Chapter Two. Kinetic theory of gases
The kinetic theory of gases is of great importance for understanding the
physical vapor deposition. Although it is a classic model, its predictions
are remarkably accurate and it affords a physical picture of the
behavior of a gas.
The fundamental assumptions are:
 The gas contains an enormous number of classical particles.
 The particles are infinitely hard spheres.
 All collisions are elastic.
 There are no forces on the particles except during collisions with
each other or with the container.
2.1 The Boltzmann distribution
The distribution in energy of the particles (f) is given by the classical
Boltzmann distribution function:
f ( E )  exp
E
kT
II-1
where k is Boltzmann’s constant, T is the absolute temperature, and E is
the energy. It is suitable for identical, distinguishable (thus classical)
particles that do not obey the Pauli exclusion principle. The number of
particles having a kinetic energy between E and E+dE is therefore given
by dN=Nf(E)dE/kT, where N the total number of particles and 1/kT is
the necessary normalized factor.
dN/N is the ratio of the particle number in this region to the total
number. When dE is larger, the value of dN/N could be larger also, thus
dN/NdE. Actually the possibility for energy ranged from E1 to E2 is
given by
dN
1 E2
1 E2
E



f
E
dE

exp
 dE
N
kT E1
kT E1
kT
II-2
Obviously for the whole range (E from 0 to ), this ratio must be 1:
1 
f E dE  1
II-3
kT 0
The Boltzmann factor gives the fraction of the total number of particles
in the system whose energy is greater than or equal to some specified
energy Ea. It is calculated from

  E  dE
 E 
II-4
Ea exp  kT  kT  exp  kTa  .
In our theory the energy of a gas particle is entirely translational kinetic
energy, given by
2
2
2
mv 2 m(v x  v y  v z )
E

2
2
II-5
It is frequently useful to work with the subgroup of particles that have
essentially the same velocity vector. Consider those particles occupying
a certain differential volume of `velocity space’. A probability
distribution function may be constructed from the Boltzmann
distribution. It is based the probability of having the total energy
associated with the velocity v  (vx , v y , vz ) :
F (v x , v y , vz ) 

   exp  m(v

 v ) / 2kT dv dv dv
exp  m(v x2  v 2y  vz2 ) / 2kT



  
2
x
 v 2y
2
z
x
y
.
II-6
z
This is the probability of occupancy of a differential volume of velocity
space (dvxdvydvz), per unit volume of that space, by a gas particle. The
probability is F (v x , v y , v z ) dv x dv y dv z . The dominator, a normalizing factor, is
(2kT/m)3/2.
Example 2.1
Find the height distribution of air particles.
Suppose the number of air particles per unit volume at z=0 is n00, then
the number in dV=dxdydz at a height z can be expressed as
dN=n00exp(-mgz/kT) dxdydz.
Thus the number of particles per unit volume at z is
n0= dN /dV=n00exp(-mgz/kT)
2.2 Characteristic particle speeds
Speed is the magnitude of the velocity vector. To gain the probability
distribution function for speed F(v), we use a spherical coordinate. The
differential volume in velocity space, instead of an infinitesimal cube,
2
2
2
becomes that of a thin spherical shell given by 4v2dv; vx  v y  vz
becomes simply v2. The new expression is
F (v ) 

exp  mv 2 / 2kT


  

  



exp  mv 2 / 2kT v 2 sin ddv
.
II-7
The probability itself is 4 v2F(v)dv. The dominator is, again, (2 kT/m)3/2.
There are several characteristic speeds:
The most probable speed (vp) is the speed for which d[4v2F(v)]/dv is zero
(ignoring the v=0 case):
1/ 2
 2kT 
vp  

 m 
.
II-8
3/ 2
2
2
The average speed is equal to m / 2kT  0 v  exp  mv / 2kT  4v dv

1/ 2
 8kT 
vav  

 m 
.
II-9
The root-mean-square speed (vrms) is found from the square root of
m / 2kT 3 / 2 0 v 2  exp  mv2 / 2kT  4v 2dv :

1/ 2
vrms
 3kT 


 m 
.
II-10
The average kinetic energy (Eav) of the particle is
2
mvrms
3kT
Eav 

.
2
2
II-11
It is always true for the Boltzmann distribution that vp<vav<vrms.
Example 2.2
Calculate the rms speed and the average kinetic energy of a nitrogen
molecule at room temperature.
1/ 2
 3  (1.38  10 23 J / K )  298K 
vrms  
  515m / s
 27
28

1
.
66

10
kg


28  1.66  1027 kg  (515m / s )2
Eav 
 6  10 21 J  0.04eV .
2
2.3 The ideal gas law
The ideal gas law follows directly from the kinetic theory of gases. We
can derive it by compute the pressure as the rate of transfer of
momentum, per unit area, to the wall of the container.
The pressure due to particles with velocity component between vy and
vy+dvy is equal to the number of particles striking the wall per unit area,
times momentum transferred per collision, per unit time:
dP(v y ) 
nFy dv y  v yt  2mv y
t
II-12
The total pressure then is
P  2mn 

0
1/ 2
 m 


 2kT 
  mv 2y  2
v y dv y  nkT
exp 

2
kT


II-13
It relates the thermodynamic state variables P and T (through n).
The more popular form of the ideal gas law is
M
PV
RT
 C or PV 

T
II-14
where V is the volume of gas, T is absolute temperature (unit: K), C is a
constant, M the mass of gas (unit: gram),  the mass of gas molecular,
and R is the Molar gas constant:
R=8.31441 Jmol-1K-1.
For mixture of gases
The partial pressure of a mixture of gases may be calculated by
separating the preceding integrals into individual integrals for each
gaseous specie. The total pressure is the sum of the partial pressures.
P   Pn
n
II-15
Thus
M
M
M
M 
PV  ( Pn )V   1  2  3  ...  n  RT
 2 3
n 
n
 1
Example 2.3
The height dependence of the pressure.
From the ideal gas law (Eq. II-13),
P=n0kT
II-16
According to the example 2.1, we have
P=n00kTexp(-mgz/kT)
=P0exp(-mgz/kT)
=P0exp(-gz/RT),
where P0= n00kT is the pressure at the sea level (z=0),  is the mass of air.
Mean free path
The mean free path () is the distance a particle travels, on average,
before experiencing a collision. To estimate , we can calculate the
survival probability Q(x) that a given particle travels a distance x, from
the point of previous collision to the next collision. 1-Q(x) is the
probability to have a collision after traveling a distance x. Q decreases
with x according to the following expression for the probability (d[1Q(x)]/dx) of particles experiencing a collision while traveling between x
and x+dx:
 dQ
 QPc
dx
II-17
where Pc is the average number of collisions per unit distance traveled,
a constant. Thus we have
Q(x)=exp(-Pcx).
II-18
The mean free path is found by calculate the average distance traveled:


1
   x ( dQ / dx )dx   xPc exp(  Pc x )dx  .
II-19
0
0
Pc
Thus the mean free path is the reciprocal of the probability of a
collision per unit distance traveled.
The collision section
We use the assumption that the particles are hard sphere of diameter d.
Therefore a particle’s collision section is d2. On average, a particle
sweeps out a volume of space at the rate given by d2vav and thus
encounters other particles at a rate given by d2vavn.
Here we assumed the target particles are stationary. But this is not true
and we should use the average relative speed of particles, e.g.
Hence the correct collision rate r is
r = 2  vav d2n
2  vav .
II-20
On the other hand, the collision rate is also the average speed multiplied
by the collision probability per unit distance traveled:
r=vavPc
II-21
Thus
Pc=
2 d
2
n
II-22
1
2d 2n
II-23
and

The mean free path is typically determined
experimentally from viscosity
measurements. Use eq.II-23, the effective
particle diameter can be found. It is in the
range 2-5 Å for most common gases.
d
Example 2.3
Estimate the mean free path in N2 gas at one atmosphere and room
temperature.
The density under these conditions, from ideal gas law, is
n

760torr
25  3

2
.
46

10
m
1.38  10 23 J / K  298K
1
2
o

2   3.75    2.46  1025 m 3


o
 650 
Pressure
Suppose the gas container with a volume V contains N molecular of
mass m. Thus the number per unit volume is n0=N/V. To simplify the
discussion, we separate the gas into several groups, in which the gas
particles have the same speed with the same direction.
n0   n0i
II-24
i
Consider the pressure on a small area dA perpendicular to the x-axis. A
particle with a speed vi collision with dA. Since the collision is fully
elastic, vix changes to –vix while viy and viz keep unchanged. The variation
of the momentum is –2mvix, which is the impulse within a period dt, the
number of particles can reach dA is n0ivixdtdA. The total impulse of gas
particles with speed vi on dA:
2n0imvix2dtdA.
Thus the total impulse of all gas particles on dA is
dK 
2n



0i
v ix  0
mvix2 dAdt .
II-25
The condition for vix0 is necessary as the particle with vix<0 will not
reach dA. In average, the particle with vix0 should be a half of the total
particles, therefore we have
dK 
n



v ix  0
2
mv
0i
ix dAdt
II-26
It shows the interaction of gas to dA in time dt.
dK  PdAdt 
2
n
mv
 0i ixdAdt
vix  0 
II-27
The pressure P is then:
P
1
n0imvix2 dAdt   mn0i vix2

dAdt vix  0 
vix  0 
II-28
There are various systems for the pressure units, including bar, torr, Pa,
psi (lb/ft2), kg/cm2, etc.
2.4 Adsorption and condensation
The most important factors in physical vapor deposition are vapors and
gases. Indeed, the key of PVD is to produce a vapor that will condense
on a solid surface (substrate).
For vapor creation, if the condensed phase is a liquid, we speak of
evaporation; is a solid, of sublimation. The two together are referred to
as vaporization and the reverse, condensation.
Adsorption of gases
Gas particles adsorb because there are stable positions for them on
surface. The adsorption energy, EdA, is the depth of the potential well
with respect to infinite separation of the molecule from the surface.
If Ed is less than ~0.4 eV, the particle (whether a diatomic molecule or
something else) is said to be in a state of physisorption, which is an
energy range typical of van der Waals bonding. The noble gases
physisorb on all surfaces; a molecule probably also physisorb if it is able
to adsorb without dissociating. If Ed is large than ~1eV, the particle is
said to be chemisorbed, which is a state of relatively strong chemical
bonding. Single oxygen atoms chemisorb on virtually all solid surfaces.
Mean residence time
The mean residence time indicates how long will an individual particle
stay adsorbed. It can be estimated as the inverse of its desorption rate:

1
  exp  Ed / kT  ,
II-29
where  is a frequency of vibration perpendicular to the surface, almost
always assumed to be ~1013 Hz.
Example 2.4
A particle physisorbed with a desorption energy of 0.4 eV has a mean
residence time at a room temperature:
1
  13
 0.58s
10 Hz  exp  0.4eV /{( 8.62  10 5 eV / K )  298K }
A desorption energy of 1.2 eV gives a room temperature  of 1.9107 s;
at 500 K this  is reduced to 0.12 s.
Condensation of vapors
A film will condense on a substrate when there exists a supersaturated
vapor above the substrate. For film deposition of pure elements by
condensation from a vapor, there is a critical incident flux at the
substrate below which no film accumulates. The value of this critical
flux increases with the substrate temperature. The following principles
are illustrated:
 For a given substrate temperature, there is critical incident
flux above which a film will form, but below which no
deposition is obtained.
 The greater the substrate temperature, the greater the
critical incident flux.
For compounds whose vapors consists of particles having the
stoichiometric composition, stoichiometric films may be obtained by
direct vaporization of these compounds. An important thing is to obtain
a stoichiometric vapor, which is technique dependent.