25Section 1.8Summarythe relative velocity8kT<U,>=2<U>(1.40)Tuand the mean free path,1U>入(1.47)ZiV2ad'n;These concepts form thebasis forfurtherinvestigation into transport properties andchemical reactionkineticsappendix 1.1The Functional Form of the Velocity DistributionWe demonstrate in this appendix that the exponential form used in equation 1.23is the only function that satisfies the equation f(a + b + c) = f(a)f(b)f(c). Considerfirst the simpler equation(1.50)f(z) =f(a)f(b),where z = a + b. Taking the derivative of both sides of equation 1.50 with respectto a we obtain(2) = f(a)(b),(1.51)dzdaOn the other hand, taking the derivative of both sides of equation 1.50 with respecttob,weobtaindf(z) dz = f(a)f(b).(1.52)dzdbSince z = a + b, dz/da = dz/db = 1. Consequently,df(z)(1.53)= f(a)f(b) = f(a)f(b)dzDivision of both sides of theright-hand equality byf(a)f(b)yieldsf(a)_ f(b)(1.54)f(a)f(b)Now the left-hand side of equation 1.54 depends only on a, while the right-handside depends only on b. Since a and b are independent variables, the only way thatequation 1.54 can be true is if each side of the equation is equal to a constant, ±k,wherek is defined as nonnegative:f(a)f(b)(1.55)±K±K.f(a)f(b)
Section 1.8 Summary the relative velocity, and the mean free path, These concepts form the basis for further investigation into transport properties and chemical reaction kinetics. appendix 1.1 The Functional Form of the Velocity Distribution We demonstrate in this appendix that the exponential form used in equation 1.23 is the only function that satisfies the equation f(a + b + c) = f(a)f(b)f(c). Consider first the simpler equation where z = a + b. Taking the derivative of both sides of equation 1.50 with respect to a we obtain On the other hand, taking the derivative of both sides of equation 1.50 with respect to b, we obtain Since z = a + b, dzlda = dzldb = 1. Consequently, Division of both sides of the right-hand equality byf(a)f(b) yields f'(4 - f'o - f(a> f(b) Now the left-hand side of equation 1.54 depends only on a, while the right-hand side depends only on b. Since a and b are independent variables, the only way that equation 1.54 can be true is if each side of the equation is equal to a constant, t~, where K is defined as nonnegative:
26Chapter1KineticTheoryof GasesSolution of these differential equations using x to represent either a orb leads todf(x)f'(α)±kdx.(1.56)±korf(x)f(x)Integration showsthatf(x) =Ke+x(1.57)where K is related to the constant of integration. Equation 1.23 is obtained byreplacing x with u?.appendix 1.2Spherical CoordinatesMany problems in physical chemistry can be solved more easily using sphericalrather than Cartesian coordinates. In this coordinate system, as shown in Figure1.14, a point P is located by its distance r from the origin, the angle between thez axis and the line from the point to the origin, and the angle between the x axisand the linebetween the origin and a projection of the point onto the x-y plane.Anypoint can be described by a value of r between O and oo, a value of e between O andT,and a value of between O and 2.The Cartesian coordinates arerelated to thespherical ones by the following relationships: x = r sin cos p, y = r sin sin ,and z=rcose.The volume element in spherical coordinates can be calculated with the helpof Figure1.15.As the variable is increased forfixed r, the position of thepointdescribed by (r,o,) moves along a longitudinal line on the surface of a sphere,while if is increased at fixed r, the position of the point moves along a latitudi-nal line.Starting at a point located at (r,,Φ),if r is increased by dr, is increasedby do, and is increased by do,then the volume increase is the surface area on theFigure 1.14Spherical coordinates
Chapter 1 Kinetic Theory of Gases Solution of these differential equations using x to represent either a or b leads to f '(4 - df (4 -+K or - - +-K dx. f (4 f (4 Integration shows that f(x) = KekKX, where K is related to the constant of integration. Equation 1.23 is obtained by replacing x with u,2. Spherical Coordinates Many problems in physical chemistry can be solved more easily using spherical rather than Cartesian coordinates. In this coordinate system, as shown in Figure 1.14, a point P is located by its distance r from the origin, the angle 8 between the z axis and the line from the point to the origin, and the angle 4 between the x axis and the line between the origin and a projection of the point onto the x-y plane. Any point can be described by a value of r between 0 and m, a value of 8 between 0 and T, and a value of 4 between 0 and 2~. TheCartesian coordinates are related to the spherical ones by the following relationships: x = r sin 8 cos 4, y = r sin 8 sin 4, and z = r cos 8. The volume element in spherical coordinates can be calculated with the help of Figure 1.15. As the variable 8 is increased for fixed l; the position of the point described by (r,8,+) moves along a longitudinal line on the surface of a sphere, while if 4 is increased at fixed l; the position of the point moves along a latitudinal line. Starting at a point located at (r,8,4), if r is increased by dl; 8 is increased by d8, and 4 is increased by d4, then the volume increase is the surface area on the X Figure 1.14 Spherical coordinates
27Appendix 1.3rsinedsrsinerdedAFigure 1.15The volume element in spherical coordinates.sphere times the thickness dr (for clarity,the thickness dr is not shown in the dia-gram).The surface area is given by the arc length on the longitude, r do, times thearc length on the latitude, r sin do.Thus, the volume element is dV=sin dodo dr.appendix 1.3The Error Function and Co-Error FunctionIt often occurs that we need to evaluate integrals of the form of those listed inTable1.1but for limits less than the range of 0 to infinity.For such evaluations itisuseful to definethe errorfunction:2e-rduerf(x)(1.58)VJoFrom Table 1.1 we see that for x = o, the value of the integral is V /2, so thaterf(co) = 1. Note that if we “complement" the error function by 2/V^ times theintegral fromx to o, we should getunity:2Le- du+" du = erf(x) +e-"duoVTJVTVT(1.59)udu=1
Appendix 1.3 I rsin Bd4 r sin 0 1 II Figure 1.15 The volume element in spherical coordinates. sphere times the thickness dr (for clarity, the thickness dr is not shown in the diagram). The surface area is given by the arc length on the longitude, r do, times the arc length on the latitude, r sin 0 d4. Thus, the volume element is dV = r2sin 0 d0 d4 dr. appendix 1.3 The Error Function and Co-Error Function It often occurs that we need to evaluate integrals of the form of those listed in Table 1.1 but for limits less than the range of 0 to infinity. For such evaluations it is useful to define the error function: erf(x) = - e-"- du. Go I" From Table 1.1 we see that for x = m, the value of the integral is G/2, so that erf(m) = 1. Note that if we "complement" the error function by 2/G times the integral from x to m, we should get unity:
28Chapter1 KineticTheoryof Gases1.00.80.6(g) ja0.40.2006110.00.51.01.52.02.5xFigure1.16Values of the error function.Consequently,it is also useful to define the co-errorfunction, erfc(x),as the complement to the errorfunction:C1erfec(x) = 1 - erf(x) =e-idu(1.60)VJxTables of the error function and co-error function are available,but the pervasiveuse of computers has made them all but obsolete. For calculational purposes, theintegrand in equation 1.58 or equation 1.60 can be expanded using a series,erf(c) =2受(-1)2+1(1.61)V=0n! (2n +1)and then the integration can be performed term by term. Figure 1.16 plots erf(x) asa function of x.appendix 1.4The Center-of-Mass FrameWe show in this appendix that the total kinetic energy of two particles of veloc-ities Vi and v, is given by uu? + IMvcom, where v, - V2 - Vi, and where Vcomthe vector describing the velocity of the center of mass, is defined by the equa-tion (mi+m2)Vcom=miVi+m2V2,andM=m+m2.Figure1.17 shows thevectorrelationships
Chapter 1 Kinetic Theory of Gases I Figure 1.16 Values of the error function. Consequently, it is also useful to define the co-error function, erfc(x), as the complement to the error function: Tables of the error function and co-error function are available, but the pervasive use of computers has made them all but obsolete. For calculational purposes, the integrand in equation 1.58 or equation 1.60 can be expanded using a series, and then the integration can be performed term by term. Figure 1.16 plots erf(x) as a function of x. appendix 1.4 The Center-of-Mass Frame We show in this appendix that the total kinetic energy of two particles of velocities v, and v, is given by ipv; + ~MV:~, where v, = v, - v, and where v, the vector describing the velocity of the center of mass, is defined by the equation (m, + m2)vCom = m,v, + m,v2, and M = m, + m,. Figure 1.17 shows the vector relationships
Appendix1.429mVeomVNV2m2Figure1.17Vector diagram for center-of-mass conversion.Thevirtue of this transformation is that the total momentum of the systemp=m,y, + m2vz is also equal to the momentum of the center of mass, defined as MVeomBecause we assume that no external forces are acting on the system,F = Macom (dpcom/dt) = 0, so that the momentum of the center of mass does not change dur-ing the interaction between the two particles.Notethatsince(m/M)+(m/M)=1 wecanwritemmV2-VcomV2VeomMM(1.62)mim2IV2+FV2-Veom'MMHowever,(1.63)m,V,+mV2=MVcomso thatm2V2miVi-Voom'MMConsequently,mimiV2VcomV2MyM(1.64)miVrMIn a similar way,wefind thatm2(1.65)Veom - Vj =My.1We now note an important point, that the velocities of the particles with respect tothe center of mass are just given by the two pieces of the vector v: u,=-(mg/M)y,.and u, = (m,/M)vr, as shown in Figure 1.18. Note also that in the moving frame ofthe center of mass, there is no net momentumfor the particles; that is, m,u,+ mzu,O. This important property enables us to calculate the velocity of one particle in thecenter-of-mass frame given just the mass and the velocity of the other particle
Appendix 1.4 II Figure 1.17 Vector diagram for center-of-mass conversion. The virtue of this transformation is that the total momentum of the system p = mlvl + m2v2 is also equal to the momentum of the center of mass, defined as Mv,. Because we assume that no external forces are acting on the system, F = Ma, = (dp,ldt) = 0, so that the momentum of the center of mass does not change during the interaction between the two particles. Note that since (mllM) + (m21M) = 1 we can write However, so that Consequently, In a similar way, we find that We now note an important point, that the velocities of the particles with respect to the center of mass are just given by the two pieces of the vector v,: ul = -(m21M)v, and u, = (m,lM)v, as shown in Figure 1.18. Note also that in the moving frame of the center of mass, there is no net momentum for the particles; that is, mlu, + m2u2 = 0. This important property enables us to calculate the velocity of one particle in the center-of-mass frame given just the mass and the velocity of the other particle