Drs. Jaslin Ikhsan, M.App.Sc., Ph.D.
Chemistry Ed. Department
State University of Yogyakarta
Kinetic Theory of Gases I:
 Gas Pressure
 Translational Kinetic Energy
 Root Mean Square Speed
GASES
 Gases are one of the most pervasive aspects of our
environment on the Earth. We continually exist with
constant exposure to gases of all forms.
 The steam formed in the air during a hot shower is
a gas.
 The Helium used to fill a birthday balloon is a gas.
 The oxygen in the air is an essential gas for life.
GASES
A windy day or a still day is a result of the difference in
pressure of gases in two different locations. A fresh breeze on
a mountain peak is a study in basic gas laws.
Mixture of gases
The Kinetic Molecular Model for Gases




Gas consists of large number of small
individual particles with negligible size
Particles in constant random motion and
collisions
No forces exerted among each other
Kinetic energy directly proportional to
temperature in Kelvin
3
KE 
2
 R T
The Ideal Gas Law
PV  nRT
in K
n: the number of moles in the ideal gas
N
n
NA
total number
of molecules
Avogadro’s number: the number of
atoms, molecules, etc, in a mole of
a substance: NA=6.02 x 1023/mol.
R: the Gas Constant: R = 8.31 J/mol · K
Pressure and Temperature
Pressure: Results from collisions of molecules
on the surface
Pressure:
F
P
A
Force:
dp
F
dt
Force
Area
Rate of momentum
given to the surface
Momentum: momentum given by each collision
times the number of collisions in time dt
Only molecules moving toward the
surface hit the surface. Assuming the
surface is normal to the x axis, half the
molecules of speed vx move toward
the surface.
Only those close enough to the
surface hit it in time dt, those within
the distance vxdt
The number of collisions hitting an area A in
time dt is
1 N 
 A vx  dt
2 V 
Average density
The momentum given
by each collision to
the surface 2mvx
Momentum in time dt:
1 N 
dp  2mv x  
 Av x dt
2 V 
Force:
dp
1 N 
F
 2mvx  
 A v x


dt
2 V
Pressure:
F N 2
P   mv x
A V
Not all molecules have the same v x  average v2x
N 2
P  mv x
V

2
vx
1 2 1 2 2 2
= v  v x  v y  vz
3
3
2
vx
1 2 1 2
= v  vrms
3
3

vrms is the root-mean-square speed
2
vrms  v 
2
vx
2
+ vy
2
+ vz
3
1 N 2 2 N 1 2 
Pressure: P 
mv 
mv




3V
3 V 2
Average Translational Kinetic Energy:
1 2 1 2
K  mv  mvrms
2
2
Pressure:
2 N
P   K
3 V
2
From PV   N  K and PV  nRT
3
Temperature:
3 nRT 3
K 
  k BT
2 N
2
R
23
Boltzmann constant: k B 
 1.38  10
J/K
NA
1
2
From PV   N  mvrms
3
N
and PV  nRT  N RT
A
vrms
3RT

M
Avogadro’s number
N  nN A
Molar mass
M  mN A
Pressure  Density x Kinetic Energy
Temperature  Kinetic Energy
(a) Compute the root-mean-square speed of a nitrogen
molecule at 20.0 ˚C. At what temperatures will the root-meansquare speed be (b) half that value and (c) twice that value?
vrms
(a)
vrms
3RT

M
3RT
38.31 J/mol  K  293 K


 511 m/s
-3
M
28.0 10 kg/mol
2
vrms

2
vrms
(b) Since vrms  T
for 0.5 vrms
for 2 vrms
T 

T

2
T
 0.5 T  73.3K = -200 C

2
3

T
 2 T 1.17 10 K = 899 C

Please estimate the root mean square
mean velocity of Hydrogen gas.
A. 2000
B. 1000
E. 250
D. 100
C. 500
m/s
What is the average translational kinetic energy of nitrogen
molecules at 1600K, (a) in joules and (b) in electron-volts?
(a)


3
3
K  k BT  1.38 10 23 J/K 1600K 
2
2
 3.3110 20 J
(b) 1 eV = 1.60 x 10-19 J
3.31 10 20 J
K
 0.21 eV
19
1.60 10 J/eV
3
K   kB T
2