29_562ln08

29_562ln08 - MIT OpenCourseWare http:/ocw.mit.edu 5.62...

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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 .
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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 vessel and into a vacuum size “d” slit oven walls at definite T We assume that: (1) d is so small that the pressure in the vessel is unchanged; (2) the effusion does not perturb the velocity of the gas in the vessel; (3) there are no collisions when 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. dA vdt ! 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 gas Volume 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 = ρ vcos θ dAdt (number density times volume) #molecules dAdt = ρ vcos ! =FLUX. We must integrate this expression over the distribution of velocities of the gas to obtain the average flux J. The Maxwell-Boltzmann distribution (from Lecture #28) is revised 4/22/08 5:38 PM
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5.62 Spring 2008 Lecture #29 Page 2 F ! v ( ) d ! v = F(v, ! , " )v 2 dvsin ! d ! d " = m 2 # kT $ % & ( ) 3 2 e * mv 2 kT v 2 dvsin ! d ! d " . Thus, for the average flux: The result is: J = ! m 2 " kT # $ % & ( 3 2 e ) mv 2 kT v 3 dv cos * sin * d * o " 2 + 0 , + d - 0 2 " + ! /2 rather than ! to obtain forward direction only Factor of v 3 : v 2 from volume element, v from flux J = ! m 2 " kT # $ % & ( 3/2 1 2 2kT m # $ % & ( 2 ) * + , - . 1 2 ) * + , - . 2 " [ ] = ! 4 8kT " m # $ % & ( 1/2 = ! v 4 Note : This result could have been obtained in an alternative way. Consider a volume of gas behind the hole that contains ρ dV 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=v z dt Only the molecules with v z > 0 can exit the hole. J = ! dv x "# + # $ dv y dv z 0 + # $ dv z m 2 % kT & ( ) * + 3 2 e " mv 2 2kT Angular Distribution of Flux (or Effusing Molecules) The velocity distribution of j(v, ! , " ) leaving the hole is. ! , " )dvd # = $ m 2kT % & ( ) * 3 2 v 3 e + mv 2 2kT cos ! dvd # solid angle d # =sin ! d ! d " revised 4/22/08 5:38 PM
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5.62 Spring 2008 Lecture #29 Page 3 The angular distribution of the flux is cos ! j angle ( " )d " = dv 0 # $ j v, ! , % ( ) = & v 4 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.
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29_562ln08 - MIT OpenCourseWare http:/ocw.mit.edu 5.62...

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