Distribution of Molecular Speeds



Within an ideal gas,
molecules moves with a
diversity of speeds
A mathematical rigorous way
to describe this statistically is
thru a distribution function
f(v).
Properties:
 f(v) ~ 0, v ~ 0
 f(v)  0, v  large
 f(v) largest at mid-range
Maxwell-Boltzmann Distribution
 m 
f    4 

 2 kT 
3/ 2
2  m 2 / 2 kT
e


f(v)dv gives the probability of
finding molecules with speed in
range [v,v+dv].
Diff averages with respect to the
distribution of molecular speeds
can be calculated using f(v):
1.

vav   vf (v)dv
(avg of v)
0
2.
(v2 )
av 

2 f (v)dv (avg of v2)
v

0
Notes: Weighted Average
7 Test Scores:
si  65,80,80,95,95 , i  1, ,5
Simple Avg:
1

av N
s
N
 si 
i 1
1
 65  80  80  95  95
5
sav  15  65  52 85  52  95
sav   f (s)s

vav   f (v)vdv
0
(Weighted Avg)
where n(s) = fraction of occurrence of s
Notes: Weighted Average
7 Test Scores:
si  65,80,80,95,95 , i  1, ,5
Simple Avg:
1

av N
s
N
 si2 
i 1
1
652  802  802  952  952 

5
sav  15  65   52 85   52  95 
2
sav   f (s)s
(v2 )
av 


0
2
2
2
(Weighted Avg)
where n(s) = fraction of occurrence of s
f (v)v2dv
Statistical Description of Molecular Speed

Average speed (mean value):

Root Mean Square (RMS) speed:
vav  8kT
m


kT
vrms   v2   3m

av
Note: vav does not equal vrms !
In addition to simple “mean” and “median”, there are other
ways to statistically describe the “average” values for a
distribution of molecules moving at different speeds!

Most Probable Speed - the maximum value of the distribution
function f(v):
2kT
vmp  m
Ddrinking
Mean Free Path for Gas Molecules
r
r
v
  2r 
Effective collision area =
2
2r
 (2r ) 2
Assumptions:
• Molecule with finite radius r (note:
point-like particles do not collide)
• The number density (# per unit vol) is
given by N/V.
• Molecules move at an average speed v
(assuming this average to be vrms).
For simplicity, let assume that the redder shaded
molecule is the only one which is moving and to
the right with speed v.
Then within a time interval dt, the number of
molecules that it might collide will be given by:
 effective volume 
dn  
  number density 
 indicated (blue) 
   4r 2  vdt   N V  (we take v  vrms )
Thus, the number of collisions/unit time is,
dn 4 r 2 vN

dt
V
Mean Free Path for Gas Molecules
To take into account that other molecules besides the red one are also moving, the
estimated collision rate should be higher and it can be shown that this mean collision
rate will be larger by 2 . So,
dn 4 2r 2 vN

dt
V
note
The average time between collisions (mean free time) is then the reciprocal of this
value,
tmean 
V
4 2r 2 vN
And, the mean free path will be,
  vtmean 
V
4 2r 2 N
or
(we used PV  NkT )
kT

4 2r 2 P
Derivation of the
2
Factor
Key Idea: If all the molecules are moving, the relevant quantity for
consideration should be the relative velocity between a pair of molecules.
v1
v 2  v1
v2
v 2 v1
Again, we take the average of this relative velocity to be the root-mean-square
value:
v 
2
rel av
   v 2  v1  v 2  v1  av
  v 2  v 2  2 v 2  v1  v1  v1 av
  v 2  v 2 av  2  v 2  v1 av   v1  v1 av
Note: v1 and v 2 are in
random relative directions !
 v 2  v1 av  0
Derivation of the
2
Factor
2
This gives,  vrel
av   v 2  v 2 av  2  v 2  v1 av   v1  v1 av
  v22    v12   2  v 2 
av
av
av
Again, random direction assumption
implies these two averages to be the same !
Taking the square root of the above equation,
 vrel rms 
Thus,
v 
2
rel av
 2  v 2   2vrms
av
dn 4 r 2 2vN
dn 4 r 2 vN

should be written as

dt
V
dt
V
back
vrms and  Example: 18.6 & 18.8
18.6: a) What is the average translational KE of an ideal gas molecule at 27oC?
Kinetic Theory  KE per molecule 
3
3
kT  1.38  1023 J / K   273.15  27  K  6.211021 J
2
2
b) What is the root-mean-square speed vrms of O2 at this T?
vrms 
3kT
mO2
mO2  5.311026 kg (molecular mass)
3(1.38 1023 J / K )(300 K )

 484 m s
26
5.3110 kg
18.8: a) Evaluate the mean free path of an air molecule at 27oC.
1.38  1023 J / K   300 K 

kT
8


5.8

10
m

2
2

10
5
4 2r p 4 2  2.0  10 m  1.01 10 Pa 
 r  2 10
10
m
Dperfume
Heat Capacities of Gases (at constant V)
Using the Kinetic-Molecular model, one can
calculate heat capacity for an Ideal Gas!
 For point-like molecules (monoatomic gases),
molecular energy consists only of translational
kinetic energy Ktr
 We just learned that Ktr is directly proportional
to T.
 When an infinitesimal amount of heat dQ
enters the gas, dT increases, and dKtr increases
accordingly,
dKtr  3 NkdT (or  3 nRdT )
2
2
Heat Capacities of Gases

From definition of molar heat
capacity, we also know:
dQ  nCvdT

From energy conservation,
requiring dQ=dKtr (no work) gives,
nCvdT  3 nRdT
2
Monoatomic ~ Ideal Gas
(matches well with prediction)

Cv  3 R (12.47 J / mol  K )
2
Heat Capacity (diatomic)
A diatomic molecule can absorb energy in its translational
motion, its rotational motion and in its vibrational motions.
Equipartition of Energy
This principle states that each degree of freedom (“separate
mechanisms in storing energy”) will contribute (½ kT) to the total
average energy per molecule.

Monoatomic: 3 translational dofs  3 (½ kT)
This give Etot= 3/2 NkT (same as before).

Diatomic (without vibration): 3 trans dofs + 2 rotational dofs
This give Etot= 5/2 NkT or = 5/2 nRT.
Again, consider an infinitesimal energy change, we have
nCvdT  5 nRdT, and this gives Cv = 5/2 R.
2
Heat Capacities (real gases, e.g., H2)



At low T, only 3
translational dofs can
be activated
At higher T, additional
rotational dofs can be
activated
At higher T still,
vibrational dofs might
also get activated
Heat capacity for a H2 gas
 For normal T range, one take Cv 
5
R for H2 gas
2
Heat Capacities of Ideal Solids




Atoms are connected
together by springs
Assume harmonic motions
for these springs (Hooke’s
law)
For each spatial direction,
there are two dofs
(vibrational KE, vibrational
PE) and we have 3 dims
For the entire solid, we
have 3/2 kT (KE) + 3/2 kT (PE)
 Etot = 3kT
Cv = 3R
Dulong Petit Prediction and Improvements
Einstein: treating
atoms as QM
HMOs…
Cv  0
Debye: extension
by including
low-f phonons…
Cv  T 3