Maxwell-Boltzmann velocity distribution
f  v   T 3/ 2 exp  mv 2 / 2kT 
T=1
T=2
T=4
T=8
T = 16
1.00
T=1
T=2
T=4
T=8
T = 16
0.75
2
exp(v /T)
0.75
2
exp(v /T)
1.00
f  vi   T 3/ 2 exp  mvi2 / 2kT 
T
3/2
0.50
0.50
0.25
0.25
0.00
-6
-4
-2
0
v
0.00
-6
-4
-2
0
v
2
4
6
2
4
6
Maxwell-Boltzmann speed distribution
 m 
f  v   4 

2

kT


3/ 2

v 2 exp mv 2 / 2kT

Maxwell-Boltzmann speed distribution
 m 
f  v   4 

 2 kT 
1/ 2
 2kT 
vm  

m


3/ 2
v 2 exp  mv 2 / 2kT  ;
N (v )  N  f (v )
1/ 2

 8kT 
v 


m


1/ 2

vrms
 3kT 


m


;
Gas pressure and the ideal gas law
Kinetic theory provides a natural interpretation of the
absolute temperature of a dilute gas. Namely, the
temperature is proportional to the mean kinetic energy (e )
of the gas molecules.
•
The mean kinetic energy is independent of pressure,
volume, and the molecular species, i.e. it is the same for
all molecules.
1
2
1 2 2
2
PV  Nmv   N  mv   N e  NkT
3
3
2
 3
The probability density function
• The random motions of the molecules can be
characterized by a probability distribution function.
• Since the velocity directions are uniformly distributed, we
can reduce the problem to a speed distribution function
which is isotropic.
• Let f(v)dv be the fractional number of molecules in the
speed range from v to v + dv.
• A probability distribution function has to satisfy the
condition

 f  v  dv  1
0
•We can then use the distribution function to compute the
average behavior of the molecules:

v   vf  v  dv
0

v   v 2 f  v  dv
2
0
vrms  v 2