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FEDERAL UNIVERSITY OF TECHNOLOGY MINNA
SECOND SEMESTER
PHYSIC 123 HANDOUT

DIGITALLY PREPARED BY:
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Matter:
THis is anything that has mass and occupies space
...


Electron:
This is the outer skin of matter, when electron are charged i
...
rubbed
one’s palm together, heat is generated
...
The
intermolecular force is high and this makes solid very difficult to
break
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Solid can vibrate at a foxed position transforming energy from points
to point
...

Gas:
The molecules in a gas are scattered
...
Energy in gas is transfer from molecules to
molecules
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V P = Saturated Vapour Pressure
Q = Mct
Q = Ct
Q = ML
Q = mgh = ½ mv2 = IVt

EFFECT OF HEAT

1
...


Melting ( it involves change in temperature, is applicable to
solid only)

3
...


It causes change in state (It is applicable to liquid)

5
...


It causes thermal agitation in molecules

7
...


it causes boiling

SATURATED VAPOR PRESSURE
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This is a pressure/degree of how hot or cold a body is
...


ZENOTH’S LAW
For instance, when A = B = C
...

Zenoth’s law states if the temperature of three bodies after
separation remains the same they are said to be in thermodynamic
equilibrium

Frictionless Piston

F

P

Distance (s)

Consider a mass of gas enclosed in a cylinder by frictionless piston of
cross sectional area (A) which is in equilibrium under the action of an
external force (F) ,acting to the left and a force view to be pressure (P) of
a gas acting on the right as shown on the diagram
...
dv
Final

W = P(V2 – V1) where V2 > V1
W = PdV
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NB: dV = change in pressure so that
P
...
dV
From the first law of thermodynamics,
dq = du + dw
Hence
dq = du + PdV

-

-

-

2

Remember that PV = RT (Gas law)
Hence,

V =

RT
P

HEAT CHANGE
Principle of heat capacity of gas

Molar heat capacity at constant volume (Cv) is the heat required to produce a rise in
temperature in one mole of a gas when volume is kept constant
...

From equation (2) above, dq = du + Pdv
NB
dq = change in quantity of heat
du = change in internal energy
PdV = change in external energy

dq = No
...
e
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of moles × Cv
du = nCvdt -

-

-

-

-

-

-

4

-

-

-

5

Substitute dq and du in equation (2)
From equation (2) dq = du + PdV
nCpdt = nCvdt + PdV -

-

However, from ideal gas equation
PV = RT
P(V + dv) = R(T + dT)

-

-

-

-

6

PdV = RdT

-

-

-

-

7 (when differentiated)

-

-

Substitute equation (7) in equation (5)
NB: from equation (5) CpdT = RdT + PdV
CpdT = CvdT + RdT - -

-

-

-

-

8

-

-

-

-

9

(Cp – Cv)dT = RdT
Cp – Cv = R

-

-

NB: R is called universal constant (R = 0
...
40

QUESTION

Why is Cp greater than Cv
Ans: Cp is greater than Cv because at constant volume, heat energy is higher
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properties of

matter
Remember from ideal gas law; PV = RT
Also remember that an ideal gas is a gas in which its volume is negligible compare
with the volume of the container
A real gas is a gas that deviates from Raull’s Law (Remember the Raul’s Law?)
From Boyle’s law PV = K
Charles Law V/T = K
From Pressure Law P/T = K
Now show that

PV
=K
T

(remember all of that? Great)
NB: K = constant

Therefore, PV = KT

QUESTIONS

1
...
A piston is lowered into the cylinder decreasing the
volume occupied by the gas to 80L and raising the temperature to 25oC
...
1atm

QUESTION:

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Calculate Cp and Cv of oxygen if their ratio is 1
...
43Kgm-3 at STP, the pressure at 1atm = 1
...
40

-

-

1

NB: Density of O = 1
...
01 × 105Nm-2 and molar mass O = 32 ×
10-3
But from the formulas state above; R = Cp - Cv

-

2

So, from equation (1) Cp = 1
...
8205 = 0
...
05
Substitute the value of Cv in equation (3); Cp = 1
...
40(2
...
87
NB: Check that

Cp 2
...
40 correct!
Cv 2
...
What will
be the temperature of the gas when the pressure increases by 1/3atm?

Solution
From the equation
P 1 / P 2 = T1 / T2
NB: P1 = 1500N/m2, P2 = 1/3 × 1500 = 500 + 1500 = 2000Nm-2
T1 = 28 + 273 = 301K and T2 = ?
Hence,
301 / 200 = 301 / T2
Therefore, T2 = 401
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Examples of microscopic properties of gas are;
-

speed

-

mass

NB: Macroscopic Properties of gas are
-

Pressure

-

Volume

-

Temperature

ASSUMPTION OF KINETIC THEORY OF MATTER
1
...
The molecules of the gas moves in random motion, still obeying Newton’s law
of motion
3
...
The molecules are the solid

L
c

m

u

x

v
y
5
...
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We
can resolve that (c) in three components (u, v, w) in x, y and z axis respectively
NB: A cubic box is used because it has three dimensional surfaces and matter has
three states
...
e
...
e
...
un2 ) L

- (3)

Where u1, u2, u3 …un are the different “Ox” components of the velocities of the
molecules of (1, 2, 3, …n)
Since the gas molecule exerts pressure on the container
Hence P = F / A

ρP=

M 2
(u1 + u22 + u32 +
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un2 )
N

__

Ox direction then

u

2

=

-

-

(5)

(6)

From equation No
...
un2

-

-

-

-

-

-

-

(7)

From equation (4)
_

ρ=

NM u 2

-

L2

-

NB: Resolving “c” the velocity into three dimensional surfaces, hence
C2 = u2 + y2 + w2 -

-

-

-

(8)

2

The Mean Square Value of the velocity (C ) can be given as
−

−

−

−

C 2 = u 2 + v 2 + w2
(9)
Since N (number of molecules ( ) is large and the molecules moves randomly the
−

mean square velocity C 2 are equal
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ofMass
(i
...
e
...
(Can you figure out why?)
Recall also that,
PV = RT
- (4)
(from General gas law)
So that,
2
1 −2
PV = N A m C = RT (5)
3
2
Hence

PV =

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1 −2 3 R
mC =
T - (6)
2
2 NA
(Try rearrange equation (5) and confirm to yourself that equation (6) is perfectly
correct)
NB: ½ mc2 is known as kinetic energy or translational motion
So we can see that R α N A
The ration R : NA = Boatman’s constant = k
By introducing this constant to equation (6) therefore,
1 −2 3
mC = kT
- (7)
2
2

K
...


DEDUCTION FROM KINETIC THEORY OF MATTER
1
...
e
...
This statement is known as Avogadro’s
Law
2
...

The kinetic theory attributes gas pressure to bombardment of the walls of
the containing vessels by molecules
...
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It determines grahams law of diffusion
Diffusion is the movement of gas molecules in air from one place to another
GAS1

GAS2

Ratio of diffusion of Gas1
Ratio of diffusion of Gas2
− 2

C1

=

−

C2

2

− 2

NB:

C

is the mean square velocity (Crms)

P =1
− 2

C1
_

C2

=
2

d1
d2

Rate of diffusion of Gas 1
Rate of diffusion of Gas 2
d1
=
d2
Therefore, the rate of diffusion of gas is inversely proportional to the square root of its
density
QUESTION
Calculate the root mean square speed of the molecules of hydrogen at (a) 273K (b)
373K
Density of hydrogen at s
...
p = 9
...
01 x 105N/m2

Solution
From equation above
−

dC 2
P=
3
Where P = pressure, C 2 = mean square velocity and d = density

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3P = dC2
NB: Crms = C
But P = 1atm = 1
...
8205 × 273
1
...
8205)
V
so solving the above therefore V = 2
...
8205 × 273
5
=
9
×
10
C
2
...
4 x 102m/s
...
Therefore
Cr
373
=
Where C = 1
...
150 x 103m/s
QUESTION
From the following speed below
0
...
5, 1
...
75, 3
...
72
Compute
_

(i)

Average speed (mean speed value) ( C )
_ 2

(ii)
(iii)

Mean Speed square ( C )
Root Mean Square (Crms)

Solution
_ 2

0
...
5 + 1
...
75 + 3
...
72
= 1
...
5) 2 + (2
...
6) 2 + (2
...
2) 2 + (0
...
56m / s
6

_ 2

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_ 2

(iii)

Root Mean Square Speed (Crms) =

C

=

4
...
1m / s
_ 2

_ 2

NB: This example shows clearly how to compute C , C and Crms

ADIABATIC EXPANSION

P
Adiabatic Expansion

Isothermal Expansion
V-1

An adiabatic system is an isolated system in which no heat leaves the system or enters
the system
NB: dq = du + PdV
When dq = 0, then
0 = du + PdV
du = -PdV
When there is increase in thermal energy by the system it means work is been done on
the system
...

EXAMPLES OF ADIABATIC PROCESS
i
...
Expansion of hot gases in an internal combustion engine
iii
...

The calorimeter is lagged with cotton wool or silk
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This means that a small increase in volume is accompanied by increase in
pressure, hence there is a spontaneous expansion towards C, that also implies that a
small decrease in volume will lead to a spontaneous contracti8on towards “b”
dW = RT
...

V1
P2

In reversible adiabatic process, we move the piston extremely slowly from an ideal
γ −1
T2
V1
gas, an adiabatic expansion gives a temperature drop while the ratio
=
T1
V2
Cp
NB γ =

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Therefore, by substitution into equation (7) we have
PdV + VdT
dT =
(viii)
Cp − Cv
PdV + VdT
PdV
=−
(ix)
Cp − Cv
Cv
PdV
NB: dT =
from equation (5)
Cv
PdVCp + VdPCv = 0 (x)
Dividing through by V
PdVC p
+ dPC v = 0 (xi)
V
Dividing through by P
dpC v
dVC p +
=0
P
This process constitutes the initial separation in the two phases gaseous at “c” and
liquid at “b”
...
The liquid can be made in this region
...
So therefore, the region c is also known as unstable region because a gas ca
be made at super cooling
...
e
...
This also
occurs in liquid phase
...

(i
...
Critical Isotherm at Z)
...

2
2
d P/dV = 0 at z
multiplying V with Van der Waal equation;
a
V P + 2 (v − b ) = RT
V

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a
− Pb − ab = RT
V
dP
a
dP
2ab
P +V
T − −b
T+ 2 =0 dV
v
dV
V

3

PV +

dP
T =0
dV
a 2ab
P= 2 − 3
V
V
dP
2a 6ab
=− 3 + 4 =0
dV
V
V
a
Vc = 3b,
Pc =
27h 2
8a
T c=
27 Rb
RT c
for a single gas;
PcV c
Coefficient lies between 3 – 3
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TYPES OF MODULUS
1
...
Shear Modulus ( ε )
3
...
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E)
NB: s is distance covered while h is the height
Hence,
1
W = Fe
2
NB:
1
Fe
1 F e
1 Tstress
2
×
=
=
2 A L
2 Tstrain
AL

When stress is applied to a material we have dW/W
While strain applied to a material is e/L
Hence,
dW e
e
α = −ϕ
W
L
L
dW
W = dW × L
- ϕ = poison ratio =
e
W
e
L
Example:
A material 4m long and 2cm2 in cross section stretched 2mm ,when a mass of 20Kg is
placed at the lower end
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8
4
E=
×
= 1960 x 107Nm-2
−4
2 × 10
2 × 10− 3

SHEAR MODULUS

F|| (Force line is tangential

a

c

b

F|| = Force line is tangential
A = Area
dL = change in Length
L = Original length
Ts α T||s

d
Share Modulus for a particular material is 50Pa
...
Find the sheer
stress
(2) How far will the surface be displaced NB: material under consideration is cube
...
No working was provided
to this problem as at the time of typing this note)
EXAMPLE
Two parallel and opposite forces each 4000N are applied tangentially to the upper and
lower phases of a cubical metal box which is 25cm on the side
Find the angle of shear and the displacement of the upper surface relative to the lower
surface shear modulus (G = 80gPa
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dP = 5mPa, V = 10mm = 10-6m3 Vy = ?
Vy = 5 x 10-6 / 103 =
5 x 10-9m3
Cohesion

Change of
contact

•
•
•
•
•

The force of cohesion is greater than the force of adhesion e
...
mercury
...
Also you should know that surface tension is the elastic tendency
of liquids which makes them enquires the least surface area possible
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NB: Surface tension = γ =

F
2L

ADHESION: This involves the molecules of different substances
...

COHESION: This involves the forces acting between the same molecules
...
g
...


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EXAMPLE:
The excess pressure “P” within a bubble of a radius 0
...
Take surface
tension T = 72
...
7 x 10-3, r = 0
...
7 × 10 −3
=
= 1454 N / m
r
1 × 10 − 4

CAPILLARITY
The narrower the capillary tube, the faster the movement of the liquid above
the tube, while the wider the capillary tube the slower the movement of the liquid
above the capillary tube
...

A body weighs 0
...
12N when fully immersed in water and 0
...
Calculate ( i) It lost weight (ii) it
relative density
2
...
5gcm3 until ¾
of its volume in water is immersed in the liquid
...

Differentiate between cohesive and adhesive forces
...

If adhesive forces are greater than cohesive forces, what happens to the
liquid in capillary tube?
5
...

If a surface is wetted by water, what will be the angle of contact?
7
...
60; relative density of water to be 1000kg/m3
NB: Relative density of substance = RD

VISCOSITY AND COEFFICIENT OF VISCOSITY
Viscosity is the inherent property of all fluids and may be called the internal
friction offered by a fluid to the flow
...
The motion subside after sometimes which can happen only in the presence of a
resisting force acting on the liquid
...
Viscosity is that property of fluid due to which they
opposes relative motion between its adjacent layers
...
which flows readily are said to be mobile,
while liquid such as tar, pitch which do not flow readily are said to be viscous
...

Consider a liquid contained in between two plate A and B, a small distance
apart as shown below
...
B
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Thin parallel horizontal layer and
the upper plate are polished with a small force (F) in the direction indicated
...

The next layer also moves with a constant velocity but its velocity is slightly
less than the velocity of the upper layer
...
In fact, all the layers moves but the velocity keep
decreasing towards B and the velocity of the layer in contact with B is zero
...
Each layers of the liquid like each card drags with the next lower
layer
...

Similarly, due to sliding of one area over the other, forces are called into
action within the liquid which resist the motion of layers
...

Consider a layer AB of a liquid moving with a velocity (V) ant a parallel layer
(CA) at a distance (DS) from it moving with a velocity of (V + DV)
...
Then DV/DS is known as velocity gradient
...
In the absence of this force the relative motion between the two layer cease
after some time
...


Velocity gradient (DV/DS) in a direction perpendicular to the layer

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2
...
A
DS
Therefore,
DV
F =− A
DS
Fα −

-

-

-

(i)

This is called Newton’s law of viscous flows in straight line motion
...

symbol
The –ve sign indicates that air is a backward dragging force
...
e
...

Now if A = 1, DV/DS = 1, then
F=
(coefficient of viscosity)
Hence, the

coefficient of viscosity of a liquid may be defined as the tangential

force per unit are which resist the flow of two parallel layer having unit velocity
gradient perpendicular to the streamline flow
...
The coefficient of velocity of a liquid
is equal to the tangential force per unit area that is required to maintain unit relative
velocity between its two layer unit distances

STREAMLINE AND TURBULENT
A streamline flow is that in which all particles of the liquid moves in an orderly
pattern and the part of every particles is the same as that of the liquid as a whole
...
In such a flow the velocity of every point within the liquid
remains constant both in magnitude and direction the flow remains streamline so long
as the velocity is not large and remain the load in a certain velocity is called critical
velocity beyond which the flow becomes turbulent and is hardly regular
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On the basis of series of experiment,
Reynolds concluded that the constant K which is the combination of four factors
determines whether the flow is streamline or turbulent that is, Reynolds number
determines the nature of the flow of liquid through a fluid on the basis of the
experiment, it is observed that for a tube of 1cm, flow is laminar if K is 2000 and in
general when K is less than about 2000, the flow is laminar, where as above 3000, the
flow is turbulent
...


POISEUILE EQUATION FOR THE FLUID OF LIQUID THROUGH A TUBE
When a liquid is flowing under a constant pressure between the end of a capillary, the
tube of radius (A) and length (m) it is assumed that;
The flow of liquid is parallel to the axis of the tube everywhere and to the streamline
...
e
...

The velocity of the liquid layer in contact with the wall is zero and increases regularly
and continuously towards the interior becoming maximum along the axis of the tube
...
3
Consider a cylinder layer of liquid to coaxial to the tube and radius shown in fig
above
...
And hence,
will exert a backward tangential force on the inner faster moving liquid
...
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Title: Properties of Matter
Description: Concise note material on properties of matter in physics.

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