10ChapterIKineticTheoryofGasesTABLE1.1Integralsof UseintheKineticTheoryof Gasesxne-Bx dx -2ne-Br dxx2n+le-8e dx m 0[e-μr dx=1Vuβ-1/2 xe-β, dx = 1β-1xe-rdx=1V1p-3/2Xe-ae dx = 1β2oxe-A dx = β-3Xe-8e dx=↓ Vrβ-52Pe-rdt =+ Vr(2n)1B-a)x2n+le-Be dx = 2(n!)β-(n+1)22nn!As a consequence, the average of the square of the total speed, <u?> = <u?>+<U2>+<u2>>=3<u2>issimply3<u>=(1.28)2kFromequation1.6wehavethat<v2>=3kT/mforagreementwiththeidealgaslaw, so that 3kT/m = 3/(2k), or k = m/(2kT). The complete one-dimensional dis-tribution function is thusImum(1.29)F(u.)du,du2kT2TkTThis equation is known as the one-dimensional Maxwell-Boltzmanndistributionformolecular velocities.Plots of F(u,)are shown in Figure1.2.Note that equation 1.29 is consistent with the Boltzmann distribution law,which states that the probability of finding a system with energy e is proportionaltoexp(-e/kT).Since e,=mu?isequal tothetranslational energyofthemole-cule in the x direction, the probability of finding a molecule with an energy exshould beproportional toexp(-e,/kT),asit isin equation1.29.InSection1.5.1we ensured F(u,) to be even by choosing it to depend on the square of the velocity,F(u,)=f(u).Had we chosen some other evenfunction, sayF(u,)=f(u),thefinalexpression for the one-dimensional distribution would not have agreed with theBoltzmann distribution law.Equation 1.29 provides the distribution of velocities in one dimension. In threedimensions, because F(u,Uyu,) = F(v,)F(u,)F(u), and because u2 = v + vg + v2,we find that the probabilitythat the velocity will have components u,between u,andU,+du.,betweenu,and u.+du.,andu,between v,andu,+du,isgivenbyF(ur,Uy,u,)du,du,du,=F(ux)F(u,)F(u,)du,dudu(1.30)mu2mdu.d,d,2kT2kT
Chapter 1 Kinetic Theory of Gases As a consequence, the average of the square of the total speed, <v2> = <v:> + <v;> + <u:> = 3<v:>, is simply From equation 1.6 we have that <v2> = 3kTlm for agreement with the ideal gas law, so that 3kTlm = 3/(2~), or K = ml(2kT). The complete one-dimensional distribution function is thus du,. This equation is known as the one-dimensional Maxwell-Boltzmann distribution for molecular velocities. Plots of F(v,) are shown in Figure 1.2. Note that equation 1.29 is consistent with the Boltzmann distribution law, which states that the probability of finding a system with energy E is proportional to exp(-~lkT). Since E, = kmu: is equal to the translational energy of the molecule in the x direction, the probability of finding a molecule with an energy E, should be proportional to exp(-~,lkT), as it is in equation 1.29. In Section 1.5.1 we ensured F(v,) to be even by choosing it to depend on the square of the velocity, F(v,) =flu:). Had we chosen some other even function, say F(u,) =flu:), the final expression for the one-dimensional distribution would not have agreed with the Boltzmann distribution law. Equation 1.29 provides the distribution of velocities in one dimension. In three dimensions, because F(u,u,v,) = F(v,)F(v,)F(u,), and because v2 = v: + u; + v;, we find that the probability that the velocity will have components v, between v, and vx + dv, v, between v, and v, + dv, and u, between u, and v, + dv, is given by F(vx, u, vZ) dv, dv, du, = F(v,) F(vy)F(vZ) dv, dv, dv, (1.30) = (~yex~(-~) 2rrkT 2kT dv,dv,dv
11Section 1.5TheMaxwellDistributionofSpeeds1.4-1.2TTTTTT= 300 K1.01.TTTTTTTTT=1000K0.40.20.020001000010002000Ux (m/s)Figure1.2One-dimensional velocity distribution for a mass of 28 amu and two temperatures1.5.4TheDistributionDependsOnly ontheSpeedNote that the right-hand side of equation 1.30 depends on u? and not on the direc-tional property of v. When we have a function that depends only on the length of thevelocity vector, u = Ivl, and not on its direction, we can be more precise by sayingthatthe function depends on the speed and not on thevelocitySinceF(u,uy,u,)=f(u2) depends on the speed, it is often more convenient to know the probability thatmolecules havea speed in a particular range than toknow theprobability that theirvelocity vectors will terminate in a particular volume.As shown in Figure 1.3, theprobabilitythat the speed will bebetweenu and +du is simplytheprobabilitythat velocity vectors will terminate within the volume of a spherical shellbetween theradius u and the radius u + dv. The volume of this shell is dv, du, du, = 4u? dv, sothat the probability that speed will be in the desired range isffAn alternatemethod for obtaining equation1.31 is to note that du,du,du,can be written as v'singde dp du in spherical coordinates (see Appendix 1.2) and then to integrate over the angular coordinates. Sincethe distribution does not depend on the angular coordinates,the integrals over d and dp simply give 4r andwe are left with the factor 2du.mmUF(o)du =sindudedplexp(2mkT2VT[( (-)32kT-4m()(-) a.2kTA more complete description of spherical coordinates is found in Appendix 1.2
Section 1.5 The Maxwell Distribution of Speeds II Figure 1.2 One-dimensional velocity distribution for a mass of 28 amu and two temperatures. 1.5.4 The Distribution Depends Only on the Speed Note that the right-hand side of equation 1.30 depends on v2 and not on the directional property of v. When we have a function that depends only on the length of the velocity vector, v = Ivl, and not on its direction, we can be more precise by saying that the function depends on the speed and not on the velocity. Since F(v,v,v,) = f(v2) depends on the speed, it is often more convenient to know the probability that molecules have a speed in a particular range than to know the probability that their velocity vectors will terminate in a particular volume. As shown in Figure 1.3, the probability that the speed will be between v and u + dv is simply the probability that velocity vectors will terminate within the volume of a spherical shell between the radius v and the radius v + dv. The volume of this shell is dux dv, dv, = 4.rrv2 dv, so that the probability that speed will be in the desired range isf 'An alternate method for obtaining equation 1.31 is to note that dux du, du, can be written as uZsinO dB d+ du in spherical coordinates (see Appendix 1.2) and then to integrate over the angular coordinates. Since the distribution does not depend on the angular coordinates, the integrals over dO and d+ simply give 4.rr and we are left with the factor u2 du. F(u)du = IT IT ( exp ( - - mu2)sinOdudOd+ +=, ,=, 2~kT 2kT A more complete description of spherical coordinates is found in Appendix 1.2
Chapter1KineticTheoryofGases12Area=4u2Figure1.3The shell between and u + du has a volume of 4ru2 du. The thickness of the shell here isexaggerated forclarity.2.0F(u/sgoD(a)d1.51.00.0.0250010001500U(m/s)2000Figure1.4Maxwell-Boltzmann speed distribution as a function of temperature for a mass of 28 amu.13/2mum(1.31)duF(u)du=4uexp2kT2㎡kTBy analogy to equation 1.29, we will call equation 1.31 the Maxwell-Boltzmannspeed distribution. Speed distributions as a function of temperature are shown inFigure 1.4.We often characterize the speed distribution by a single parameter, for exam-ple, the temperature.Equivalently,we could specify one of several types of"average" speed, each of which is related to the temperature. One such average iscalled the root-mean-squared (rms) speed and can be calculated from equation
Chapter 1 Kinetic Theory of Gases z 4 Area =4 nu2 II Figure 1.3 The shell between u and v + dv has a volume of 4rrv2 dv. The thickness of the shell here is exaggerated for clarity. Figure 1.4 Maxwell-Boltzmann speed distribution as a function of temperature for a mass of 28 amu. ( m )-exp( mu2) F(u) du = 471-u2 - - du. 2rkT 2kT By analogy to equation 1.29, we will call equation 1.31 the Maxwell-Boltzmann speed distribution. Speed distributions as a function of temperature are shown in Figure 1.4. We often characterize the speed distribution by a single parameter, for example, the temperature. Equivalently, we could specify one of several types of "average" speed, each of which is related to the temperature. One such average is called the root-mean-squared (rms) speed and can be calculated from equation
13Section1.5TheMaxwellDistribution of Speeds1.6: Cms = <u2>1/2 = (3kT/m)/2. Another speed is the mean speed defined byusing equation 1.16 to calculate <v>:UF(u)du<U>(1.32)8kTmudexp2kTTm2TkTwhere the integral was evaluated using Table 1.1 as described in detail in Example1.4.Finally, the distribution might also be characterized by the most probable speed,c*, the speed at which the distribution function has a maximum (Problem 1.8):(1.33)example 1.4Using the Speed DistributionObjectiveThe speed distribution can be used to determine averages.Forexample,find theaveragespeed,<u>.MethodOnce one has the normalized distribution function, equation 1.16givesthemethodforfindingtheaverageofanyquantity.IdentifyingQas thevelocityandP(Q)dQas thevelocitydistributionfunctiongiven in equation 1.3l, we see that we need to integrate vF(v) dufrom limits u = o to u = co.Solution<>vF(u)du=exp(-au)du4TU(1.34)1a'v'exp(-a)dv,VJowhere a = (m/2kT)/2. We now transform variables by letting x =au. The limits will remain unchanged, and du = dxla. Thus theintegralinequation1.34becomes441xexp(-x2)dx :aV.2aVn(1.35)18KT2k(mTmVTwhere we have used Table 1.1 to evaluate the integral.The molecular speed is related to the speed of sound, since sound vibrationscannottravelfasterthan themoleculescausing thepressurewaves.Forexample,inExample 1.5 we find that the most probable speed for O, is 322 m/s, while the
Section 1.5 The Maxwell Distribution of Speeds 1.6: c, = <v2>lR = (3kTlm)'". Another speed is the mean speed defined by using equation 1.16 to calculate <v>: where the integral was evaluated using Table 1.1 as described in detail in Example 1.4. Finally, the distribution might also be characterized by the mostprobable speed, c*, the speed at which the distribution function has a maximum (Problem 1.8): example 1.4 Using the Speed Distribution Objective The speed distribution can be used to determine averages. For example, find the average speed, <v>. Method Once one has the normalized distribution function, equation 1.16 gives the method for finding the average of any quantity. Identifying Q as the velocity and P(Q) dQ as the velocity distribution function given in equation 1.31, we see that we need to integrate vF(v) dv from limits v = 0 to v = m. Solution <v> = Ipuqv) dv = dv where a = (m/2kT)1'2. We now transform variables by letting x = av. The limits will remain unchanged, and dv = dxla. Thus the integral in equation 1.34 becomes where we have used Table 1.1 to evaluate the integral. The molecular speed is related to the speed of sound, since sound vibrations cannot travel faster than the molecules causing the pressure waves. For example, in Example 1.5 we find that the most probable speed for 0, is 322 rnls, while the
14Chapter1KineticTheoryof Gases2.0T = 300 K(u)1.5(u/sSn(a)s1.00.510001500500v (m/s)Figure 1.5Maxwell-Boltzmann speed distribution for a mass of 28 amu and a temperature of 300 K.Thevertical lines mark c,<u>,and Crmsspeed of sound in O, is measured to be 330m/s.For an ideal gas the speedof soundcan be shown to be (ykT/m)/2, where is the ratio of heat capacities, = CJCy.The Mach number is defined as the ratio of the speed of an object in a medium tothe speed of sound through the same medium, so that when an aircraft "breaks thesound barrier"(or exceeds"Mach 1")itis actually traveling faster than the speedof themolecules inthe medium.Figure1.5 shows the shape of the distribution function forT=300K and thelocationsof thevariouslydefinedspeeds.example 1.5Comparison of the Most Probable Speeds for Oxygen and HeliumObjectiveCompare the most probable speed for O, to that for He at 200 K.MethodUseequation1.33withT=200Kandm=2amuorm=32amu. Note that the relative speeds should be proportional to m-12.Solutionc(He) = (2kT/m)1/2 = [2(1.38 × 10-23 J K-1)(200 K)(6.02 ×1023 amu/g)(1000g/kg)/(2amu))/2=1290m/s.Asimilarcal-culation substituting 32 amu for 2 amu gives c(O2) = 322 m/s.CommentThe escape velocity from the Earth's gravitational field is roughlyU=1.1 × 104m/s, only about 10 times themostprobable speedfor helium.Because the velocity distribution shifts so stronglytoward high velocities as themass decreases, the fraction of helium
Chapter 1 Kinetic Theory of Gases II Figure 1.5 Maxwell-Boltzmann speed distribution for a mass of 28 amu and a temperature of 300 K. The vertical lines mark ?, <u>, and c,. speed of sound in 0, is measured to be 330 m/s. For an ideal gas the speed of sound can be shown to be (ykTlm)l", where y is the ratio of heat capacities, y = CdC, The Mach number is defined as the ratio of the speed of an object in a medium to the speed of sound through the same medium, so that when an aircraft "breaks the sound barrier" (or exceeds "Mach 1") it is actually traveling faster than the speed of the molecules in the medium. Figure 1.5 shows the shape of the distribution function for T = 300 K and the locations of the variously defined speeds. example 1.5 Comparison of the Most Probable Speeds for Oxygen and Helium Objective Compare the most probable speed for 0, to that for He at 200 K. Method Use equation 1.33 with T = 200 K and rn = 2 amu or m = 32 mu. Note that the relative speeds should be proportional to m-'I2. Solution c*(He) = (2kTlm)lD = [2(1.38 x J KP1)(200 K)(6.02 X amu/g)(1000 g/kg)/(2 amu)l1" = 1290 mls. A similar calculation substituting 32 amu for 2 amu gives ~'(0,) = 322 mls. Comment The escape velocity from the Earth's gravitational field is roughly u, = 1.1 X lo4 mls, only about 10 times the most probable speed for helium. Because the velocity distribution shifts so strongly toward high velocities as the mass decreases, the fraction of helium