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