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Lecture 2/3.
Kinetic Molecular Theory of
Ideal Gases
Last Lecture ….
IGL is a purely empirical law - solely the
consequence of experimental
observations
Explains the behavior of gases over a
limited range of conditions.
IGL provides a macroscopic explanation.
Says nothing about the microscopic
behavior of the atoms or molecules that
make up the gas.
1
Today ….the Kinetic Molecular Theory (KMT) of gases.
KMTG starts with a set of assumptions
about the microscopic behavior of matter
at the atomic level.
KMTG Supposes that the constituent
particles (atoms) of the gas obey the laws
of classical physics.
Accounts for the random behavior of the
particles with statistics, thereby
establishing a new branch of physics statistical mechanics.
Offers an explanation of the macroscopic
behavior of gases.
Predicts experimental phenomena that
suggest new experimental work (MaxwellBoltzmann Speed Distribution).
Kotz, Section 11.6, pp.532-537
Chemistry3, 1st edition: Section 7.4, pp.316-319; Section 7.5, pp.319-323.
2nd edition: Section 8.4, pp.354-357, section 8.5,
pp. 358-368.
Kinetic Molecular Theory (KMT) of Ideal Gas
•
•
•
•
•
•
•
Gas sample composed of a large number of
molecules (> 1023) in continuous random
motion.
Distance between molecules large
compared with molecular size, i.e. gas is
dilute.
Gas molecules represented as point
masses: hence are of very small volume so
volume of an individual gas molecule can
be neglected.
Intermolecular forces (both attractive
and repulsive) are neglected. Molecules do
not influence one another except during
collisions. Hence the potential energy of
the gas molecules is neglected and we only
consider the kinetic energy (that arising
from molecular motion) of the molecules.
Intermolecular collisions and collisions
with the container walls are assumed to
be elastic.
The dynamic behaviour of gas molecules
may be described in terms of classical
Newtonian mechanics.
The average kinetic energy of the
molecules is proportional to the absolute
temperature of the gas. This statement in
fact serves as a definition of
temperature. At any given temperature
the molecules of all gases have the same
average kinetic energy.
Air at normal conditions:
~ 2.7x1019 molecules in 1 cm3 of air
Size of the molecules ~ (2-3)x10-10 m,
Distance between the molecules ~ 3x10-9 m
The average speed - 500 m/s
The mean free path - 10-7 m (0.1 micron)
The number of collisions in 1 second - 5x109
2
Random trajectory of
individual gas molecule.
Assembly of ca. 1023 gas molecules
Exhibit distribution of speeds.
Gas pressure derived from KMT analysis.
Pressure
The pressure of a gas can be explained by KMT
as arising from the force exerted by gas molecules
impacting on the walls of a container (assumed to be
a cube of side length L and hence of Volume L3).
We consider a gas of N molecules each of mass m
contained in cube of volume V = L3.
When gas molecule collides (with speed vx) with wall of
the container perpendicular to x co-ordinate axis and
bounces off in the opposite direction with the same speed
(an elastic collision) then the momentum lost by the particle and
gained by the wall is px.
p x  mvx  ( mvx )  2mvx
t 
2L
vx
y
The force F due to the particle can then be computed
as the rate of change of momentum wrt time (Newtons Second
Law).
F
L
z
The particle impacts the wall once every 2L/vx time units.
Derivation:
Box 8.2 pp.356-357.
x
vx
-vx
p 2mvx mv x2


t 2 L v x
L
3
Force acting on the wall from all N molecules can be computed
by summing forces arising from each individual molecule j.
F
m N 2
 vx, j
L j 1
The magnitude of the velocity v of any particle j can also be
calculated from the relevant velocity components vx, vy, and vz.
v 2  v x2  v 2y  v z2
The total force F acting on all six walls can therefore be
computed by adding the contributions from each direction.
F 2
N
N

m N 2
m N 2
m N 2
2
2
2
2
 v x , j   v y , j   v z , j   2  v x , j  v y , j  v z , j  2  v j
L  j 1
L j 1
L j 1
j 1
j 1



Assuming that a large number of particles are moving randomly then the force on each of
the walls will be approximately the same.
F
1  m N 2 1 m N 2
2  v j  
vj
6  L j 1  3 L j 1
2
v 2  vrms

The force can also be expressed in terms of the average velocity
v2rms where vrms denotes the root mean square velocity of the
collection of particles.
F
1
N
N
v
j 1
2
j
2
Nmvrms
3L
The pressure can be readily determined once the force is known using the definition
P = F/A where A denotes the area of the wall over which the force is exerted.
P
2
2
F Nmvrms
Nmvrms


A
3 AL
3V
AL  V
The fundamental KMT result for the gas pressure P can then be stated in a number
of equivalent ways involving the gas density , the amount n and the molar mass M.

PV 
1
2  v 2  1  M  2
1 N  2
1
2
2
Nmvrms
  Mvrms
 nMvrms
  rms   N  vrms
3
3  2  3  NA 
3  NA 
3
m
Avogadro Number
= 6 x 1023 mol-1
Nm
V
M
NA
Using the KMT result and the IGEOS we can derive a
Fundamental expression for the root mean square
Velocity vrms of a gas molecule.
PV  nRT
IGEOS
KMT result
PV 
1
1
2
2
Nmvrms
 nMvrms
3
3
1
2
nMvrms
 nRT
3
2
Mvrms
 3RT
vrms 
3RT
M
4
Gas
103 M/kg mol-1
Vrms/ms-1
H2
2.0158
1930
H2O
18.0158
640
N2
28.02
515
O2
32.00
480
CO2
44.01
410
Internal energy of an ideal gas
We now derive two important results.
The first is that the gas pressure P is proportional to the average
kinetic energy of the gas molecules.
The second is that the internal energy U of the gas, i.e. the mean
kinetic energy of translation (motion) of the molecules is directly
proportional to the temperature T of the gas.
This serves as the molecular definition of temperature.
Average kinetic
Energy of gas
molecule
E 
P
Nmv
3V
2
rms

2N 1 2  2N
E
 mvrms  
3V  2
 3V
Boltzmann Constant
n
P
nRT NRT Nk BT


V
N AV
V
1 2
mvrms
2
N
NA
kB 
R 8.314 J mol1 K 1

 1.38x1023J K 1
6.02x1023 mol1
NA
nR  Nk B
Nk BT
2N
E 
V
3V
3
E  k BT
2
3
3
U  N E  Nk BT  nRT
2
2
5
Maxwell-Boltzmann velocity distribution
•
•
•
•
In a real gas sample at a given temperature T, all
molecules do not travel at the same speed. Some
move more rapidly than others.
We can ask : what is the distribution (spread) of
molecular velocities in a gas sample ? In a real
gas the speeds of individual molecules span wide
ranges with constant collisions continually
changing the molecular speeds.
Maxwell and independently Boltzmann analysed
the molecular speed distribution (and hence
energy distribution) in an ideal gas, and derived a
mathematical expression for the speed (or
energy) distribution f(v) and f(E).
This formula enables one to calculate various
statistically relevant quantities such as the
average velocity (and hence energy) of a gas
sample, the rms velocity, and the most probable
velocity of a molecule in a gas sample at a given
temperature T.
James Maxwell
1831-1879
3/ 2
 m 
F (v)  4 v 2 

 2 k BT 
F (E)  2
E
  k BT 
3
 mv 2 
exp  

 2k BT 
 E 
exp  

 k BT 
http://en.wikipedia.org/wiki/Maxwell_speed_distribution
http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution
Maxwell-Boltzmann velocity
Distribution function
 m 
F (v)  4 v 

 2 k BT 
3/ 2
2
m = particle mass (kg)
kB = Boltzmann constant
= 1.38 x 10-23 J K-1
Gas molecules exhibit
a spread or distribution
of speeds.
F (v)
 mv 2 
exp 

 2 k BT 
Ludwig Boltzmann
1844-1906
• The velocity distribution curve
has a very characteristic shape.
• A small fraction of molecules
move with very low speeds, a
small fraction move with very
high speeds, and the vast majority
of molecules move at intermediate
speeds.
• The bell shaped curve is called a
Gaussian curve and the molecular
speeds in an ideal gas sample are
Gaussian distributed.
• The shape of the Gaussian distribution
curve changes as the temperature
is raised.
• The maximum of the curve shifts to
higher speeds with increasing
temperature, and the curve becomes
broader as the temperature
increases.
•
A greater proportion of the
gas molecules have high speeds
at high temperature than at
low temperature.
v
6
Properties of the
Maxwell-Boltzmann
Speed Distribution.
Maxwell Boltzmann (MB) Velocity Distribution
vmax = vP
0.0025
T = 300 K
T = 400 K
T = 500 K
T = 600 K
T = 700 K
T = 800 K
T = 900 K
T = 1000K
0.0020
Ar M = 39.95 kg mol-1
F(v)
0.0015
0.0010
vP (300K) = 353.36 ms-1
vrms (300K) = 432.78 ms-1
‹v› (300K) = 398.74 ms-1
0.0005
0.0000
0
200
400
600
800
1000
v / ms-1
Features to note:
The most probable speed is at the peak of the curve.
The most probable speed increases as the temperature increases.
The distribution broadens as the temperature increases.
1200
1400
1600
1800
Relative mean speed
(speed at which one molecule
approaches another.
vrel  2 v
7
MB Velocity Distribution Curves : Effect of Molar Mass
T = 300 K
0.005
He
Ne
Ar
Xe
0.004
M = 4.0 kg/mol
M = 20.18 kg/mol
M = 39.95 kg/mol
M = 131.29 kg/mol
F(v)
0.003
0.002
0.001
0.000
0
500
1000
1500
2000
v / ms-1
Determining useful statistical quantities from MB
Distribution function.
Average velocity of a gas molecule

v   v F v  dv
MB distribution of velocities
enables us to statistically
estimate the spread of
molecular velocities in a gas
0
 m 
F (v)  4 v 2 

 2 k BT 
3/ 2
 mv 2 
exp 

 2 k BT 
Maxwell-Boltzmann velocity
Distribution function
Some maths !
v 
Most probable speed, vmax or vP derived from differentiating
the MB distribution function and setting the result equal to
zero, i.e. v = vmax when dF(v)/dv = 0.
vmax  vP 
Derivation of these formulae
Requires knowledge of Gaussian
Integrals.
8k BT
8 RT

m
M
Mass of
molecule
Molar mass
vP  v  vrms
2 k BT
2 RT

m
M
Root mean square speed
Yet more maths!
1/ 2


vrms   v 2 F (v)dv 
0


3k BT
3RT

m
M
8
vP  v  vrms
Gas
103 M/
kg mol-1
vrms/ms-1
‹v›/ms-1
Vrel/ms-1
vP/ms-1
H2
2.0158
1930
1775
2510
1570
H2 O
18.0158
640
594
840
526
N2
28.02
515
476
673
421
O2
32.00
480
446
630
389
CO2
44.01
410
380
537
332
2200
Typical molecular velocities
Extracted from MB distribution
At 300 K for common gases.
2000
H2
1800
1400
v 
1200
1000
8k BT
8 RT

m
M
H2O
800
N2
O2
600
vmax  vP 
CO2
400
2 k BT
2 RT

m
M
200
0
10
20
30
3
10 M/kg mol
40
50
-1
vrms 
vrel  2 v
3k BT
3RT

m
M
Maxwell Boltzmann Energy Distribution
0.00025
T = 300 K
T = 400 K
T = 500 K
T = 600 K
T = 700 K
T = 800 K
T = 900 K
T = 1000 K
Ar
0.00020
0.00015
F(E)
vrms/ms
-1
1600
0.00010
0.00005
0.00000
0
2000
4000
6000
E /J
F (E) 

2

E   E F ( E ) dE 
0

E 

k
 BT 
k BT 3 / 2 E1/ 2 exp 
2



E 
3
 dE  k BT
k
2
 BT 
k BT 3 / 2  E 3 / 2 exp 
0
9

T
v

L

2L
v
Apparatus used to measure gas molecular speed distribution.
Rotating sector method.
Chemistry3
1st edition: Box 7.3, pp.322-323.
2nd edition: Box 8.3, pp.360-361.
10
Effusion and Diffusion of Gases.
Effusion is the process by which gas molecules pass through a small hole such as a pore in a membrane.
Diffusion occurs when two or more gases come together and mix. Diffusion arises due to the presence
of a concentration gradient.
Grahame’s Law of effusion:
At a given temperature and Pressure the rate of effusion (number of molecules
Passing through hole per second) is inversely proportional to square root of molar mass M of gas.
Reffusion 
K
M
K = constant
For a mixture of two gases A and B
With molar masses MA and MB the
Relative effusion rate is given by:

RA

RB
MB
MA
We conclude that gases with different molar masses
Will effuse at different rates.
A gas with a low molar mass (e.g. He) will effuse
Faster than a gas with a higher molar mass (e.g. N2).
Nitrogen passes through membrane 2.6 times more
Slowly than helium.

1
MN2
RHe
28gmol


 7  2.6
RN2
MHe
4gmol1
Gas mixing takes time. Rate of diffusion given
by Diffusion coefficient D (units: cm2s-1). Mean
distance L travelled by a diffusing molecule is:
L  6 Dt
Molecular collisions in gases.
Chemical reactions occur when molecules collide
With each other.
We can use KTG to estimate collision frequency in gases.
Molecular size will affect probability of collision.
2 molecules A,B will collide if the center of one (molecule B)
Comes within a distance of two radii – one diameter- of
Molecule A. This area is the target presented by A.
The area of the target for molecule B to hit
Molecule A is a circle of area s called the
Collision Cross Section .
The collision cross section is related to
Molecular size.
  d2
11
Collision frequency & mean free path.
Collision frequency Z = mean number of collisions that a molecule undergoes per second.
Number of collisions per second depends on distance travelled, and number of molecules
per unit volume (concentration). This is the pressure.
In a given time period, a molecule will collide with any other molecule whose centre lies
within a cylinder with cross sectional area .
Using Kinetic theory we can show that the collision frequency is:
Z  2 N v   P RT 
Knowing average speed that molecules move and number of collisions they undergo, the
distance travelled between collisions can be determined. This is called the mean free path .

v
Z

v
2 N v   P RT 

RT
2 N P
Example: N2 gas at SATP. Collision Cross Section  = 0.43 nm2. Mean speed <v> = 475 ms-1.
Collision Frequency Z:
Z = 21/2 x(6.02x1023 mol-1)x(475 ms-1)x(0.43 x 10-18 m2)x{(1x105Pa)/(8.314JK-1mol-1)x(298K)}
Hence collision frequency Z = 7.0 x 109 s-1.
Mean free path:  = RT/21/2 NP,
Hence  ={ (8.314 Jmol-1K-1)x (298K)}/{21/2x(6.02x1023mol-1)x(0.43x10-18m2x(1x105Jm-3)}
So mean free path  = 6.7x10-8 m = 67 nm.
Note significant orders of magnitude for N2: molecular size typically 0.3 nm.
Mean free path typically 67 nm. At atmospheric pressure gas molecules collide every 0.1 ns.
12