Chapter 1 Kinetic Theory of Gases20will occur. The collision rate is proportional to the number of possible kid-adultpairs, which is proportional to the number density of adults times the number den-sityofkids.But the collision rate depends on other factors as well. If all the skaters followthe rules and skate counterclockwise around the rink at the same speed, then therewillbenocollisions.Moreoften,thekidswillskateatmuchfasteror slowerspeeds,and they will rarely move uniformly.The rate at which they collide with theadults is proportional to the relative speed between the adults and kids.Finally, consider the dependence of the collision rate on the size of the adultsandkids.Peoplearetypicallyabout40cmwide.Whatwouldbetheeffectofincreasingor decreasing this diameterby a factor of 10?If thediameter weredecreased to 4 cm, the number of collisions would go down dramatically;if thediameterwereincreasedto4m,itwouldbedifficulttomovearoundtherinkatall.Thus, simple considerations suggest that the collision rate between moleculesshould beproportional tothe relative speed of themolecules,totheirsize,and tothe number of possible collision pairs.Let us assume that the average of the magnitude of the relative velocity betweenmoleculesoftypes1and2is<u,>and thatthemoleculesbehavelikehard spheres;there are no attractive forces between them, and they bounce off one another like bil-liard balls when they collide.h Let the quantity b, shown in Figure 1.11, be definedasthedistanceofalineperpendiculartotheeachoftheinitialvelocitiesoftwocol-lidingmolecules,one oftype1and theother oftype2.Thisdistance is oftenreferredto as the impactparameter If theradii of the twomolecules are ri and r2,then, asshown in Figure 1.1l, a “collision" will occur if the two molecules approach oneanother so that their centers arewithin thedistancebmax=ri+r2.Thus,bmaxis themaximumvalueof theimpactparameterforwhichacollisioncanoccur.Fromthepoint of view of one type of molecule striking a molecule of the other type, the tar-get area for a collision is then equal to (r, + r2)? = bmax*Figure1.11A collision will occur if the impactparameter is less than bmax,thesumof thetwomolecularradi.hWe consider only the relative velocity between the molecules. Appendix 1.4 shows that the total veloc-ity of each molecule can be written as a vector sum of thevelocity of the center of mass of the pair of mole-cules and the relative velocity of the molecule with respectto the center of mass.The forces between molecules depend on the relative distance between them and do not change the velocity of their center of mass,which must be conserved during the collision
Chapter 1 Kinetic Theory of Gases will occur. The collision rate is proportional to the number of possible kid-adult pairs, which is proportional to the number density of adults times the number density of kids. But the collision rate depends on other factors as well. If all the skaters follow the rules and skate counterclockwise around the rink at the same speed, then there will be no collisions. More often, the kids will skate at much faster or slower speeds, and they will rarely move uniformly. The rate at which they collide with the adults is proportional to the relative speed between the adults and kids. Finally, consider the dependence of the collision rate on the size of the adults and kids. People are typically about 40 cm wide. What would be the effect of increasing or decreasing this diameter by a factor of lo? If the diameter were decreased to 4 cm, the number of collisions would go down dramatically; if the diameter were increased to 4 m, it would be difficult to move around the rink at all. Thus, simple considerations suggest that the collision rate between molecules should be proportional to the relative speed of the molecules, to their size, and to the number of possible collision pairs. Let us assume that the average of the magnitude of the relative velocity between molecules of types 1 and 2 is <u,> and that the molecules behave like hard spheres; there are no attractive forces between them, and they bounce off one another like billiard balls when they collide." Let the quantity b, shown in Figure 1.11, be defined as the distance of a line perpendicular to the each of the initial velocities of two colliding molecules, one of type 1 and the other of type 2. This distance is often referred to as the impact parametel: If the radii of the two molecules are r, and r, then, as shown in Figure 1.11, a "collision" will occur if the two molecules approach one another so that their centers are within the distance b, = r, + r,. Thus, b, is the maximum value of the impact parameter for which a collision can occur. From the point of view of one type of molecule striking a molecule of the other type, the target area for a collision is then equal to ~(r, + r,), = rrb;,. Figure 1.11 A collision will occur if the impact parameter is less than b, the sum of the two molecular radii. hWe consider only the relative velocity between the molecules. Appendix 1.4 shows that the total velocity of each molecule can be written as a vector sum of the velocity of the center of mass of the pair of molecules and the relative velocity of the molecule with respect to the center of mass. The forces between molecules depend on the relative distance between them and do not change the velocity of their center of mass, which must be conserved during the collision
21Section 1.7Collisions:MeanFreePath and CollisionNumberA=Tbmm(ot)ArFigure 1.12Molecule 1 sweeps out a cylinder of area rb2max.Any molecule of type 2whose center is withinthe cylinder will be struck.Consider a molecule of type1 movingthrough a gas witha speed equal to theaverage magnitude of the relative velocity <u,>. Figure 1.12 shows that any mole-culeoftype2located inacylinderofvolumerbmax<u,>Atwill thenbestruckinthetime Ati Iif the density of molecules of type 2 is nz, then the number of collisions onemolecule of type 1 will experience with molecules of type 2 per unit time isZ, = Tbmax<u,>n2(1.42)Of course,for a molecule of type1 moving through other molecules of the same type,Z,=mbmx<u,>ni=Td?<u,>ni,(1.43)where bmax has been replaced by d? since r + r2 = 2r, = d. The quantity rbmax isknown as thehard-sphere collision cross section.Cross sections aregenerallygiventhe symbol .Equation 1.42 gives the number of collisions per unit time of one molecule oftype 1 with a density n2 of molecules of type 2. The total number of collisions ofmolecules of type 1 with those of type 2per unit time and per unit volume is foundsimply bymultiplyingbythedensity of typel molecules:Zi2=Zan=bmax<u,>nin2(1.44)Note that the product nin,is simply proportional to thetotal numberofpairs ofcol-lisionpartners.By a similar argument, if there were only one type of molecule, the number ofcollisions per unit time perunit volumeis given byTbmx<u,>(ni)2.(1.45)Zu =2Zini=2The factor of is introduced for the following reason.The collision rate should beproportional to the number of pairs of collision partners. If there are n molecules,then the number of pairs is n(n - 1)/2, since each molecule can pair with n - 1others and thefactor of2in thedenominator corrects for having counted eachpairtwice.If n is a largenumber,then we can approximate n(n --1)as n2,and since thenumber of molecules is proportional to the number density, we see that the numberof pairs goes as (n)2/2It remains for us to determine the value of the relative speed, averaged over thepossibleangles of collision and averaged over the speed distribution foreachmolecule. One way to arrive quickly at the answer for a very specific case is shown inBecauseofthecollisions, themolecule undrconsideration wllactuallytravelalongazigzagpath, butthe volume swept out per unit time will be the same
Section 1.7 Collisions: Mean Free Path and Collision Number II Figure 1.12 Molecule 1 sweeps out a cylinder of area rb2-. Any molecule of type 2 whose center is within the cylinder will be struck. Consider a molecule of type 1 moving through a gas with a speed equal to the average magnitude of the relative velocity Or>. Figure 1.12 shows that any molecule of type 2 located in a cylinder of volume .rrb2,<ur>At will then be struck in the time At.' If the density of molecules of type 2 is n;, then the number of collisions one molecule of type 1 will experience with molecules of type 2 per unit time is Of course, for a molecule of type 1 moving through other molecules of the same type, where biax has been replaced by d2 since r, + r2 = 2r, = d. The quantity rb;, is known as the hard-sphere collision cross section. Cross sections are generally given the symbol a. Equation 1.42 gives the number of collisions per unit time of one molecule of type 1 with a density n; of molecules of type 2. The total number of collisions of molecules of type 1 with those of type 2 per unit time and per unit volume is found simply by multiplying by the density of type 1 molecules: Note that the product nyn,* is simply proportional to the total number of pairs of collision partners. By a similar argument, if there were only one type of molecule, the number of collisions per unit time per unit volume is given by 1 1 Z, = l~ln; = - rbi, <u,> (n;)2. 2 (1.45) The factor of is introduced for the following reason. The collision rate should be proportional to the number of pairs of collision partners. If there are n molecules, then the number of pairs is n(n - 1)/2, since each molecule can pair with n - 1 others and the factor of 2 in the denominator corrects for having counted each pair twice. If n is a large number, then we can approximate n(n - 1) as n2, and since the number of molecules is proportional to the number density, we see that the number of pairs goes as (12;)~/2. It remains for us to determine the value of the relative speed, averaged over the possible angles of collision and averaged over the speed distribution for each molecule. One way to arrive quickly at the answer for a very specific case is shown in 'Because of the collisions, the molecule under consideration will actually travel along a zigzag path, but the volume swept out per unit time will be the same
22Chapter1KineticTheoryofGases(or) =0(or) = 2 (0)(ur) = V2 (0)(u)()(u)()(o)Figure1.13In a hypothetical collision where two molecules eachhave a speed equal to the average≤u>the relative velocity between two molecules, averaged over all collision directions, is V2<u>Figure1.13.Suppose that the twotypesofmoleculeshave the samemass,m.Letusassumeforthemomentthatwecanaccomplishtheaverageof the speeddistributionby assuming that the two molecules each have a speed equal to the average of theirdistribution.Sincethetwomoleculesareassumedtohavethe samemass (andtemperature), they will also have the same average speed, <u>. We now consider theaverage over collision angles.If the molecules are traveling in the same direction,then the relative velocity between them will have zero magnitude, v, O, while ifthey aretraveling in opposite directions alongthe same linethe relative velocitywillhave a magnitude of u, = 2<u>. Suppose that they are traveling at right angles toone another. In that case, which is representative of the average angle of collision,the relative velocity will have a magnitude of u, = <u,> = V2 <u>. Recallingfrom equation1.32that<u>=(8kT/m)1/2,wefind that<U>=V2<U>(1.46)8kT8kV(T(m/2)TUwhere we have introduced the reduced mass, μ, defined as μ = m,m/(m, + m).When the masses m; and m are the same, μ = m?/2m = m/2. If the masses are dif-ferent, then the mean velocities will not be the same, and the simple analysis of Figure 1.13 is not adequate. However, as shown for the general case in Appendix 1.4 andProblem 1.12, the result for <u,> is the same as that given in equation 1.46.Theappendix also shows why the definition of μ as m,m/(m, + m) is a useful one.example 1.6TheCollisionRateof NOwithO3ObjectiveFind the collision rate of NO with O, at 300 K if the abundances at1 atm total pressure are each 0.2ppm and if the molecular diametersare 300 and 375 pm, respectively. Reactive collisions between thesetwo species areimportant inphotochemical smogformation
Chapter 1 Kinetic Theory of Gases In a hypothetical collision where two molecules each have a speed equal to the average <v>, the relative velocity between two molecules, averaged over all collision directions, is A<v>. Figure 1.13. Suppose that the two types of molecules have the same mass, m. Let us assume for the moment that we can accomplish the average of the speed distribution by assuming that the two molecules each have a speed equal to the average of their distribution. Since the two molecules are assumed to have the same mass (and temperature), they will also have the same average speed, <v>. We now consider the average over collision angles. If the molecules are traveling in the same direction, then the relative velocity between them will have zero magnitude, v, = 0, while if they are traveling in opposite directions along the same line the relative velocity will have a magnitude of v, = 2<v>. Suppose that they are traveling at right angles to one another. In that case, which is representative of the average angle of collision, the relative velocity will have a magnitude of u, = <u,> = %b <u>. Recalling from equation 1.32 that <v> = (8kTl~m)~'~, we find that (1.46) 3 n-m where we have introduced the reduced mass, p, defined as p = m,ql(m, + q). When the masses m, and m, are the same, p = m2/2m = ml2. If the masses are different, then the mean velocities will not be the same, and the simple analysis of Figure 1.13 is not adequate. However, as shown for the general case in Appendix 1.4 and Problem 1.12, the result for <v,> is the same as that given in equation 1.46. The appendix also shows why the definition of p as mlql(ml + m,) is a useful one. example 1.6 1 The Collision Rate of NO with O3 Objective Find the collision rate of NO with O3 at 300 K if the abundances at 1 atm total pressure are each 0.2 ppm and if the molecular diameters are 300 and 375 pm, respectively. Reactive collisions between these two species are important in photochemical smog formation
23Section-1.7Collisions:MeanFreePathand CollisionNumberMethodUse equation 1.44,remembering to convert the abundances tonumberdensities at 300Kand calculating the average relativevelocity by use of equation 1.46.SolutionFirst find the total number density n" at 1 atm: n"= (n/V)NA =(p/RT)NA = (1 atm)(6.02× 1023 molec/mole)/[(0.082 L atmmo1-1K-1)(300K)]=2.45×1022molec/L.Nextdeterminethenumberdensities of NOand O,each being thetotal densitytimes 0.2X10-6: n(NO)=n(O)=(0.2×10-6)(2.45×1022)= 4.9 ×1015molec/L.Theaveragerelativevelocityis<u,>=(8kT/w)i/2=[8(1.38 × 10-23 J K-1)(300 K)(6.02 × 1023 amu/g)(1000 g/kg)/((48× 30/78)amu)j/2586m/s.The averagediameter is(300 + 375 pm)/2 = 337.5 pm. Then Zi2 = π(337.5 × 10-12 m)2(586 m/s)(4.9 × 1015 molec/L)(1 L/10-3 m3)2 = 5.0 × 1021collisions s-1 m-3. If every collision resulted in a reaction, thiswouldbe thenumber of reactions per unit second per cubic meter.A quantity related to Z, is the mean free path, A. This is the average distance amolecule travels before colliding with another molecule.If we divide the averagespeed<u>inmeterspersecondbythecollisionnumberZ,incollisionspersec-ond,weobtainthemeanfreepathinmeterspercollision:<U><U>入：ZTdV2<U>ni(1.47)1V2ad'niNote that the mean free path is inversely proportional to pressure.The mean free pathwill be important in Chapter 4, where we will see that the transport of heat, momen-tum, and matter are all proportional to the distance traveled between collisions.example 1.7nThe Mean Free Path of NitrogenObjectiveFind Z, and the mean free path of N, at 300 K and 1 atm given thatthemoleculardiameteris218pmMethodUse equation 1.46 to calculate <v,>, equation 1.43 to calculateZ, and equation 1.47 to calculate 入.SolutionWe start by calculating <u,> = (8kT/μ)1/2, where μ = 28 ×28/(28+28)=14amu.1/8(1.38 × 10-23 J k-l)(300 K)(6.02 × 1023 amu/g)(1000 g/kg)(3.1415× 14amu)(1.48)= 673 m/s
Section 1.7 Collisions: Mean Free Path and Collision Number Method Use equation 1.44, remembering to convert the abundances to number densities at 300 K and calculating the average relative velocity by use of equation 1.46. Solution First find the total number density n* at 1 atm: n* = (nIV)N, = (pIRT)N, = (1 atm)(6.02 X molec/mole)/[(0.082 L atm mol-l K-')(300 K)] = 2.45 X lo2, molecL. Next determine the number densities of NO and O, each being the total density times 0.2 X lop6: n*(NO) = n*(03) = (0.2 X 1OV6)(2.45 X lo2,) = 4.9 X 1015 molecL. The average relative velocity is <vr> = (8kTl~p)"~ = [8(1.38 X J K-l)(300 K)(6.02 X amu/g)(1000 g/kg)/ (~(48 X 30178) amu)l1/& 586 m/s. The average diameter is (300 + 375 pm)/2 = 337.5 pm. Then Z12 = ~(337.5 X 10-l2 rn), (586 m/s)(4.9 X 1015 m~lecL)~(l L/10p3 m3)2 = 5.0 X 1021 collisions s-' m-3. If every collision resulted in a reaction, this would be the number of reactions per unit second per cubic meter. A quantity related to Z, is the mean free path, A. This is the average distance a molecule travels before colliding with another molecule. If we divide the average speed <v> in meters per second by the collision number Z, in collisions per second, we obtain the mean free path in meters per collision: Note that the mean free path is inversely proportional to pressure. The mean free path will be important in Chapter 4, where we will see that the transport of heat, momentum, and matter are all proportional to the distance traveled between collisions. example 1.7 - The Mean Free Path of Nitrogen Objective Find Z, and the mean free path of N, at 300 K and 1 atm given that the molecular diameter is 218 pm. Method Use equation 1.46 to calculate <vr>, equation 1.43 to calculate Z, and equation 1.47 to calculate A. Solution We start by calculating <vr> = (8kTl~p)"~, where p = 28 X 28/(28 + 28) = 14 mu. 8(1.38 X J ~-')(300 K)(6.02 X amu/g)(1000 g/kg) <vr> = (3.1415 X 14 amu) = 673 m/s
24Chapter1 KineticTheory of GasesNext, we calculate Z, noting that the density(1 atm)(6.02 × 1023 molec/mole)pn;RT(0.082Latmmole-/k-/)(10~3m/L)(300K)(1.49)= 2.45 × 1025 molec/m2.Then,Z, = π(218 ×10-12 m)2(673m/s)(2.45_× 1025molec/m3) =2.46_×109 collision/s.Finally,<u,>/(V2z)=(673m/s)/(V2×2.46×10°collision/s)=1.93×10-m.1.8SUMMARYBy considering the pressure exerted by ideal gas molecules on a wall, we deter-mined that, for agreement with the observed ideal gas law, the average energy of amolecule must be given by3<E>kT(1.5)2To learn how to perform averages, we discussed distribution functions of a continu-ousvariable.Theaverageof someobservablequantityQwasfoundtobegivenby<Q>P(e)Q dQ,(1.16)where P(Q) is the distribution function for the quantity Q.We then made the fol-lowing observations about the molecular speed distribution: (1) the speed distribu-tionmustbean evenfunctionofu,(2)the speeddistribution inanyparticulardirec-tion is independent from and uncorrelated with that in orthogonal directions, (3) thevalue of <u2> must be equal to 3kT/m to agree with the ideal gas law, and (4) thedistribution depends only on the magnitude of u.Thesefour considerations allowedus to determine the Maxwell-Boltzmann distribution of speeds:mw2mF(u)du=4udu(1.31)exp2kT2TkTCalculations using this distribution gave us an equation for the average speed of amolecule,(8kT)1/2<Uy(1.32)and the most probable speed,(1.33)A simple transformation of variables in the speed distribution led to the Maxwell-Boltzmann energy distribution:G(e)de = 27(1.37)deFinally, for molecules behaving as hard spheres, we determined the collision rate,Z,=Tbmax<u,>ni,(1.42)
Chapter 1 Kinetic Theory of Gases I Next, we calculate 2, noting that the density * P (1 atm)(6.02 X molec/mole) nl=-= RT (0.082 L atrn mole-' K-')(~o-~ m3/~)(300 K) Then, 2, = ~(218 X 10-l2 m)2(673 rn/s)(2.45 X molec/m3) = 2.46 X lo9 collision/s. Finally, <u,>/(fiZ1) = (673 m/s)/ (fi X 2.46 X lo9 collision/s) = 1.93 X lo-' m. 1.8 SUMMARY By considering the pressure exerted by ideal gas molecules on a wall, we determined that, for agreement with the observed ideal gas law, the average energy of a molecule must be given by 3 <e> = -kT. 2 (1.5) To learn how to perform averages, we discussed distribution functions of a continuous variable. The average of some observable quantity Q was found to be given by where P(Q) is the distribution function for the quantity Q. We then made the following observations about the molecular speed distribution: (1) the speed distribution must be an even function of v, (2) the speed distribution in any particular direction is independent from and uncorrelated with that in orthogonal directions, (3) the value of <v2> must be equal to 3kTlm to agree with the ideal gas law, and (4) the distribution depends only on the magnitude of u. These four considerations allowed us to determine the Maxwell-Boltzmann distribution of speeds: Calculations using this distribution gave us an equation for the average speed of a molecule, and the most probable speed, .* (x)'". A simple transformation of variables in the speed distribution led to the MaxwellBoltzmann energy distribution: 3/2 G(e)de=2~(~) T~T Finally, for molecules behaving as hard spheres, we determined the collision rate