Chapter 33 Kinetic Theory of Gases
33.1 Kinetic Theory of Gas Motion and Pressure
33.2 Velocity Distribution in One Dimension
33.3 The Maxwell Distribution of Molecular Speeds
33.4 Comparative Values for Speed Distributions:
vmp, vave, and vrms
33.5 Gas Effusion
33.6 Molecular Collisions
33.7 The Mean Free Path
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.1 Kinetic Theory of Gas Motion and Pressure
General Chemistry
Kinetic Molecular Theory (KMT): Explain ideal gas behavior.
1.
The particles are so small compared with the distances between them that the
volume of the individual particles can be assumed to be negligible (zero).
2.
The particles are in constant motion. The collisions of the particles with the
walls of the container are the cause of the pressure exerted by the gas.
3.
The particles are assumed to exert NO forces on each other; they are assumed
neither to attract nor to repel each other.
4.
The average kinetic energy of a collection of gas particles is assumed to be
directly proportional to the Kelvin temperature of the gas.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.1 Kinetic Theory of Gas Motion and Pressure
Gas kinetic theory
(1) The distance between particles is very large in comparison to their size.
Example: One mole argon at 298 K and 1 atm.
PV = nRT
V = 24.4 L = 0.0244 m3
NA = 6.02 x 1023 particles
dAr = 2.9 Å = 0.29 nm
VAr =
4 3
πrAr
3
= 0.013 nm3
Free space/particle = 40.5 nm3
L free speace = 3.43 nm
Diameter/distance ~ 1/12
Volume/free space ~ 0.03%
The distance between particles in the gas phase is substantially far away.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.1 Kinetic Theory of Gas Motion and Pressure
Gas kinetic theory
(2) The pressure (exerted by a gas on the container confining the gas) arises from
collisions of gas particles with the container walls.
(3) The collisions with the wall are elastic such that the translational energy of the
particle is conserved.
-- Each particle travels through space as a separate, unperturbed entity until a
collision occurs with another particle or with the walls of the container confining
the gas.
-- Particle motion is described using Newton’s law of motion.
The energy spacing between translational states is very small relative to kT such
that a classical description of translational motion is appropriate.
-- Linear momentum is imparted to the wall, which results in pressure.
(a hallmark 標記 of kinetic theory)
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
The Macroscopic Variables - PVT
** The total momentum change ∝ Pressure
∆pTotal = (∆p molecule )× ( number of molecules )
** Collisional volume:
V = A×
×Δx
** For a given velocity vx at a given time period Δt:
Δx = vx×Δt
~
** The number density of the particles within the container is N = nNA/V.
The number of particle collisions to the container wall in the time interval Δt:
Directionality of particle motion
** Δptotal of the container wall imparted by particle collisions:
∆pTotal = (∆p molecule )× ( number of molecules )
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
Average
value
1.2
The Macroscopic Variables - PVT
Total momentum change:
** In general chemistry: Gas law for ideal gases:
½ m<vx2> = ½ kT
nN A kT
P =
V
(empirical)
(From Statistic thermodynamics, P. 898: ½ m<vx2> = ½ kT)
(4) The root-mean-squared speed of the gas particles increases as the square
root of temperature and decreases as the square root of the particle mass.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
1
3
2
K .E. = m v = kT
2
2
33.2 Velocity Distribution in One Dimension
** Velocity distribution function – The probability of a gas particle having a
velocity within a given range.
-- Energy-level spacings of translation are small.
Continuous distribution.
** Assume the speed distribution function as, Ω(vx, vy, vz) = f(vx)f(vy)f(vz).
** The velocity distribution along a
single direction,
Appendix A: the chain rule
v = (vx2 + v y2 + v z2 )
y = f(u) and u = g(x)
df ( u ) df ( u ) ∂u
=
dxof Applied du
∂xNational University of Kaohsiung
Grace H. Ho, Department
Chemistry,
12
⇒
vx
∂v
=
∂vx
v x2 + v 2y + v z2
(
)
12
vx
=
v
33.2 Velocity Distribution in One Dimension
** The velocity distributions along each direction are equivalent:
d ln Ω ( v ) d ln f ( v x ) d ln f ( v y ) d ln f ( v z ) d ln f ( v j )
=
=
=
=
= −γ
vdv
vx dvx
v y dv y
v x dvz
v j dv j
-- γ is a constant and a positive quantity to ensure that f(vx,y,z) does not diverge
as vx,y,z ∞.
∫ d ln f ( v
j
) = − ∫ γv j dv j
1 2
⇒ ln f ( v j ) = − γv j
2
 1 2
∫ d ln f (v j ) = ∫ d  − γv j 
 2

⇒ f (v j ) = Ae
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
− γv 2j / 2
Normalization
constant
33.2 Velocity Distribution in One Dimension
** Normalization:
f (v j ) = Ae
−γv 2j / 2
∞
= 2A ∫ e
-( γ/2)v 2j
0
12
 γ 

A= 
 2π 
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
∞ − ax 2
dx
∫0 e
π
dv j = 2A
 2γ
12
π 
= 
 4a 
1/ 2



 2π
= A
 γ
1/ 2



33.2 Velocity Distribution in One Dimension
** Evaluate the value of γ:
Gas kinetic theory:
γ 1
π
1
=
=
3
2π 2 (γ 2)
γ
Maxwell – Boltzmann velocity distribution in one dimension:
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.2 Velocity Distribution in One Dimension
33.1
Average velocity = 0
Average speed?
(Check Appendix A)
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.2 Velocity Distribution in One Dimension
298 K
Ar
One dimensional
speed distribution:
** The width of the distribution increases with temperature.
= The increasing probability of higher energy translational states.
** The velocity distribution is mass dependent with a given temp. : Kr < Ar < Ne.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.3 The Maxwell Distribution of Molecular Speeds
** The relation between particle speed, v, and the one-dimensional velocity:
v = (v x2 + v y2 + v z2 )1/ 2
** Particle speed distribution F(v) is defined as follows:
Velocity
Space:
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
2
4πv dv
33.3 The Maxwell Distribution of Molecular Speeds
** Maxwell speed distribution
-----
K. E.
= ½ m <v2>
= 3RT/2
v2 dependency.
The range of the speed distribution is from 0 to infinity.
The speed distribution is not symmetric.
The v2 term dominates the distribution curve at low speeds, and the exponential
term is important at high speed region.
-- The maximum of the distribution shifts to higher speed as temperature increases.
-- The curvature of the distribution changes, and the change is pronounced on the
high speed side.
An increase in kT will increase the probability of occupying higher energy
translational states.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.3 The Maxwell Distribution of Molecular Speeds
Experimental apparatus to
verify the Maxwell speed
distribution by Miller and
Kusch [Phys. Rev. 99
(1955), 1314].
Experimental results:
Potassium at 466±2 K.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.4 Comparative Values for Speed Distribution
** The most probable speed, vmp:
= 0
P. 933, What is the most probable speed for Ne and Kr at 298 K?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.4 Comparative Values for Speed Distribution
** The average speed, vave:
∫
∞
0
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
x
2 n +1 − ax 2
e
n!
dx = n +1
2a
33.4 Comparative Values for Speed Distribution
** The root-mean-squared speed, vrms:
< v 2 >= ∫
∞
0
3/ 2
3/ 2


2
m
m



 ∞ 4 −mv 2 / 2 kT
2
2 − mv / 2 kT 
v 4π 
ve
dv = 4π 
dv
 ∫0 v e
  2πkT 

 2πkT 


 m 
= 4π 

 2πkT 
3/ 2
3  2kT 

3 
2  m 
2
π 2kT
m
= 3kT / m
∫
∞
0
2 n − ax 2
x e
(2n − 1)!! π
dx = n +1 n
2 a
a
n!!
n(n-2)…5．3．1, n>0 odd
n(n-2)…6．5．2, n>0 even
-1!! = 0!! =1 by definition
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.4 Comparative Values for Speed Distribution
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.5 Gas Effusion
** Gas effusion:
-- Some finite pressure within a container is separated from a vacuum by a thin wall.
-- The pressure of the gas and size of the aperture is such that molecules do not
undergo collisions near or when passing through the aperture.
-- Produced a “beam” of gas particles.
Application: molecular beam techniques to study chemical dynamics.
** Graham’s Law (1831):
Early experimental observation by
Thomas Graham:
Effusion rate for gas 1
Effusion rate for gas 2
=
√ M2
√ M1
Graham’s law of effusion:
The rate of effusion is inversely proportional to the square root of the
mass of its particles.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.5 Gas Effusion
** Effusion rate, dNc/dt:
-- Proportional to the area being struck, A.
-- Proportional to particle speed.
~
-- Proportional to the particle density, N.
~
∞ N ( Av dt ) × f ( v )dv
∞
~
dN c
x
x
x
=∫
= NA∫ v x f ( vx )dv x
0
0
dt
dt
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
∫
∞
0
x
2 n +1 − ax 2
e
n!
dx = n +1
2a
33.5 Gas Effusion
Effusion rate:
** Collisional flux, Zc –
The number of collisions
per unit time and
per unit area.
-- Particle density:
1
= N • vave
4
)
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.5 Gas Effusion
** Effusion to vacuum
Pressure drop as a function of time.
N: number of particles in the container
--
1
2
P
t  A  kT 
dP
∫P0 P = ∫P0 d ln P = − ∫0  V  2πm  dt
P
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
Example 33.5 (P. 936)
33.6 Molecular Collisions
** Simplified model = Hard-sphere collisions (no intermolecular interaction)
Assumption:
• Collision cross section = σ = π(r1 + r2)2
• The particle of interest is
moving and that all other
molecules are stationary.
(A concept of “relative”)
-- Effective speed, #1 speed with respect to #2’s
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
<v12> = <v1> - <v2>
33.6 Molecular Collisions
< v12 >=
** The number of collisional partner = (N2/V) × Vcyl
8kT
πµ
~
= N2σ<v12>dt
m1m2
µ=
m1 + m2
** Particle collisional frequency, z12: The number of collisions an
individual molecules (#1) undergoes with other collisional partners (#2) per unit
time (dt).
N 2  Vcyl

z12 =
V  dt
• number density
• cross section
• relative velocity
~
-- If
~
z1 X = N X σ 1 X vavg ,rel = N X σ 1 X
 N  σ < v12 > dt  N 2
8kT
 = 2 
=
σ

V
dt
πµ

 V

#1 = #2, μ = m1/2
8 RT
πµ
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.6 Molecular Collisions
~
z1 X = N X σ 1 X
8 RT
πµ
** One CO2 molecule undergoes ~7x109 collisions per second at 298 K and 1 atm.
** The time between molecular collisions ~ 150 x10-12 second.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.6 Molecular Collisions
** Total collisional frequency: Total number of collisions that occurs for all gas
particles. = Total number of collisions per unit volume.
collisions m-3 s-1
π(r1+r2)2
1A-1B = 1B -1A
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
Collisional parameters
for various gases
33.6 Molecular Collisions
(NAr/V) =
(NKr/V) =
0.040 × 0.084
0.040 + 0.084
= 0.027 kg mol-1
µ=
8RT/πµ = 484 m s -1
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.7 The Mean Free Path
** Mean free path – The average distance a gas particle travels between successive
collisions.
-- In a given time dt, the distance a particles will travel = vave dt
-- In a given time dt, the overall number of collisions for a particle = (z11 + z12)dt
-- The mean free path = the average distance traveled / the number of collisions:
For a container with two types of particles:
A gas with one type of particle, N2 = 0:
kT
kT
=
=
2
2σP
2πd P
-- The higher the number density (higher the pressure)
shorter λ .
-- The greater the particle size (higher collision probability)
shorter λ.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.7 The Mean Free Path
Example: Ar at a pressure of 1 atm and temperature of 298 K, (1)λ = ?
σ = 0.36 nm2 = 3.6 x 10-19 m2
 RT  1
kT

=
= 8 ×10 −8 m = 80 nm
λ = 
2σPAr
 PAr N A  2σ
-- dAr = 0.29 nm
cf.
λAr = 80 nm
-- An Ar atom travels an average distance ~ 275 times dAr between collisions.
KMT: The distance between particles is very large in comparison to their size.
** The time scale of collision = 1/ number of collisions per second = ?
1
z Ar − Ar
=
λ
vave
8 × 10 −8 m
−10
=
=
2
×
10
s
−1
397 m s
** Collisional frequency = 1/time scale of collision = vave/λ
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
33.7 The Mean Free Path
Exercise
a. What is z11 (single-particle collisional frequency) of air under 1.0 atm and 298K?
12
 101325 × N A 
 8 × 8.314 × 298 
−18
z11 = 
×
0
.
43
×
10
×
2
×



π
8
.
314
×
298
×
0
.
0288




= 7.0 ×109 s −1
b. At the tropopause (11 km in altitude), z11 = 3.16x109 s-1. T = 220 K, P=?
P
 1 atm 
9
7.0 × 10 : 
 = 3.16 ×10 :
220
 298 
9
Ptropopause ~ 0.38 atm
c. Mean free path of air at the tropopause = ?
 8RT 
-1
9 -1
−7
z
=
402
m
s
/(
3
.
16
×
10
s
)
=
1
.
27
×
10
m
= 
 11
 πM 
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
台大-101
中山-100
成大-100
7. Show that l Latm ~ 10l J
中山-99
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung