•
In-class Exam #1, Wednesday, Feb 1 (50 minutes)

Covers chapters 1-3, HW_A and lecture material.

Formula sheets/notes (2 pages single sided) allowed.

Hand held calculators allowed.
•
HW_B due Friday, Feb 10

Maxwell-Boltzmann velocity distribution
  
T
  i

T

3/ 2

3/ 2 exp exp



 mv
2
/ 2 kT mv
2
/ 2 kT i


1.00
0.75
T = 1
T = 2
T = 4
T = 8
T = 16
1.00
0.75
T = 1
T = 2
T = 4
T = 8
T = 16
0.50
0.50
0.25
0.25
0.00
-6 -4 -2 v
0 2 4 6
0.00
-6 -4 -2 v
0 2 4 6

Maxwell-Boltzmann speed distribution

4

2
 m kT

3/ 2 v
2 exp

 mv
2
/ 2 kT


Maxwell-Boltzmann speed distribution

4

2
 m kT
3/ 2 v
2 exp

 mv
2
/ 2 kT

; ( )

N

( ) v m


2 kT
 1/ 2
 m 
 v


8 kT
 1/ 2
  m 
 v rms


3 kT
 1/ 2
;
 m 

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

0
   
1
•We can then use the distribution function to compute the average behavior of the molecules: v
 
0
 v
2  
0

2
  v rms
 v
2