5Section1.4Distributions,Mean Values,and Distribution Functions3 (1.38×10-23J/K)(300 K)D[(28/6.02 × 1026)]= 2.67 × 105 (m/s)2 = (516 m/s)2To summarizethe discussion sofar,wehave seen from equation1.2that pVisproportional to <u2> and that the ideal gas law is obtained if we take the defini-tion of temperature to be that embodied in equation 1.5. Since <e> =m<u2>,both temperature and pV are proportional to the average of the square of the veloc-ity.The use of an average recognizes that not all the molecules will be moving withthe same velocity.In the next few sections we consider the distribution of molecu-larspeeds.Butfirst wemust consider what wemean bya distribution.DISTRIBUTIONS,MEANVALUES,AND1.4 DISTRIBUTIONFUNCTIONSSuppose that five students take a chemistry examination forwhich the possiblegrades are integers in the range from 0 to 100. Let their scores be S, = 68, S, = 76,S, = 83, S4 =91, and S,=97.The average scorefor the examination is then-2<S>=(1.7)NTwhere N = 5 is the number of students. In this case, the average is easily calcu-latedtobe83Now suppose thatthe class had500 studentsratherthan5.Ofcourse,the aver-age grade could be calculated in a manner similar to that in equation 1.7 with anindex i running from 1 to N=500.However,another method will be instructive.Clearly, if the examination is still graded to one-point accuracy, it is certain thatmorethanone studentwill receivethe same score.Supposethat, instead of sum-ming over the students, represented by the index i in equation 1.7, we form theaverage by summing over the scores themselves, which range in integer possibili-tiesfromj=Oto100.Inthis case,toobtaintheaverage,wemustweighteach scoreS, by the number of students who obtained that score, N:ANSN.(1.8)<SV=NT oNotethat thedefinition of N,requires thatZN,=Nr-Thefactor1/Nin equation1.8 is included for normalization, since,for example,if all the studentshappenedtogetthesamescoreS,=Sthens1Zs,N,=EN, = s.(1.9)<S>:NTNTNow let us define the probability of obtaining score S, as the fraction of stu-dents receiving that score:N(1.10)PiNt
Section 1.4 Distributions, Mean Values, and Distribution Functions To summarize the discussion so far, we have seen from equation 1.2 that pV is proportional to <v2> and that the ideal gas law is obtained if we take the definition of temperature to be that embodied in equation 1.5. Since <E> = $rn<v2>, both temperature and pV are proportional to the average of the square of the velocity. The use of an average recognizes that not all the molecules will be moving with the same velocity. In the next few sections we consider the distribution of molecular speeds. But first we must consider what we mean by a distribution. 1.4 DISTRIBUTIONS, MEAN VALUES, AND DISTRIBUTION FUNCTIONS Suppose that five students take a chemistry examination for which the possible grades are integers in the range from 0 to 100. Let their scores be S, = 68, S, = 76, S, = 83, S, = 91, and S, = 97. The average score for the examination is then where NT = 5 is the number of students. In this case, the average is easily calculated to be 83. Now suppose that the class had 500 students rather than 5. Of course, the average grade could be calculated in a manner similar to that in equation 1.7 with an index i running from 1 to N, = 500. However, another method will be instructive. Clearly, if the examination is still graded to one-point accuracy, it is certain that more than one student will receive the same score. Suppose that, instead of summing over the students, represented by the index i in equation 1.7, we form the average by summing over the scores themselves, which range in integer possibilities from j = 0 to 100. In this case, to obtain the average, we must weight each score Sj by the number of students who obtained that score, Nj: Note that the definition of N, requires that ZNj = NT. The factor l/NT in equation 1.8 is included for normalization, since, for example, if all the students happened to get the same score Sj = S then Now let us define the probability of obtaining score Sj as the fraction of students receiving that score:
6Chapter 1Kinetic Theory of GasesThen another way to write equation 1.8 is<SV=Es,Pj(1.11)where ,P, = 1 from normalization.Equation 1.11 provides an alternative to equation 1.7 for finding the averagescore for the class. Furthermore, we can generalize equation 1.11 to provide amethod forfinding the average of anyquantity,<Q>:(1.12)Pojwhere P, is the probability of finding the jth result.example 1.2Calculating Averages from ProbabilitiesObjectiveFind the average throw for a pair of diceMethodEach die is independent, so the average of the sum of the throwswill be twice the average of the throw for one die. Use equation1.12 to find the average throw for one die.SolutionThe probability for each of the six outcomes, 1-6, is the same,namely, 1/6. Factoring this out of the sum gives <T> =(1/6)>T, where T, = 1,2,3,4,5,6 for i = 1-6. The sum is 21, so that theaverage throw for one die is <T> = 21/6= 3.5.For the sum oftwo dice, the average would thus be 7.Themethod can be extended to calculate more complicated averages.Letf(Q)be some arbitrary function of the observation Q, Then the average value of thefunctionf(Q)isgiven by<f(0)>= ZP,f(0)(1.13)For example, ifQ were the square of a score,then<SV=Zp,s?(1.14)Suppose now that the examination is a very good one, indeed, and that the tal-ented instructor can grade it not just to one-point accuracy (a remarkable achieve-ment in itself!)but to an accuracy of ds, whereds is a very small fraction of a pointLet P(S) d be the probability that a score will fall in the range between S and S+dS, and letds become infinitesimally small.The fundamental theorems of calculustell us that we can convert the sum in equation 1.11 to the integralP(s)s ds,<SV(1.15)
Chapter 1 Kinetic Theory of Gases Then another way to write equation 1.8 is where C .P. = 1 from normalization. 1 1. Equation 1.11 provides an alternative to equation 1.7 for finding the average score for the class. Furthermore, we can generalize equation 1.11 to provide a method for finding the average of any quantity, where Pi is the probability of finding the jth result. example 1.2 Calculating Averages from Probabilities Objective Find the average throw for a pair of dice. Method Each die is independent, so the average of the sum of the throws will be twice the average of the throw for one die. Use equation I 1.12 to find the average throw for one die. Solution The probability for each of the six outcomes, 1-6, is the same, namely, 116. Factoring this out of the sum gives <T> = (116) 2 Ti, where Ti = 1,2,3,4,5,6 for i = 1-6. The sum is 21, so that the average throw for one die is <T> = 2116 = 3.5. For the sum of two dice, the average would thus be 7. The method can be extended to calculate more complicated averages. Let f(Qj) be some arbitrary function of the observation Qj. Then the average value of the functionflQ) is given by For example, if Q were the square of a score, then Suppose now that the examination is a very good one, indeed, and that the talented instructor can grade it not just to one-point accuracy (a remarkable achievement in itself!) but to an accuracy of dS, where dS is a very small fraction of a point. Let P(S) dS be the probability that a score will fall in the range between S and S + dS, and let dS become infinitesimally small. The fundamental theorems of calculus tell us that we can convert the sum in equation 1.11 to the integral
7Section 1.4Distributions,Mean Values,andDistributionFunctionsor,moregenerallyforanyobservablequantity<0>=P(Q)odo(1.16)Equation 1.16 will form the basis for much of our further work. The probabil-ity function P(Q) is sometimes called a distribution function, and the range of theintegral is over all values of Q where the probability is nonzero. Note that normal-ization of the probability requiresP(Q)dQ = 1(1.17)The quantity l(x)P dx is simply a specific example of a distribution function.Although knowledge of quantum mechanics is not necessary to solve it, you mayrecognize a connection to the particle in the box in Problem I.7, which like Exam-ple 1.3isanexercisewith distributionfunctions.example 1.3Determining Distribution FunctionsObjectiveBees likehoney.A sphere of radius rois coated with honey andhanging in a tree. Bees are attracted to the honey such that theaveragenumber of bees perunitvolumeisgiven byKr-5,whereK is a constant and r is the distance from the center of the sphere.Derive the normalized distribution function for thebees.They canbe at any distance from the honey,but they cannot be inside thesphere. Using this distribution, calculate the average distance of abeefromthecenterofthesphere.MethodFirst we need to find the normalization constant K by applyingequation 1.17,recalling that we have a three-dimensional problemand that in spherical coordinates the volume element for a problemthat does not depend on the angles is 4rrrdrThen, to evaluate theaverage, we apply equation 1.16.SolutionRecall that, by hypothesis, there is no probability for the beesbeing at r < ro, so that the range of integration is from ro toinfinity.TodetermineKwerequire(1.18)(Kr-5)4mdr = 1,or4元Kr-3dr= 1 = 4mF(1.19)4元K2r2so thatrK :2㎡
Section 1.4 Distributions, Mean Values, and Distribution Functions or, more generally for any observable quantity, Equation 1.16 will form the basis for much of our further work. The probability function P(Q) is sometimes called a distribution function, and the range of the integral is over all values of Q where the probability is nonzero. Note that normalization of the probability requires The quantity 1+(x)I2 dx is simply a specific example of a distribution function. Although knowledge of quantum mechanics is not necessary to solve it, you may recognize a connection to the particle in the box in Problem 1.7, which like Example 1.3 is an exercise with distribution functions. example 1.3 - Determining Distribution Functions Objective Bees like honey. A sphere of radius ro is coated with honey and hanging in a tree. Bees are attracted to the honey such that the average number of bees per unit volume is given by Kr-5, where K is a constant and r is the distance from the center of the sphere. Derive the normalized distribution function for the bees. They can be at any distance from the honey, but they cannot be inside the sphere. Using this distribution, calculate the average distance of a bee from the center of the sphere. Method First we need to find the normalization constant K by applying equation 1.17, recalling that we have a three-dimensional problem and that in spherical coordinates the volume element for a problem that does not depend on the angles is 45-9 dr: Then, to evaluate the average, we apply equation 1.16. Solution Recall that, by hypothesis, there is no probability for the bees being at r < ro, so that the range of integration is from ro to infinity. To determine K we require so that
8Chapter1KineticTheoryofGasesHavingdeterminedthenormalization constant,we nowcalculate theaveragedistance:4m2dr(1.20)= 2ro(-r=2ro=2ro"ro1.5THEMAXWELLDISTRIBUTIONOFSPEEDSWeturnnowtothedistribution of molecular speeds.We will denotetheprobability offinding u,in therange from ,to u+ du,u,in the range from u,to u,+duy,andu,in therangefromu,to ,+ du,byF(uyuy.u,)du,du,du,Theobjectofthis section is to determine the function F(u,uyu,). There are four main points inthe derivation:1.In each direction,the velocity distribution must be an evenfunction ofu2.The velocity distribution in any particular direction is independent from anduncorrelated with the distributions in orthogonal directions3.Theaverageof thesquareof thevelocity<u2>obtained usingthedistributionfunction should agree with the value required by the ideal gas law: <u2>3kT/m.4.The three-dimensional velocity distribution depends only on the magnitude of (i.e., the speed) and not on the direction.We now examine these four points in detail.1.5.1TheVelocityDistributionMustBeanEvenFunction of yConsider the velocities v,of molecules contained in a box.The number of molecules moving in the positive x direction must be equal to the number of moleculesmoving in the negative x direction.This conclusion is easily seen by examining theconsequences of the contrary assumption.If the number of molecules moving ineach direction were not the same,then the pressure on one side of the box wouldbe greater than on the other.Aside from violating experimental evidence that thepressure isthe same wherever it is measured in a closed system,our common observation is that the box does not spontaneously move in either the positive or nega-tive x direction, as would be likely if the pressures were substantially different. Weconclude that the distribution function for the velocity in the x direction, or moregenerally in any arbitrary direction,must be symmetric;i.e.,F(u,)=F(-v,).Functions possessing the property that f(x) = f(-x) are called even functions, while-f(-x)arecalled oddfiunctions.Wecanthosehavingthepropertythatf(x)=ensure that F(u) be an even function by requiring that the distribution functiondepend on the square of the velocity: F(v) = f(u2). As shown in Section 1.5.3, thiscondition isalsoin accordwiththeBoltzmann distribution law.ecOtherevenfunctions,for example,=fu) would bemathematicallyacceptable,but would not satisfy the requirement of Section 1.5.3
Chapter 1 Kinetic Theory of Gases Having determined the normalization constant, we now calculate the average distance: 1.5 THE MAXWELL DISTRIBUTION OF SPEEDS We turn now to the distribution of molecular speeds. We will denote the probability of finding v, in the range from u, to v, + dux, u, in the range from v, to v, + dv, and v, in the range from v, to u, + du, by F(v,v,v,) dux dv, dv,. The object of this section is to determine the function F(v,v,u,). There are four main points in the derivation: 1. In each direction, the velocity distribution must be an even function of u. 2. The velocity distribution in any particular direction is independent from and uncorrelated with the distributions in orthogonal directions. 3. The average of the square of the velocity <v2> obtained using the distribution function should agree with the value required by the ideal gas law: <u2> = 3kTlm. 4. The three-dimensional velocity distribution depends only on the magnitude of u (i.e., the speed) and not on the direction. We now examine these four points in detail. 1.5.1 The Velocity Distribution Must Be an Even Function of v Consider the velocities u, of molecules contained in a box. The number of molecules moving in the positive x direction must be equal to the number of molecules moving in the negative x direction. This conclusion is easily seen by examining the consequences of the contrary assumption. If the number of molecules moving in each direction were not the same, then the pressure on one side of the box would be greater than on the other. Aside from violating experimental evidence that the pressure is the same wherever it is measured in a closed system, our common observation is that the box does not spontaneously move in either the positive or negative x direction, as would be likely if the pressures were substantially different. We conclude that the distribution function for the velocity in the x direction, or more generally in any arbitrary direction, must be symmetric; i.e., F(v,) = F(-v,). Functions possessing the property that Ax) = A-x) are called even functions, while those having the property that f(x) = -f(-x) are called odd functions. We can ensure that F(v,) be an even function by requiring that the distribution function depend on the square of the velocity: F(v,) = f(v,2). As shown in Section 1.5.3, this condition is also in accord with the Boltzmann distribution law.e eOther even functions, for example, F =flu:) would be mathematically acceptable, but would not satisfy the requirement of Section 1.5.3
9Section1.5TheMaxwellDistributionofSpeeds1.5.2The Velocity Distributions Are Independent and UncorrelatedWenow considertherelationshipbetweenthedistribution of x-axisvelocities andy- or z-axis velocities. In short, there should be no relationship. The three compo-nents of the velocity are independent of one another since the velocities are uncor-related.An analogy might be helpful.Consider the probability of tossing three hon-est coins and getting "heads" on each. Because the tosses t, are independentuncorrelated events, the joint probability for a throw of three heads,P(t, =heads,t =heads,t,=heads),is simply equal to theproduct of the probabilitiesfor thethree individual events, P(t = heads) × P(t2 = heads) × P(t, = heads) × × . In a similar way,because the x-,y-,and z-axis velocities are independentand uncorrelated, we can write that(1.21)F(Ur,Uy,vz) = F(ux)F(u,)F(uz).Wecan now use the conclusion of theprevious section.We can write,for exam-ple, that F(u,) = f(u2) and similarly for the other directions. Consequently,F(ux,Uy,) =F(u)F(u,)F(uz) =f()f()f(u2),(1.22)What functional form has the property that f(a + b + c) =f(a)f(b)f(c)? A lit-tle thought leads to the exponential form, since exp(a + b + c) = eaebec. It can beshown, in fact, that the exponential is the only form having this property (seeAppendix1.1),so thatwecanwrite(1.23)F(ux) = f(u) = K exp(±ku),where K and k are constants to be determined. Note that although k can appearmathematically with either a plus or a minus sign, we must require the minus signon physical grounds because we know from common experience that the probabil-ityof veryhighvelocities shouldbe small.The constantK can be determined from normalization since,using equation1.17,the total probability that u,lies somewhere in the range from -oo to +ooshould be unity:(1.24)F(u.)du, = 1.Substitution of equation 1.23 into equation 1.24 leads to the equationexp(-ku)du, = K(1.25)wherethe integral was evaluated using Table1.1.The solution is then K= (k/r)1/21.5.3<u2>ShouldAgreewith theIdeal Gas LawThe constant k is determined by requiring <u2> to be equal to 3kT/m, as in equa-tion1.6.Fromequation1.16wefindF(ur)dr= exp(ku2)dur.<U>(1.26)The integral is a standard one listed in Table 1.1, and using its value we find that()()"-<U>=(1.27)
Section 1.5 The Maxwell Distribution of Speeds 1.5.2 The Velocity Distributions Are Independent and Uncorrelated We now consider the relationship between the distribution of x-axis velocities and y- or z-axis velocities. In short, there should be no relationship. The three components of the velocity are independent of one another since the velocities are uncorrelated. An analogy might be helpful. Consider the probability of tossing three honest coins and getting "heads" on each. Because the tosses ti are independent, uncorrelated events, the joint probability for a throw of three heads, P(tl = heads, t, = heads, t3 = heads), is simply equal to the product of the probabilities for the three individual events, P(tl = heads) X P(t2 = heads) X P(t3 = heads) = $ X $ X $ . In a similar way, because the x-, y-, and z-axis velocities are independent and uncorrelated, we can write that F(ux7uy9uz) = F(ux)F(vy)F(uz). (1.21) We can now use the conclusion of the previous section. We can write, for example, that F(v,) = f(u2) and similarly for the other directions. Consequently, What functional form has the property that f(a + b + c) = f(a)f(b)f(c)? A little thought leads to the exponential form, since exp(a + b + c) = eaebec. It can be shown, in fact, that the exponential is the only form having this property (see Appendix 1.1), so that we can write F(vx) = f(u;) = K exp(?~u:), (1.23) where K and K are constants to be determined. Note that although K can appear mathematically with either a plus or a minus sign, we must require the minus sign on physical grounds because we know from common experience that the probability of very high velocities should be small. The constant K can be determined from normalization since, using equation 1.17, the total probability that u, lies somewhere in the range from -m to +w should be unity: 00 J-/(ux)dux = 1. (1.24) Substitution of equation 1.23 into equation 1.24 leads to the equation where the integral was evaluated using Table 1.1. The solution is then K = (~l.rr)l". 1.5.3 <v2> Should Agree with the Ideal Gas Law The constant K is determined by requiring <u2> to be equal to 3kTlm, as in equation 1.6. From equation 1.16 we find The integral is a standard one listed in Table 1.1, and using its value we find that