CHAPTER 14:
Kinetic Theory of Gases
(3 Hours)
1
Learning Outcome:
14.1 Ideal gas equation (1 hour)
At the end of this chapter, students should be
able to:
• Sketch
•
–
P-V graph at constant temperature
–
V-T graph at constant pressure
–
P-T graph at constant volume of an ideal gas.
Use the ideal gas equation:
PV  nRT
2
14.1 Ideal Gas Equation
14.1.1
Boyle’s law
• states : “The pressure of a fixed mass of gas at constant
temperature is inversely proportional to its volume.”
1
P
V
if
PV  constant
where
3
T  constant
OR
P1V1  P2V2
P1 : initial pressure
P2 : final pressure
V1 : initial volume
V2 : final volume
• Graphs of the Boyle’s law.
b. PV
a. P
T2 > T1
T2
T1
0
c. P
T2
T1
V
T2
0
d. PV
T1
4
0
P
T2
T1
1
V
0
V
14.1.2
Charles’s law
• states : “The volume of a fixed mass of gas at constant
pressure is directly proportional to its absolute
temperature.”
V T
If
P  constant
V
 constant
T
where
6
V1 V2

T1 T2
T1 : initial absolute temperatu re
T2 : final absolute temperatu re
V1 : initial volume
V2 : final volume
• Graphs of the Charles’s law.
7
a.
V
273.15
0
b. V
T(C)
0
T(K)
14.1.3
Gay-lussac’s (pressure) law
• states : “The pressure of a fixed mass of gas at constant
volume is directly proportional to its absolute
temperature.”
P  T If V  constant
P
 constant
T
where
P1 P2

T1 T2
T1 : initial absolute temperatu re
T2 : final absolute temperatu re
P1 : initial pressure
P2 : final pressure
9
• Graphs of the Gay-lussac’s (pressure) law.
a.
P
273.15
0
10
b. P
T(C)
0
T(K)
14.1.4 Equation of state for an ideal gas
• An ideal gas is defined as a perfect gas which obeys the
three gas laws (Boyle’s, Charles’s and Gay-Lussac’s)
exactly.
• Consider an ideal gas in a container changes its pressure
P, volume V and temperature T as shown in Figure 15.1.
P1
V1
T1
12
1st
stage
P2
V'
T1
2nd
Figure 14.1
stage
P2
V2
T2
– In 1st stage, temperature is kept at T1 ,
Using Boyle’s law :
P1V1
P2V '  P1V1
V '
P2
nd
– In 2 stage, pressure is kept constant at P2 ,
Using Charles’s law :
V ' V2
V2T1

V '
T1 T2
T2
– Equating eqs. (1) and (2), thus
P1V1 P2V2

T1
T2
Initial
13
Final
OR
PV
 constant
T
(1)
(2)
(3)
• Consider 1 mole of gas at standard temperature and pressure
(S.T.P.), T = 273.15 K, P = 101.3 kPa and Vm = 0.0224 m3
 From equation (3),


PVm 101.3 103 0.0224
R

T
273.15
1
1
R  8.31 J K mol
where R is called molar gas constant and its value is the same
for all gases.
 Thus PV
m
T
R
where
Vm : volume of 1 mole gas
PVm  RT
• For n mole of an ideal gas, the equation of state is written as
PV  nRT
 where n : the number of mole gas
m
n
M
where
m : mass of a gas
M : molar mass of a gas
OR
N where
n
N : number of molecules
N A N : Avogadro' s constant
A
 6.02 1023 mol 1
– If the Boltzmann constant, k is defined as
R
k
 1.38 10 23 J K 1
NA
then the equation of state becomes
15
PV  NkT
Example 14.1 :
The volume of vessel A is three times of the volume vessel B. The
vessels are filled with an ideal gas and are at a steady state. The
temperature of vessel A and vessel B are at 300 K and 500 K
respectively as shown in Figure 15.2.
A
B
(300 K)
(500 K)
Figure 14.2
If the mass of the gas in the vessel A is m, obtain the mass of the
gas in the vessel B in terms of m.
16
Solution :
17
Solution :
18
Example 14.2 :
connecting tap
B

A
Figure 14.3
Refer to Figure 15.3. Initially A contains 3.00 m3 of an ideal gas at
a temperature of 250 K and a pressure of 5.00  104 Pa, while B
contains 7.20 m3 of the same gas at 400 K and 2.00  104 Pa.
Calculate the pressure after the connecting tap has been opened
and the system reached equilibrium, assuming that A is kept at
250 K and B is kept at 400 K.
19
Solution :
20
Exercise 14.1 :
Given R = 8.31 J mol1 K1 and NA = 6.0  1023 mol1
1. Estimate the number of molecules in a flask of volume
5.0  104 m3 which contains oxygen gas at a pressure of
2.0  105 Pa and temperature of 300 K.
ANS. : 2.41  1022 molecules
2. A cylinder contains a hydrogen gas of volume 2.40  103 m3 at
17 C and 2.32  106 Pa. Calculate
a. the number of molecules of hydrogen in the cylinder,
b. the mass of the hydrogen,
c. the density of hydrogen under these conditions.
(Given the molar mass of hydrogen = 2 g mol1)
ANS. : 1.39  1024 molecules; 4.62 g; 1.93 kg m3
21
Learning Outcome:
14.2 Kinetic theory of gases (1 hour)
At the end of this chapter, students should be able to:
•
State the assumptions of kinetic theory of gases.
•
Apply the equations of ideal gas,
PV 
1
Nm  v 2 
3
and pressure ,
P
•
22
1
  v2 
3
in related problems.
Explain and use root mean square (rms) speed,
kT
2
 v  3
m
of gas molecules.
14.2 Kinetic theory of gases
• The macroscopic behaviour of an ideal gas can be describe by using
the equation of state but the microscopic behaviour only can be
describe by kinetic theory of gases.
14.2.1 Assumption of kinetic theory of gases
• All gases are made up of identical atoms or molecules.
• All atoms or molecules move randomly and haphazardly.
• The volume of the atoms or molecules is negligible when
compared with the volume occupied by the gas.
• The intermolecular forces are negligible except during collisions.
• Inter-atomic or molecular collisions are elastic.
• The duration of a collision is negligible compared with the time
spent travelling between collisions.
• Atoms and molecules move with constant speed between
collisions. Gravity has no effect on molecular motion.
23
14.2.2
Force exerted by an ideal gas
• Consider an ideal gas of N molecules are contained in a
cubical container of side d as shown in Figure 15.4.
Wall B
Figure 14.4
Wall A
• Let each molecule of the gas have the mass m and
velocity v.
• The velocity, v of each molecule can be resolved into
their components i.e. vx, vy and vz.
24
• Consider, initially a single molecule moving with a velocity vx
towards wall A and after colliding elastically , it moves in the
opposite direction with a velocity vx as shown in Figure 15.5.
Wall
A
Wall
B
Wall
Figure 14.5
A
Wall
B
• Therefore the change in the linear momentum of the molecule is
given by
P  mv  mv
x
25
x
Px  2mvx

x

• The molecule has to travel a distance 2d (from A to B
and back to A) before its next collision with wall A.
The time taken for this movement is
2d
t 
vx
• If Fx1 is the magnitude of the average force exerted by
a molecule on the wall in the time t, thus by applying
Newton’s second law of motion gives
Px 2mvx
m 2
Fx1 

Fx1   v x
t  2d 
d
 
v 
 x 
m 2 m 2
m 2
Fx   vx1   vx 2  .......   vxN
d
d
d
• For N molecules of the ideal gas,
26

m 2
2
2
Fx  vx1  vx 2  .......  vxN
d

where vx1 is the x component of velocity of molecule 1, vx2
is the x component of velocity of molecule 2 and so on.
• The mean (average ) value of the square of the velocity in
the x direction for N molecules is
vx1  vx 2  .......  vxN
2
 vx 
N 2
2
2
2
vx1  vx 2  .......  vxN  N  vx 
2
2
2


• Thus, the x component for the total force exerted on the
wall of the cubical container is
m
2
Fx  N  vx 
d
• The magnitude of the velocity v is given by
v  vx  v y  vz
2
2
2
2
 v  vx    v y    vz 
2
then
27
2
2
2
• Since the velocities of the molecules in the ideal
gas are completely random, there is no
preference to one direction or another. Hence
 vx  v y  vz 
2
2
 v  3  vx 
2

v

2
 vx 
3
2
2
2
• The total force exerted on the wall in all direction,
F is given by
m
2
F  N  vx 
d
28
where
m   v2  

F  N 
d  3 
N  m  v2  

F  
3
d

 v : mean square velocity of the molecule
2
14.2.3 Pressure of an ideal gas
• From the definition of pressure,
F
P
A
where
A  d2
and
N  m  v2  

F  
3
d

1  Nm  v 2  
3
and
d

V


P 
3

3
d

1  Nm

2
P 
 v 
3 V



(14.1)
1
(14.2)
PV  Nm  v 2 
3
where Nm : mass of an ideal gas in the container
29
• Since the density of the gas,  is given by
Nm

V
• hence the equation (14.1) can be written as
1
2
P  v 
3
where
(14.3)
P : pressure by the gas
 : density of the gas
 v 2 : mean square velocity of the gas molecules
30
14.2.4 Root mean square velocity ( vrms)
• is defined as
vrms   v 2 
• From the equation of state in terms of Boltzmann
constant, k : PV  NkT
(14.4)

1
NkT  Nm  v 2 
3
3kT
2
 v 
m

• By equating the eqs. (14.4) and (14.2), thus
vrms
31
3kT

m
OR
vrms
3RT

M
where vrms
: root mean square velocity (speed)
m : mass of a molecule gas
M : molar mass of a gas
T : absolute temperatu re
3P
1
2
2
P    v  thus  v 

3
• Since
32
vrms 
3P

therefore the equation of root mean square
velocity of the gas molecules also can be written
as
Example 14.3 :
Eight gas molecules chosen at random are found to have speeds
of 1,1,2,2,2,3,4 and 5 m s1. Determine
a. the mean speed of the molecules,
b. the mean square speed of the molecules,
c. the root mean square speed of the molecules.
Solution :
a.
33
Solution :
b.
34
Example 14.4 :
A cylinder of volume 0.08 m3 contains oxygen gas at a temperature
of 280 K and pressure of 90 kPa. Determine
a. the mass of oxygen in the cylinder,
b. the number of oxygen molecules in the cylinder,
c. the root mean square speed of the oxygen molecules in the
cylinder.
(Given R = 8.31 J mol1 K1, k = 1.38  1023 J K1, molar mass of
oxygen, M = 32 g mol1, NA = 6.02  1023 mol1)
Solution :
a.
35
Solution :
.
36
Exercise 14.2 :
Given R = 8.31 J mol1 K1, Boltzmann constant, k = 1.381023 K1
1. In a period of 1.00 s, 5.00  1023 nitrogen molecules strike a
wall with an area of 8.00 cm2. If the molecules move with a
speed of 300 m s1 and strike the wall head-on in the elastic
collisions, determine the pressure exerted on the wall.
(The mass of one N2 molecule is 4.68  1026 kg)
ANS. : 17.6 kPa
2. Initially, the r.m.s. speed of an atom of a monatomic ideal gas is
250 m s1. The pressure and volume of the gas are each
doubled while the number of moles of the gas is kept constant.
Calculate the final translational r.m.s. speed of the atoms.
ANS. : 500 m s1
3. Given that the r.m.s. of a helium atom at a certain temperature
is 1350 m s1, determine the r.m.s. speed of an oxygen (O2)
molecule at this temperature.
(The molar mass of O2 is 32.0 g mol1 and the molar mass of He
is 4.00 g mol1)
ANS. : 477 m s1
37
Learning Outcome:
14.3 Molecular kinetic energy and internal energy
At the end of this chapter, students should be able to:
• Explain and use translational kinetic energy of gases,
3 R 
3
T  kT
K tr  
2  NA 
2
•
•
•
State the principle of equipartition of energy.
Define degree of freedom.
State the number of degree of freedom for monoatomic,
diatomic and polyatomic gas molecules.
Explain internal energy of gas and relate the internal energy to
the number of degree of freedom.
•
1
U  fNkT
2
Explain and use internal energy of an ideal gas
•
38
3
U  NkT
2
14.3.1 Molecular kinetic energy
• From equation (14.1), thus
1  Nm
P 
 v2
3 V



2  N  1
P    m  v 2
3  V  2
This equation shows that
P increases () When
• Rearrange equation (14.5), thus
2 1

2
PV  N  m  v  
39
3 2




(14.5)
N
  increases
V 
1
2
m

v

2
and

 increases

PV  NkT
40
2 1

2
NkT  N  m  v  
3 2

1
1
3
2
2
and
m  v  K tr
m  v  kT
2
2
2
3
3 R 
T
K tr  kT  
2
2  NA 
where K : average translati onal kinetic energy
tr
of a molecule
T : absolute temperatu re
k : Boltzmann constant
R : molar gas constant
N A : Avogadro constant
• For N molecules of an ideal gas in the cubical container,
the total average (mean) translational kinetic energy, E is
given by
E  NKtr
3 
E  N  kT 
2 
3
E  NkT
2
OR
3
E  nRT
2
41
Principle of equipartition of energy
• States : “the mean (average) kinetic energy of
1
every degrees of freedom of a molecule is kT .
2
Therefore
f
 K  kT
2
Mean (average) kinetic
energy per molecule
OR
where
f
Mean (average) kinetic
 K  RT
energy per mole
2
f : degrees of freedom
T : absolute temperatu re
42
Degree of freedom ( f )
• is defined as a number of independent ways in
which an atom or molecule can absorb or
release or store the energy.
Monoatomic gas (e.g. He, Ne, Ar)
• The number of degrees of freedom is 3 i.e. three
direction of translational motion where
contribute translational kinetic energy as shown
y
in Figure 15.6.

vy
He
43
z

vz

vx
Figure 15.6
x
Diatomic gas (e.g. H2, O2, N2)
• The number of degrees of freedom is
y
H
Translational kinetic energy
Rotational kinetic energy

vy
x
H


vz v x
3
2
5
z
y
Polyatomic gas (e.g. H2O, CO2, NH3)
Figure 14.8

• The number of degrees of freedom is v y
Translational kinetic energy
3
O

Rotational kinetic energy
3
x

 v
6
H
vz H x
z 
Figure 14.7
44

• Table 14.1 shows the degrees of freedom for various molecules.
Degrees of Freedom ( f )
Molecule
Example
Translational Rotational
Monatomic
Diatomic
Polyatomic
He
3
0
3
H2
3
2
5
H 2O
3
3
(At temperature of 300 K)
Table 14.1
45
Total
6
Average kinetic
energy per
molecule,<K>
3
kT
2
5
kT
2
6
kT  3kT
2
• Degrees of freedom depend on the absolute temperature of the
gases.
– For example : Diatomic gas (H2)
H
H
vibration
Figure 15.9
– Hydrogen gas have the vibrational kinetic energy (as shown in
Figure 15.9) where contribute 2 degrees of freedom which
correspond to the kinetic energy and the potential energy
associated with vibrations along the bond between the atoms.
when the temperature,
f

3
At 250 K
f 5
At 250 – 750 K
f 7
At >750 K
46
Example 14.5 :
A vessel contains hydrogen gas of 2.20  1018 molecules per unit
volume and the mean square speed of the molecules is
4.50 km s1 at a temperature of 50 C. Determine
a. the average translational kinetic energy of a molecule for
hydrogen gas,
b. the pressure of hydrogen gas.
(Given the molar mass of hydrogen gas = 2 g mol1,
NA= 6.02  1023 mol1 and k = 1.38  1023 J K1)
Solution :
47
Solution :
48
14.3.2 Internal energy, U
• is defined as the sum of total kinetic energy and total
potential energy of the gas molecules.
• But in ideal gas, the intermolecular forces are assumed to
be negligible thus the potential energy of the molecules
can be neglected. Thus for N molecules,
U NK
f
U  NkT
2
and
R
k
NA
OR
f
U  nRT
2
• For N molecules of monoatomic gas ,
3
U  NkT
2
OR
3
U  nRT
2
where
49
U : internal energy of the gas
Example 14.7 :
Neon is a monoatomic gas. Determine the internal energy of 1 kg
of neon gas at temperature of 293 K. Molar mass of neon is 20 g.
Solution :
50
THE END…
Next Chapter…
CHAPTER 15 :
Thermodynamics
51