4. 1.Which law describes the Pressure-
Volume relationship?
A. Avogadro’s Law
B. Boyle’s Law
C.Charles’ Law
D.Gay-Lussac’s Law
5. 2. Who is the proponent of Boyle’s Law?
A. Amadeo Avogadro
B. Jacques Charles
C. Joseph Gay-Lussac
D. Robert Boyle
6. 3. Which of the following diagrams best
describes Boyle’s Law?
A.↓ V → ↑ P, ↑ V → ↓ P
B.↓ V → ↓ P, ↑ V → ↓ P
C.↓ V → ↑ P, ↑ V → ↑ P
D.↓ V → ↓ P, ↑ V → ↑ P
7. 4. Which of the following is the
corresponding relationship if Volume
increases?
A. increased V
B. increased P
C. decreased V
D. decreased P
8. 5. What are the constant values
needed to demonstrate Boyle’s
Law?
A. T and mole
B. P and V
C. V and T
D. P and T
18. POP GOES THE BUBBLE!
Are you one of those people who can’t resist
popping the bubbles of bubble wrap? Before the
bubbles pop, the air pressing against the inside of
the bubbles inflates them like tiny balloons. When
you pop the bubbles, most of the air rushes out.
19. What does popping bubble
wrap have to do with science?
Actually, it demonstrates an
important scientific law called
Boyle’s law. Like other laws in
science, this law describes
what always happens under
certain conditions.
21. Procedure:
1. The students will place a few marshmallows in the
syringe.
2. Then push the plunger until it touches the
marshmallows.
3. Covering the tip of the syringe with their finger,
students pull the plunger up and watch as the
marshmallows expand. Students then return the
marshmallows to their original position, pull the
plunger to the top of the syringe, cover the tip and
push the plunger down.
4. Students observe how the marshmallows "shrink.“
5. Students will write their observation in their notes.
Procedure:
1. The students will place a few marshmallows in the
syringe.
2. Then push the plunger until it touches the
marshmallows.
3. Covering the tip of the syringe with their finger,
students pull the plunger up and watch as the
marshmallows expand. Students then return the
marshmallows to their original position, pull the
plunger to the top of the syringe, cover the tip and
push the plunger down.
4. Students observe how the marshmallows "shrink.“
5. Students will write their observation in their notes.
22. ANALYSIS
1. What happen to the marshmallows when
you push the plunger of the syringe?
When the plunger of the syringe is push, the
marshmallows are compressed.
23. 2. What happen to the marshmallows when
you pull the plunger of the syringe?
When the plunger of the syringe is pull, the
marshmallows expand.
25. Robert Boyle and Robert Hooke
used a J-tube to measure the
volume of a sample of gas at
different pressures.
They trapped a sample of air in
the J-tube and added mercury to
increase the pressure on the gas.
They observed an inverse relationship between
volume and pressure.
Hence, an increase in one causes a decrease in
the other.
26.
27. Boyle’s Law leads to the mathematical
expression: *Assuming temp is constant
P1V1=P2V2
Where P1 represents the initial pressure
V1 represents the initial volume,
And P2 represents the final pressure
V2 represents the final volume
28. Example Problem:
A 200 ml sample of hydrogen gas is collected when the pressure is 800 mmHg.
What volume will the gas occupy at 760 mmHg?
Solution:
Given:
P1= 800 mmHg
V1= 200 ml
P2= 760 mmHg
V2=?
Since final volume is asked. We will derive the formula of final volume from this
equation
P1V1=P2V2
V2= P1V1
P2
Substitute.
V2= 800mmHg(200ml)
760mmHg
V2= 160000ml
760
V2= 210.53 ml
29. Molecular Interpretation of
Boyle’s Law
As the volume of a gas sample is decreased, gas molecules
collide with surrounding surfaces more frequently, resulting in
greater pressure.
30. Boyle’s Law and Diving
For every 10 m of
depth, a diver
experiences
approximately one
additional atmosphere
of pressure due to the
weight of the
surrounding water.
At 20 m, for example,
the diver experiences
approximately 3 atm
of pressure.
31. If a diver holds his or her breath and rises to
the surface quickly, the outside pressure drops
to 1 atm.
According to Boyle’s law, what should
happen to the volume of air in the lungs?
Because the pressure is decreasing by a factor
of 3, the volume will expand by a factor of 3,
causing damage to internal organs.
Always exhale when rising!
32. ABSTRACTION
Based on the graph below, how does pressure
affects the volume of a gas if temperature is
held constant?
33. Based on the graph below, what is the
relationship between pressure and volume of
a gas if temperature is held constant?
The graph
represents variables
that have an inverse
relationship, if
the temperature of a
gas is held constant,
as the volume of the
gas increases, the
pressure decreases
and vice versa
34. APPLICATION
Imagine a container of gas molecules like the one in
the Figure below. The container in the sketch has a lid
that can be pushed down to shrink the volume of the gas
inside. What happens as the lid is lowered?
35. Imagine a container of gas molecules like the one in
the Figure below. The container in the sketch has a lid
that can be pushed down to shrink the volume of the gas
inside. What happens as the lid is lowered?
The gas molecules crowd closer
together because there is less space
for them to occupy and they have
nowhere else to go. Gas molecules
have a lot of energy. They are always
moving and bouncing off each other
and anything else in their path. When
gas molecules bump into things, it
creates pressure. Pressure is greater
when gas molecules occupy a smaller
space, because the greater crowding
results in more collisions. In other
words, decreasing the volume of a gas
increases its pressure.
36. EVALUATION
Direction: Solve the following
1. A gas occupies 25.3 mL at a pressure of 790.5
mm Hg. Determine the volume if the pressure is
reduced to 0.804 atm
2. Convert 77.0 L at 18.0 mm of Hg to its new
volume at standard pressure.
37. 1. A gas occupies 25.3 mL at a pressure of 790.5
mm Hg. Determine the volume if the pressure is
reduced to 0.804 atm
P1= 790.5 mm Hg
V1 = 25.3 ml
P2= 0.804 atm. x 760 mm Hg = 611.04 mm Hg
1 atm.
V2=?
P1V1=P2V2
Substitute:
V2= P1V1
P2
V2= 790.5 mm Hg (25.3 ml )
611.04 mm Hg
V2= 19999.65 ml.
611.04
V2= 32.73 ml
38. 2. Convert 77.0 L at 18.0 mm of Hg to its new volume
at standard pressure.
Solution:
P1= 18.0 mm Hg
V1 = 77.0 L
P2= 760 mm Hg
V2=?
P1V1=P2V2
Substitute:
V2= P1V1
P2
V2= 18.0 mm Hg ( 77.0 L )
760 mm Hg
V2= 1386 L
760
V2= 1.82 L
