3.
SOLIDS
• Fixed shape and volume
• Normally hard and rigid
• Large force needed to
change shape
• High density
• Incompressible
4.
Model of Solids
• Closely packed
together
• Occupy minimum space
• Regular pattern
• Vibrate about fixed
position
• Not free to move about
5.
LIQUIDSLIQUIDS
• Fixed volume but no fixed
shape
• High density
• Not compressible
6.
Model of Liquids
• Occur in clusters
with molecules
slightly further apart
as compared to solids
• Free to move about
within confined
vessel
7.
GASESGASES
•No fixed shape or
volume
•Low density
•Compressible
8.
Model of GasesModel of Gases
• Very far apart
• Travel at high speeds
• Independent and
random motion
• Negligible forces of
attraction between
them
9.
Brownian MotionBrownian Motion
• Movement of
smoke cells
were observed
under the
microscope
• Random
motion
10.
Pressure in Gases
(Ideal Gases)
Air molecules in a container are in
as state of continuous motion.
11.
Pressure in Gases
(Ideal Gases)
Air molecules in a container are in
as state of continuous motion.
When they collide with the wall of
a container, they exert a force, F on
the wall.
F
12.
Pressure in Gases
(Ideal Gases)
Air molecules in a container are in
as state of continuous motion.
When they collide with the wall of
a container, they exert a force, F on
the wall.
F
The force per unit area is the
pressure exerted on the wall.
13.
Pressure-volume (p-V)
relationship of a gas
Air molecules in a container will
exert a certain amount of pressure.
14.
Pressure-volume (p-V)
relationship of a gas
Air molecules in a container will
exert a certain amount of pressure.
If the volume of this container was
to decrease, the air molecules will
have less space to move about. This
will result in the molecules
colliding with the walls more
frequently.
15.
Pressure-volume (p-V)
relationship of a gas
Therefore, when we decrease the volume of
the container, the pressure exerted by the
air molecules on the container increases.
V
1
αp
16.
To form an equation,
p = k/V
pV = k (k is a constant)
p1V1 = p2V2
Where p1 and V1 are the initial pressure and volume,
And p2 and V2 are the final pressure and volume.
17.
Example:
The volume of a fixed mass of gas at 600 Pa is
1500cm3
. What is the pressure if the volume is
reduced to 1000 cm3
at constant temperature?
Solution:
Using the formula: p1V1 = p2V2
(600)(1500) = p2(1000)
p2=
p2= 900 Pa
(1000)
)(600)(1500
18.
P-T Relationship
Now we will keep the
volume of the
container constant.
We will investigate to
see how the pressure
will vary with
temperature of the
gas.
19.
P-TRelationshipP-TRelationship
From the applet, we can see that
Pressure increases as the temperature increases.
TP ∝
when the volume is kept constant
20.
Example
Air is being trapped in a container of fixed volume. At room
temperature of 300 K, the pressure exerted by the gas is 100 Pa.
If the air in the container was heated to 600 K, what is the new
pressure exerted by the gas now?
Solution:
Since pressure is proportional to temperature, when temperature
increases, pressure should also increase.
Temperature increases by 2 times, so pressure should increase
by 2 times.
New pressure = 100 x 2 = 200 Pa
21.
V-T Relationship
This is the most commonly occurring relationship.
When gas gets heated, the amount of space that it
occupies expands.
So when temperature increase, volume would also
increase. Temperature is proportional to volume.
TV ∝
at constant pressure
22.
Example
A balloon is filled with gas, at a temperature of 300 K, to a
volume of 50cm3
. If I want to expand the balloon to a volume of
150cm3
, what is the temperature of the gas now? Assuming that
the pressure exerted by the gas does not change.
Solution:
Volume is proportional to temperature.
Since the volume has to be increased by 3 times, the
volume should also be increased accordingly.
Required temperature = 300 x 3 = 900 K