CHAPTER 9: KINETIC THEORY OF GASES & THERMODYNAMICS

CHAPTER 9

KINETIC THEORY OF
GASES
&

THERMODYNAMICS

KMM

TOPIC 9
KINETIC THEORY OF GASES AND THERMODYNAMICS

9.1 Kinetic Theory of Gases
9.2 Molecular kinetic energy and internal energy
9.3 First Law of Thermodynamics
9.4 Thermodynamic processes
9.5 Thermodynamics work

1

9.1 Kinetic theory of gases

Learning Outcome :

a) State the assumptions of kinetic theory of gases. vrms  v 2
b) Discuss root mean square (rms) speed of gas molecules,

c) Solve problems related to root mean square (rms) speed of gas molecule

vrms  3kT  3RT
m M

d) Solve problem related to the equation, PV  1 Nmvr 2 and pressure , P  1  v2
3 ms 3 rms

Microscopic Behaviour

• 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.

a) Assumption of kinetic theory of gases

3

1. All gases are made up of identical atoms or molecules.
2. All atoms or molecules move randomly and haphazardly.
3. The volume of the atoms or molecules is negligible when compared with the

volume occupied by the gas.
4. The intermolecular forces are negligible except during collisions.
5. Inter-atomic or molecular collisions are elastic.
6. The duration of a collision is negligible compared with the time spent

travelling between collisions.
7. Atoms and molecules move with constant speed between collisions. Gravity

has no effect on molecular motion.

4

b) Root Mean Square (rms) Speed of Gas Molecules, vrms  v2

vrms is defined as the square root of the mean squares of speed of the individual

molecules in a gas.

Measuring the velocities of particles at a given time results in a large distribution of
values where some particles may move very slowly while others move very quickly, and
they are constantly moving in different directions. To properly assess the average
velocity, average the squares of the velocities and take square root of that value. It is
represented as follows:

vrms  v2  v12  v22  v32  ....vN2
N

From the equation of state in terms of
Boltzmann constant, k :

PV  NkT    (9.1)

 NkT  1 Nm v2
3

v2  3kT    (9.2)
m

By equating the equations. (9.1) and (9.2), vwhere ;
rms: root mean square velocity (speed)
thus, m : mass of molecule gas

vrms  3kT or vrms  3RT M : molar mass of gas
m M
T : absolute temperature

 k  R then k  R 
NA m M

How fast molecules move is directly proportional to their absolute temperature and
inversely proportional to their masses. At a given temperature the molecules of lighter
mass move faster on an average than the molecules with heavier masses.

Vrms and pressure of an ideal gas.

From the definition of pressure, P  F
A

Given that: F  N  m  v2   and A d2
3 d

Where F is the total force of the molecules exerted on the wall in all direction and A is

area. Thus,  so 1
3
P  1  Nm  v2   and d3 V PV Nmvrms2    (9.3)
3 d3

Where Nm : mass of an ideal gas in the container

Since the density of the gas,  is given by   Nm , hence the equation (9.3) can
V,

be written as P  1  vrm 2    (9.4) .
3 s

Other related vrms equation in terms of pressure and density can be write as below:

v2  3P or  v2  3P vrms  3P
rms
  therefore 

c) Solve problems related to root mean square (rms) speed of gas molecules,

vrms  3kT  3RT
m M

d) Solve problems related to the equations,

 PV1 2 and pressure P  1  v2
3 Nmvr m s 3 rms

10

11

12

P  1 vrm 2
3 s

13

14

Example 9.1.6

Determine the temperature at which rms speed of a helium molecule is 1.3 times greater
than the rms speed at the temperature of 310 K.

Solution :
The rms speed is given by the equation

vrms  3RT
M
3RT2
At T = 320 K, vrms  3RT1 Therefore, vrms 2  M
M vrms1
3RT1
3RT2 M
M
At T = T2, vrms   vrms 2  2 T1
vrms1
T2 

T2 = 523.90 K

15

9.2 Molecular Kinetic Energy And Internal Energy

Learning Outcome :

a) Explain and use translational kinetic energy of a molecule,

K tr  3  R  T  3 kT
2 NA 2

b) Define degree of freedom.

c) Identify number of degrees of freedom, f for monoatomic, diatomic and

polyatomic gas molecules. U  1 fNkT
d) State the principle of equipartition of energy. 2

e) Discuss internal energy of gas.

f) Solve problems related to internal energy,

P  1  Nm  v 2  NkT  2 N 1 m  v2 
3 2 
3 V  1 m  v2  3 kT

22

P  2  N  1 m  v2  3  R  T  3 kT
3V  2  2 NA 2
K tr 

Example 9.2.1

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.
(Molar mass of hydrogen gas = 2 gmol1, NA= 6.02  1023 mol1 and k =1.38  1023 JK1)

Solution :

= 2.2 × 1018; < 2 >= 4.50 × 1018 −1; = 323.15


3 3 1.38 × 10−23 323.15
= 2 =2

= 6.69 × 10−21

18

Solution :

(b) The pressure of hydrogen gas is given by

= 1 < 2 > ℎ =
3

= 1 < 2 >
3 4.50 × 103
= 1 2.2 × 1018 0.002
6.02×1023
3
= 1.1 × 10−5

19

Example 9.2.2
The volume of air in a fully expanded pair of human lungs is 5.0 × 10–3 m3. The pressure
of the air in the lungs is 1.0 × 105 Pa and its temperature is 310 K. Calculate the number

of moles of air in the lungs. the average kinetic energy of an air molecule in the lungs.

Solution :

Use equation pV=nRT

(1.0 × 105)(5.0 × 10−3)
n = =
(8.31)(310)

n = 0.194 moles

= 3 = 3 (1.38 × 10−23)(310)
2 2
= 6.4 × 10−21

20

Example 9.2.3

a) Calculate the average kinetic energy of a gas molecule of an ideal gas at a
temperature of 20 °C.

b) Two different gases at the same temperature have molecules with different mean
square speeds. Explain why this is possible.

Solution :

a) = 3 = 3 1.38 × 10−23 293
2 2
= 6.1 × 10−21

b) As = 3 , the average kinetic energy of both gases are the same since they are
2

at the same temperature. Kinetic energy is related to the mass and mean square of

speed, = 1 < 2 > , then, 1 < 2 >= 3 , < 2 >= 3
2
2 2

therefore if the masses of the molecules are different, the mean square speeds must be

different.

b) Define degree of freedom ( f )

c) Identify number of degrees of freedom, f for monoatomic, diatomic and
polyatomic gas molecules

Degree of freedom 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) (a)
The number of degrees of freedom is 3 i.e.
three direction of translational motion
where contribute translational kinetic energy
as shown in Figure (a). In an monatomic
molecule, there are three different ways in
which they can translate (along each axis).

• Diatomic gas (e.g. H2, O2, N2) (b)
On the other hand, any diatomic gas molecule can (c)
translate in those three same ways and also rotate about
two different planes. A molecule of diatomic gas has
five degrees of freedom where translational kinetic
energy is 3 and rotational kinetic energy is 2. The
number of degree of freedom can be referred in Figure
(b).

• Polyatomic gas (e.g. H2O, CO2, NH3)
A molecule of polyatomic gas has six degrees of
freedom where translational kinetic energy is 3 and
rotational kinetic energy is 3. The number of degree of
freedom can be referred in Figure (c).

Table below shows the degrees of freedom for various molecules.
Degrees of freedom
depend on the absolute
temperature of the gases.
For example : Diatomic
gas (H2)

Hydrogen gas have the vibrational kinetic energy (as shown in figure above) 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 at 250 K, f = 3, at 250 - 750 K, f = 5 and at >750 K, f = 7

(d) The Principle of Equipartition of Energy
• Equipartition of energy, law of statistical mechanics stating that, in a system in

thermal equilibrium, on the average, an equal amount of energy will be associated
with each independent energy state.

• The mean (average) kinetic energy of every degrees of freedom of a molecule is

. That’s means each independent degree of freedom has an equal amount of


energy equal to .  K  f kT Mean (average) kinetic energy per
2 molecule.

 K  f RT Mean (average) kinetic energy per
2 mole.

(e) Internal Energy of Gas, U
• Internal energy of gas 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 

U  f Nk The equation can rewrite as; U  f nRT
22

Let say for N molecules of monoatomic gas ,

U  3 nRT or U  3 NkT
2 2

f) Solve problems related to internal energy, U  f NkT
2

Example 9.2.4
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 :

U  3 nRT ,
2

U  3 m RT ,
2M

U  3 1 (8.31)(293)
2 0.02

1.83105 J

Example 9.2.5

Determine the total internal energy for 4 mol of ideal gas at the following temperatures:

a) 200 K b)303 K

Solution : b) U  3 nRT ,
a) U  3 nRT ,
2
2  3 (4)(8.31)(303)
 3 (4)(8.31)(200)
2
2  1.51104 J
 9.97 103 J

9.3 First Law of Thermodynamics

Learning Outcome :
a) State the First Law of Thermodynamics,

U  Q  W

b) Solve problem related to First Law of Thermodynamics.

a) First law of Thermodynamics (FLOT)

The heat (Q) supplied to a system is equal to the increase in the internal energy
(U) of the system plus the work done (W) by the system on its surroundings.

Q = ΔU + W Signs convention for Q, ΔU and W
HEAT (Q)

heat change in work done heat added/ heat removed/
internal supplied/ flow into released/ flow out
from the system
energy the system

CHANGE IN INTERNAL ENERGY (ΔU)

temperature temperature temperature

increase constant decrease
Tf > Ti Tf = Ti Tf < Ti

Internal energy, Change in internal energy.
U  f nRT
2 U  f nRT  f nR(T f  Ti )
2 2

WORK DONE (W)

work done by the work done on the
system system

(gas expansion) (gas compression)
(volume increase) (volume decrease)

expansion initially
compression

b) Solve problem related to First Law of Thermodynamics.

Example 9.3.1
In each of the following situations, find the change in internal energy of the system.
(a) A system absorbs 2090 J of heat and at the same time does 400 J of work.
(b) A system absorbs 1255 J of heat and at the same time 420 J of work is done on it.
(c) 5020 J of heat is removed from a gas held at constant volume.
Solution:
(a) “heat absorb” Q is positive, Q=2090 J, “does” W is positive, W=400 J

U  Q W  2090  400  1690J

(b) “heat absorb” Q is positive, Q=1255 J, “done on it” W is negative, W=-420 J

U  Q W  1255   420  1675J

(c) “heat removed” Q is negative, Q=-5020J, “constant volume” no work done, W=0

U  Q W  5020  0  5020J

9.4 Thermodynamic processes

Learning Outcome :

a) Define the following thermodynamic processes:
i. Isothermal;
ii. Isochoric;
iii. Isobaric and
iv. Adiabatic

b) Analyse P-V graph for all the thermodynamic processes.

a) Thermodynamics Process and and Isothermal Isothermal
b) Analyse P-V graph expansion compression

1. ISOTHERMAL

• Thermodynamic process that occurs at constant

temperature.

• when temperature is constant, there is no change

in internal energy, ΔU = 0.

• FLOT: Q = ΔU + W become Q = W

• related gas law: Boyle’s Law • volume increase: isothermal

Notes: expansion
• volume decrease: isothermal
• T2 > T1
• the curve of isothermal graph must follow compression

• the same temperature curve (temperature is constant)

2. ISOCHORIC / ISOVOLUMETRIC isochoric isochoric
• Thermodynamic process that occurs at constant (pressure (pressure
increase) decrease)
volume.
• when there is no change in volume (no expansion/

compression), there is no work done by/on the

system, W = 0 J
• FLOT: Q = ΔU + W become Q = ΔU
• related gas law: Gay Lussac’s Law/ Pressure Law

Notes:
• T2 > T1
• vertical line graph

i. arrow upward: pressure increase

ii. arrow downward: pressure decrease

3 ISOBARIC Isobaric Isobaric
• Thermodynamic process that occurs at constant expansion compression

pressure.
• when there is no change in volume (no expansion/

compression), there is no work done by/on the

system, W = 0 J
• FLOT: Q = ΔU + W
• related gas law: Charles’s Law

Notes:
• T2 > T1
• horizontal line graph (bar graph)

i. arrow to the right: volume increase

ii. arrow to the left: volume decrease

4 ADIABATIC Adiabatic Adiabatic
• Thermodynamic process that occurs without the expansion compression

transfer of heat (into or out of the system).
• heat flow can be prevented by thermally

insulating material.
• when no heat is transferred, Q = 0
• FLOT: Q = ΔU + W become ΔU = -W or

- ΔU = W
• related gas law: Charles’s Law

Notes:
• T2 > T1
• the curve of adiabatic graph is steeper compare than the curve of isothermal graph

i. volume increase: adiabatic expansion

ii.volume decrease: adiabatic compression

9.5 Thermodynamics work

Learning Outcome :

a) Derive equation of work done in isothermal, isochoric and
isobaric processes from P-V graph.

b) Solve problem related to work done in:

i. isothermal process; W  nRT ln V2 and W  nRT ln p1
V1 p2

ii. isobaric process; W  PV and W  P(V2 V1)

iii. isochoric process; W 0

a) Derive equation of work done in Generally, W  Vf p dV and
isothermal, isochoric and isobaric • Vi
processes from P-V graph.
W  area under p V graph

Example 9.5.1 In isothermal process, T is constant:
Ti =Tf = 481.35 K
Three moles of an ideal gas do 3500 J of
work to its surrounding as it expands from the formula of work done (for
isothermally to a final pressure of 4 x 105 isothermal):
Pa and volume 0.030 m3. Determine the
initial volume of the gas. W  nRT ln V f
Vi
Solution:
Given: n = 3, W= 3500J, 4 x 105 Pa, 3500  38.31481.35ln 0.03
Vf = 0.03 m3.
Vi
p f V f  nRTf Vi  0.022 m3

 4 105 0.03  38.31Tf

Tf  481.35K

Example 9.5.2 U  f nRT
In an isochoric process, the temperature of 2
1 mole of ideal monoatomic gas increase
from 30 ℃ to 83 ℃. How much heat is  3 18.31356.15  303.15
transferred in the process?
Solution: 2
. n  1, f  3,  660.65J

Ti  30  273.15  303.15K Q  U  660.65J
Tf  83  273.15  356.15K

in an isochoric process, W = 0

FLOT : Q  U  W
Q  U

Example 9.5.3 (a) Isobaric compression

A gas is compressed at a constant pressure (b) Work done for isobaric process:
of 1 atm from 8 L to 3 L. In the process,
400 J of heat is leaving the gas.  W  pV  p V f Vi

(a) State the type of thermodynamic   1.013105 3103  8 103
process
 506.5J
(b) Calculate the work done on the gas
(negative sign show that the work is done
(c) Calculate the change in internal energy
of the system. on the gas)

(d) Sketch the p-V graph. (c) U  Q W

 400  (506.5)  106.5J

(Given: 1 atm = 1.013 x 105 Pa, (d)

1 L = 1 dm3)

Solution:
Given : p  1 atm  1.013105 Pa

Vi  8 L  8 103 m3, V f  3 L  3103 m3

Example 9.5.4 FLOT :
1 mole of helium gas expanding under
adiabatic conditions does 480 J of work. Q  U  W
Calculate the change of temperature of the U  W  480
gas.
Solution: f nRT  480
2
given: n=1, helium (monoatomic gas), f=3
and W=480 J 3 18.31T  480
adiabatic process, Q = 0
2
T  38.5K  38.5o C

Example 9.5.5 (b) the change in internal energy in the
process of ACD.
In the process of AC, 160 J of heat is added
to the system while in the process of CD, (c) the total heat added in the process of
550 J of heat is added to the system. ABD.
Calculate
(a) the change in internal energy in the Solution:
process of AC.
Given : QAC = 160 J, QCD = 550 J
a) AC: isochoric process, W=0

U AC  QAC  WAC
 160  0  160J

b) change in internal energy, U AC

U ACD  QACD  WACD
QACD  QAC  QCD  160  550  710J
WACD  WAC  WCD  0  pV

  7 105 6 104  3104  210J

c) total heat added (ABD), QABD = ?

QABD  U ABD  WABD
U ABD  U ACD  500J
WABD  WAB  WBD

  pV  0  2 105 6 104  3104  60J

QABD  500  60  560J

THE END…

Next Chapter…

CHAPTER 15 :
Thermodynamics

47