Physics Chapter 14- Kinetic Theory of Gases

Physics DF025 Chapter 14

CHAPTER 14:
Kinetic theory of gases
(3 Hours)

Learning Outcome:
14.1 Ideal gas equations (1 hour)
At the end of this chapter, students should be able to:

a. Sketch and interpret b. Use the ideal gas equation:
– P-V graph at constant
temperature PV nRT
– V-T graph at constant
pressure
– P-T graph at constant
volume
of an ideal gas.

14.1 Ideal gas equation • The related equations to the
Boyle’s law are
14.1.1 Boyle’s law
• states : “The pressure PV constant
of a fixed mass of gas
at constant OR
temperature is
inversely proportional P1V1 P2V2
to its volume.”
OR
where P : initial pressure
1
1 P2 : final pressure
P if T constant
V1 : initial volume
V
V2 : final volume

• Graphs of the Boyle’s law.
a. P
The pressure of a fixed
mass of gass at constant
temperature is inversely
T2 propertional to its volume.
T1
0 V
b. P
T2

T1

1
0 V

14.1.2 Charles’s law The related equations to the Charles’ law are

• states : “The volume V
of a fixed mass of constant
gas at constant T
pressure is directly OR
proportional to its V1 V2
absolute
temperature.” T1 T2
OR where
T1 : initial absolute temperatu re
T2 : final absolute temperatu re
V T If P constant V1 : initial volume
V2 : final volume

Graphs of the Charles’s law.

b. V
a V
.

273.15 0 T( C) 0 T(K)

14.1.3 Gay-lussac’s The related equations to the Gay-
lussac’s law are
(pressure) law
P
states : “The pressure constant
of a fixed mass of gas T
OR
at constant
volume is directly P1 P2
proportional to its T1 T2
absolute where
temperature.” T1 : initial absolute temperatu re
OR
T2 : final absolute temperatu re
P : initial pressure
1

P T If V constant P2 : final pressure

Graphs of the Gay-lussac’s (pressure) law

a. P b. P

273.15 0 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 14.1.

P1 P2 P2
2nd stage
V1 1st stage V' V2
T1 T1 T2
Figure 14.1

– In 1st stage, temperature is
kept at T1 , – Equating eqs. (1) and (2), thus
Using Boyle’s law :

PV '
2 PV1
1
PV1
1 P2V2
OR
PV1 T1 T2
V' 1 (1)
P2
– In 2nd stage, pressure is kept Initial Final
constant at P2 ,
Using Charles’s law : PV
constant (3)
V' V2 T
T1 T2
V2T1
V' (2)
T2

• 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),
3
PVm 101.3 10 0.0224
R
T 1
273.15
1
R 8.31 J K mol
where R is called molar gas constant and its value is the
same for all gases.
– Thus PVm
R
T
PVm RT where Vm : volume of 1 mole gas
• 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 – If the Boltzmann constant, k
is defined as
M
where R 23 1
m : mass of a gas k 1.38 10 JK
M : molar mass of a gas NA
OR
then the equation of state
N becomes
n
NA
PV NkT
where
N : number of molecules
N A : Avogadro' s constant
6.02 1023 mol 1

Example 1 : Solution :
VA 3VB ; mA m; T0A 300 K;T0B 500 K
The volume of vessel A is three
times of the volume vessel B. The Since the vessels A and B are
vessels are filled with an ideal gas connected by a narrow tube thus the
pressure for both vessels is same,
and are at a steady state. The finally i.e.
temperature of vessel A and vessel PA PB P
B are at 300 K and 500 K
The system is in the steady state, thus
respectively as shown in Figure
14.2. T0A TA 300 K;T0B TB 500 K
By applying the equation of state for an
A B ideal gas,
(300 K) (500 K)
m
Figure 14.2 PV nRT and n
m M
If the mass of the gas in the vessel PV RT
A is m, obtain the mass of the gas M
in the vessel B in terms of m.

Vessel B :
Therefore,
VA 3VB ; mA m; T0A 300 K;T0B 500 K
Vessel A :

mA m
PVB 100 R (1)
PAVA RTA M
M
mB
PVB 500 R (2)
M
m
P 3VB R 300 By equating the eqs. (1) and (2) hence
M
m mB
100 R 500 R
M M
m
mB
5

Example 2 :
connecting tap

B A

Figure 14.3
Refer to Figure 14.3. 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.

Solution : V0A V1A 3.00 m3 ; T0A T1A 250 K;
P0A 5.00 104 Pa;V0B V1B 7.20 m3 ;
T0B T1B 400 K; P0B 2.00 104 Pa
After the connecting tap has been opened and the system reached
equilibrium, thus
P1A P1B P
By using the equation of state for ideal gas, P0V0 PV1
1
T0 T1
P0AV0A P0BV0B V1A V1B
P
T0A T0B T1A T1B
5.00 104 3.00 2.00 104 7.20 3.00 7.20
P
250 400 250 400
4
P 3.20 10 Pa

Exercise 14.1 :
Given R = 8.31 J mol 1 K 1 and NA = 6.0 1023 mol 1

1. A gas has a volume of 60.0 cm3 at 20 C and 900 mmHg.
What would its volume be at STP?
(Given the atmospheric pressure = 101.3 kPa and the density
of mercury = 13600 kg m 3)
ANS. : 66.2 cm3
2. 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
3. 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

Learning Outcome: and pressure ,
14.2 Kinetic theory of
gases (1 hour) 1
P v2
3
At the end of this chapter,
students should be able in related problems.
to:
c. Explain and use root
a. State the assumptions mean square (rms)
of kinetic theory of speed,
gases.
b. Apply the equations of 2 kT
ideal gas, v 3
m
1 of gas molecules.
PV Nm v 2
3

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.

14.2.2 Force exerted by an ideal • Let each molecule of the gas have the
gas mass m and velocity v.
• The velocity, v of each molecule can be
• Consider an ideal gas of N resolved into their components i.e. vx, vy
molecules are contained in a and vz.
cubical container of side d as • Consider, initially a single molecule
shown in Figure 14.4. 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 14.5.

Figure Wa
Wall B ll A
14.5 Wa
ll B

Wall A Wa
ll A
Wa
Figure 14.4 ll B

Therefore the change in the If Fx1 is the magnitude of the
linear momentum of the average force exerted by a
molecule is given by molecule on the wall in the time
t, thus by applying Newton’s
Px mvx mvx second law of motion gives
Px 2mvx Px 2mvx m 2
Fx1 Fx1 vx
The molecule has to travel a t 2d d
distance 2d (from A to B and vx
back to A) before its next
collision with wall A. The time For N molecules of the ideal gas,
taken for this movement is
m 2 m 2 m 2
2d Fx vx1 vx 2 ....... vxN
d d d
t
vx Fx
m 2 2
vx1 vx 2 ....... vxN
2

d

where vx1 is the x Thus, the x component for the
component of velocity of total force exerted on the wall of
molecule 1, vx2 is the x the cubical container is
component of velocity of
molecule 2 and so on. m 2
Fx N vx
• The mean (average ) value d
of the square of the velocity
in the x direction for N • The magnitude of the velocity
molecules is v is given by
2 2 2 2 2 2
vx1 vx 2 ....... vxN 2
vx
2
v vx vy vz
N
2 2 2
v x1 vx 2 ....... vxN then
2 2 2
2 v2 vx vy vz
N vx

• Since the velocities of the molecules in the ideal gas are
completely random, there is no preference to one direction
or another. Hence 2 2 2
v x v y v z
2 2
v 3 vx
2 v2
vx
3
• The total force exerted on the wall in all direction, F is given by

F
m
N vx
2 m v2
F N
d d 3
N m v2
F
3 d
where v 2 : mean square velocity of the molecule

14.2.3 Pressure of an ideal gas
• From the definition of pressure,
F N m v2
P where A d2 and F
A 3 d
1 Nm v 2
P 3
and d3 V
3 d
1 Nm 2
P v (14.1)
3 V
1 2
PV Nm v (14.2)
3
where Nm : mass of an ideal gas in the container

• Since the density of the 14.2.4 Root mean square
gas, is given by
velocity ( vrms)

Nm • is defined as vrms v2
V
hence the equation • From the equation of state in terms
(15.1) can be written as of Boltzmann constant, k :

1 2
PV NkT (14.4)
P v (14.3)
3 • By equating the eqs. (15.4) and
(14.2), thus
where
P : pressure by the gas 1
NkT Nm v 2
: density of the gas 3
v 2 : mean square velocity 2 3kT
v
of the gas molecules m

• Therefore • Since 1
P v2
3
3kT 3RT
vrms OR vrms
m M
2 3P
thus v

where
therefore the equation of root
vrms : root mean square velocity (speed) mean square velocity of the
gas molecules also can be
m : mass of a molecule gas written as
M : molar mass of a gas 3P
vrms
T : absolute temperatu re

a. The mean speed of the
Eight gas molecules chosen molecules is given by
at random are found to have
speeds of 1,1,2,2,2,3,4 and N 8
5 m s 1. Determine N
a. the mean speed of the vi
i 1
molecules, v
b. the mean square speed of N
the molecules, 1 1 2 2 2 3 4 5
v
c. the root mean square 8
speed of the molecules.
1
v 2.5 m s

Solution :
N 8
b. The mean square speed of c. The root mean square speed
the molecules is given by of the molecules is
N
2
vi
v 2 i 1 vrms v2
N

v 2 1
2
1
2
2
2
2
2
2
2
3
2
4
2
5
2
vrms 8
8

1
v 2
8m s 2 2
vrms 2.83 m s

Example 4 : (Given R = 8.31 J mol 1 K 1, k = 1.38
10 23 J K 1, molar mass of oxygen, M = 32
A cylinder of volume 0.08 m3 g mol 1, NA = 6.02 1023 mol 1)
contains oxygen gas at a
Solution :
temperature of 280 K and
pressure of 90 kPa. V 0.08 m3 ; T 280 K;
Determine
P 90 103 Pa
a. the mass of oxygen in the
cylinder, a. By using the equation of
b. the number of oxygen state, thus
molecules in the cylinder, m
c. the root mean square
PV nRT and n
M
speed of the oxygen
m
molecules in the cylinder. PV RT
M

Solution : V 0.08 m3 ; T 280 K; P 90 103 Pa
a.
3 m
90 10 0.08 8.31 280
0.032
m 9.90 10 2 kg
b. The number of oxygen molecules in the cylinder is
m N m
n N NA
M NA M
9.90 10 2 23
N 6.02 10
0.032
24
N 1.86 10 molecules
c. The root mean square speed of the oxygen molecules is

3RT 3 8.31 280 1
vrms vrms 467 m s
M 0.032

Exercise 15.2 :
Given R = 8.31 J mol 1 K 1, Boltzmann constant, k = 1.38 10 23K1
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

Learning Outcome:
14.3 Molecular kinetic energy and internal energy

a. Explain and use c. Define degree of freedom
translational kinetic
energy of gases,
d. State the number of
3 R 3 degree of freedom for
K tr T kT monoatomic, diatomic,
2 NA 2 and polyatomic gas
molecules.
b. State the principle of
equipartition of energy.

Learning Outcome:

e. Explain internal energy f. Explain and use
of gas and relate the internal energy of an
internal energy to the ideal gas
number of degree of
freedom,

14.3.1 Translational kinetic energy of molecule
• From equation (14.1), thus

1 Nm 2
P v
3 V
2 N 1
P m v2 (14.5)
3 V 2
This equation shows that N
increases
V
P increases ( ) When
1
m v2 increases
2

14.3.1 Translational kinetic energy of molecule

This equation shows that
N
increases
P increases ( ) When V

1
m v2 increases
2

Rearrange equation (14.5), thus

2 1
PV N m v2 and PV NkT
3 2


2 1 2 3 3 R
NkT N m v K tr kT T
3 2 2 2 NA
1 2 3 where
m v kT
2 2 K tr : average translati onal
kinetic energy of a
and
molecule
1 T : absolute temperatu re
m v2 K tr k : Boltzmann constant
2 R : molar gas constant
N A : Avogadro constant


E NKtr
• For N molecules of
3
an ideal gas in the E N kT
cubical container, 2
the total average 3
(mean) translational E NkT
2
kinetic energy, E is
given by OR

3
E nRT
2

14.3.2 Principle of equipartition of energy
• States : “the mean (average) kinetic energy of every degrees
of freedom of a molecule is
1
kT .
Therefore 2
f Mean (average) kinetic
K kT energy per molecule
2
OR

f Mean (average) kinetic
K RT energy per mole
2
where f : degrees of freedom
T : absolute temperatu re

N
A vessel contains hydrogen gas of 7.50 1017 ; vrms 2.50 103 m s 1 ;
7.50 1017 molecules per unit V
volume and the root mean T 303.15 K
square speed of the molecules is
2.50 km s 1at a temperature of 30 a. The average translational
C. Determine kinetic energy of a molecule
a. the average translational kinetic is
energy of a molecule for 3
hydrogen gas, K tr kT
b. the pressure of hydrogen gas. 2
(Given the molar mass of hydrogen
gas = 2 g mol 1, NA= 6.02 1023
3 23
K tr 1.38 10 303.15
mol 1and k = 1.38 10 23 J K 1) 2
21
K tr 6.28 10 J

N
Solution : 7.50 1017 ; vrms 2.50 103 m s 1 ; T 303.15 K
V
b. The pressure of hydrogen gas is given by
1 N 2 M 2 2
P m v where m and v vrms
3 V NA
1 N M 2
P vrms
3 V NA
1 17 0.002 3 2
P 7.50 10 23
2.50 10
3 6.02 10
3
P 5.19 10 Pa

14.3.3 Degree of freedom ( f )
• is defined as a number of
independent ways in which y
an atom or molecule can
absorb or release or store 
the energy. vy
He
 x
Monatomic gas (e.g. He,Ne,Ar)
 vx
vz
• The number of degrees of freedom
z
Figure 14.6
is 3 i.e. three direction of
translational motion where
contribute translational kinetic
energy as shown in Figure 14.6.

Diatomic gas (e.g. H2, O2, N2) y
• The number of degrees of freedom is

Translational kinetic energy 3 vy
H x
Rotational kinetic energy 2  H 
5 vz v x
z
Figure 14.7

y
Polyatomic gas (e.g. H2O, CO2, NH3)
• The number of degrees of freedom is 
Translational kinetic energy 3 vy
Rotational kinetic energy 3
O
6
 x
H  vx
vz H
z Figure 14.8

• Table 14.1 shows the degrees of freedom for various molecules.

Degrees of Freedom ( f ) Average kinetic
Molecule Example energy per
Translational Rotational Total molecule,<K>
3
Monatomic He 3 0 3 2
kT
5
Diatomic H2 3 2 5 2
kT
6
Polyatomic H 2O 3 3 6 2
kT 3kT

(At temperature of 300 K)
Table 14.1

• Degrees of freedom depend on
the absolute temperature of
the gases.
H H
– For example : Diatomic gas
(H2) vibration
Figure 14.9
– Hydrogen gas have the
vibrational kinetic energy
(as shown in Figure 14.9)
where contribute 2 degrees
of freedom which when the temperature,
correspond to the kinetic
energy and the potential At 250 K f 3
energy associated with
vibrations along the bond At 250 – 750 K f 5
between the atoms. At >750 K f 7

Solution :
Example 6 :
T1 293.15 K; E1 3.00 10 6 J;
A vessel contains an ideal E2 9.00 10 6 J ; m2 2m1
diatomic gas at temperature of By applying the equation of the total
20 C. The total translational translational kinetic energy, thus
kinetic energy of the gas 3
molecules is 3.00 10 6 J. E NkT
2
The mass of the gas is then
m and k
R
doubled and the total where N NA
translational kinetic energy M NA
of the molecules becomes 3 m R
9.00 10 6 J. Determine the E NA T
2 M NA
new temperature of the gas.
3 m
E RT
2 M

Solution : T1293.15 K; E1 3.00 10 6 J; E2 9.00 10 6 J
m22m1 3 m1
For temperature T1 : E RT (1)
1 1
2 M
3 m2
For temperature T2 : E2 RT2
2 M
3 2m1
E2 RT2 (2)
E2 2T2 2 M
(2) (1) :
E1 T1
6
9.00 10 2T2
6
3.00 10 293.15
T2 440 K OR 167  C

Learning Outcome:

• Explain internal energy • Define molar specific heat
of gas and relate the at constant pressure and
internal energy to the volume.
number of degree of • Use equations,
freedom.
• Explain and use C P CV R
internal energy of an
ideal gas and

f CP
U NkT
2 CV

14.3.4 Internal energy of gas Thus for N molecules,
and relate the internal
energy to the number U N K
of degree of freedom.
f R
• is defined as the sum of total U NkT and k
2 NA
kinetic energy and total
potential energy of the gas OR
molecules.
f
• But in ideal gas, the U nRT
intermolecular forces are 2
assumed to be negligible thus
the potential energy of the where U : internal energy
molecules can be neglected.
of the gas

Table 14.2 shows the properties for 1 mole of an ideal gas.

Table 14.2

Monatomic Diatomic Polyatomic

Degrees of freedom, f 3 5 6
Average kinetic energy per 3 5 6
molecule, <K> kT kT kT 3kT
2 2 2
3 5 6
Internal energy, U RT RT RT 3RT
2 2 2

Exercise 14.3 :

Given R = 8.31 J mol 1 K 1, Boltzmann constant, k = 1.38 10 23 K 1

1. One mole of oxygen has a mass of 32 g. Assuming oxygen
behaves as an ideal gas, calculate
a. the volume occupied by one mole of oxygen gas
b. the density of oxygen gas
c. the r.m.s. speed of its molecules
d. the average translational kinetic energy of a molecule
at 273 K and pressure of 1.01 105 Pa.
ANS. : 2.25 10 2 m3; 1.42 kg m 3; 461 m s 1; 5.65 10 21 J


THE END…
Next Chapter…
CHAPTER 15 :
Thermodynamics

Physics Chapter 14- Kinetic Theory of Gases

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