?126755?10.43T ??15246?1.255T ?Hvap?R(15246?1.255?Hvap?126755?10.43T?114.4?103J/mol?114.4kJ/mol

?15755?0.755ln(T)?19.25T?Hfus1157550.755 d(lnP)??d() d(lnP)?(2?)dTRTTT1?(?15755?0.755)(?2)dTT1?(?15755?0.755)d()TlnP???HfusR??15755?0.755T

?Hfus?R(15755?0.755T)

At triple point, Ttr = 695.4K

?Hfus?8.314?(15755?0.755T)?126.6?103J/mol?126.6kJ/mol

(e) d(?Hfus)??CPdT

?CP?d(?Hfus)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 of the gas. Find the fraction of material in the vapor phase in the initial and final states. (4.9, P118) Solution: In the initial state,

P1V1?n1RT,5T1?300K,P1?1atm?1.013?10Pa,V1?1L

P1V110130?10?3n1???4.06?10?3mol

RT18.314?300mol(vapor)%?4.06%

In the final state, P2=2 atm, V2 = 1L According to Clayperon equation:

?11??T?T2?2 2500011ln??(?)18.314T2300RTlnT2?458.6K?HvapP2??P1R????

P2V2?n2RT2,P2V22?10130?10?3n2???5.3?10?3molRT28.314?458.6

mol(vapor)%?5.3%

8. The melting point of gold is 1336K, and vapor pressure of liquid gold is given by:

lnP(atm)?23.716?43522?1.222lnT(K). (a) Calculate the heat of vaporization of gold at its Tmelting 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(lnP)?(435221.222?)dTTT2

?Hvap??R(?43522?1.222T)?361841?10.19T

1)dTT21?(?43522?1.222)d()T?(?43522?1.222)(?(a) At 1336K, ?Hvap?361841?10.19?1336??348?103J??348kJ (b) “Solution not possible”; (c) “Solution not possible”. 9.

(a) At 298K, what is the Gibbs free energy change for the following reaction?

Cgraphite?Cdiamond

(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

?Ho) Ho.K) f(298)(kJ/molf(298)(J/molDiamond 1.897 2.38 Graphite 0 5.74

Solution: (a)

o?H??HokJ/mol f,diamond??Hf,graphite?1897Cgraphite?Cdiamond

o?S??So.K) f,diamond??Sf,graphite??5.74?2.38??3.36J/(mol?G??H?T?S?1897?298?(?3.36)?2898.28J/mol

(b) No, diamond is not thermodynamically stable relative to graphite at 298K.

?612?10(c ) ?G?Vdiamand?P?3.51?99?10130?34.29J/mol

(c ) Assuming N atm , ?G = 0, reversible processes as following can be designed to realize this,

?G(4)＝?G(1)＋?G(2)＋?G(3)＝Vgraphite?P?2898.28?Vdiamond(??P)?(Vgraphite?Vdiamond)(1?N)?10130?2898.28?12?10?612?10?6??N??2.25?3.51???0.194N?2898.28?0N?14939(atm)?Cp?0,??CpdT?0,?TT'T'T（4） graphite, 298K, N atm diamond, 298K, N atm （1）（2） (3) diamond, 298K,1atm graphite, 298K,1atm

????10130?2898.28??CpTdT?0

?H900??H298?1897J/mol ?S900??S298??3.36J/mol.K

?G??H?T?S?1897?900?3.36?4921J/mol

Chemical Equilibrium

1. Calculate the partial pressure of monatomic hydrogen in hydrogen in hydrogen gas at 2000K and 1atm.

For

1H2(g)?H(g) 2o?217990J ?H298o?S298?49.35J/KFor the reaction :

1H2(g)?H(g) 2131?Cp?CP,H(g)?CP,H2??8.314??31??3.035

222oo?H2000??H298??2000o?CPdT??H298??CP?(2000?298)

298?217990?3.035?1702?212824J?Ho2000??Ho298??2000298o?CPdT??H298??CP?(2000?298)

?217990?3.035?1702?212824Joo?S2000??S298??2000298o?CPdT??S298??CP?ln2000298

?49.35?3.035?ln2000?43.57J298000 ?G2000??H2000?T?S2000?212824?2000?43.57?12568J 4?Go2000??RTlnK??RTln1/2PH(g)PH2PH21/21/2?RTln?PH21/2PH(g)

RTlnPH2PH(g)?125684?lnPH(g)125684?7.568.314?2000PH2(g)?PH2(g)?1,PH2?1ln1PH(g)?7.562?PH?0.0005atmCo(s)?

2. For the reaction :

1O2(g)?CoO(S) 2

?Go??59850?19.6T, where ?Gois 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:

?Go??RTlnK??RTlnlnPO2??27.6PO2?1.07?10?12atm1PO21/2?1RTlnPO2??145879J2

(b) uncertainty in ?Ho = 500cal/mol = 2090J/mol So uncertainty in ?Go = 500cal/mol = 2090J/mol That means:

11'RTlnPORTlnPO2?20902?22'PO1 RTln2?20902PO2'POln2?0.25PO2'PO?P2?1.286??28.6%PO2PO2 Similarly, uncertainty in ?Ho =- 500cal/mol =- 2090J/mol ?Go = -2090J/mol

11'RTlnPORTlnPO2??20902?22' PO2ln??0.25PO2'PO?P2?0.779???22.1%PO2PO23. 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

Standard Entropy at 298K

?HfforAg2O??7300cal/mol

Assume that ?Cp = 0 for the decomposition reaction. Solution: (a) Ag2O = 1/2O2 + 2Ag

?Ho???Hocal/mol?30514f,AgO?7300?So?2?SAg,298??2?10.2?1?SO2,298??SAg2O,2982

1?49?29.1?66.044J/mol.K2?Go??Ho?T?So?30514?T?So?30514?66.044T

when Ag2O begins to decompose,

o ?G??G?RTlnJ?0 ie30514?66.044T?RTlnPO2?0(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

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)?1O2(g)?Cu2O(s) 2(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)?1o

O2(g)?Cu2O(s), ?G (in calories ) is –39850+15.06T. (5.4, p145) 2Solution: (a) When the equilibrium is reached,

1?G??Go?RTlnJ??Go－RTlnPO2?0

2lnPO2?4.18?(?39850?15.06T)

1RT2 T = 500?C = 773K

lnPO2?PO24.18?(?39850?15.06?773)??36.69 1?8.314?7732?1.14?10?26atm(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

2 (b) ?H??Ka2PH211 PH2?1atm,?H??7.34cm3/100gCu

Ka2 is a constant,

1/2PH[H][H]'0.1'1/22[H]'??(P)???0.0136 H21/2'1/2[H]7.34PH2(PH2)'(PHatm?18.56?10130?18.8Pa2)?0.000191 (c ) The amount of H2 needed to be brought out by Ar is:

P?V10130?(0.5?0.1)?10?6n???1.6?10?6mol

RT8.314?298This amount of H2 is in equilibrium with the melt in the bubble, ie. The partial pressure of H2 in the bubbles is 18.8Pa.

'?2?PH2Vbubble?nRT?4.05?10

'H2?4.05?10/18.8?0.00215m?2.15L?22.15L Ar is needed to be bubbled into the melt.

6. The following equilibrium data have been determined for the reaction:

NiO(s)?CO(g)?Ni(s)?CO2(g)T(?C) K?10-3 663 4.535 716 3.32(a) Plot the data using appropriate axes and find ?Ho, K and ?Go at 1000K;

(b) Will an atmosphere of 15%CO2, 5%CO, and 80%N2 N2 oxidize nickel at 1000K? (5.6, p145)

o?H1Solution: (a) dlnKa??d() RT Plot

lnKa~1/T

8.68.48.28.0 Kduishu Linear Fit of Data1_KduishulnKa7.87.67.47.20.880.900.920.940.960.981.00-3lnKa =2.01+6003(1/T)1.021.041.061.081/T, 10

dlnKa?Ho???6003??Ho??R?6003??49909J dTRAt T=1000K, lnKa =8.01, Ka = 3010

o?G1000??RTlnKa??8.314?1000?8.01?66600J?66.6kJ

o(b)?G??G?RTlnJ??RTlnKa?RTlnJ?RTlnJKa

J?15%?3?Ka5%So the atmosphere will oxidize Ni.

7. At 1 atm pressure and 1750?C, 100 g of iron dissolve 35cm3 (STP) of nitrogen. Under the same conditions, 100 g of iron dissolves 35 cm3 of hydrogen. Argon is insoluble in molten iron. How much gas will 100 g of iron dissolve at 1750?C and 760 mm pressure under an atmosphere that consists of: (a) 50% nitrogen and 50% hydrogen? (b) 50% argon and 50% hydrogen? (c) 33% nitrogen, 33 hydrogen, and 34 argon? (5.7, p145) Solution: N2 =2N, H2 = 2H

?N??K2a,N1P2N2

1，

?H??K2a,H12PH2,1

[N][N]'?For N2 dissolving : 1/2'1/2 PN2(PN2)[H][H]''1/2For H2 dissolving：1/2?PH2(PH2)

（a）For dissolving N2, PN2 = 1 atm, [N]=35cm3/100g melt,

'1/2(PN[N]2)[N]??35?(0.5)1/2?24.75cm3/100gmelt1/2PN2‘ similarly: [H]’ =24.75cm3/100g melt

total gas : [H]?＋[N]? = 49.5 cm3/100g melt (b) [H]? =24.75 cm3/100g melt

(c ) [H]?＋[N]? = [N](0.33)1/2 /1+[H](0.33)1/2 /1=20.10+20.10 = 40.2cm3/100g melt

8. Solid silicon in contact with solid silicon dioxide is to be heated to a temperature of 1100 K in a vaccum furnace. The two solid phases are not soluble in each other, but is known that silicon and silicon dioxide can react to form gaseous silicon monoxide. For the reaction: Si(s)?SiO2?2SiO(g)the Gibbs free energy change (J) is ? G o ? ? 25 . 0 T ln. T (a) Calculate the equilibrium 667000?510Tpressure of SiO gas at 1100K; (b) For the reaction above, calculate ?Ho and ?So at 1100K; (c) Using the Ellingham chart (Figure 5.7), estimate the pressure of oxygen (O2) in equilibrium with the materials in the furnace. (5.8, P146)

2(a) ?Go??RTlnK??RTlnPSiO??2RTlnPSiO

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

66700025?lnT?510RTR 66700025?66700025?HodlnK?（?)dT?(?T)d(1/T)??d(1/T)RTRRRRT2?Ho25????667000?TRRo?H?667000?25TlnK?? T = 1100K, ?Ho = 639500J

?Ho??Go667000?298584?S???334.9J/K

T1100o (c ) PO2 =10-30 atm

9. What is the pressure of uranium (gas) in equilibrium with uranium dicarbide DATA: At 2263K, ? for UCGo2 is –82,000 cal/mol

f Vapor pressure of pure uranium is: (5.9,p146)

Solution:

lnP(atm,uranium)?25.33?100000(TinK)T?Gocal/mol??342760Jf??82000U(g)?C(s)?UC2(s)

1Pu(g))?RTlnPu(g)?Gof,UC2??RTlnK??RTln(RTlnPu(g)??342760lnPu(g)?1.2?10?8(atm)

vapor pressure of uranium: dicarbide.

10. The direct reduction of iron oxide by hydrogen maybe represented by the following equation:

lnPu(g)(atm,uranium)?25.33?Pu(g)?0.6?10?8(atm)100000100000?25.35??18.89T2263the vapor pressure is lower than the one determined by chemical reaction. It is the one in equilibrium with

Fe2O3?3H2?2Fe?3H2O What is the enthalpy change, in joules, for the reaction? Is it exothermic

3O2?Fe2O3or endothermic? 21H2?O2?H2O22Fe?Solution:

?Go??810250?254.0T?Go??246000?54.8T (5.10)

2Fe?3O2?Fe2O321H2?O2?H2O2(1)(2)?G1o??810250?254.0To?G2??246000?54.8T

Fe2O3?3H2?2Fe?3H2O?72250?89.6T?H3o?72250Jo (3)?G3o?G3o?3?G2??G1o?3??24600?54.8T??(?810250?254.0T)

The reaction is an endothermic one.

11.Calcium carbonate decomposes into calcium oxide and carbon dioxide according to the reaction

CaCO3?CaO?CO2

DATA for the pressure of carbon dioxide in equilibrium with CaO and CaCO3:

Temperature (K) 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)

?Ho?1?dlnK??d??,K?PCO2R?T??Ho?1??dlnPCO2??d??R?T? oPCO2,2?H?11????ln???PCO2,1R??T2T1?ln0.01?Ho?11??????0.1R?9211030??Ho?166528Jthe reaction is endothermic

(b) PCO2 =1atm

1?Ho11ln??(?)0.1RT1030T?1168K

At 1168K, the equilibrium pressure of CO2 equals one atmosphere.

12. In the carbothermic reduction of magnesium oxide, briquettes of MgO and and carbon are heated at high temperature in a vacuum furnace to form magnesium (gas) and carbon monoxide(gas).

(a) write the chemical reaction for the process; (b) What can you say abou the relationship between the pressure of magnesium gas and the pressure of carbon monoxide? (c) Calculate the temperature at which the sum of the pressures of Mg(gas) and CO reaches on atmosphere. With T in Kelvin, the free energies of formation, in calories, of the relevant compounds are:

MgOCO?Go?48.7T f??174000?Gof??28000?20.2T(a). The reaction is:

MgO(s)?C(s)?CO(g)?Mg(g)

(b).

o?Go??Go?68.9Tf,CO??Gf,MgO?146000

?G??RTlnK??RTln(PCO(g)PMg(g))?(146000?68.7T)PCO?PMg

(c ) Ptotal = 1 atm, PCO = 0.5 atm, PMg =0.5 atm ?RTln(0.5?0.5)?(146000?68.7T)?4.18

T = 2037 K

13. Metallic silicon is to be heated to 1000?C. To prevent the formation of silicon dioxide (SiO2), it is proposed that a hydrogen atmosphere be used. Water vapor, which is present as an impurity in the hydrogen, can oxidize the silicon. (a) Write the chemical equation for the oxidation of silicon to dioxide by water vapor; (b)Using the accompanying data, where ?Go is in joules, determine the equilibrium constant fro the reaction at 1000?C (1273K); (c) What is the maximum content of water in the hydrogen (ppm) that is permitted if the oxidation at 1000?C is to be prevented ? (d) Check the answer to part c on the Ellingham diagram (Figure 5.7) DATA H2(g)?1O2(g)?H2O(g)2Si?O2?SiO2(s)?Go??246000?54.8T ?Go??902000?174To5.13, P147

Solution: (a) Si(s)?2H2O(g)?SiO2(s)?2H2(g)(b)

H2(g)?O2(g)?H2O(g)(1)(2)12Si?O2?SiO2(s)?G(o?54.8T 1)??246000?G(o2)??902000?174To ?G3

Si(s)?2H2O(g)?SiO2(s)?2H2(g)(3)o?G3o??G2?2?G1o??902000?174T?(?246000?54.8T)?2

??410000?64.4T?G??RTlnK??410000?64.4TAtT?1273K,lnK??K?2.9?1013(c )

2o3

?410000?64.4?1273?318.314?1273?PH2(g)?13K????2.9?10??PH2O(g)?? PH2(g)?5.38?106PH2O(g)PH2O(g)PH2(g)?1?0.186?10?6?0.186ppm65.38?10

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).

atmosphere? ( 5.14, P147 )

DATA

Solution: (a) (b) ?CP = 0

oo?H??Hof,CO2(298)??Hf,BaO(298)??Hf,BaCO3(298)Thermodynamic [KCAL/(g.mol)]

Properties

BaCO3(s)?BaO(s)?COo2(g) ?Gof(298) ?Hf(298)

??94?133?(?219)?64kcal?267.52kJ the reaction is endothermic (c ) At 298K,

ooo?G298??Gof,CO2,298??Gf,CaO,298??Gf,CaCO3,298??94?126?272?52kcal?217.36kJ

ooo?G298??H298?T?S298oo?H298??G298(267.52?217.36)?S???168J/mol.KT298ooo??GT??HT?T?STo298

when PCO2=1 atm,

o?GT?0,ie2675201?168TT?1592K

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:

MgCl2(l)?H2(g)?Mg(g)?2HCl(g)

(a) Calculate the equilibrium pressure of H2(g), Mg(g) and HCl(g) if the total pressure is maintained constant at

1 atm.

(b) Calculate the maximum vapor pressure of H2O that can be tolerated in the hydrogen without causing the

oxidation of the Mg vapor. DATA

（5．15， p48）

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 Solution: (a) MgCl2(l)?H2(g)?Mg(g)?2HCl(g)(1)?G1o

Mg(g)+Cl2(g) = MgCl (l) (2) ?Go2 －425484 J

H2 (g) + Cl2(g) = 2HCl(g) (3) ?Go3 －207856 J

oo?G1o??G3??G2??207856?425484?217628J

2PHClPMg(g)(g)2PHClPMg(g)(g)?G??RTlnK??RTln?217628ln2PHClPMg(g)(g)o1PH2(g)??8.314?1473lnPH2(g)

PH2(g)PH2(g)＝17.782PHClPMg(g)(g)?5.27?107PH2(g)?PMg(g)?PHCl?1,PHCl?2PMg(g)let PMg(g) =x, PHCl = 2x, PH2 = 1-3x

1?3x?5.27?1072 x(2x)

x?1.6?10?3(atm)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

ooo?G4??G5??G6??437185?(?165280)??271905Jo?G4??RTlnK??RTlnPH2(g)PH2O(g)PMg(g)??8.314?1473lnPH2(g)PH2O(g)PMg(g)??271905lnPH2(g)PH2O(g)PMg(g)＝22.2

1.6?10?3?22.2PH2O(g)?1.6?10?3lnPH2O(g)??22.2PH2O(g)?2.28?10?10(atm)16. A common reaction for the gasification of coal is:

H2O(g)?C(s)?H2(g)?CO(g)

(a) Write the equilibrium constant for this reaction and compute its value at 1100K;

(b) If the total gas pressure is kept constant at 10 atm, calculate the fraction of H2O that reacts;

your answer. Use the data in Table 5.1. (5.16, 148)

Solution:

H2O(g)?C(s)?H2(g)?CO(g)

(a)

K?PCO(g)PH2(g)PHO2(g)

?Go??Gf,CO??Gf,H2O??111710?87.65?1100?(?246740?54.81?1100)??21676J ?Go??RTlnK?lnK?2.3?K?9.97 (b)

letPH2(g)?xatm,PCO(g)?xatm,PH2O(g)?(10?2x)atmx2?9.971?2xx?4.14atmK?

(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:

2, where xCu is the mole fraction of copper. Calculate the partial RTln?Zn??38300xCupressure 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:

22?38300xCu?38300?0.6Culn?Zn????1.38RT8.314?1200 ?Zn?0.25aZn??ZnxZn?0.25?0.4?0.1pPZn?PZnaZn?0.1?1.17?0.117(atm)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:

xZnd(lnaZn)?xCud(lnaCu)?0xZnd(ln?Zn)?xZnd(lnxZn)?xCud(lnaCu)?xCud(lnxCu)?0?xZnd(lnxZn)＋xCud(lnxCu)?dxZn?dxCu?dxZn?d(1?xZn)?0?xZnd(ln?Zn)＋xCud(ln?Cu)?0xd(ln?Cu)??Znd(ln?Zn)xCu

?ln?0d(ln?Cu)????1xCu1?xZn?38300??2xCudxCunxCuRTxCuxZn3830038300?2xZndxCu????2xZndxZn1RTRTln?Cu??3.

383002xZnRT(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.

is the solubility of graphite in ?-iron at 900K?

DATA AT 900K,

3Fe?C(graphite)?Fe3C?Go??3463J

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 NHg aHg (7.4,196) Solution: NHg aHg ?Hg

Plot ln?Hg ~xHg

0.949 0.961 1.013

0.893 0.929 1.04

0.851 0.908 1.06

0.753 0.840 1.12

0.653 0.765 1.17

0.537 0.650 1.21

0.437 0.542 1.24

0.330 0.432 1.31

0.207 0.278 1.34

0.063 0.092 1.46

0.949 0.961

0.893 0.929

0.851 0.908

0.753 0.840

0.653 0.765

0.537 0.650

0.437 0.542

0.330 0.432

0.207 0.278

0.063 0.092

0.400.350.300.25 lnB Linear Fit of Data1_lnBln?Hg0.200.150.100.050.000.00.20.40.60.81.0ln?Hg=0.395-0.391xHgxHg

d(ln?Bi)??xHgxBid(ln?Hg)??xBixHgxBi?(?0.391)dxHg

?ln?0d(ln?Bi)??0.391?(11?1)dxBixBiln?Bi??0.391(lnxBi?xHg)

when xBi = 0.4, xHg =0.6

?Bi?1.107ln?Bi??0.391(ln0.4?0.6)?0.1

7.5 For a given binary system at constant T and P, the liquid molar volume of the solution (cm3/mol) is given by :

V?100xA?80xB?2.5xAxB

(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:

V?100xA?80xB?2.5xAxA

??V?VA???100?2.5xB ??x???A?T,P,xB??VVB????x?B105???80?2.5xA?82.5?2.5xB ??T,P,xB100 partial volume of A partial volume of B molar volume of the solution95cm/mol90385800.00.20.40.60.81.0xB

(b)

VA?100,VB?80VM?V?(100(1?xB)?80xB?V?(100?xB)?2.5xAxB4. For an ideal binary solution of A and B atoms, plot schematically the chemical potential of both species as a function of the composition of the solution. Indicate on the plot the molar Gibbs free energy of pure A and B. (7.6,196)

5. At 473?C, the system Pb-Sn exhibits regular solution behavior, and the activity coefficient of Pb is given by:

log??Pb???0.32?1?xPb?at 473?C. (7.7, p196)

2. Write the corresponding equation of the variation of

?Sn with composition

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:

?GMix?2RT(xMgCl2lnxMgCl2?xMgF2lnxMgF2)?xMgCl2xMgF2(?2556?25(xMgF2?xMgCl2)).

Compute the maximum solubility of MgF2 in liquid MgCl2 at 900?C. MgCl2 does not dissolve in solid MgF2. (7.8, 197)

7. The thermodynamic properties of Al-Mg solution at 1000K are given in accompanying table. (a) If one mole of pure liquid aluminum and one mole of pure liquid magnesium, each at 1000K, are mixed adiabatically, what will be the final temperature of the solution that is formed ? (b) What is the total change in entropy for the process ? DATA Quantities of Mixing Liquid Alloys at 1000K

xMg

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

?GM(cal/mol)

-800 -1250 -1550 -1700 -1800 -1700 -1550 -1250 -800

.K)] ?HM(cal/mol) ?SM(cal/mol) Cp[cal/(mol-300 -600 -750 -850 -900 -850 -750 -600 gf-300

0.5 0.65 0.8 0.85 0.9 0.85 0.8 0.65 0.5

7.1 7.18 7.26 7.34 7.42 7.5 7.58 7.66 7.74

The problems of the phase rule

8.1 Zinc sulfide (ZnS) is reacted in pure oxygen to form zinc sulfate (ZnSO4) (a) write the chemical reaction representing the process (b) how many solid phases may exist in equilibrium if pressure and temperature are arbitrarily fixed? (c) if the temperature is fixed, will the pressure be determined if ZnS and ZnSO4 exist in equilibrium? Solution: (a)ZnS?2O2(b)two

8.2 an Fe-Mn solid solution containing 0.001 mole fraction Mn is in equilibrium with an FeO-MnO solid solution and a gaseous atmosphere containing oxygen at 1000K. How many degree of freedom does the equilibrium have? What is the composition of the equilibrium oxide solution, and what is the oxygen pressure in the gas phase? Assume that both solid solutions are ideal? Data: for Fe Fe(s)?1O2?FeO(s)??ZnSO4

2?G??259000?62.55T For Mn Mn(s)?O2?MnO(s)

12??G??384700?72.8TSolution: (a) F=(5-2-1-1)-2+1=0

1RTlnPO220 (b)P1?exp(2?G)?3.0?10?21

O2RT2?G0PO22?exp()?2.6?10?33RTP?0.999PO21?0.001PO22?3.0?10?21?G0?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)

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 1300 1000+ 1000-

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

Phase Liquid α β Liquid α β Liquid α β Composition 60 8 99 70 9 98 _ 7 98 Fraction 61.5 38.5 0 50.8 49.2 0 0 63.7 36.3 ?Ag??Cusolution (b)F=C-P=(2-1)-1=0 (c) ?Agphase in the solution increases and ?Cu decreases

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

degrees of freedom are therr in this equilibrium?

(a) 2900℃, α(12%) (b) 2300℃, liq(95%) (c) 8.2%α(composition is 24% )+91.8%β(85%)

9.10 the accompanying diagram represents the liquidus surface in the ternary phase diagram for BaCl2-NaCl-KCl-CaCl2 assume for the purpose of this problem that there is no solid solubility of the compounds in one another.

(a) on the diagram trace the path of the liquid composition when a material consisting of BaCl2 60%, NaCl 20%,

KCl-CaCl220% is cooled from 900℃. Assume that equilibrium is maintained at all temperature. (b) Sketch two blank ternary diagram and draw in the isothermal section at 750 and 650℃

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 (b)

750℃

BaCl2

20% NaCl

20%

BaC20Soli750KCl, CaCl2

750Na

SoliLiqKCl, CaCl2

600℃

BaCl2

20% Solid Liquid NaCl

20%

KCl, CaCl2

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