大学物理(热学) 1) Mean Free Path 2)Diffusion, Thermal conductivity, and viscosity 2006-10-12
大学物理（热学） 1) Mean Free Path 2) Diffusion, Thermal conductivity, and Viscosity 2006-10-12
Equilibrium distribution v2n1n2+n22+ ∑(舞)一 n -the number of molecules of dn is the number of molecules in vx>vx+dvx near v V,→V,+dv ν→)1+dv
Equilibrium distribution N v v v v N 2 2 2 2 2 1 + + + = dn is the number of molecules x x x v →v + dv vy vy + dv → y z z z v → v + dv in → = N n v N n v i i i 2 2 d N n v + n v + = 2 2 2 2 1 1 ni ⎯ the number of molecules of i v near v
Meaning of dn/n among N molecules, how many are near probability that one molecule is near v proportional to avx fly dv dn(vr,vy,v =/()如yv,y,fvk N
among N molecules, how many are near v probability that one molecule is near v proportional to dvx ... ( ) x x f v dv Meaning of dn/N ( )d ( )d ( )d d ( , , ) x x y y z z x y z f v v f v v f v v N n v v v =
Isotropic assumption (v,/(v,)(2)=m2)=V2+y+n2) The solution is (v,)=Cexp(2/a2) The probability should be normalized dn
Isotropic assumption ( ) ( ) ( ) ( ) ( ) 2 2 2 2 x y z x y z f v f v f v = v = v + v + v ( ) ( ) 2 2 f vx =Cexp − vx The solution is The probability should be normalized = N dn 1
∫=c=/) 2/a2 1=Cl explvx ly each normalized √兀 a exp C C
( ) 3 3 2 2 exp d d 1 = − = x x C v v N n − = − x x v v C exp d 1 2 2 ( ) = − x x 1 C exp v dv 2 2 12 = 1 each normalized
dn( +13+12 d v. av. av a√r exp 3k T Known v C dn(vr, vy, v)3 whule C N 2
( ) exp d d d 1 d ( , , ) 2 2 2 2 3 x y z x y z x y z v v v v v v N n v v v + + − = = ? m k T v 3 2 B Known = 2 2 2 2 d ( , , ) 3 = = N n v v v v v while x y z
dr mlv+.+ exp d v..av 2R B 2KT For unit volume dnlvx,y7 m(2+n2+2) =no 2kBT expI d v dv.av ckpT Maxwell velocity distribution
( ) ( ) x y z x y z x y z v v v k T m v v v k T m N n v v v d d d 2 exp 2 d , , B 3 2 2 2 2 B + + − = For unit volume d ( , , ) x y z n v v v Maxwell velocity distribution ( ) x y z x y z v v v k T m v v v k T m n d d d 2 exp 2 B 3 2 2 2 2 B 0 + + − =
dn(v) exp 4πv2dv 2TRB nof(v)vdv Maxwell speed distribution Most probable speed For one molecule, what speed is most possible? Among N molecules, what is the speed most molecules have in unit interval?
v v k T m v k T m n v n 4 d 2 1 exp 2 d ( ) 2 B 2 3 2 B 0 − = Maxwell speed distribution For one molecule, what speed is most possible? Most probable speed n f (v)4 v dv 2 0 Among N molecules, what is the speed most molecules have in unit interval?
dy exp-2 kBT 0 dy kOt Maxwellian can be written as 3/2 n(v 4 2 exp d m
0 2 1 exp d d B 2 2 = − k T mv v v m k T v B m 2 = v v v v N v n v exp d 1 4 d ( ) 2 2 m 2 3 2 2 m − = Maxwellian can be written as
The average speed v=4 vexp(-v2/1mh2dv 0 m 8kT 几1
The average speed v ( v v )v v v v exp d 1 4 2 2 m 2 0 3 2 2 m − = m k T = 8 B v x x x e d 4 2 0 3 m − =
