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MIT 5 62 - Study Notes

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MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.62 Spring 2008 Lecture #29 Page 1 Kinetic Theory of Gases: Effusion and Collisions EFFUSION Consider the process by which molecules escape through a hole in a vesseland into a vacuum size “d”slitoven walls atdefinite TWe assume that: (1) d is so small that thepressure in the vessel is unchanged; (2) theeffusion does not perturb the velocity of thegas in the vessel; (3) there are no collisionswhen the molecules pass through the slit. Molecules that would have been incident on the portion of the wall where the hole is, now pass through the hole. This creates a flux of particles defined as the number of particles per unit area per unit time that leave the vessel. dAvdt!Consider a square hole of area dA. A particle that is a distance vdt from the hole!moves with speed v and at angle from the surface normal toward the hole. Draw a parallelepiped around the hole with length equal to vdt, at angle θ from the normal. All molecules within this volume moving toward the hole (i.e., with the correct !, "angle) with speed v will pass through the hole in time interval dt. ρ = density of gasVolume of parallelepiped = v cos !dA dt (Note that, at grazing angles, θ ~ π/2, the volume is small) # of molecules crossing through dA in dt = ρvcos θdAdt (number density times volume) # moleculesdAdt= ρvcos !=FLUX. We must integrate this expression over the distribution of velocities of the gas to obtainthe average flux J. The Maxwell-Boltzmann distribution (from Lecture #28) is revised 4/22/08 5:38 PM5.62 Spring 2008 Lecture #29 Page 2 F!v( )d!v = F(v, !, ")v2dvsin !d!d" =m2#kT$%&'()32e*mv2kTv2 dvsin!d!d" . Thus, for the average flux: The result is: J = !m2"kT#$%&'(32e)mv2kTv3dv cos* sin*d*o"2+0,+d- 02"+!/2 rather than ! to obtainforward direction onlyFactor of v3: v2 from volumeelement, v from fluxJ = !m2"kT#$%&'(3/ 2122kTm#$%&'(2)*+,-.12)*+,-.2"[ ]=!48kT"m#$%&'(1/ 2=!v4Note : This result could have been obtained in an alternative way. Consider a volume of gas behind the hole that contains ρdV molecules. Then J simply is: particles exiting through the hole are the same as the velocity distribution of particles hitting the hole. As a result the angular distribution and speed distribution of flux Volume dV=vzdtOnly the molecules with vz > 0 can exit the hole.J = ! dvx"#+#$dvydvz0+#$dvzm2%kT&'()*+32e"mv22kTAngular Distribution of Flux (or Effusing Molecules) The velocity distribution of j(v, !, ")leaving the hole is. j(v, !, ")dvd# = $m2kT%&'()*32v3e+mv22kTcos !dvd#solid angled#=sin!d!d"revised 4/22/08 5:38 PM5.62 Spring 2008 Lecture #29 Page 3 The angular distribution of the flux is cos !jangle(")d" = dv0#$j v,!,%( )=&v4'cos !d"0 < ! <'2, 0 < % < 2'Effusion is an important mechanism for creating molecular beams that have practical use (for example in molecular beam epitaxy used in the manufacture of electronic devices) and for studying collisions and gas phase chemical reactions. One important application is time of flight verification of the Maxwell Boltzmann distribution (1955!). The experiment is described in the figure below. Puffs of gas are released by opening the shutter. The gas spreads as a result of the spread in speed. {velocity distributions can also be inferred from Doppler shift measurements on spectral lines}. q(t) is the number of molecules thathits the detector between t and t + dttimeL = vtgassourceshutterdetectort1t2t3There is a relation between the distribution of arrival times q(t) and the distribution ofspeeds h(v): . q(t)dt = h (v)dvrevised 4/22/08 5:38 PM5.62 Spring 2008 Lecture #29 Page 4 Dynamics determines the exact relation between distance, speed, and time L=vt. Using the Dirac delta function to set this relation gives: dvh(v)! v "Lt#$%&'(0)*= hLt#$%&'(. Thus measurement of q(t) allows us to infer the functional form of h according to: hLt!"#$%&= q(t)dtdv= q(t)t2L. The flux is obtained by multiplying the amplitude in the arrival time distribution at thecorresponding time t with a factor t2/L. The flux of molecules with large v arrive at the detector early and have not spread out in arrival times as much as the later arriving ones. Molecular Collisions Goal: To calculate collision frequency between pairs of molecules in a gas. We begin by defining terms. Z = average number of collisions of a single particle per unit time. Collision event occurs when centers of 2 molecules approach within distance d. Distance d is the hard sphere diameter Collision cross section ! "d2=area of circle of radius d (surrounding particle 1), if the center ofparticle 2 enters (or touches) this circle, a collision occurs. A collision occurs when the relative velocity permits an encounter between two particles. The relative velocity is defined as: !vr=!v1!!v2. In a time dt, a collision volume dV is swept out by a particle. This volume is: dV = !d2!vrdt; !vr=!vr"!vr[ ]12. revised 4/22/08 5:38 PM5.62 Spring 2008 Lecture #29 Page 5 The number of encounters in time dt at relative velocity vr is ρdV. The number of encounters per unit time is # of encounters/time=!dVdt= "d2!!vr . The collision frequency is the average of this quantity: Z=!d2"!vr Since !vr = vr, the average of the relative velocity, is vr= d!v1d!v2!!F(!v1)F(!v2)vrIn this expression, we have assumed that the velocity distributions for the two molecules are independent of each other; this is true. But, at high density, intermolecular interactions will influence the spatial arrangement of the colliding pairs and this effect is not taken into account in the above expression. Thus, the average we are calculating is only valid for dilute gas conditions. The two distribution functions have the form: F(!v1) =m12!kT"#$%&'32exp (m1v122kT"#$%&' and F(!v2) =m22!kT"#$%&'32exp (m2v222kT"#$%&'. Thus the integral to be done is: vr=m12!kT"#$%&'32m22!kT"#$%&'32d!v1d!v2((exp )m1v12+ m2v222kT"#$%&'*+,-./ vrIn order to do the integrals we must change variables from !U !v1 and !v2to relative and center of mass velocity : !vr=!v1!!v2 and M!U=m1!v1+ m2!v2 where M=m1+ m2Solving for the particle velocities gives: !v1=!U +m2M!vr and !v2=!U !m1M!vr . revised 4/22/08 5:38 PM5.62 Spring 2008 Lecture #29 Page 6 A bit of algebra shows that the kinetic energy term is: m1v12+ m2v22= (m1+ m2)U2+m2m1m1+ m2vr2= MU2+


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