Transcript Chapter 4 Notes

Chapter 4
Physics of Matter
Matter: Phases, Forms &
Forces
 We classify matter into four categories:

Solid: rigid; retain its shape unless distorted by a
force.

Liquid: flows readily; conforms to the shape of a
container; has a well-defined boundary; has higher
densities than gases.

Gas: flows readily; conforms to the shape of a
container; does not have a well-defined surface; can
be compressed readily.

Plasma: has gaseous properties but also conducts
electricity; interacts strongly with magnetic fields;
commonly exists at higher temperatures.
2
Matter: Phases, Forms &
Forces, cont’d
 The chemical elements represent the
simplest and purest forms of everyday matter.


There are currently 114 different elements.
110 of them have accepted names.
 Each element is composed of incredibly small objects
called atoms.



There are 114 different atoms, one for each of the
known elements.
Only about 90% of the elements exist naturally on
Earth.
The others are artificially produced in laboratories.
3
Matter: Phases, Forms &
Forces, cont’d
 The atom is not indivisible.

It has its own internal structure.
 Every atoms has a very dense, compact core called
the nucleus.

The nucleus is composed of two kinds of particles:
 Protons: have a positive electric charge.
 Neutrons: have no electric charge.
 The nucleus is surrounded by one or more particles
called electrons.

Electrons have the same electric charge as protons but
are negatively charged.
4
Matter: Phases, Forms &
Forces, cont’d
 Every atom associated with a particular
element has a fixed number of protons.

The number of protons distinguishes the element.
 Atoms with two protons are helium atoms.
 The atomic number of an element specifies the
number of protons.

The atomic number of helium (He) is 2 because it
has two protons.
 Each element is given its own chemical symbol.
 This is a one- or two- letter abbreviation.
5
Matter: Phases, Forms &
Forces, cont’d
 Atoms can have various numbers of neutrons.
 Atoms with different numbers of neutrons for a
certain elements are called isotopes.

More on this in chapter 11.
6
Matter: Phases, Forms &
Forces, cont’d
 Chemical compounds are the next simplest form
of everyday matter.

Examples: water, salt, sugar, etc.
 Compounds are made from building blocks
called molecules.

Every molecule of a particular compound consists
of the same unique combination of two or more
atoms.

Each water molecule consists of two hydrogen atoms
and one oxygen atom.
7
Matter: Phases, Forms &
Forces, cont’d
 Each compound can be represented by a
chemical formula.





water is H2O;
salt is NaCl;
carbon dioxide is CO2;
sugar is C12H22O11;
ethyl alcohol is C2H5OH.
8
Matter: Phases, Forms &
Forces, cont’d
 Many substances are composed of two or more
different compounds that are physically mixed
together called mixtures and solutions.


Air is a mixture of
several gases.
The actual composition
varies widely from day
to day and place to
place.
9
Behavior of atoms and
molecules
 The constituent particles of atoms and
molecules exert electrical forces on each
other.

“Static cling” is an example of an electrical force.
 The forces depend upon the configuration of
the atoms in each atom.
10
Behavior of atoms and
molecules, cont’d
 Solids: Attractive forces between particles are very
strong; the atoms or molecules are rigidly bound to
their neighbors and can only vibrate.
 Liquids: The particles are bound together, though
not rigidly; each atom or molecules move about
relative to the others but is always in contact with
other atoms or molecules.
 Gases: Attractive forces between particles are too
weak to bind them together; atoms or molecules
move freely with high speed and are widely
separated; particles are in contact only when they
collide.
11
Behavior of atoms and
molecules, cont’d
 Atoms or molecules in a solid arranged in a
regular geometric pattern are called crystals.
 Solids that do not have a regular crystal
structure are called amorphous solids.
12
Behavior of atoms and
molecules, cont’d
 Carbon has two common crystalline forms.


Graphite forms crystalline sheets with little
bonding between
sheets.
Diamond forms very
strong bonds between
adjacent carbons
— it’s the hardest
known natural
substance.
13
Behavior of atoms and
molecules, cont’d
 In liquids, the inter-atomic forces are
insufficient to bind the atoms rigidly.

The atoms are free to move and vibrate.
 In gases, the inter-atomic forces are virtually
negligible unless the atoms are very close.

Gaseous atoms have rather high speeds: ~1,000 mph.
14
Behavior of atoms and
molecules, cont’d
 Whenever a high-speed gas atom impacts a
large force on a container.

It is this force that produces a pressure on the
container.
 As the air molecules strike the inside of a tire, they
produce the pressure that inflates the tire.
 If the molecules are stationary, there is no pressure.
15
Behavior of atoms and
molecules, cont’d
 Gases are easily compressed because the
majority of their volume is the space between
the atoms.

If you compress it enough, you force the atoms
close enough together that you form a liquid.
16
Pressure
 Pressure is the force per unit area for a force
acting perpendicular to a surface.

Since we use the perpendicular component of
the force, pressure is a scalar.
force
pressure 
area
F
p
A
17
Pressure, cont’d
 The units of pressure are:

Metric:



pascal (Pa; 1 Pa = 1 N/m2) — SI unit;
millimeters of mercury (mm Hg).
English:



pound per square foot (lb/ft2);
pound per square inch (lb/in2 or psi);
inches of mercury (in. Hg).
18
Pressure, cont’d
 Some pressure conversions:

1 psi = 6,890 Pa.
 We also use an atmosphere (atm) as a
pressure unit:

One atmosphere is the average pressure
exerted by air at sea level:


1 atm = 1.01×105 Pa;
1 atm = 14.7 psi.
19
Pressure, cont’d
20
Example
Example 4.1
A 160-pound person stands on the floor. The
area of each show that is in contact with the
floor is 20 square inches. What is the
pressure on the floor?
Assume the person’s weight is shared equally
between the two shows.
21
Example
Example 4.1
ANSWER:
The problem gives us:
F  160 lb
A  20 in 2
Since the shoes support the weight equally,
each show must support a weight of 80 lb.
The pressure is
F 80 lb
p 
 4 psi.
2
A 20 in
22
Example
Example 4.1
DISCUSSION:
If the person stands on only one foot:
F 160 lb
p 
 8 psi.
2
A 20 in
If the person wore high heels and stood on
only one heel (0.5 in by 0.5 in):
F
160 lb
p 
 480 psi.
2
A 0.5  0.5 in
23
Example
Example 4.2
In the late 1980s, there were several
spectacular aircraft mishaps involving rapid
loss of air pressure in the passenger cabins.
The cabin pressure of a passenger jet cruising
at high altitude (25,000 ft) is about 6 psi (0.41
atm) greater than the pressure outside. What
is the outward force on a window measuring
1 foot by 1 foot and on a door measuring 1
meter by 2 meters?
24
Example
Example 4.2
ANSWER:
The force on the window is:
F  pA   6 psi 12 in 12 in 
 864 lb.
The force on the door is
F  pA   6 psi  39.4 in  39.4 in 
 9,310 lb.
25
Example
Example 4.2
DISCUSSION:
The force on a window is approximately the
weight of four adults.
The force on the door is nearly the weight of
five pickups.
26
Pressure, cont’d
 Pressure is a relative quantity.
 When you measure a pressure, you are
measuring it relative to some other pressure.

When you measure your
tire pressure, you are
measuring the pressure
in the tire above the
atmospheric pressure.
27
Pressure, cont’d
 The gauge pressure is the pressure relative
to the current atmospheric pressure.
 The absolute pressure is the gauge
pressure plus the atmospheric pressure.
pabsolute  pgauge  patmospheric
28
Pressure, cont’d
 The standard pen-shaped tire gauge
compares the tire pressure against a spring
and the atmospheric pressure.
29
Pressure, cont’d
 An important statement about gases.
 Consider a gas held at constant temperature.

Increasing the temperature increases the
motion of the gas atoms and therefore the
pressure.
 Under these conditions:
pressure  volume  constant
pV  constant
30
Pressure, cont’d
 This means:

Decreasing the volume increases the
pressure.


Imagine squeezing part of an inflated balloon.
Decreasing the pressure increases the
volume.

Imagine letting the air out of an inflated balloon.
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Density
 Mass density is the mass per unit volume of
a substance.

It is the ratio of the mass to the volume of the
substance.
mass
mass density 
volume
m
D
V
32
Density, cont’d
 Units of mass density:

Metric:



kilogram per cubic meter (kg/m3);
gram per cubic centimeter (g/cm3).
English:

slug per cubic foot (slug/ft3).
33
Density, cont’d
 Measure density by finding the mass of a
sample and dividing by the volume of the
sample.
 The actual size of the sample is irrelevant.

If you use a sample with twice the volume you
will twice the mass.
34
Density, cont’d
 You can measure the freezing point of your
car’s coolant by measuring the density.

The density of water and antifreeze are
different.



Water: 1,000 kg/m3.
Antifreeze: 1,100 kg/m3.
The overall density depends on the ratio of the
amount of water and amount of antifreeze.
35
Example
Example 4.3
The dimensions of a rectangular aquarium are
0.5 meters by 1 meter by 0.5 meters. The
mass of the aquarium is 250 kilograms larger
when it is full of water than when it is empty.
What is the density of the water?
36
Example
Example 4.3
ANSWER:
The problem gives us:
l  0.5 m
w 1 m
h  0.5 m
m  250 kg
The volume is:
V  l  w  h   0.5 m   1 m    0.5 m 
 0.25 m .
3
37
Example
Example 4.3
ANSWER:
The mass density is then
m 250 kg
kg
D =
=1,000 3 .
3
V 0.25 m
m
38
Example
Example 4.3
DISCUSSION:
If the width of the tank was doubled, the
amount of water would be doubled — the
density would remain the same.
If filled with gasoline, the mass is 170 kg. The
density of gasoline is
m 170 kg
kg
D =
=680 3 .
3
V 0.25 m
m
39
Density, cont’d
 If you know the volume and what type of
substance you have, you can find the mass:
m V D
40
Example
Example 4.4
The mass of water needed to fill a swimming
pool can be computed by measuring the
volume of the pool. Let’s say a pool is going
to be guilt that will be 10 meters wide, 20
meters long, and 3 meters deep. How much
water will it hold?
41
Example
Example 4.4
ANSWER:
The problem gives us:
l  20 m
w  10 m
h3m
The volume is:
V  l  w  h   20 m   10 m    3 m 
 600 m .
3
42
Example
Example 4.4
ANSWER:
Since the density of water is 1,000 kg/m3, the
amount of water is:
m  D V

 
 1000 kg/m3  600 m3

 600, 000 kg.
43
Example
Example 4.4
DISCUSSION:
Since a common bathtub is about 0.25 m3
(let’s say 1 m by 0.5 m by 0.5 m), this
swimming pool is about 2400 bathtubs of
water.
44
Density, cont’d
 Weight density is the weight per unit volume
of a substance.

It is the ratio of the object’s weight and its
volume.
weight
weight density 
volume
W
DW 
V
45
Density, cont’d
 Units or weight density:

Metric:


newton per cubic meter (N/m3).
English:


pound per cubic foot (lb/ft3);
pound per cubic inch (lb/in3).
46
Example
Example 4.5
A college dormitory room measures 12 feet
wide by 16 feet long by 8 feet high. What is
the weight of the air in it under normal
conditions?
47
Example
Example 4.5
ANSWER:
The problem gives us:
l  16 ft
w  12 ft
h  8 ft
The volume is:
V  l  w  h  16 m   12 m   8 m 
 1,536 m .
3
48
Example
Example 4.5
ANSWER:
The weight of the air in the room is
W  DW  V


 0.08 lb/ft 3 1,536 ft 3

 123 lb.
49
Example
Example 4.5
DISCUSSION:
Notice that as the temperature increases, the
density would decrease.
So the air in the room would weigh less on
hotter days.
Because there is less air (fewer air molecules) in the
room.
50
Fluid pressure and gravity
 The law of fluid pressure states that the
(gauge) pressure at any depth in a fluid at
rest equals the weight of the fluid in a column
extending from that depth to the “top” of the
fluid divided by the cross-sectional area of the
column.
51
Fluid pressure and gravity, cont’d
 Mathematically,
F W
weight of liquid
p  
A A area of liquid column
 The force is:
F  W  DW V  DW  l  w  h
 The area is:
A  lw
52
Fluid pressure and gravity, cont’d
 We can write then write the pressure as:
F DW  l  w  h
p 
A
lw
 DW h
 Dgh

The pressure depends on:



the fluid’s density,
the gravitational acceleration, and
the depth of the fluid
53
Fluid pressure and gravity, cont’d
 Note that this is the gauge pressure at the
depth h.

To get the absolute pressure, we need to add
the atmospheric pressure pushing down at the
top of the fluid column.
54
Example
Example 4.6
Let’s calculate the gauge pressure at the
bottom of a typical swimming pool — one that
is 10 feet (3.05 meters) deep.
55
Example
Example 4.6
ANSWER:
The problem gives us:
h  10 ft
D  62.4 lb/ft 3
The gauge pressure is:
p  Dgh  DW h


 62.4 lb/ft 3 10 ft 
 624 lb/ft .
2
56
Example
Example 4.6
DISCUSSION:
To convert this to psi:
lb
1 ft 1 ft
p  624 2 
ft 12 in 12 in
624 lb

2
144 in
 4.33 psi.
57
Fluid pressure and gravity, cont’d
 The general result for pressure increase
under water is:
p  0.433 psi/ft  h


The pressure increases
4.33 psi for every 10 ft.
It increases by 1 atm
every 10 meters, or
about 35 feet.
58
Example
Example 4.7
At what dept in pure water is the gauge
pressure 1 atmosphere?
59
Example
Example 4.7
ANSWER:
The problem gives us: p  1 atm  14.7 psi
The pressure depends on depth according to:
p  0.433 psi/ft  h
Rearranging for the depth:
p
14.7 psi
h

0.433 psi/ft 0.433 psi/ft
 34 ft.
60
Example
Example 4.7
DISCUSSION:
If you went swimming in mercury, which is
13.6 times as dense as water:
p
h
13.6  0.433 psi/ft
14.7 psi

5.89 psi/ft
 2.5 ft.
61
Fluid pressure and gravity, cont’d
 The atmosphere exerts a pressure on
everything since there is a very high column
of air above Earth’s surface.

The column height for
air, water and mercury
are different because
the mass densities are
increasingly larger.
62
Fluid pressure and gravity, cont’d
 A device to measure air pressure is a
barometer.



As the air pressure increases, it exerts a larger
pressure on the fluid’s surface.
The forces the fluid
farther up the tube.
The height of the
column in the tube
indicates the air
pressure.
63
Archimedes’ principle
 We’ve found that the force of gravity on a fluid
causes the pressure to increase with depth.

This means that the mass density of the fluid
actually decreases with depth.

We typically neglect this fact to keep things
simple.
 This produces something we call a buoyant
force that exerts an upward force on an
object placed in a fluid.
64
Archimedes’ principle, cont’d
 The buoyant force is an upward force
exerted by a fluid on a substance partly or
completely immersed in the fluid.

The buoyant
force depends
on the density
of the fluid and
the substance.
65
Archimedes’ principle, cont’d
 Archimedes’ principle states that the
buoyant force acting on an object in a fluid at
rest is equal to the weight of the fluid
displaced by the object.
FB  weight of displaced fluid

This means the object will sink until the weight
of the object equals the weight of the water
displaced by the object.
66
Archimedes’ principle, cont’d
 Consider weighing an object in air and then
partly immersed in water.



In air, the scale must
support the entire
weight.
In water, the buoyant
force adds an
additional upward
force.
The scale reads less.
67
Archimedes’ principle, cont’d
 Mathematically, we can write the buoyant
force as:
FB  DW  fluid  V  displaced fluid 
68
Example
Example 4.8
A contemporary Huckleberry Finn wants to
construct a raft by attaching empty, plastic, 1gallon milk jugs to the bottom of a sheet of
plywood. The raft and passengers will have a
total weight of 300 pounds. How many jugs
are required to keep the raft afloat on water?
69
Example
Example 4.8
ANSWER:
The problem gives us:
F1  300 lb
DW  62.4 lb/ft 3
The buoyant force is:
FB  DW V
Rearranging for the necessary volume:
FB
V
DW
70
Example
Example 4.8
ANSWER:
The necessary volume to just float is:
300 lb
3
V

4.8
ft
62.4 lb/ft 3
7.48 gal
3
 4.8 ft 
1 ft 3
 36 gal.
71
Example
Example 4.8
DISCUSSION:
So Huck needs at least 36 jugs.
Note that he would just float.
For a safety margin, you should probably have
many more than just 36 jugs — just in case
a wave rolls through or you want to take a
friend along.
72
Pascal’s Principle
 Pascal’s Principle states that the pressure
applied to an enclosed fluid is transmitted
undiminished to all parts of the fluid and to
the walls of the container.

This principle is used in hydraulic systems.

hydraulic lifts, braking systems,etc.
73
Pascal’s Principle, cont’d
 The piston on which the force is applied (left) has a
small area:
p
Finput
Ainput
 The piston that applies the force (right) has a larger
area:
p
Foutput
Aoutput
74
Pascal’s Principle, cont’d
 Since the pressures are the same, the output force is:
Finput
Ainput


Foutput
Aoutput
 Foutput 
Aoutput
Ainput
Finput
With the output piston’s area larger than the input
piston’s area, the output force is larger than the input
force.
75
Pascal’s Principle, cont’d
 A common example is the braking system in
your car.
76
Bernoulli’s principle
 So far we spoke only of static fluids.
 We now talk about moving fluids.
 Bernoulli’s principle states that for a fluid
undergoing steady flow, the pressure is lower
where the fluid is flowing faster.

Steady flow means the fluid flow is smooth.

No random swirling, eddies, or “white water.”
77
Bernoulli’s principle, cont’d
 One idea behind the Bernoulli principle is:

The amount of mass passing through an area
during an amount of time must be the same
throughout the pipe.

There additional fluid enters or leaves the pipe so
what goes in must come out.
78
Bernoulli’s principle, cont’d
 In order for the same amount of mass to pass
through a cross-section of the wide and
narrow sections in a certain amount of time,
the fluid must flow faster in the narrow
section.
79
Bernoulli’s principle, cont’d
 The other idea behind the Bernoulli principle
is energy conservation.
 We can consider the fluid to have a pressure
potential energy.

The greater the pressure, the greater the PE.
80
Bernoulli’s principle, cont’d
 Just as with a falling ball, as the PE
decreases the KE increases.
 So as the pressure decreases, the pressure
PE decreases.
 By energy conservation, the KE must
increase.
81
Bernoulli’s principle, cont’d
 Another example is an atomizer.




When you squeeze the bulb, you increase the
pressure in the nozzle.
This lowers the pressure in the nozzle.
The higher pressure in
the bottle draws the
fluid up the tube.
The moving air carries
the fluid out of the
nozzle.
82