Chapter 21 The Kinetic Theory of Gases. Molecular Model of an Ideal Gas The model shows that the...

Chapter 21Chapter 21

The Kinetic Theory of GasesThe Kinetic Theory of Gases

Molecular Model of an Ideal GasMolecular Model of an Ideal Gas

The model shows that the pressure that a gas The model shows that the pressure that a gas exerts on the walls of its container is a exerts on the walls of its container is a consequence of the collisions of the gas molecules consequence of the collisions of the gas molecules with the wallswith the walls

It is consistent with the macroscopic description It is consistent with the macroscopic description developed earlierdeveloped earlier

Assumptions for Ideal Gas TheoryAssumptions for Ideal Gas Theory The gas consist of a very large number of identical The gas consist of a very large number of identical

molecules each with mass molecules each with mass mm but with negligible size (this but with negligible size (this

assumption is approximately true when the distance between assumption is approximately true when the distance between

the molecules is large compared to the size) the molecules is large compared to the size)

The consequence The consequence rarrrarr for negligible size molecules we can for negligible size molecules we can

neglect the intermolecular collisions neglect the intermolecular collisions

The molecules donrsquot exert any action-at-distance forces The molecules donrsquot exert any action-at-distance forces

on each other This means there are no potential energy on each other This means there are no potential energy

changes to be considered so each molecules kinetic energy changes to be considered so each molecules kinetic energy

remains unchanged This assumption is fundamental to the remains unchanged This assumption is fundamental to the

nature of an ideal gasnature of an ideal gas

Assumptions for Ideal Gas TheoryAssumptions for Ideal Gas Theory

The molecules are moving in random The molecules are moving in random directions with a distribution of speeds that is directions with a distribution of speeds that is independent of directionindependent of direction

Collisions with the container walls are elastic Collisions with the container walls are elastic conserving the moleculersquos energy and conserving the moleculersquos energy and momentummomentum

Pressure and Kinetic EnergyPressure and Kinetic Energy

Assume a container Assume a container is a cubeis a cube

Edges are length Edges are length dd Look at the motion Look at the motion

of the molecule in of the molecule in terms of its velocity terms of its velocity componentscomponents

Look at its Look at its momentum and the momentum and the average forceaverage force

Pressure and Kinetic EnergyPressure and Kinetic Energy Since the collision is elastic the Since the collision is elastic the y y - -

component of moleculersquos velocity component of moleculersquos velocity remains unchanged while the remains unchanged while the xx - - component reverses sign Thus the component reverses sign Thus the molecule undergoes the momentum molecule undergoes the momentum change of magnitude change of magnitude 2mv2mvxx

After colliding with the right hand wall After colliding with the right hand wall

the the x x - component of moleculersquos - component of moleculersquos velocity will not change until it hits the velocity will not change until it hits the left-hand wall and its left-hand wall and its x x - velocity will - velocity will again reverses again reverses

ΔΔt = 2d vt = 2d vxx

Pressure and Kinetic EnergyPressure and Kinetic Energy The average force due to the The average force due to the

each molecule on the walleach molecule on the wall

To get the total force on the To get the total force on the wall we sum over all wall we sum over all N N molecules Dividing by the wall molecules Dividing by the wall area area AA then gives the force per then gives the force per unit area or pressure unit area or pressure

d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22

Since is just the average of the squares of is just the average of the squares of xx ndash ndash components of velocities components of velocities

2xvV

mNP


2xv 2

yv 2zv

Since the molecules are moving in random directions the Since the molecules are moving in random directions the

average quantities and must be average quantities and must be

equal and the average of the molecular speed equal and the average of the molecular speed

and and vv22 = 3v = 3vxx22 or or vvxx

22 = v = v223 3 Then the expression for Then the expression for

pressurepressure

2222zyx vvvv

2

3v

V

mNP


The relationship can be writtenThe relationship can be written

This tells us that pressure is proportional to the This tells us that pressure is proportional to the number of molecules per unit volume (number of molecules per unit volume (NNVV) and ) and to the average translational kinetic energy of the to the average translational kinetic energy of the moleculesmolecules

___22 1

3 2

NP mv

V


This equation also relates the macroscopic This equation also relates the macroscopic quantity of pressure with a microscopic quantity quantity of pressure with a microscopic quantity of the average value of the square of the of the average value of the square of the molecular speedmolecular speed

One way to increase the pressure is to increase One way to increase the pressure is to increase the number of molecules per unit volumethe number of molecules per unit volume

The pressure can also be increased by The pressure can also be increased by increasing the speed (kinetic energy) of the increasing the speed (kinetic energy) of the moleculesmolecules

A A 200-mol 200-mol sample of oxygen gas is confined to a sample of oxygen gas is confined to a 500-L 500-L vessel at a pressure of vessel at a pressure of 800 atm800 atm Find the Find the average translational kinetic energy of an oxygen average translational kinetic energy of an oxygen molecule under these conditions molecule under these conditions


223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2

3 800 atm 1013 10 Pa atm 500 10 m32 2 2 2 mol 602 10 molecules mol

505 10 J molecule

mv PVK N nN N

N

PVK

N

K

Molecular Interpretation of Molecular Interpretation of TemperatureTemperature

We can take the pressure as it relates to the kinetic We can take the pressure as it relates to the kinetic energy and compare it to the pressure from the equation energy and compare it to the pressure from the equation of state for an ideal gasof state for an ideal gas

Therefore the temperature is a direct measure of the Therefore the temperature is a direct measure of the average molecular kinetic energyaverage molecular kinetic energy

TnkmvV

NP B

2

2

1

3

2


Simplifying the equation relating temperature and Simplifying the equation relating temperature and kinetic energy giveskinetic energy gives

This can be applied to each direction This can be applied to each direction

with similar expressions for with similar expressions for vvyy and and vvzz

___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T

A Microscopic Description of A Microscopic Description of TemperatureTemperature

Each translational degree of freedom Each translational degree of freedom contributes an equal amount to the contributes an equal amount to the energy of the gasenergy of the gas

A generalization of this result is called A generalization of this result is called the the theorem of equipartition of energytheorem of equipartition of energy

Theorem of Equipartition of EnergyTheorem of Equipartition of Energy

Each degree of freedom contributes Each degree of freedom contributes frac12frac12kkBBTT to the energy of a system where to the energy of a system where

possible degrees of freedom in addition to possible degrees of freedom in addition to those associated with translation arise those associated with translation arise from rotation and vibration of moleculesfrom rotation and vibration of molecules

Total Kinetic EnergyTotal Kinetic Energy The total kinetic energy is just The total kinetic energy is just NN times the kinetic energy times the kinetic energy

of each moleculeof each molecule

If we have a gas with only translational energy this is the If we have a gas with only translational energy this is the internal energy of the gasinternal energy of the gas

This tells us that the internal energy of an ideal gas This tells us that the internal energy of an ideal gas depends only on the temperaturedepends only on the temperature

___2

tot trans B

1 3 3

2 2 2K N mv Nk T nRT

Root Mean Square SpeedRoot Mean Square Speed The root mean square (The root mean square (rmsrms) speed is the square root of ) speed is the square root of

the average of the squares of the speedsthe average of the squares of the speeds Square average take the square rootSquare average take the square root

Solving for Solving for vvrmsrms we findwe find

MM is the molar mass and is the molar mass and MM = = mNmNAA

Brms

3 3k T RTv

m M

Some Example Some Example vvrmsrms ValuesValues

At a given At a given temperature lighter temperature lighter molecules move molecules move faster on the faster on the average than average than heavier moleculesheavier molecules

Molar Specific HeatMolar Specific Heat

Several processes can Several processes can change the temperature of change the temperature of an ideal gasan ideal gas

Since Since ΔΔTT is the same for is the same for each process each process ΔΔEEintint is also is also the samethe same

The heat is different for the The heat is different for the different pathsdifferent paths

The heat associated with a The heat associated with a particular change in particular change in temperature is temperature is notnot unique unique


We define specific heats for two processes that We define specific heats for two processes that frequently occurfrequently occurndash Changes with constant volumeChanges with constant volumendash Changes with constant pressureChanges with constant pressure

Using the number of moles Using the number of moles nn we can define we can define molar specific heats for these processesmolar specific heats for these processes

Molar Specific HeatMolar Specific Heat Molar specific heatsMolar specific heats

QQ = = nCnCVV ΔΔTT for constant-volume processes for constant-volume processes QQ = = nCnCPP ΔΔTT for constant-pressure processes for constant-pressure processes

QQ (constant pressure) must account for both the (constant pressure) must account for both the increase in internal energy and the transfer of increase in internal energy and the transfer of energy out of the system by workenergy out of the system by work

QQconstant constant PP gt Q gt Qconstant constant VV for given values of for given values of nn and and ΔΔTT

Ideal Monatomic GasIdeal Monatomic Gas

A monatomic gas contains one atom per A monatomic gas contains one atom per moleculemolecule

When energy is added to a monatomic gas in When energy is added to a monatomic gas in a container with a fixed volume all of the a container with a fixed volume all of the energy goes into increasing the translational energy goes into increasing the translational kinetic energy of the gaskinetic energy of the gas There is no other way to store energy in There is no other way to store energy in

such a gassuch a gas


Therefore Therefore

ΔΔEEintint is a function of is a function of TT only only At constant volumeAt constant volume

QQ = = ΔΔEEintint = = nCnCVV ΔΔTTThis applies to all ideal gases not just monatomic onesThis applies to all ideal gases not just monatomic ones

nRTTNkKE Btranstot 2

3

2

3int

Monatomic GasesMonatomic Gases

Solving Solving

for for CCVV gives gives CCVV = 32 = 32 RR = 125 Jmol = 125 Jmol K K for all for all

monatomic gasesmonatomic gases This is in good agreement with experimental This is in good agreement with experimental

results for monatomic gasesresults for monatomic gases

TnRTnCE V 2

3int

Monatomic GasesMonatomic Gases In a constant-pressure process In a constant-pressure process ΔΔEEintint = = QQ + + WW and and

Change in internal energy depends only on temperature Change in internal energy depends only on temperature for an ideal gas and therefore are the same for the for an ideal gas and therefore are the same for the constant volume process and for constant pressure constant volume process and for constant pressure process process

CCPP ndash C ndash CVV = R = R

VPTnCWQE P int

TnRTnCTnC PV


CCPP ndash ndash CCVV = = RR This also applies to any ideal gasThis also applies to any ideal gas

CCPP = 52 = 52 RR = 208 Jmol = 208 Jmol K K

Ratio of Molar Specific HeatsRatio of Molar Specific Heats

We can also define We can also define

Theoretical values of Theoretical values of CCVV CCPP and and are in are in excellent agreement for monatomic gasesexcellent agreement for monatomic gases

But they are in serious disagreement with the But they are in serious disagreement with the values for more complex moleculesvalues for more complex moleculesndash Not surprising since the analysis was for Not surprising since the analysis was for

monatomic gasesmonatomic gases

5 2167

3 2P

V

C R

C R

Sample Values of Molar Specific HeatsSample Values of Molar Specific Heats

A A 100-mol 100-mol sample of air (a diatomic ideal gas) sample of air (a diatomic ideal gas) at at 300 K300 K confined in a cylinder under a heavy confined in a cylinder under a heavy piston occupies a volume of piston occupies a volume of 500 L500 L Determine Determine the final volume of the gas after the final volume of the gas after 440 kJ440 kJ of of energy is transferred to the air by heatenergy is transferred to the air by heat

A A 100-mol 100-mol sample of air (a diatomic ideal gas) at sample of air (a diatomic ideal gas) at 300 K300 K confined in a cylinder under a heavy piston occupies a confined in a cylinder under a heavy piston occupies a volume of volume of 500 L500 L Determine the final volume of the gas Determine the final volume of the gas after after 440 kJ440 kJ of energy is transferred to the air by heat of energy is transferred to the air by heat

nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(

The piston moves to keep pressure constant The piston moves to keep pressure constant

nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005

Molar Specific Heats of Other MaterialsMolar Specific Heats of Other Materials

The internal energy of more complex gases The internal energy of more complex gases must include contributions from the rotational must include contributions from the rotational and vibrational motions of the moleculesand vibrational motions of the molecules

In the cases of solids and liquids heated at In the cases of solids and liquids heated at constant pressure very little work is done since constant pressure very little work is done since the thermal expansion is small and the thermal expansion is small and CCPP and and CCVV are approximately equalare approximately equal

Adiabatic Processes for an Ideal GasAdiabatic Processes for an Ideal Gas

Assume an ideal gas is in an equilibrium state Assume an ideal gas is in an equilibrium state and so and so PVPV = = nRTnRT is valid is valid

The pressure and volume of an ideal gas at any The pressure and volume of an ideal gas at any time during an adiabatic process are related by time during an adiabatic process are related by PV PV = constant= constant

= = CCPP CCVV is assumed to be constant during the is assumed to be constant during the processprocess

All three variables in the ideal gas law (All three variables in the ideal gas law (PP VV TT ) ) can change during an adiabatic processcan change during an adiabatic process

Equipartition of EnergyEquipartition of Energy

With complex With complex molecules other molecules other contributions to internal contributions to internal energy must be taken energy must be taken into accountinto account

One possible way to One possible way to energy change is the energy change is the translational motion of translational motion of the center of massthe center of mass


Rotational motion Rotational motion about the various axes about the various axes also contributesalso contributes

We can neglect the We can neglect the rotation around the rotation around the yy axis since it is negligible axis since it is negligible compared to the compared to the xx and and zz axesaxes


The molecule can The molecule can also vibratealso vibrate

There is kinetic There is kinetic energy and potential energy and potential energy associated energy associated with the vibrationswith the vibrations


The translational motion adds three degrees of The translational motion adds three degrees of freedomfreedom

The rotational motion adds two degrees of The rotational motion adds two degrees of freedomfreedom

The vibrational motion adds two more degrees The vibrational motion adds two more degrees of freedomof freedom

Therefore Therefore EEintint = 72 = 72 nRTnRT and and CCVV = 72 = 72 RR

Agreement with ExperimentAgreement with Experiment Molar specific heat is a function of temperatureMolar specific heat is a function of temperature At low temperatures a diatomic gas acts like a At low temperatures a diatomic gas acts like a

monatomic gas monatomic gas CCVV = 32 = 32 RR

Agreement with ExperimentAgreement with Experiment

At about room temperature the value increases At about room temperature the value increases to to CCVV = 52 = 52 RR

This is consistent with adding rotational energy but This is consistent with adding rotational energy but not vibrational energynot vibrational energy

At high temperatures the value increases At high temperatures the value increases to to CCVV = 72 R = 72 R

This includes vibrational energy as well as rotational This includes vibrational energy as well as rotational and translationaland translational

Complex MoleculesComplex Molecules

For molecules with more than two atoms the For molecules with more than two atoms the vibrations are more complexvibrations are more complex

The number of degrees of freedom is largerThe number of degrees of freedom is larger The more degrees of freedom available to a The more degrees of freedom available to a

molecule the more ldquowaysrdquo there are to store molecule the more ldquowaysrdquo there are to store energyenergy

This results in a higher molar specific heatThis results in a higher molar specific heat

Quantization of EnergyQuantization of Energy

To explain the results of the various molar To explain the results of the various molar specific heats we must use some quantum specific heats we must use some quantum mechanicsmechanics

Classical mechanics is not sufficientClassical mechanics is not sufficient

In quantum mechanics the energy is In quantum mechanics the energy is proportional to the frequency of the wave proportional to the frequency of the wave representing the frequencyrepresenting the frequency

The energies of atoms and molecules are The energies of atoms and molecules are quantizedquantized


This energy level diagram This energy level diagram shows the rotational and shows the rotational and vibrational states of a vibrational states of a diatomic moleculediatomic molecule

The lowest allowed state is The lowest allowed state is the the ground stateground state


The vibrational states are separated by larger The vibrational states are separated by larger energy gaps than are rotational statesenergy gaps than are rotational states

At low temperatures the energy gained during At low temperatures the energy gained during collisions is generally not enough to raise it to collisions is generally not enough to raise it to the first excited state of either rotation or the first excited state of either rotation or vibrationvibration


Even though rotation and vibration are Even though rotation and vibration are classically allowed they do not occurclassically allowed they do not occur

As the temperature increases the energy of the As the temperature increases the energy of the molecules increasesmolecules increases

In some collisions the molecules have enough In some collisions the molecules have enough energy to excite to the first excited stateenergy to excite to the first excited state

As the temperature continues to increase more As the temperature continues to increase more molecules are in excited states molecules are in excited states


At about room temperature rotational energy is At about room temperature rotational energy is contributing fullycontributing fully

At about At about 1000 K1000 K vibrational energy levels are vibrational energy levels are reachedreached

At about At about 10 000 K10 000 K vibration is contributing fully vibration is contributing fully to the internal energyto the internal energy

Molar Specific Heat of SolidsMolar Specific Heat of Solids

Molar specific heats in solids also demonstrate a Molar specific heats in solids also demonstrate a marked temperature dependencemarked temperature dependence

Solids have molar specific heats that generally Solids have molar specific heats that generally decrease in a nonlinear manner with decreasing decrease in a nonlinear manner with decreasing temperaturetemperature

It approaches zero as the temperature It approaches zero as the temperature approaches absolute zeroapproaches absolute zero

DuLong-Petit LawDuLong-Petit Law

At high temperatures the molar specific heats At high temperatures the molar specific heats approach the value of approach the value of 3R3R

This occurs above This occurs above 300 K300 K

The molar specific heat of a solid at high The molar specific heat of a solid at high temperature can be explained by the equipartition temperature can be explained by the equipartition theoremtheorem

Each atom of the solid has six degrees of freedomEach atom of the solid has six degrees of freedom

The internal energy is The internal energy is 33nRTnRT and and CCvv = 3 = 3 RR


As As TT approaches approaches 00 the molar specific heat the molar specific heat approaches approaches 00

At high temperatures At high temperatures CCVV becomes a becomes a

constant at constant at ~3~3RR

Specific Heat and Molar Specific Heat of Some Solids and LiquidsSpecific Heat and Molar Specific Heat of Some Solids and Liquids

SubstanceSubstance c kJkgc kJkgKK c JmolKc JmolK

AluminumAluminum 09000900 243243

BismuthBismuth 01230123 257257

CopperCopper 03860386 245245

GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264

SilverSilver 02330233 249249

TungstenTungsten 01340134 248248

ZinkZink 03870387 252252

Alcohol (ethyl)Alcohol (ethyl) 2424 111111

MercuryMercury 01400140 283283

WaterWater 418418 752752

The molar mass of copper is 635 gmol Use the Dulong-Petit law to calculate the specific heat of copper


The molar specific heat is the heat capacity per mole

The Dulong-Petit lawgives c in terms of R

c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3

This differs from the measured value of 0386 kJkgThis differs from the measured value of 0386 kJkgK given K given in Table by less than 2in Table by less than 2

The specific heat of a certain metal is measured to The specific heat of a certain metal is measured to be be 102 kJkg102 kJkgKK (a) Calculate the molar mass of (a) Calculate the molar mass of this metal assuming that the metal obeys the this metal assuming that the metal obeys the Dulong-Petit law (b) What is the metalDulong-Petit law (b) What is the metal

The specific heat of a certain metal is measured to be The specific heat of a certain metal is measured to be 102 102 kJkgkJkgKK (a) Calculate the molar mass of this metal assuming (a) Calculate the molar mass of this metal assuming that the metal obeys the Dulong-Petit law (b) What is the that the metal obeys the Dulong-Petit law (b) What is the metalmetal

(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3

(b) The metal must be magnesium which has a molar mass of 2431 gmol

Boltzmann Distribution LawBoltzmann Distribution Law

The motion of molecules is extremely chaoticThe motion of molecules is extremely chaotic Any individual molecule is colliding with others at Any individual molecule is colliding with others at

an enormous ratean enormous rateTypically at a rate of a billion times per secondTypically at a rate of a billion times per second

We add the We add the number densitynumber density nnV V ((E E ))

This is called a distribution functionThis is called a distribution function

It is defined so that It is defined so that nnV V ((E E ) ) dEdE is the number of is the number of

molecules per unit volume with energy between molecules per unit volume with energy between EE and and EE + + dE dE

Number Density and Boltzmann Number Density and Boltzmann Distribution LawDistribution Law

From statistical mechanics the number density From statistical mechanics the number density is is

nnV V ((E E ) = ) = nnooe e ndashndashE E kkBBTT

This equation is known as the Boltzmann This equation is known as the Boltzmann distribution lawdistribution law

It states that the probability of finding the It states that the probability of finding the molecule in a particular energy state varies molecule in a particular energy state varies exponentially as the energy divided by exponentially as the energy divided by kkBBTT

Distribution of Molecular SpeedsDistribution of Molecular Speeds

The observed speed The observed speed distribution of gas distribution of gas molecules in thermal molecules in thermal equilibrium is shown equilibrium is shown at rightat right

NNVV is called the is called the

Maxwell-Boltzmann Maxwell-Boltzmann speed distribution speed distribution functionfunction

Distribution FunctionDistribution Function

The fundamental expression that describes the The fundamental expression that describes the distribution of speeds in distribution of speeds in NN gas molecules is gas molecules is

mm is the mass of a gas molecule is the mass of a gas molecule kkBB is is Boltzmannrsquos constant and Boltzmannrsquos constant and TT is the absolute is the absolute temperaturetemperature

2

3 2

22

B

42

Bmv k TV

mN N v e

k T

Most Probable SpeedMost Probable Speed

The average speed is somewhat lower than the The average speed is somewhat lower than the rmsrms speed speed

The most probable speed The most probable speed vvmpmp is the speed at is the speed at

which the distribution curve reaches a peakwhich the distribution curve reaches a peak

B Bmp

2141

k T k Tv

m m

Speed DistributionSpeed Distribution The peak shifts to the right as The peak shifts to the right as TT increases increases

This shows that the average speed increases with This shows that the average speed increases with increasing temperatureincreasing temperature

The asymmetric shape occurs because the lowest The asymmetric shape occurs because the lowest possible speed is possible speed is 00 and the highest is infinity and the highest is infinity

Speed DistributionSpeed Distribution

The distribution of molecular speeds depends The distribution of molecular speeds depends both on the mass and on temperatureboth on the mass and on temperature

The speed distribution for liquids is similar to that The speed distribution for liquids is similar to that of gasesof gases

EvaporationEvaporation Some molecules in the liquid are more energetic Some molecules in the liquid are more energetic

than othersthan others Some of the faster moving molecules penetrate Some of the faster moving molecules penetrate

the surface and leave the liquidthe surface and leave the liquidThis occurs even before the boiling point is reachedThis occurs even before the boiling point is reached

The molecules that escape are those that have The molecules that escape are those that have enough energy to overcome the attractive forces enough energy to overcome the attractive forces of the molecules in the liquid phaseof the molecules in the liquid phase

The molecules left behind have lower kinetic The molecules left behind have lower kinetic energiesenergies

Therefore evaporation is a cooling processTherefore evaporation is a cooling process

Mean Free PathMean Free Path

A molecule moving through a A molecule moving through a gas collides with other gas collides with other molecules in a random molecules in a random fashionfashion

This behavior is sometimes This behavior is sometimes referred to as a referred to as a random-walk random-walk processprocess

The The mean free pathmean free path increases as the number of increases as the number of molecules per unit volume molecules per unit volume decreasesdecreases


The molecules move with constant speed The molecules move with constant speed along straight lines between collisionsalong straight lines between collisions

The average distance between collisions is The average distance between collisions is called the called the mean free pathmean free path

The path of an individual molecule is randomThe path of an individual molecule is randomThe motion is not confined to the plane of the The motion is not confined to the plane of the paperpaper


The The mean free pathmean free path is related to the is related to the diameter of the molecules and the density of diameter of the molecules and the density of the gasthe gas

We assume that the molecules are spheres of We assume that the molecules are spheres of diameter diameter dd

No two molecules will collide unless their No two molecules will collide unless their paths are less than a distance paths are less than a distance dd apart as the apart as the molecules approach each othermolecules approach each other


The mean free path The mean free path ℓℓ equals the average equals the average distance distance vvΔΔtt traveled in a time interval traveled in a time interval ΔΔtt divided by the number of collisions that occur in divided by the number of collisions that occur in that time intervalthat time interval

22

1

VV

v t

d nd v t n

Collision FrequencyCollision Frequency

The number of collisions per unit time is the The number of collisions per unit time is the collision frequencycollision frequency

The inverse of the collision frequency is the The inverse of the collision frequency is the collision collision mean free timemean free time

2ƒ Vd vn

Molecular Model of an Ideal GasMolecular Model of an Ideal Gas

The model shows that the pressure that a gas The model shows that the pressure that a gas exerts on the walls of its container is a exerts on the walls of its container is a consequence of the collisions of the gas molecules consequence of the collisions of the gas molecules with the wallswith the walls

It is consistent with the macroscopic description It is consistent with the macroscopic description developed earlierdeveloped earlier




























d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




























d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn

















d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn














d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn









d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




d

mv

vd

mv

t

pF x

x

xi

2

)2(

2

V

vm

Ad

vm

Ad

mv

A

F

A

FP xx

x

i 22

2


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


N

vx 2

N

v

V

mN

V

vmP xx

22


2xvV

mNP


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


2xv 2

yv 2zv






pressurepressure

2222zyx vvvv

2

3v

V

mNP




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




___22 1

3 2

NP mv

V







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn







223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn



223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


223 2N mv

PV

2

av A A

5 3 3

av 23A

21av

3 where 2

2 2


505 10 J molecule

mv PVK N nN N

N

PVK

N

K




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




TnkmvV

NP B

2

2

1

3

2





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn





___2

B

1 3

2 2mv k T

___2

B

1 1

2 2xmv k T











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn











___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn








___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn





___2

tot trans B

1 3 3






Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn





Brms

3 3k T RTv

m M




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


















Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn













Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn










Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn






Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


Therefore Therefore




3

2

3int


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


Solving Solving




TnRTnCE V 2

3int




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




VPTnCWQE P int

TnRTnCTnC PV









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn









5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn






5 2167

3 2P

V

C R

C R




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn



nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


nR

Q

RR

n

Q

RCn

Q

nC

QT

VP 5

2

2

3)(


nR TV

P

P VQ nC T n C R T

nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


nRT

QV

P

Q

nR

Q

P

nRT

P

nRV

5

2

5

2

5

2

LK

Kmol

Jmol

LJV 533

)300(3148)1(

)5)(10404(

5

2 3

LLLVVV if 538533005










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn










































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn







































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn































































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




























































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn

























































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




















































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn


















































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn













































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn








































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn



































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn
































GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn





























GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn
























GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn




















GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn
















GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

1

VV

v t

d nd v t n




2ƒ Vd vn










GoldGold 01260126 256256

Ice (-10Ice (-1000C)C) 205205 369369

LeadLead 01280128 264264



ZinkZink 03870387 252252








c= 3R

Mcn

mc

n

Cc

KkgkJKgJmolg

KmolJ

M

cc

39203920

563

)3148(3




(a)(a)

gmolkgmolKkgJ

KmolJ

c

RM

RMc

Rc

Mcc

452402445010021

)3148(33

3

3

3





















2

3 2

22

B

42

Bmv k TV

mN N v e

k T





B Bmp

2141

k T k Tv

m m



























22

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Date post:	18-Dec-2015
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Chapter 21 The Kinetic Theory of Gases. Molecular Model of an Ideal Gas The model shows that the...

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