Module_SP015

In general for fixed quantity of gas of mole occupy a volume at
pressure and absolute temperature , we have :

Volume (unit in m3) Molar gas constant
(R: 8.31 J K−1 mol−1)
1L = m3

Pressure (unit in Pa) Number of mole Absolute
(unit in mol) Temperature
1 atm = Pa (unit in K)
( )
( )

IDEAL GAS EQUATION

The number of moles of gas can also be expressed as :

where : mass of gas ( in kg )
: molar mass of gas ( in kg mol–1)
: number of gas molecules
: Avogadro’s constant (6.02 1023)

Hence, the ideal gas equation can also be written in terms of number of

molecules present: Ideal Gas is a

hypothetical perfect

or gas which obeys all the

gas laws over all pressure

and temperature.

where is called Boltzmann constant and has the value



. / =
. × /

13.2 Kinetic Theory of Gases

Why we need Kinetic Theory of Gases ?

It relates

the macroscopic

property of gases

(pressure-
temperature-volume-

internal energy)

to

the microscopic

property – the motion of
atoms or molecules

(speed)

Remember the
Assumptions of kinetic Theory of Gases

1) All gases are made up of identical atoms or molecules.
2) All atoms or molecules are in constant, move randomly,

colliding each other and with the walls of the container. By
“randomly” mean that any molecule can move in any direction
with any speed.
3) All collisions are perfectly elastic collision.
4) The volume of the atoms or molecules is negligible, when
compared with the volume occupied by the gas.
5) Intermolecular forces are negligible except during the
collision. (There are no interactions-attraction/repulsion between
atoms or molecules)
6) The duration of collision is negligible compared with the time
spent travelling between collisions.

Why take rms speed ?

According to Kinetic theory of gases, all gas molecules are in a
state of constant random motion; individual particles move at
different speed, constantly colliding and changing directions.

We use velocity to describe the movement of gas molecules (considering both
speed and direction).

Gas molecules moving in opposite directions have velocities of opposite signs

Velocities of gas molecules are + and , by assuming that all particles are moving
equally in different directions, so an average would be zero ! This value is
unhelpful !

Squaring first gives all positive velocities

Taking the root of the square of the average velocities give the average speed

without direction.  

Mean squared speed of gas molecules,

The root-mean-square speed, is the measure of the speed of

molecules in a gas, defined as the square root of the average (mean) velocity-

squared of the molecules in a gas.

  For example, five gas molecules are found to
have speeds of 500,600,700,800, and 900 m s─1

, then the rms speed of the gas molecules would

  be

=   ( ) +( ) +( ) +( ) +( )



To compute the rms speed, we m s─1
square each molecular speed,
add, divide by the number of Of course, in a room, there are a lot more than 5
molecules and take the particles of gas!
square root.

Pressure of gases on the basis of kinetic theory of gases

According to the kinetic theory of gasses,

gas pressure is the result of the
collisions of large number of
molecules on the wall of the
container.

As gas molecules collide the walls of its
container, it will change direction, when
this happens the particle exerts force on
the wall of the container

Refer : pressure

The force exerted over a unit area by all
gas molecules as they hitting the walls of
the container around them is pressure.

Pressure can relate to the root mean squared speed of the gas

Rearrange : we obtain an important
result that relates the
Density of gas is
We have : macroscopic quantity
(pressure) to a
microscopic quantity
(root mean square
molecular speed) Thus,

we have established
a key link between the
molecular world and the
large-scale world.

Pressure of gases on the basis of kinetic theory of gases Extra note for better understanding

Consider a cubical box of side
length containing molecules of
gas each of mass, .

Focus our attention on one of
these molecules and assume that
it is moving
so that its component of velocity in
the direction is

Perfectly For identical
elastic molecules:
collision
with wall Extra note for better understanding

+ +


Force of molecular collision with wall


But this can related to the average
(mean) square speed:

The time for a “round trip” is
So the average force is



so total force on the wall due to
molecules colliding with it is :

Consider one molecule with velocity Total pressure exerted on the wall:

components and apply < >

Pythagorean theorem

motion is completely random Knowing :   Extra note for better understanding
Thus

Thus from eq. (1), total force on the wall Density of gas is
due to molecules colliding with it

We have :

rms speed of gas molecules and temperature

From the ideal gas equation Can also be written as

 

Compare it with

: Mass of a molecule (in kg)

: Molar mass of the gas

( in )

: Absolute temperature ( in K )
: Boltzmann constant

( )
  : molar gas constant

(8.31 J K−1 mol−1 )

This indicates that the root – mean – square speed , is directly proportional to
  and inversely proportional to  

o rms speed of gas molecule increase with increasing temperature.

o At a given temperature, lighter molecules moves faster, on average, than heavier
ones.

On average, the nitrogen
molecules ( = 28 g/mol )
in the air around us are
moving faster than the
oxygen molecules ( = 32
g/mol). Hydrogen molecules
( = 2 g/mol ) are fastest of
all.

13.3 Molecular kinetic energy and internal
energy

Average translational kinetic energy of a molecules

Equation

Can be written as

NOTE : (average translational kinetic energy)






This indicates that the pressure of a gas, is proportional to the number of

molecules per unit volume ( ) and to the average translational kinetic

energy of the molecules (
)


Since Therefore
Compare (1) and (2)
Average translational
kinetic energy of a
gas molecules

Average translational This result tells us that temperature is a direct
kinetic energy of a measure of average translational molecular
gas molecules kinetic energy. At a given temperature T, all ideal
gas molecules—no matter what their mass—have
the same average translational kinetic energy

The total translational kinetic energy
of molecules of gas is simply
times the average kinetic energy of a
molecule, which is given by Equation



Since

Molecules of gas that move randomly
can have translational (linear) kinetic
energy which is :

Beside translational (linear) KE, gas
molecules can also have kinetic energy
due to rotational and vibrational motion

Degree of freedom, Translational
motion of the center
Degree of freedom, of mass

An independent mode of motion or an rotational motion about
the various axes
independent ways in which a gas molecule
Vibrational motion along
acquire or store energy. the molecular axis.

3 type of orientation mode in molecule :
1) Translational (Linear),
2) Rotational and
3) Vibrational (can only be present at very high
temperature above 1000K

Number of degree of freedom depends on whether
the molecule is monoatomic, diatomic or
polyatomic.

Type of molecule Degree of freedom,

Monatomic gas Translational Rotational Total

(e.g. He, Ne, Ar) 3
5
30
6
Diatomic gas

(e.g. H2, O2, N2)
32






Polyatomic gas

(e.g. H2O, CO2, NH3)
33






Principle of equipartition of energy

Principle of equipartition of energy

The principle of equipartition of energy states that the average (mean) kinetic
energy stored in each every degree of freedom of a molecule is

Remember In general, for a gas with degrees
of freedom,
Applying the principle of equipartition of energy, The average kinetic energy per
the average molecular kinetic energy for : molecule can be expressed as:

Monoatomic gas

Diatomic gas Average kinetic energy of a gas molecules
Polyatomic gas

Internal energy of gas,

o The internal energy of gas is the total of the kinetic energy and potential
energy of all molecules of gas.

o The potential energy is associated with the forces between 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

o The internal energy of an ideal gas is simply the sum of average
kinetic energy of its molecules.

o Average kinetic energy of a molecule is given by:
o If there are molecules in the ideal gas, its internal energy is simply multiply

with average kinetic energy of a molecule:

Internal energy of gas

Since (



Therefore Internal energy of gas

: degree of freedom (depends on types of gas)

: numbers of mole

: number of molecules

: Boltzmann constant ( )

: molar gas constant (8.31 J K−1 mol−1 )

This result implies that the internal energy of an ideal gas depends only
on the temperature.

For example if the internal energy of a gas system decreases, the temperature
will drops. REMEMBER : As temperature change, internal energy of gas
change.

TOPIC 14 THERMODYNAMICS

14.1 FIRST LAW OF THERMODYNAMICS MODE Face to face Non Face
14.2 THERMODYNAMIC PROCESSES SLT to face SLT
14.3 THERMODYNAMICS WORK Lecture
Tutorial 1.5 1.5

33

LEARNING OUTCOMES

At the end of this topic, students should be able to :

14.1 First law of thermodynamics b) Discuss p-V graph for all the thermodynamic
a) State the first law of thermodynamics,
processes.
Q  U W
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) (Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

b) Solve problem related to first law of 14.3 Thermodynamics work
thermodynamics
a) Discuss work done in : V
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6) V
(i) Isothermal, W  nRT ln[ f ]
14.2 Thermodynamic processes (ii) Isochoric, W  0 i
a) Define the following thermodynamics Vf
processes :
(i) Isothermal  PdV 
(ii) Isovolumetric Vi
(iii) Isobaric (iii) Isobaric, Vf Vf
(iv) Adiabatic
W   PdV  P dV  p(Vf Vi )
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) Vi Vi

(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)

b) Solve problem related to work done in

isothermal, isobaric and isochoric

process.

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

14.1 FIRST LAW OF THERMODYNAMICS

Thermodynamics is the study of how

energy can be exchanged between a system
and its surroundings.

Any system can exchange energy with its
surroundings in two general ways, as heat
or work.

This will affect the macroscopic properties of the system like
pressure ( ), volume ( ) and temperature ( ).

LO14.1 (a)

First law of Thermodynamics Outline the relationship between internal energy, work and heat

is a statement of the conservation of energy for a thermodynamic system.

First law of thermodynamics

states the heat ( ) supplied to a closed system is equal to the increase in
the internal energy ( ) of the system plus the work done ( ) by the
system on its surroundings”

: The change in internal energy
()

: Heat added into the system
: Work done by the system

CAUTION! Sign convention of

Positive ( ) Negative ( )
System absorb heat or System loses heat or
Heat flow into the system Heat flow out of the system
increases ( increases) decreases ( decreases)

Work done by expansion Work done compression

the system on the system

Make sure
you use the
correct sign !

LO14.1 (b) Let’s see how to solve problem related to first law of thermodynamics

100 J of work is done on a system
Internal energy increases by 74 J
Does the gas system gain or lose
heat and how much ?

J

System lose heat

Non Face-to-face SLT

Example 1 (LO 14.1 (b))

2500 J of heat is added to a system, and 1800 J of work is done on
the system What is the change in internal energy of the system?

Heat added → = +2500 J ; work done on system → = −1800 J

= ∆ +
∆ = −
∆ = 2500 − −1800 = J

Example 2 (LO 14.1 (b))

A gas expands under constant temperature condition and does
work of 30 J against the surrounding.

a) What is the change in the internal energy of the gas?
b) Find the amount of heat absorbed or lost by the gas.

(a) Constant temperature  ∆ =
(b) Find =? Work against surrounding  = +30 J

= ∆ +
= 0 + 30
= J

The heat, Q add to a gas system is used to How to know
(i) Change internal energy of the gas there is a change
(ii) Do work on gas system in internal
(iii) Both (i) and (ii) energy, or
not ?
How to know is there
any work done on or Refer to the change in
by the gas system ? temperature ( )

Refer to the change in
volume ( )

Change in internal energy

Recall Topic 13 :

temperature change, internal energy change.

 ( : initial ; : final) , also Initial state
 : increasing in internal energy (
final state
indicates increasing in temperature of the gas Referring to Fig. above,
(1) ∆ for all the path is same because all
 : decreasing in internal energy ( , also
have same initial and final point.
indicates decreasing in temperature of the gas. (2) ∆ = −∆

 If temperature constant, no change of internal

energy ( =0)

 value of only depends on initial and final state. It
does NOT depends on path taken.

W If the gas is compressed...

If the gas expands ...

...volume must increase and ...volume must
work done by gas ( ) decrease and work
done on the gas ( )

Work done on or by a gas depends on the path taken because work is represented
by the area under the line or curve in graph (will be explained more detail in subtopic 14.3)

LO14.2 (a)

14.2 THERMODYNAMIC PROCESSES

A thermodynamics process is a process This can happen due to either:

where one or more of the parameters of a a) Exchange of heat between system and surrounding.
system such as pressure, volume and b) Mechanical work being done on or by the system.
temperature undergo change. c) Both of (a) and (b).

4 ways in which thermodynamic process can be carried out in a controlled manner:

1. Isothermal process: A process with no change in temperature ( constant).
2. Isochoric (Isovolumetric) process: A process with no change in volume (

constant).

3. Isobaric process: A process with no change in pressure ( constant).
4. Adiabatic process: A process with no heat transfer into or out of the system (Q = 0).

LO14.2 (b)

How can we represent a thermodynamic process ?

graph Initial state
1

• A graph is a plot of pressure

vs. volume

• The initial state is a point on the final state
2
graph
V
• The process is the line or curve 0

• The final state is another point

• Work done during the process

shown by the graph can be

determine by looking at the area

under the line or curve Increasing temperature

Remember this is same as integral

expression for the work.

LO14.2 (b)

Thermo- Condition First law of − graph Ideal gas law
dynamic (What is Thermodynamic apply
Process constant ? ) Expansion Compression
= ∆ +
Isobaric constant On a − graph, an isobaric process would show in accordance
There are
(changing in typically internal up as a horizontal line, since it takes with Charles
and ) energy changes. ' law - the
Work is done by place under a constant pressure.
(or on) the volume of a gas
system, and heat Isobaric expansion Isobaric compression is proportional
is transferred (in ( > ) ( ) ( < ) to its
or out), so none (Pa) temperature
of the quantities
in the first law of = constant
thermodynamics < 0
readily reduce
to zero. > 0

( 3) 0 ( 3) =

0

∆

 For expansion, Δ and work are positive
 For compression, Δ and work are

negative

LO14.2 (b)

Thermo- Condition First law of − graph Ideal gas law
dynamic (What is Thermodynamic apply
Process constant ? ) On a − graph, isovolumetric process would
= ∆ + in accordance
Isovolu- constant makes show up as a vertical line, since it takes
metric with pressure
a.k.a (changing in thus place under a constant volume. law - the
Isochoric and )
pressure of a gas
(Pa) (Pa) is proportional to
its temperature

all of the heat = constant
either comes
from internal
energy or goes
into increasing 0 (m3) 0 (m3) =
the internal
energy.
(a) Increased pressure (b) Decreased pressure

volume is constant, the system does no work

LO14.2 (b)

Thermo- Condition First law of − graph Ideal gas law
dynamic (What is Thermodynamic apply
Process constant ? ) Expansion Compression
= ∆ + in accordance
Isother constant On a − graph, an isothermal process would
mal makes with Boyle’ s
(changing in show up as a curved line law - the
and ) internal energy
pressure of a gas
of an ideal is inversely
proportional to
gas depends its volume
solely on the
temperature, so = constant

the change in internal =
energy during an
isothermal process In general one curve on − graph
for an ideal gas is O. represent one temperature. Curve closer to
the origin, has lower temperature.
thus

All heat added

to a system (of

gas) performs
work

LO14.2 (b)

Thermo- Condition First law of − graph Ideal gas law
dynamic (What is Thermodynamic apply
Process constant ? ) Expansion Compression
= ∆ + not
Adiabatic No heat On a − graph, an adiabatic process would applicable
flowing thus
into or out show up as a curved line.
from This can be
system accomplished by:
(i) surrounding the
= entire system with
a strongly
insulating A system that expands under adiabatic conditions
material or does positive work, so the internal
(ii) performing the energy decreases, is a cooling process
process very A system that compress under adiabatic conditions
rapidly -- there is does negative work, so the internal energy
no time for a increases, is a warming process.
significant heat
transfer to take
place.

CAUTION ! Quick check 1

Note that the − graph for isothermal and A : .......................................................
adiabatic processes look similar (curved line). B : .......................................................
But actually the adiabatic curve is steeper than C : .......................................................
isothermal curve and adiabatic process D : .......................................................
involves 2 different temperature curves.

→ ( = )
→

( < )

→

( > )

→

( = )

LO14.3 (a)

14.3 THERMODYNAMICS WORK

How does a thermodynamic system do work?

Figure shows a gas confined to a cylinder that
has a movable piston at one end.

If the gas expands against the piston, it
exerts a force through a distance and does
work on the piston.

If the piston compresses the gas as it is
moved inward, work is also done—in this
case, on the gas.

LO14.3 (a)

The work associated with such volume changes can be determined
as follows:

Let the gas pressure on the piston face be .

Then the force on the piston due to the gas is , where A is the

area of the face.

When the piston is pushed outward an infinitesimal distance , the

magnitude of the work done by the gas is

Since the change in volume of the gas is , this becomes

LO14.3 (a)

For a finite change in volume from to , we can integrate this
equation from to to fin  d the wo  rk :

  

The integral is interpreted graphically as )

the area under the curve (the )

shaded area of Figure below).

Work done by the gas is positive for
expansion and negative for
compression.

LO14.3 (a) final Work Done in
isovolumetric process
Work Done in isobaric
process In the isovolumetric process, volume is
constant. Therefore,
In the isobaric process, the pressure
remain constant, so final

initial

initial

Since

Work done is also represented on
the diagram by the area

under the process line

LO14.3 (a)

Work Done in isothermal process

In the isothermal process, temperature, T is

constant.

Work done is = ∫ initial final
For an ideal gas, = final initial

=


= 1 Since temperature is constant, apply Boyle’s law :
=

= ln rearrange :

= (ln − ln ) Therefore,




FORMAT KERTAS PSPM FIZIK SP015

Paper 1 Paper 2

Topic (1 hour) (2 ½ hour)

1 PHYSICAL QUANTITIES AND MEASUREMENTS 40 MCQ answer ALL 13 questions answer ALL
2 KINEMATICS OF LINEAR MOTION Mark Allocated Mark Allocated
3 MOMENTUM AND IMPULSE 1 2
4 FORCES 3 10
5 WORK, ENERGY AND POWER 1 6
6 CIRCULAR MOTION 2 7
7 GRAVITATION 3 10
8 ROTATIONAL OF RIGID BODY
9 SIMPLE HARMONIC MOTION (SHM) 2 5
10 MECHANICAL WAVES AND SOUND 8
11 DEFORMATION OF SOLIDS 2
12 HEAT ASSIGNMENT
13 GAS LAW AND KINETIC THEORY 2 12
14 THERMODYNAMICS 16
2 4
TOTAL 6 6
6 8
2 6
3 100
3
40

CAN I SCORE ?

Let’s find out :

Assignment 9.0 36/40 Grade C : 50%
Practical Test 14.0 Only 14 marks more to pass the course !
Lab report 13.0
PSPM 44.0 Paper 1 Paper 2
{15} {45}
Aku dapat You are
Fizik A !!! what
you
think ;)

Fake it until 3 days (18, 19, 20 Oct) just focus and prepare for 1 subject !
you make it You have no excuses that you can’t score for Physics !

oIt’s the little things
that lead to a BIG
slump.
1×1×1×1×1×1×1×1
×1×1….. = 1

But if add in 0.1
everyday ….

1.1× 1.1 × 1.1 × 1.1
× 1.1 × 1.1 × 1.1
= 2.14358881

THE END

SP015

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