Unformatted text preview: Kinetic Theory of Gases(Chapter 27.13) When we think of think of temperature and heat flow, we (well at least I!) think of thermal energy and motion. For this discussion, we will consider how this relates to the motion of gases. We will consider the gas to be quasiideal with a finite volume and hard sphere potential (but no attractive interactions). Thermal energy, we always associate with motion and with temperature. Let's start to quantify these relationships and properties based on the above assumptions. What causes pressure? Collisions of gas molecules with walls. Here in the xdirection, with an elastic (no energy transfer) collision, u1x Fig 27.1 changes direction with a headon collision. dp ( m u1x ) F= = p  momentum = m u1x dt t 2a ( m x ) = 2mu1x (reversal of velocity); t= u1 u1x
Lecture 29 1 2mu1x mu1x 2 so F1 = = 2a / u1x a for gas molecule #1 F1 mu1x 2 / a mu1x 2 mu1x 2 Now P = = = = 1 A bc abc V If we sum the effect of all the molecules: m N 2 P = i = ix P u V i =1 i =1 1 N 2 2 but uix = u x 2 = average of the squared velocity (not u x ) 1 N i =1 so PV = mN u x 2 We could have picked any direction. It was arbitrary, so: ux 2 = u y 2 = uz 2
Lecture 29 2
N u 2 = ux 2 + u y 2 + uz 2 = ux 2 + u y 2 + uz 2 1 2 1 so u x = u and PV = mN u 2 3 3 3 1 For an ideal gas the average kinetic energy (translation) = k BT = m u 2 2 24 3 1 24
2 classical K.E. 3 1 1 2 so RT = mN A u = M u 2 2 2 2 3RT = M u 2 1 1 1 2 2 so for one mole: N =, N A PV = mN A u = M u = ( 3RT ) 3 3 3 PV = RT ( n=1) ideal gas equation and u
2 3RT = M 3 RT so u = 1 2 3 M 2 1/ 2 RMS vel. 1/ 2 Lecture 29 3 For Ar at 300K 1 For N 2 at 300K 2 u u 2 1/ 2 Ar 2 1/ 2 N2 3 8.314 300 = = 432 m / sec! 0.040 = 515 m / sec = 3.2 101 miles / sec 1/ 2 rms speed is very fast gases transfer energy quickly 5 RT note: usound = 3M This gives us a way to relate speeds to temperature, but what about the velocity distribution? What is it like? Inquiring minds want to know! Let h ( u x , u y , u z ) du x du y du z represent the fraction (probability) of molecules
having velocities between u x and u x + du x , u y and u y + du y , u z and u z + du z .
1/ 2 In 1860(!) Maxwell posited that ux,uy, and uz are completely independent (proved more elegantly by Boltzmann) in gas sample.
Lecture 29 4 If this is the case: h ( u x , u y , u z ) = distributions in each independent direction f ( ux ) f ( u y ) f ( uz ) 1 4 44 2 4 4 4 3 Since motion of the gas is isotropic, there is no preferred direction so all that is important is the magnitude of the velocity u 2 = u x 2 + u y 2 + u z 2 so h ( u x , u y , u z ) = h ( u ) now ln h ( u ) = ln f ( u x ) + ln f ( u y ) + ln f ( u z ) ln h ( u ) 1 ln f ( u x ) ln h ( u ) and = = ux u 1u x ,uz uy u 1 2 from above: ( u ) = 2u u = 2ux 1u x x , u u
y z u 1 u x , u z uy or u ux = ux u d ln h ( u ) u x d ln f ( u x ) so = du u du x d ln h ( u ) udu d ln f ( u x ) d ln f ( u y ) d ln f ( u z ) = = = Lecture u x du x u y du y 29 u z du z 5 This is like Q.M. separation of variables and must be equal to some constant since ux,uy and uz are independent variables. d ln h ( u ) d ln f ( u x ) Let = =  udu u x du x d ln f ( u j ) = . d ln f ( u j ) =  u j du j u j du j so ln f ( u j ) =  f ( u j ) = Ae u j2 2 +C  u j 2 / 2 f ( u j ) is a probability distribution so ( Prof. Lisy's copy of McQ&S doesn't have the 2 )
du j = 2 A e du j 0 1 4 2 44 4 3 1 /2 4 1/ 2 f ( u ) du
 j j = 1 = A e
  u j 2 / 2  u j 2 / 2 1 = 2A 2 1/ 2 A = 2 1/ 2 Lecture 29 6  u j 2 / 2 So our normalized velocity distribution is: f ( u j ) = e 2 uj
2
2 RT  u j / 2 1 2 2  u j / 2 = =j e u du j = 2 j e u du j =  0 M 2 2 1 4 4 2 4 43 2 1/ 2 1/ 2 1/ 2 2 2 1 2 2 1/ 2 M so = RT f ( uj ) M  Mu j 2 / 2 RT = e 2 RT As T decreases, velocity distribution narrows close to 0. Note <ui>=0 valid for all three components x,y, and z. 1/ 2 Fig 27.3 Lecture 29 7 h ( u ) = f ( ux ) f ( u y ) f ( uz ) M = e 2 RT 3/ 2 Mu 2  2 RT M  2M ( ux2 +u y 2 +uz 2 ) = e RT 2 RT 3/ 2 the distribution depends on the scalar speed, u M  Mu 2 / 2 RT h ( u ) du = du x du y du z e 2 RT We assume that we have an isotropic distribution of velocity components (why is this valid?) so du x du y du z = u 2 du sin d d and 2 M  Mu 2 / 2 RT 2 h ( u ) du = u du d d sin e 0 2 1 4 4 2 4 04 3 RT 4 3/ 2 3/ 2 Lecture 29 8 M  Mu 2 / 2 RT 2 u du 1 1 4 e 2 RT 1 F ( u ) du 1 MaxwellBoltzmann vel. dist. 3/ 2 1 24 1 4 3 m  mu 2 / 2 kBT 2 McQ & S u du 11 4 k T e 2 B 1 This is a probability distribution for speeds, u, in this gas kinetic model. We can do quite a bit with this: m 3  mu 2 / 2 kBT 8 k BT u = ( u ) du = 4 uF du = ue 0 0 2 k BT m 1/ 2 3/ 2 1/ 2 3/ 2 3k BT 3 k BT 2 2 1/ 2 u = , u = m m One last useful speed is the most probable maximum in the speed distribution:
dF ( u ) du =0 solve for ump 2 k T ump = B m Lecture 29
1/ 2 9 We can also express our distribution in terms of the molecular 1 kinetic energy, , i.e. we can find F ( ) d where = mu 2 2 We need to change variables: 2 1 m 2 d u = , du = = d 1/ 2 m 2 2 m ( 2m ) 2 d so u du = = d 1/ 2 3/ 2 m ( 2m ) m
2 1/ 2 1/ 2 ( 2 ) 1/ 2 m  / k BT F ( u ) du 2 F ( ) d = 4 e 2 k BT 3/ 2 3/ 2 ( 2 )
m 1/ 2 3/ 2 d 1 1/ 2  / kBT F ( ) d = 2 d e k BT This is useful for readily determining moments of the energy: 3 = F ( ) d = k BT 0 2 Lecture 29 10 ...
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This note was uploaded on 04/13/2008 for the course CHEM 444 taught by Professor Jameslis during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
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