ContentsX8.7.31Potential Energy Surfaces from Spectroscopic Information:.305vanderWaalsComplexes.8.8.307Summary...309SuggestedReadingsProblems.310AnswersandSolutionsto315SelectedProblemsIndex.319
Contents 8.7.3 Potential Energy Surfaces from Spectroscopic Information: van der Waals Complexes . 305 8.8 Summary . 307 Suggested Readings . 309 Problems . 310 Answers and Solutions to Selected Problems . 315 Index . 319
CopyrightedMaterialsKinetic Theoryof GasesChapter Outline1.1Introduction1.2Pressure of an ldeal Gas1.3Temperature and Energy1.4Distributions,Mean Values,and Distribution Functions1.5TheMaxwellDistributionofSpeeds1.6EnergyDistributions1.7Collisions:MeanFreePath and Collision Number1.8SummaryThe Functional Form of the Velocity DistributionAppendix 1.1Appendix 1.2SphericalCoordinatesThe Error Functionand Co-Error FunctionAppendix1.3Appendix 1.4TheCenter-of-MassFrame1.1INTRODUCTIONThe overall objective of this chapter is to understand macroscopic properties suchas pressure and temperature on amicroscopic level.We willfind that the pressureof an ideal gas can be understood by applying Newton's law to the microscopicmotion of themoleculesmakingup thegas and thata comparisonbetweentheNewtonian prediction and the ideal gas law can provide a function that describesthe distribution of molecular velocities.This distribution function can in turn beused to learn aboutthefrequency of molecular collisions.Sincemolecules canreactonly as fast as they collide with one another, the collision frequency provides anupper limit on the reaction rate.The outline of the discussion is as follows. By applying Newton's laws to themolecular motion we will find that the product of the pressure and the volume isproportional totheaverageof thesquareofthemolecularvelocity,<u2>,orequiv-alently to the average molecular translational energy e. In order for this result to beconsistent with the observed ideal gas law, thetemperature T of thegas must alsobeproportional to<u2>or<e>.Wewillthenconsiderindetailhowtodeterminethe average of the square of the velocity from a distribution of velocities, and wewill use the proportionality of T with <u2>to determine the Maxwell-Boltzmanndistribution of speeds.This distribution, F(u) du, tells us the number of moleculeswith speeds between u and u + du.The speed distribution is closely related to the dis-tribution of molecular energies, G(e) de. Finally, we will use the velocity distribution
Kinetic Theory of Gases Chapter Outline 1 .I Introduction 1.2 Pressure of an Ideal Gas 1.3 Temperature and Energy 1.4 Distributions, Mean Values, and Distribution Functions 1.5 The Maxwell Distribution of Speeds 1.6 Energy Distributions 1.7 Collisions: Mean Free Path and Collision Number 1.8 Summary Appendix 1.1 The Functional Form of the Velocity Distribution Appendix 1.2 Spherical Coordinates Appendix 1.3 The Error Function and Co-Error Function Appendix 1.4 The Center-of-Mass Frame 1 .I INTRODUCTION The overall objective of this chapter is to understand macroscopic properties such as pressure and temperature on a microscopic level. We will find that the pressure of an ideal gas can be understood by applying Newton's law to the microscopic motion of the molecules making up the gas and that a comparison between the Newtonian prediction and the ideal gas law can provide a function that describes the distribution of molecular velocities. This distribution function can in turn be used to learn about the frequency of molecular collisions. Since molecules can react only as fast as they collide with one another, the collision frequency provides an upper limit on the reaction rate. The outline of the discussion is as follows. By applying Newton's laws to the molecular motion we will find that the product of the pressure and the volume is proportional to the average of the square of the molecular velocity, <v2>, or equivalently to the average molecular translational energy E. In order for this result to be consistent with the observed ideal gas law, the temperature T of the gas must also be proportional to <v2> or <E>. We will then consider in detail how to determine the average of the square of the velocity from a distribution of velocities, and we will use the proportionality of T with <v2> to determine the Maxwell-Boltzmann distribution of speeds. This distribution, F(v) dv, tells us the number of molecules with speeds between v and u + dv. The speed distribution is closely related to the distribution of molecular energies, G(E) d~. Finally, we will use the velocity distribution
2Chapter 1Kinetic Theory of Gasesto calculate the number of collisions Z that a molecule makes with other moleculesin the gas per unit time. Since in later chapters we will argue that a reaction betweentwo molecules requires that they collide, the collision rate Z provides an upper limitto the rate of a reaction.A related quantityAis the average distance a moleculetravelsbetween collisions orthe mean freepath.The history of the kinetic theory of gases is a checkered one, and serves to dis-pel the impression that science always proceeds along a straight and logical path.a In1662 Boyle found that for a specified quantity of gas held at a fixed temperature theproduct of the pressure and the volume was a constant.Daniel Bernoulli derived thislawin1738byapplyingNewton'sequationsofmotiontothemoleculescomprisingthe gas, but his work appears to have been ignored for more than a century.b A schoolteacher in Bombay, India, named John James Waterston submitted a paper to theRoyal Society in 1845 outlining many of the concepts that underlie our currentunderstanding of gases. His paper was rejected as “nothing but nonsense, unfit evenforreadingbeforetheSociety.Bernoulli'scontributionwasrediscoveredin1859and several decades later in 1892,afterJoule(1848)and Clausius (1857)had putforth similar ideas, Lord Rayleigh found Waterston's manuscript in the Royal Soci-etyarchives.Itwas subsequentlypublished inPhilosophicalTransactions.Maxwell(lllustrations of Dynamical Theoryof Gases,1859-1860)andBoltzmann(Vor-lesungen iber Gastheorie, 1896-1898) expanded the theory into its current form.1.2PRESSUREOFANIDEALGASWe start with the basic premise that the pressure exerted by a gas on the wall of a con-tainer is dueto collisions of molecules with the wall. Since the numberof moleculesin the container is large, the number colliding with the wall per unit time is largeenough so that fluctuations in the pressure due to the individual collisions are immea-surably small in comparison to thetotal pressure.The first step in the calculationis toapplyNewton's laws to the moleculesto show that theproductofthepressure and thevolume is proportional to the average of the square of the molecular velocity, <u?>.Consider molecules with a velocity component u,in thex direction and a massm. Let the molecules strike a wall of area A located in the z-y plane, as shown inFigure1.1.We would first like toknowhowmanymolecules strike thewall inatime At, where At is short compared to the time between molecular collisions.Thedistance along the x axis that a molecule travels in the time At is simply u,At, sothat all molecules located in the volume Au,At and moving toward the wall willstrike it. Let nbe the number of molecules per unit volume. Since one half of themoleculeswillbemovingtowardthewall inthe+xdirectionwhiletheotherhalfwill be moving in the -x direction, the number of molecules which will strike thewallinthetimeAtisnAu,At.The force on the wall due to the collision of a molecule with the wall is givenby Newton's law: F = ma = m du/dt = d(mu)/dt, and integration yields FtA(mu). If a molecule rebounds elastically (without losing energy) when it hits thewall, its momentum is changed from +mu, to -mu x, so that the total momentumchangeis△(mw)=2mur.Consequently,F△t=2mu,foronemolecularcollisionand FAt=(n'Au,At)(2mux)for the total number of collisions.CancelingAtfrom both sides andrecognizing that thepressure is the forceper unit area,p =FlAweobtainp=nmw?rThe history of the kinetic theory of gases is outlined by E. Mendoza, Physics Today 14, 36-39 (1961)hA translation of this paper has appeared in The World of Mathematics, J. R. Newman, Ed., Vol.2(Simon and Schuster, New York, 1956),p.774
Chapter 1 Kinetic Theory of Gases to calculate the number of collisions Z that a molecule makes with other molecules in the gas per unit time. Since in later chapters we will argue that a reaction between two molecules requires that they collide, the collision rate Z provides an upper limit to the rate of a reaction. A related quantity A is the average distance a molecule travels between collisions or the mean free path. The history of the kinetic theory of gases is a checkered one, and serves to dispel the impression that science always proceeds along a straight and logical path." In 1662 Boyle found that for a specified quantity of gas held at a fixed temperature the product of the pressure and the volume was a constant. Daniel Bernoulli derived this law in 1738 by applying Newton's equations of motion to the molecules comprising the gas, but his work appears to have been ignored for more than a ~entu1-y.~ A school teacher in Bombay, India, named John James Waterston submitted a paper to the Royal Society in 1845 outlining many of the concepts that underlie our current understanding of gases. His paper was rejected as "nothing but nonsense, unfit even for reading before the Society." Bernoulli's contribution was rediscovered in 1859, and several decades later in 1892, after Joule (1848) and Clausius (1857) had put forth similar ideas, Lord Rayleigh found Waterston's manuscript in the Royal Society archives. It was subsequently published in Philosophical Transactions. Maxwell (Illustrations of Dynamical Theory of Gases, 1859-1860) and Boltzmann (Vorlesungen iiber Gastheorie, 1896-1898) expanded the theory into its current form. 1.2 PRESSURE OF AN IDEAL GAS We start with the basic premise that the pressure exerted by a gas on the wall of a container is due to collisions of molecules with the wall. Since the number of molecules in the container is large, the number colliding with the wall per unit time is large enough so that fluctuations in the pressure due to the individual collisions are irnrneasurably small in comparison to the total pressure. The first step in the calculation is to apply Newton's laws to the molecules to show that the product of the pressure and the volume is proportional to the average of the square of the molecular velocity, <u2>. Consider molecules with a velocity component u, in the x direction and a mass m. Let the molecules strike a wall of area A located in the z-y plane, as shown in Figure 1.1. We would first like to know how many molecules strike the wall in a time At, where At is short compared to the time between molecular collisions. The distance along the x axis that a molecule travels in the time At is simply v,At, so that all molecules located in the volume Av,At and moving toward the wall will strike it. Let n* be the number of molecules per unit volume. Since one half of the molecules will be moving toward the wall in the +x direction while the other half will be moving in the -x direction, the number of molecules which will strike the wall in the time At is $~"Av,A~. The force on the wall due to the collision of a molecule with the wall is given by Newton's law: F = ma = m dvldt = d(mv)ldt, and integration yields FAt = A(mu). If a molecule rebounds elastically (without losing energy) when it hits the wall, its momentum is changed from +mu, to -mu, so that the total momentum change is A(mu) = 2mux. Consequently, FAt = 2mvx for one molecular collision, and FAt = ($n*~u,~t)(2mu,) for the total number of collisions. Canceling At from both sides and recognizing that the pressure is the force per unit area, p = FIA, we obtain p = n*mu,2. aThe history of the kinetic theory of gases is outlined by E. Mendoza, Physics Today 14,36-39 (1961). bA translation of this paper has appeared in The World of Mathematics, J. R. Newman, Ed., Vol. 2 (Simon and Schuster, New York, 1956), p. 774
Section1.2PressureofanIdeal Gas3Figure 1.1All the molecules in the box that are moving toward the z-y plane will strike the wall.Ofcourse,notallmoleculeswill betravelingwiththesamevelocityu,Wewilllearnbelowhowto characterize thedistribution ofmolecular velocities,butfornowlet us simply assume that the pressure will be proportional to the average of thesquare of the velocity in the x direction, p = n'm<u2>.The total velocity of anindividual molecule most likely contains other components along y and z.Sincev = iux + juy + kuz,d where i, j, and k are unit vectors in the x, y and z direc-tions, respectively, 2 = + + and <v2> = <u> + <2> + <u2>. Inan isotropic gas themotion ofthe molecules israndom,so there is no reasonfor thevelocity in one particular direction to differ from that in any other direction. Con-sequently,<v>=<u2>=<u2>=<u2>/3.Whenwecombinethisresultwiththecalculationaboveforthepressure,weobtain=n'm<u2>(1.1)p=3Of course, n' in equation 1.1 is the number of molecules per unit volume and canbe rewritten as nN/V, where N is Avogadro's number and n is the number ofmoles. The result isnNAm<U>(1.2)pV:2Since the average kinetic energy of the molecules is <e>=m<u?>anotherwaytowriteequation1.2is2(1.3)PV=nN<e>.Equations 1.2 and 1.3 bear a close resemblance to the ideal gas law,pV = nRT.The ideal gas law tells us that the product of p and V will be constant if the tem-perature is constant, while equations 1.2 and 1.3 tell us that the product will beconstant if <u?> or <e> is constant. The physical basis for the constancy of pvwith <u2> or <e> is clear from our previous discussion. If the volume iscIn this text, as in many others, we will use the notation <x> or x to mean “the average value of x."Throughout the text we will use boldface symbols to indicate vector quantities and normal weightsymbols to indicate scalar quantities. Thus, u lvl. Note that y? = y -y u2
Section 1.2 Pressure of an Ideal Gas Figure 1.1 All the molecules in the box that are moving toward the z-y plane will strike the wall. Of course, not all molecules will be traveling with the same velocity v,. We will learn below how to characterize the distribution of molecular velocities, but for now let us simply assume that the pressure will be proportional to the average of the square of the velocity in the x direction, p = n*m<~:>.~ The total velocity of an indivicual molecul_e most likelyA c9ntain: other components along y and z. Since v = iv, + jvy + kuz,* where i, j, and k are unit vectors in the x, y, and z directions, respectively, v2 = v: + v; + v: and <v2> = <v:> + <v;> + <v:>. In an isotropic gas the motion of the molecules is random, so there is no reason for the velocity in one particular direction to differ from that in any other direction. Consequently, <u:> = <v;> = <v:> = <v2>/3. When we combine this result with the calculation above for the pressure, we obtain 1 p = -n*m<v2>. 3 (1.1) Of course, n* in equation 1.1 is the number of molecules per unit volume and can be rewritten as nNAlv where NA is Avogadro's number and n is the number of moles. The result is Since the average kinetic energy of the molecules is <E> = irn<v2>, another way to write equation 1.2 is 2 pv = -PINA<€>. 3 (1.3) Equations 1.2 and 1.3 bear a close resemblance to the ideal gas law, pV = nRT. The ideal gas law tells us that the product of p and V will be constant if the temperature is constant, while equations 1.2 and 1.3 tell us that the product will be constant if <v2> or <E> is constant. The physical basis for the constancy of pV with <v2> or <E> is clear from our previous discussion. If the volume is CIn this text, as in many others, we will use the notation <x> or F to mean "the average value of x." dThroughout the text we will use boldface symbols to indicate vector quantities and normal weight symbols to indicate scalar quantities. Thus, v = Ivl. Note that v2 = v . v = vZ
4Chapter 1Kinetic Theory of Gasesincreased while the number, energy, and velocity of the molecules remain constant,then a longer time will be required for the molecules to reach the walls; there willthus be fewer collisions in agiven time,and thepressure will decrease.To identifyequation 1.3 with the ideal gas law, we need to consider in more detail the relationship between temperature and energy.1.3TEMPERATUREANDENERGYConsider two types of molecule in contact with one another. Let the average energyof the first type be <e>, and that of the second type be <e>2. If <e>, is greaterthan<e>2,thenwhenmoleculesoftype1 collidewiththoseoftype2,energywillbe transferred from the former to the latter. This energy transfer is a form of heatflow.From a macroscopicpoint of view,as heat flows thetemperatureof a system ofthe type 1 molecules will decrease,while that ofthe type2molecules will increase.Onlywhen<e>,=<e>,will thetemperaturesof thetwomacroscopicsystemsbethe same. In mathematical terms, we see that T, = I, when <e>, = <e>2 and thatT, > T, when <e>, > <e>2. Consequentiy, there must be a correspondencebetween <e> and T so that the latter is some function of the former: T = T(<e>).Thefunctionalformof thedependenceofTon<e>cannotbedeterminedsolely fromkinetic theory,since the temperature scale can be chosen in many pos-sible ways.In fact, one way to define the temperature is through the ideal gas law:T=pV(nR).Experimentally,this corresponds to measuring thetemperatureeitherby measuring the volume of an ideal gas held at constant pressure or by measuringthe pressureof an ideal gas held at constant volume.Division of both sides ofequa-tion 1.3 by nR and use of the ideal gas relation gives us the resultPV_2NA<e>,(1.4)T2nR3 R(1.5)<E>KT2where k, known as Boltzmann's constant, is defined as R/Na. Note that since<e>=m<u2>3kT<U>=(1.6)mexample 1.1Calculation of Average Energies and Squared VelocitiesObjectiveCalculate the average molecular energy, <e>, and the averagesquared velocity,<u?>, for a nitrogen molecule at T = 300 K.MethodUse equations 1.5 and 1.6 with m = (28 g/mole)(1 kg/1000 g)/(Nmolecule/mole)andk=1.38×10-23J/K.<e>= 3kT/2= 3(1.38 × 10-23 J/K)(300 K)/2 = 6.21 ×10-21 J.Solution3kT<u>=m
Chapter 1 Kinetic Theory of Gases increased while the number, energy, and velocity of the molecules remain constant, then a longer time will be required for the molecules to reach the walls; there will thus be fewer collisions in a given time, and the pressure will decrease. To identify equation 1.3 with the ideal gas law, we need to consider in more detail the relationship between temperature and energy. 1.3 TEMPERATURE AND ENERGY Consider two types of molecule in contact with one another. Let the average energy of the first type be <E>, and that of the second type be <E>,. If <E>, is greater than <E>~, then when molecules of type 1 collide with those of type 2, energy will be transferred from the former to the latter. This energy transfer is a form of heat flow. From a macroscopic point of view, as heat flows the temperature of a system of the type 1 molecules will decrease, while that of the type 2 molecules will increase. Only when <E>, = <E>, will the temperatures of the two macroscopic systems be the same. In mathematical terms, we see that Tl = T2 when = <E>~ and that T, > T2 when <E>, > <E>~. Consequently, there must be a correspondence between <e> and T so that the latter is some function of the former: T = T(<E>). The functional form of the dependence of T on <E> cannot be determined solely from kinetic theory, since the temperature scale can be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pVl(nR). Experimentally, this corresponds to measuring the temperature either by measuring the volume of an ideal gas held at constant pressure or by measuring the pressure of an ideal gas held at constant volume. Division of both sides of equation 1.3 by nR and use of the ideal gas relation gives us the result where k, known as Boltzmann's constant, is defined as RIN,. Note that since <E> = $m<v2>, example 1.1 Calculation of Average Energies and Squared Velocities Objective Calculate the average molecular energy, <E>, and the average squared velocity, <v2>, for a nitrogen molecule at T = 300 K. Method Use equations 1.5 and 1.6 with m = (28 g/mole)(l kg11000 g)l (NA moleculelmole) and k = 1.38 X JIK. I solution <E> = 3kTl2 = 3(1.38 x J/K)(300 K)/2 = 6.21 X J