Transcript Kinetic Theory

Kinetic theory
We consider a dilute gas of N atoms in a box in the classical limit

 a
de Broglie wave length 

p k

h
h
h


p
2mEkin
2m kBT
h
L
N
 1/ 3 with density n  3
L
N
2m kBT
N1/3 a  L
1/ 3

h
1/ 3
N
L 
n
Low density
hn
 1
2m kBT
classical limit
high temperature
Ideal gas equation of state
Momentum change during
collision with wall:
t t
t t
dt    F (t )dt 

dt
t
t
Average perpendicular force component
acting on particle during time interval t
t t
1
av
mvx (t  t )  mvx (t )  2mvx (t )  Fxav, atom t
Fx, atom 
Fx (t )dt 

t t
p 

dp
L
x
Average perpendicular force component acting on wall during time interval t
Fxav, wall  Fxav, atom
2mvx (t )  Fxav, wall t
Time of flight t between successive collisions with right wall:
av
with Fx, wall  2mvx (t ) / t
Fxav, wall
mv
 x
Lx
t 
2
2 Lx
vx
Total force exerted on the wall by N particles
N
Fxtot  
i 1
mi vx ,i
2
Lx
Pressure P on right wall of area A=LyLz
Fxtot
1
P

Ly Lz Ly Lz
N

i 1
mi vx ,i
2
Lx
1 N
2
  mi vx ,i
V i 1
Same argument for particles flying
Fytot
1 N
2
  mi vy ,i
in y-direction P 
Lx Lz V i 1
tot
z
F
1 N
2
  mi vz ,i
in z-direction P 
Lx Ly V i 1
1
P
3V
1

3V
 m v
N
i 1
N
i
 mi vi
i 1
2
2
x ,i
 v y ,i  v z ,i
2
2

1 N
2
mi vi
Further inspection of P 

3V i 1
1 N
2
m
v

i i
1
2
2
i 1
with
 mi vi
and mi  m i
N
2
arithmetic
average
21

PV  N  m v 2 
3 2

1
1
2
m v  m v2
2
2
Mean square
velocity
Compare with the ideal gas equation of state
PV  N kBT
1
3
m v 2  k BT
2
2
Correct only in the classical limit
vrms :
v2 
3kBT
m
See textbook for generalized expressions where
mi are not identical
Note: we did not derive yet the equation of state
1
3
m v 2  k BT
2
2
To do so we need to show
Calculation of statistical averages requires knowledge of the
distribution function:
f ( r , p, t ) d 3 r d 3 p :number of molecules at a given time, t, within a
volume element d3r located at r and
momentum element d3p around p=mv
Fztot
1
P

Lx Ly V
N
m v
i 1
Note that (r, p) spans the 6-dimensional µ-space where f is
defined. Later we will introduce a density function  defined over
the 6N-dimensional -space. Don’t confuse!!!
2
i z ,i
With f we are able to calculate averages such as:
v2 
2
3
3
v
f
(
r
,
p
,
t
)
d
r
d
p


f ( r , p, t ) d 3 r d 3 p
v2 
2
3
3
v
f
(
r
,
p
)
d
r
d
p


3
f ( r , p, t )  f ( r , p)  f1 ( p) f 2 ( r )
For
3
f ( r , p) d r d p

2
3
v
f
(
p
)
d
p
1


3
f1 ( p) d p

 f (r ) d
f2 (r ) d 3r
2
3
r

2
3
v
f
(
p
)
d
p
1


f1 ( p) d 3 p
How to get f ?
In equilibrium, f turns out to be the
Maxwell-Boltzmann distribution function
We will encounter various approaches for its derivation throughout the
course.
Let’s start with a heuristic consideration
P(h  h )
P(h )  P(h  h) 
(h) A h g
Mg
 P(h  h) 
A
A
 P(h  h )  (h ) h g
h
Mass of the air contained in the volume element
A
h
Remember the barometric formula P=P(h)
h
air in good approximation considered as an ideal gas
M
V
n
M PP
PVnRT
P

RT

n RT
RT
M M
1/ 
P(h )  P(h  h )  (h )
h g
P ( h  h )  P ( h )
M gP(h )

h
n RT( h )
dP(h )
M g P( h )

dh
n R T(h )
h  0
In the case T=const. and with M=nNAm and R=NAkB
dP
mg


 P k B T
P0
P
dP(h)
mg P(h)

dh
kB T
P(h)  P0 e

mgh
kB T
and
 (h)  0 e

mgh
kB T
h
 dh
0
 ( r )d 3 r = number of particles in d3r  probability of finding a particle in d3r
in accordance with our definition of f we expect

f ( r , p) d 3 p   ( r )   ( x )   0 e

mgx
kB T
 0 e

E pot ( x )
kB T
p2
 E pot ( r ))
Ansatz f ( r , p)  f ( E 
2m
d
3
p f ( E )  0 e
3
d
 p

E pot
kB T
d
dE pot
df ( E ) dE
1

0 e
dE dE pot
k BT

E pot
kB T
1
df ( E )
1
3
d
p


d
p f (E)


dE
kBT
3

1
3  df ( E )
d
p

f
(
E
)
 0
  dE kBT


1
3  df ( E )
d
p

f
(
E
)
  0 E pot
  dE kBT

df
E

 df kBT
df ( E )
1

f (E)  0
dE
k BT
f e
f e
E
k BT
e
 p2

  
 E ( r ) 
 2m

 p2

  
 E ( r ) 
 2m

e
with
 p2 
  

2
m

    E ( r )
e
 :
1
k BT
 f1 ( p 2 ) f 2 ( r )
velocity distribution
independent of position
Let’s calculate the Maxwell Boltzmann distribution f1(p2) with all constants

C d pe
3

 p2
N
2m


C d pe
3

 p2
2m


with
 dx e

 C  4 p dp e
2

 p2
2m
0
 x 2
0
1

2

 4 C  p dp e
2
0
 2m 
 2m C 2m  3/ 2  C 




f1 ( p ) 
2m
N

e
3/ 2
 2m kBT 
p2
2 mkBT
3/ 2

 p2


2m
 8m C
dp
e
 0


  2m 
N  4m C

  4m C 2m
   
2

 p2
  3/ 2 


2


When asking for the probability, f(v), of finding a particle with velocity between
v v
and
v  dv
f1 ( p 2 )d 3 p / N 
we transform from d3p to dv
1
e
3/ 2
 2m kBT 
mv 2

2 k BT
 m 
2
4  mv  d  mv   4 v 2 

2

k
T

B 
3/ 2
e
mv 2

2 k BT
dv
f(v)
Maxwell speed distribution function
 m 
f (v)  4 v 

 2 k BT 
3/ 2
2

e
mv 2
2 k BT
With, e.g., f(p) at hand we can actually derive the ideal gas equation of state


p
by calculating

3
2
3
2
2
1
1
m v2 
2
2m
d
p p f ( p)



d

3
p f ( p)
1
2m
d
ppe


d

3
pe

2m
 p2
2m

d
1
1 
m v2 
2
2m 
3
d
p p 2e
3
pe


 p2

p
2
2m

1
2m


2
p
 dp e

4
p
 dp e
0

 2m
 p2
2m
 dpp
2

0

0

e

 p2
2m
 p2
2m
0
 p2
2m
 dp p e
4
2m


dp e

 0

 p2
2m

 2m
  1 2m  1  2m 
 2m

 
  2
  4   

2
dpp
e

 0

 p2

1
1
m v2 
2
2m
d
3
p p 2e


d
3

2m
pe


m   2m 


2    
3/ 2

3/ 2
3m
1
3/ 2
2
m


 5/ 2
4

 p2
2m
p
2
2m

1 3m 3
 k BT
2m 
2

with
21

PV  N  m v 2 
3 2

PV  NkBT