Heat and theromodynaicms_engineering physics

engineering physics

1. CALORIFICVALUEDEFINED
Thecalorificvalueorheatingvalueofafoodandfuelisdefinedasthe
quantityofheatreleasedduringcombustionofaspecifiedamountofit.
Thecalorificvalueisacharacteristicforeachsubstance.
UNITS OF CALORIFIC VALUES
The units of calorific value are energy per unit of the substance, such as:
kcal/kg, kJ/kg.

2. Higher Calorific Value (HCV) or Higher Heating Value (HHV) or Gross Calorific
Value:
When 1 kg of a fuel is burnt, the heat obtained by the complete combustion after the
products of the combustion are cooled down to room temperature (usually 15 degree
Celsius) is called higher calorific value of that fuel.
Lower Heating Value (LLV) or Lower Calorific Value (LCV) or Net Calorific
Value:
When 1 kg of a fuel is completely burned and the products of combustions are not cooled
down or the heat carried away the products of combustion is not recovered and the steam
produced in this process is not condensed then the heat obtained is known as the Lower
Calorific Value.
Relation between Higher and Lower Calorific Value.
Answer: The amount of Lower Calorific Value can be obtained by subtracting the amount
heat carried away by the combustion products especially the heat carried away by the steam
LCV = HCV –Heat carried away by the steam.
Additional

3. Additional

4. Additional

5. Additional

6. Importance of Calorific Value
Itisveryimportanttohaveaknowledgeofthecalorificvalueoffueltocarryout
ourdayto-dayactivities.Thisknowledgehelpsustodeterminetheamountof
energywetransport.Thegasshippersandsuppliersrequirethisinformationto
billgasconsumers.Italsohelpstodeterminetransportationchargesofgas
shippersandsuppliers.Thehumanbodyrequirescaloriestocarryoutdaily
activities.Withoutcalories,thebodywouldstopworkingandthecellsinthebody
woulddie.But,ifpeopleconsumedonlyaspecificamountofcalorieseachday,
theywouldleadahealthylife.Toohighortoolowcalorieconsumptioneventually
leadstohealthproblems.
Additional

7. Sources of Calories in Foods
The energy content of foods primarily comes from the catabolism of carbohydrates,
proteins, and fats. Each of these macronutrients yields a different amount of
energy per gram:
•Carbohydrates and proteins yield about 4 kcal/g of energy.
•Fats provide a more significant amount of energy, approximately 9 kcal/g.
•Alcohol, found in some foods and beverages, yields about 7 kcal/g when
metabolized.
The gross calorific value of fats is 9.45 kcal/g,
for carbohydrates, it is 4.1 kcal/g and
for proteins, it is 5.65 kcal/g.
Additional

8. Calculating caloric content in foods
Ques.To determine the number of kilocalories from a specific component in a
food item, multiply the number of grams of that component by its respective
energy content.
For example, if a meal contains 37 grams of carbohydrates, 20 grams of fat, and
15 grams of protein, the caloric content can be calculated as follows:
Ans.Carbohydrates: 37 g ×4 kcal/g = 148 kcal
Fat:20 g ×9 kcal/g = 180 kcal
Protein:15 g ×4 kcal/g = 60 kcal
Total kcal in the slice of pizza: 148 kcal + 180 kcal + 60 kcal = 388 kcal.
Additional

9. Definition of Bomb Calorimeters:
The calorific value of solid and liquid fuels is determined in the laboratory by ‘Bomb
calorimeter’ It is so named shape resembles that of a Bomb. Fig shows the schematic
sketch of the bomb calorimeter.
ConstructionofBombCalorimeters:
Thecalorimeterismadeofausteniticsteel
whichprovidesconsiderableresistanceto
corrosionandenablesittowithstandhigh
pressure.Inthecalorimeteruseofa
strongcylindricalbomb inwhich
combustionoccurs.Thebombhastwo
valvesatthetoponesuppliesoxygento
thebombandtheotherreleasesthe
exhaustgases.

10. Acrucibleinwhichaweighedquantityof
fuelsampleisburntisarrangedbetween
thetwoelectrodesasshowninfig.The
calorimeterisfittedwithawaterjacket
thatsurroundsthebombToreducethe
lossesduetoradiationcalorimeteris
furtherprovidedwithajacketofwaterand
air.Astirrerforkeepingthetemperatureof
wateruniformandathermometer the
temperatureuptotheaccuracyof0.001
degreeCisfittedthroughthelidofthe
calorimeter.
Theheatreleasedbythefuelon
combustionisabsorbedbythesurrounding
waterandthecalorimeter.Fromthe
equationthecalorificvalueofthefuelcan
befound. 
M
TTwm
M
C
)(ProducedHeat
21

IfMbethemassofthesample,the
calorificvaluewillbe
wherembethemassofwater,wisthe
capacityofcalorimeter,T
1andT
2are
initialandfinaltemperatureofwater
valuewillbe

11. Apetrolengineconsume25kgofpetrolperhour.Thecalorificvalue
ofpetrolis11.4x10
6
cal/kg.Thepoweroftheengineis99.75kW.
Calculatetheefficiencyoftheengine.

12. Apetrolengineconsume25kgofpetrolperhour.Thecalorificvalue
ofpetrolis11.4x10
6
cal/kg.Thepoweroftheengineis99.75kW.
Calculatetheefficiencyoftheengine.

13. What is a bomb calorimeter?
Abombcalorimeterisadeviceusedtomeasuretheheatofcombustionofa
sample.Itconsistsofasmall,non-reactivecruciblethatholdsthesample,
whichisplacedinsideahigh-pressureoxygenatmosphere,referredtoasa
bomb.Thebombisimmersedinawaterjacket,andtheheatreleasedduring
thecombustionreactionismeasuredbythetemperaturechangeofthewater.
Additional

14. Whatistheprincipleofabombcalorimeter?
Theprincipleofabombcalorimeteristomeasuretheheatreleasedduringthe
combustionofasample.Thesampleisplacedinsideacrucibleandignitedin
anoxygenatmosphereathighpressure.Theheatreleasedduringthe
combustionisabsorbedbythewaterinthewaterjacketsurroundingthebomb,
andthetemperaturechangeofthewaterisusedtocalculatetheheatof
combustionofthesample.
Additional

15. Howaccuratearebombcalorimeters?
Bombcalorimetersaregenerallyconsideredtobeveryaccurateinmeasuring
theheatofcombustionofasample,withuncertaintiestypicallyintherangeof
1-2%.However,theaccuracyofthemeasurementcanbeaffectedbya
varietyoffactors,suchasthepurityofthesample,thecalibrationofthe
calorimeter,andtheprecisionofthetemperaturemeasurement.
Additional

16. Uses of Bomb Calorimeter
The mainuse of a bomb calorimeteris to measure the heat of combustion of a
sample, which provides important information about the energy content of the
sample. Here are some specific uses of bomb calorimeters:
1.Energy Research:Bomb calorimeters are commonly used in the field of
energy research to measure the heat of combustion of various fuels, such as
coal, oil, and natural gas.
2.Food Science:Bomb calorimeters are used in food science to determine the
caloric content of food products.
3.Pharmaceutical Research:Bomb calorimeters are also used in
pharmaceutical research to determine the heat of combustion of drug samples.
4.Environmental Testing:Bomb calorimeters can be used to measure the heat
of combustion of various materials, including waste and biomass.
Additional

21. In the solid or metal specific heat capacity is the sum of both i.e.electronlatticesolid CCC 

22. ClassicalTheoryofHeatCapacityofSolid:
TheDulong–Petitlaw,athermodynamiclawproposedin1819byFrenchphysicists
DulongandPetit,statestheclassicalexpressionforthemolarspecificheatofcertain
crystals.Thetwoscientistsconductedexperimentsonthreedimensionalsolidcrystals
todeterminetheheatcapacitiesofavarietyofthesesolids.Theassumptionofthis
theoryare:
Dulongand Petit
1.Each atom in the crystal vibrates freely constitutes a 3D
harmonic oscillator and vibrate independent to each other.
2.Each vibrational degree can be regarded as a one
dimensional harmonic oscillator
3.Each harmonic oscillator vibrates with its natural
frequency (w).

23. The total energy of a one dimensional harmonic oscillator of mass is the sum of kinetic
and potential energy i.e.)1(
2
1
2
22
2
xmw
m
p
E 
Substituting the value of E from equation (1))2(









dEe
dEEe
E
KT
E
KT
E

24. 



































































































dxee
dxeexmw
dpee
dpee
m
p
E
KT
xmw
mKT
p
KT
xmw
mKT
p
KT
xmw
mKT
p
KT
xmw
mKT
p
222
222
222
222
2
1
2
2
1
222
2
1
2
2
1
2
2
.
.
2
1
.
.
2 



































dxdpe
dxdpexmw
m
p
E
KT
xmw
m
p
KT
xmw
m
p
.
.
2
1
2
22
2
22
2
2
1
2
2
1
2
22
2 Additional

25. 































































dxe
dxexmw
dpe
dpe
m
p
E
KT
xmw
KT
xmw
mKT
p
mKT
p
22
22
2
2
2
1
2
1
22
2
2
2
2
1
2 Using Standard integral,KTxmw
KT
xmw
KT
m
p
mKT
p
Let
2222
22
2
2
2
2
2
1
,
22
,
2
,   



 






dede
22
2
,
2
Additional

26. 
























de
deKT
de
deKT
E
2
2
2
2
22 KTE
KTKT
KTKT
E


22
22



 Now, total vibrational energy in 3N one dimensional oscillator isNKTU3
Additional

27. Ris universal gas constant for a gram atom
and R=1.9856 cal. per gram atom
R=8.31J/mol.KKmolJKmolcalRNK
dT
dU
C
v
v ./93.24./96.533 






So specific heat capacity isRC
v3
Now,fromDulong&Petitlaw,thespecific
heatisindependentoftemperaturebutitis
experimentallyseenthatspecificheatat
lowertemperaturesisdirectlyproportional
tothecubeoftemperatures.Theabove
dependenceisbecauseofthefactthatthe
particlesinthecrystaloscillateasifthey
arecoupledQuantumHarmonicOscillator.

28. Limitations of DulongPetit Law
•The DulongPetit law is only relevant to the heavier elements.
•It is only applicable to elements that are in solid form.
•It cannot be applied to lighter elements having high melting points.
•It only gives a rough atomic mass.
This graph clearly shows that the law is applicable at
various higher temperatures for different elements.
Additional

29. Additional

30. Einstein’s Quantum Theory of Heat Capacity of Solid :
Themoderntheoryoftheheatcapacityofsolidsstatesthatitisduetolatticevibrationsin
thesolidandwasfirstderivedfromthisassumptionbyAlbertEinsteinin1907.
TheEinsteinsolidmodelthusgaveforthefirsttimeareasonwhytheDulong–Petitlaw
shouldbestatedintermsoftheclassicalheatcapacitiesforgases.
TheEinsteinsolidisamodelofasolidbasedontwocrudeassumptions:
1)AccordingtoEinstein’s,in0K,eachatominsolidisatrest,butthezeropointenergyis.
Asatemperatureincreases,thesolidoscillatessimpleharmonically.Theyareindependent
onotheratoms.
2)Theenergyofharmonicoscillatorisnotcontinuousbutitisdiscrete.
3)Eachatominthelatticeisanindependent3Dquantumharmonicoscillatorwithdiscrete
energylevels , wheren=0,1,2,3,…………,
histhePlanck-constantandfisthefrequency.
4)Allatomsoscillatewiththesamenaturalfrequency.hfE
2
1
 hfnE
n )
2
1
(

31. 








0
0
n
KT
E
n
KT
E
n
n
n
e
eE
EE According to Maxwell-Boltzmann Statistical distribution, the average energy of a harmonic oscillator per degree of
freedom at temperature KT is





























0
2
1
0
2
1
.
2
1
n
KT
wn
n
KT
wn
e
ewn
E


 





 wf
h
hf  

2.
2
Additional

32. 






KT
w
xLet

, 





























0
2
1
0
2
1
.
2
1
n
KT
wn
n
KT
wn
e
ewn
E


 



























0
2
1
0
2
1
.
2
1
n
xn
n
xn
e
ewn
E
 Additional

33. 


















...........
2
5
2
3
2
1
...........
2
5
2
3
2
1
2
5
2
3
2
1
2
5
2
3
2
1
xxx
xxx
eee
eeew
E
 







 ...........ln
2
5
2
3
2
1
xxx
eee
dx
d
wE )(.
)(
1
)(ln xf
dx
d
xf
xf
dx
d
  





 ...........1ln
22
1
xx
x
eee
dx
d
wE  ...........1lnln
22
1

xx
x
ee
dx
d
we
dx
d
wE  Additional

34.  ...........1lnln
22
1

xx
x
ee
dx
d
we
dx
d
wE  
x
e
dx
d
wx
dx
d
wE 





 1ln
2
1
 
 
x
x
e
e
wwE 

 0.
1
1
2
1
 
x
x
e
e
wwE


12
1
  1
1
2
1


x
e
wwE  Each atom in the lattice is an independent 3D quantum
harmonic oscillator. Total internal energy due to 3N atoms
Additional







1
3
2
3
3
KT
w
e
w
wNwNENU




35. Now, molar specific heat at constant volume
































2
1
1111
30
KT
w
KT
w
KT
w
v
e
e
dT
d
dT
d
e
wNC


 























1
1
3
2
3
KT
w
v
v
e
wNwN
dT
d
dT
dU
C

 





























1
1
3
2
3
KT
w
v
v
e
dT
d
wNwN
dT
d
dT
dU
C

 Additional

36. Now, molar specific heat at constant volume
































2
2
1
1
.0
3
KT
w
KT
w
v
e
TK
w
e
wNC


 

























2
1
3
KT
w
KT
w
v
e
e
KT
w
NKC


 K
w
RNK
E

 , 

























2
2
1
3
T
T
E
v
E
E
e
e
T
RC




37. At absolute 0K,1
KT
w
e
 2
2
.3











KT
w
KT
w
v
e
e
KT
w
RC


 












KT
wv
e
KT
w
RC

 1
.3
2 




































....
1
.3
32
2
KT
w
KT
w
KT
wKT
w
RC
v



38. At absolute 0K,0
vC 































....1
1
1
.3
KT
w
KT
w
RC
v

 







....10
1
.3RC
v

39. At high temperature,1
KT
w 2
2
.3











KT
w
KT
w
v
e
e
KT
w
RC


 




































































2
32
32
2
1.......1
........1
.3
KT
w
KT
w
KT
w
KT
w
KT
w
KT
w
KT
w
RC
v


 RC
v3 
























2
2
1
.3
KT
wKT
w
RC
v



40. At absolute 0K,0
vC 































....1
1
1
.3
KT
w
KT
w
RC
v

 







....10
1
.3RC
v

41. At high temperature,1
KT
w 










KT
w
v
e
KT
w
RC

 1
.3
2 













KT
w
v e
KT
w
RC


.3
2
At low temperature molar specific
heat decreases exponentially.

42. Degree of Freedom
Total number of independent quantities, variables or co-ordinate which are required to describe the
motion of particle are called degree of freedom.
Three types of degrees of freedom are translation, rotation, and vibration.
For a monatomic gas, degrees of freedom = 3, and all are translational: Molecules of
monoatomic gases can move linearly in any direction in space along the coordinate axis, so they
can have three independent motions and hence 3 degrees of freedom.
The example includes gases like Argon and Helium.

43. For a diatomic gas,
degrees of freedom = 5, where 3 are translational
and 2 are rotational:
In diatomic gas molecules, the centreof mass of
two atoms is free to move along three coordinate
axes. Thus, a diatomic molecule rotates about an
axis at right angles to its axis. Therefore, there are
2 degrees of freedom of rotational motion and 3
degrees of freedom of translational motion along
the three axes.
Theexampleincludesoxygenandnitrogen
molecules.

44. Diatomic molecule: There are two cases.
(i) At Normal temperature: A molecule of a diatomic gas consists of two atoms bound to each
other by a force of attraction. Physically the molecule can be regarded as a system of two point
masses fixed at the ends of a massless elastic spring. The center of mass lies in the center of
the diatomic molecule. So, the motion of the center of mass requires three translational
degrees offreedom (figure a). In addition, the diatomic molecule can rotate about three
mutually perpendicular axes (figure b). But the moment of inertia about its own axis of rotation
is negligible. Therefore, it has only two rotational degrees of freedom (one rotation is about Z
axis and another rotation is about Y axis). Therefore totally there are five degrees of freedom.
f =5

45. (ii)At High Temperature: At a very high temperature such as 5000 K, the diatomic molecules
possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of
vibration and the other is due to potential energy] (figure c ). So totally there are seven degrees of
freedom. f = 7.

46. For a non-linear triatomic gas, degrees of freedom = 6, where 3 are translational and 3
are rotational.
For a linear triatomic gas, degrees of freedom = 7, where 3 are translational, 3 are
rotational, and 1 is vibrational. Triatomic gas molecules have three atoms.
If all three atoms are aligned along a line, it is a linear molecule.
But if the three atoms are placed along the vertex of a triangle, then it is a non-
linear molecule.
•The degrees of freedom of the
system is given by the formula
f=3N–K
where,
f=degrees of freedom
N=Number of Particles in the
system.
K=Independent relation among
the particles

47. Non-linear triatomic molecule: In this case, the three atoms lie at the vertices
of a triangle. It has three translational degrees of freedom and three rotational
degrees of freedom about three mutually orthogonal axes. The total degrees of
freedom. f = 6 Example:
Triatomic molecules: There are two case.

48. At normal temperature, linear triatomic
molecule will have five degrees of
freedom. At high temperature it has two
additional vibrational degrees of
freedom. So a linear triatomic molecule
has seven degrees of freedom.

49. Law of Equipartitionof Energy:
Statement:In equilibrium, the total energy is equally distributed in all possible energy
modes, with each mode having average energy equal to
Explanation:A molecule has three types of kinetic energy a) Translational kinetic
energy,b) Rotational kinetic energy, andc) Vibrational kinetic energy. Thus total energy of a
molecule is given by
E
T= E
Translational+ E
Rotational+ E
Vibrational……… (1)

50. For translational motion, the molecule has three degrees of freedom (along the x-axis, along the
y-axis and along the z-axis).Hence
,
For rotational motion, it has two degrees of freedom along its centreof mass (clockwise and
anticlockwise).
For vibrational motion only one degree of freedom (to and fro).
Where k is the force constant and y is the vibrational coordinate.

51. Thus the total energy of the molecule is
It is to be noted that in the vibrational mode the energy has two components potential and
kinetic and by the law of equipartitionenergy, each part is equal to
Thus the total vibrational component of the energy is

52. Monoatomic Gases:
Helium and argon are monoatomic gases. The monoatomic gases have only translational
motion, hence they have three translational degrees of freedom. The average energy of the
molecule at temperature T is given by
By the law of equipartitionof energy we have

53. Thusthe ratio of specific heat capacities of monoatomic gas is 1.67
Thus the energy per mole of the gas is given by
1 moleofany
ideal
gasoccupies
exactly22.4 L
at STP.
Monoatomic Gases

54. Diatomic Gases:
Di-nitrogen, di-oxygen, and di-hydrogen are diatomic
gases. The diatomic gases have translational motion (three
translational degrees of freedom) as well as rotational
motion(rotational degree of freedom). The average energy
of the molecule at temperature T is given by
By the law of equipartitionof energy we have

55. Thus the energy per mole of the gas is given by
Thusthe ratio of specific heatcapacities of diatomic gas is 1.4
Diatomic Gases:

56. Triatomic Gas:
Trioxygen(ozone) ,water and carbondioxideare triatomic
gases. The triatomic gases have translational motion,
rotational motion as well as vibrational motion, hence has
three translational degrees of freedom and two rotational
degrees of freedom. For non-rigid molecules, there is an
additional vibrational motion.
The average energy of the molecule at temperature T is
given by

57. By the law of equipartitionof energy we have
Thus the energy per mole of the gas is given by
Thus for 7 degree of freedom,the ratio of specific heatcapacities of diatomic gas is 1.29.
Thusfor 6 degree of freedom, the ratio of specific heatcapacities of diatomic gas is 1.33.

58. Ideal Gas Law
According to the ideal gas law, when a gas is compressed into a smaller volume, the
number and velocity of molecular collisions increase, raising the gas's temperature
and pressure.
The Ideal Gas Equation

59. RepulsiveInteractions.Theoperationofrepulsiveintermolecularinteractionsimpliesthattwogasmoleculescannot
comecloserthanacertaindistanceofeachother.InsteadofbeingfreetotravelanywhereinavolumeV,theactual
volumeinwhichthemoleculecantravelisreducedbyanamountwhichisproportionaltothenumberofmolecules
presentandthevolumewhichtheyexclude.Thevolumeexcludingrepulsiveforcesaremodelledbychangingthe
volumetermVintheidealgasequationtoV-nb,wherebrepresentstheproportionalityconstantbetweenthe
reductioninvolumeandtheamountofmoleculespresentinthecontainer:itistheexcludedvolumepermole.
AttractiveInteractions.Thepresenceofattractiveinteractionsbetweenmoleculesistoreducethepressurethatthe
gasexerts.Theattractionexperiencedbyagivenmoleculeisproportionaltotheconcentrationn/Vofmoleculesin
thecontainer.Attractiveforcesslowthemoleculesdown:moleculesstrikethewallslessfrequentlyandwithless
impact.Pressuredeterminedbyimpactofmoleculesoncontainerwallsandisproportionaltorateofimpacttimes
averagestrengthofimpact.Bothofthesequantitiesareproportionaltotheconcentration.
The Van der Waals equation of state, although approximate, is based on a very simple physical model of intermolecular
interactions.

61. What Is Van der Waals Equation?
TheVanderWaalsequationisanequationrelatingtherelationshipbetweenthepressure,volume,
temperature,andamountofrealgases.Forarealgascontaining‘one’mole,theequationiswrittenas;RTbV
V
a
P 






2
where,
Pisthepressure,
Visthevolume,
Tisthetemperature,
nisthenumberofmolesofgases,
‘a’and‘b’aretheconstantsthatarespecifictoeachgas. nRTnbV
V
an
P 









2
2
For a real gas containing ‘n’ moles, the equation is written as;
J.D.VanderWaalsmadethefirst
mathematicalanalysisofrealgases.His
treatmentprovidesusaninterpretation
ofrealgasbehavioratthemolecular
level.Hemodifiedtheidealgas
equationPV=nRTbyintroducingtwo
correctionfactors,namely,pressure
correctionandvolumecorrection.

62. Pressure Correction:
Thepressureofagasisdirectlyproportionaltotheforcecreatedbythebombardmentofmoleculesonthe
wallsofthecontainer.Thespeedofamoleculemovingtowardsthewallofthecontainerisreducedbythe
attractiveforcesexertedbyitsneighbours.Hence,themeasuredgaspressureislowerthantheidealpressure
ofthegas.Hence,VanderWaalsintroducedacorrectiontermtothiseffect.
Van der Waals found out the forces of attraction experienced
by a molecule near the wall are directly proportional to the
square of the density of the gas.
where n is the number of moles of gas and V is the volume of
the container
where a is proportionality constant and depends on the nature of gas.Therefore,
Inter-molecular force of
attraction

63. VolumeCorrection:
Aseveryindividualmoleculeofagasoccupiesa
certainvolume,theactualvolumeislessthanthe
volumeofthecontainer,V.VanderWaalsintroduceda
correctionfactorV'tothiseffect.Letuscalculatethe
correctiontermbyconsideringgasmoleculesas
spheres.
whereV
misavolumeofasinglemoleculeExcluded
volumeforsinglemolecule

64. Excludedvolumefornmolecule=n(4V
m)=nb
wherebisVanderWaalsconstantwhchisequalto4V
m
V'=nb
V
ideal=V-nb
ReplacingthecorrectedpressureandvolumeintheidealgasequationPV=nRTwegetthevander
Waalsequationofstateforrealgasesasbelow,
TheconstantsaandbarevanderWaalsconstantsandtheirvaluesvarywiththenatureofthegas.It
isanapproximateformulaforthenon-idealgas.

65. EstimationofCriticalConstants
ConsiderthecriticalisothermalACBasshowninfigure.AtthecriticalpointC,the
curveishorizontal.Therefore,atpointC,0





dV
dP
Atthepointthetangentalsocrossesthecurve.Therefore,thetangentatsuchapointis
saidtobestationaryandthepointiscallthepointofinflection.Atthepintofinflection,
wehave0
2
2









dV
Pd
AtthepointC,wehave0 and 0
2
2















dV
Pd
dV
dP
A
C
B
highpressure
isotherm
gas has
disappeared
pressure of
about 62 atm

66. EstimationofCriticalConstants
TheVanderWaalsequationis2
V
a
bV
RT
P 

 ,,,
ccc VVPPTT 
AtthecriticalpointC,wehave
32
2
V
a
bV
RT
dV
dP


 
432
2
62
V
a
bV
RT
dV
Pd


 2
cc
c
c
V
a
bV
RT
P 

  
32
2
0
cc
c
V
a
bV
RT


  
43
62
0
cc
c
V
a
bV
RT


 0 and 0
2
2















dV
Pd
dV
dP

67. EstimationofCriticalConstants 
32
2
cc
c
V
a
bV
RT

  
43
62
cc
c
V
a
bV
RT


Fromaboveequations,wehave 
32
cc VbV

  
bV
bVV
VbV
c
cc
cc
3
323
23


  
32
3
2
3 b
a
bb
RT
c

  
32
2
cc
c
V
a
bV
RT

 bR
a
T
b
a
b
RT
c
c
27
8
27
2
4
32

 2
cc
c
c
V
a
bV
RT
P 

  
2
3327
8
b
a
bbbR
aR
P
c 

 
2
22
2
27
3
1
9
1
6
8
9
9227
8
b
a
P
b
a
b
a
P
b
a
bb
a
P
c
c
c








