CYCLE 4
Developing Ideas
ACTIVITY 3: Effects of Pressure Differences--
KEY
Purpose
In Activity 2 you were introduced to the definition of pressure as force per
unit area. You also learned that air exerts pressure, and at sea level air
pressure equals 1 atm or 14.7 lb/in2. (This value actually assumes that the
temperature is 20 °C. You saw in Activity 2 that the pressure of a gas depends
on the temperature of the gas.) In everyday life we don’t normally observe
the effects of air pressure because the pressure is usually the same on both
sides of every surface the air is in contact with. Thus, the air pushes equally
in opposite directions on both sides and the effects cancel each other out.
However, we can observe the effects of air pressure if we can create pressure
differences. On a large scale, differences in air pressure over large regions of
land or water can cause winds and other weather effects. In this activity you
will work on a much smaller scale and create situations where the air
pressure is different on two sides of a surface. Nevertheless, you will observe
some very interesting effects.
What happens when there are differences in air
pressure?
Initial Ideas
Your instructor will show you either a demonstration or a movie of an
interesting event involving a tin can connected to a vacuum pump.
© 2007 PSET
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Cycle 4
What happens when the air is removed from the tin can?
The tin can collapses.
Explain what you think caused the metal can to behave the way it did.
When the air is sucked out of the can, the outside air pushes in the walls of the tin
can.
Participate in a brief class discussion to share your ideas. Make a note of
any ideas that are different from those of your group.
Collecting and Interpreting Evidence
Experiment: How can you feel changes in the pressure of a gas?
You will need:
Air syringe with plunger
STEP 1. Make sure the tip of the syringe is not closed off. Push the plunger all
the way in and then pull it all the way out. It should be easy to push and pull
the plunger. We can think of the bottom of the plunger as a movable wall in
the syringe container. When you are done, leave the movable wall in the
middle of the container.
Do you think the air pressure inside the container has a value that is
greater than, equal to, or less than the value of the air pressure outside
the container (above the movable wall)? Why do you think so?
The air pressure inside the syringe is the same as the air pressure outside the syringe
since the bottom tip of the syringe is connected to the outside, so the pressures must
be the same.
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Activity 3: Effects of Pressure Differences
If we call the air pressure inside the container pin and the pressure outside the
container pout, then in this case pin = pout.
STEP 2. Hold your finger tightly over the tip of
the syringe to prevent any air from entering or
leaving the container. Push the plunger as far
in as you can. Then let go.
What happens when you try to push the
movable wall all the way down to the tip?
It gets harder and harder to push the plunger further
down.
What happened when you let go? Did the movable wall stay where it
was, or did it start moving back to the middle?
It started moving back to the middle, although it didn’t make it all the way back to the
middle before stopping.
Let’s try to understand why the wall started moving back towards the middle
after you let go. (It may not have returned all the way to the middle because
as the wall moved upwards, it rubbed against the inside walls of the
container, and the friction tended to slow it down more.)
When you pushed down on the wall, you compressed the gas, decreasing
its volume. Did the total # of air particles inside the container increase,
decrease or remain the same? Why do you think so?
The total # air particles inside the syringe remained the same since no gas was let in
or let out of the syringe.
Let us assume the temperature of the air inside the container remained the
same as you pushed down the plunger.
Did the average speed of the air particles inside the container increase,
decrease or remain the same? How do you know?
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Cycle 4
The average speed of the air particles remained the same because the temperature
remained constant (and the only way we know to change the average speed is to
change the temperature).
Fill in the following Macro/Micro Table for the air inside the container
during the time when the air was compressed. Use ⇑, ⇓, or ⇔.
Temp.
⇔
Mass
⇔
Volume
⇓
Avg.
#
#
Oomph/
Pressure
Speed
particles
collisions/sec
collision
pin
⇔
⇔
⇑
⇔
⇑
While you are compressing the air inside the container, the pressure of the air
outside the container did not change (since it is the same as the air pressure
in the room).
How does pin now compare to pout?
pin > pout
After you let go of the plunger, why do you think it began moving
upwards?
Since pin > pout, the air inside the syringe pushes upwards on the movable wall with a
greater force than the outside air pushes down on the wall. There is an unbalanced
upward force on the movable wall, and therefore, according to Newton’s Laws, the
wall begins to move in the direction of the unbalanced force. Therefore, the wall moves
upward towards the middle of the syringe.
Whenever there is a pressure difference on opposite sides of a wall, there’s an
unbalanced force on the wall and it begins moving from the region of higher
pressure to the region of lower pressure. In the above case, since
pin > pout, the force of the air in the container on the wall is greater than the
force of the air outside the container on the wall, and the wall moved
upwards, from the region of higher pressure (pin) to the region of lower
pressure (pout). Below is a force diagram for this situation. In this case, Fin >
Fout, and the unbalanced force on the wall is upwards.
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Activity 3: Effects of Pressure Differences
STEP 3. Return the movable wall to its position in the middle of the
container. Hold your finger over the tip and slowly pull the plunger out,
causing the wall to move upwards and the air in the container to expand.
Then let go of the plunger.
What happens when you try to pull the movable wall all the way out
towards the top of the syringe?
It gets harder and harder to pull the plunger upward as the wall moves upward.
What happened when you let go of the plunger?
The movable wall moved downwards, tending to return to its original position in the
middle of the syringe.
Fill in the following Macro/Micro Table during the time when the air was
expanded. Use ⇑, ⇓, or ⇔.
Temp.
⇔
Mass
⇔
Volume
⇑
Avg.
#
#
Oomph/
Speed
particles
collisions/sec
collision
⇔
⇔
⇓
⇔
Pressure
⇓
As you pull the wall out, is pin greater than, equal to, or less than pout?
pin < pout
After you pulled the wall as far out as you could and let go, why did the
wall begin moving back towards the middle of the syringe?
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Cycle 4
Since pin < pout, the air outside the syringe pushes downward on the movable wall with
a greater force than the inside air pushes up on the wall, and hence there is an
unbalanced upward force on the movable wall. According to Newton’s Laws the wall
begins moving in the direction of the unbalanced force. Therefore the wall moves
downward towards the middle of the syringe.
The vacuum pump and can
We now return to the demonstration or movie you considered in the Initial
Ideas section. To explain what happened it is important you have some
simple understanding of how the vacuum pump works.
Inside the vacuum pump there is a motor. One side of the pump is opened to
the room, where the pressure is proom. (This is also the pressure outside the
tin can.) The other side is connected to an object that is to be evacuated (in
this case, the tin can, where the pressure is pcan). When the pump is turned
on, and the motor turns, air is forced to move from the can to the room. The
air pressure in the room, however, always stays the same, at proom (which has
a value of 14.7 lb/in2).
Fill in the following Macro/Micro Table for the air inside the tin can
when the pump is first turned on. Assume the temperature of the air
inside the can remains constant, and also assume the volume of the can
has not changed. (Eventually, the can is crushed and the volume
decreases, but at first there is little change in volume, so we can ignore it.)
Temp.
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Mass
Volume
Avg.
#
#
Oomph/
Pressure
Speed
particles
collisions/sec
collision
pcan
Activity 3: Effects of Pressure Differences
⇔
⇓
⇔
⇔
⇓
⇓
⇔
⇓
After the pump is turned on, how does pcan compare to proom?
pcan < proom
Is there an unbalanced force on the walls of the tin can? Why?
Yes, there is an unbalanced force because pcan < proom, which means that the force of
the room air on the tin can walls is greater than the force of the can air (inside air) on
the same walls.
Why do the walls of the tin can eventually begin to collapse?
Since the force of the room air on the tin can walls is greater than the force of the can
air (inside air) on the same walls, there is an unbalanced force pushing the walls of
the tin can towards the inside. If this force is great enough, the walls will begin to
move inward, beginning to collapse.
Explanations involving pressure difference
When you write explanations involving the effects of pressure difference (like
the collapsing tin can), you should first construct and fill out a Macro/Micro
Table and use the SPM mechanisms for pressure to explain what happens to
the pressure either inside or outside an enclosed object. Normally, the
pressure on the other side of the object remains the same. Then, compare the
pressures on the two sides of the object, and if the pressures are different use
the idea of unbalanced forces (Newton’s Law) to explain why the wall
separating the inside and outside begins to move. Drawing a force diagram,
clearly showing the two forces, with one being greater than the other, is
usually helpful.
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Cycle 4
Summarizing Questions
S1: Imagine a glass soda bottle in
which the bottom has been
removed. A piece of rubber (e.g.
from a balloon) is tightly secured
over the open bottom.
The
mouth of the bottle is connected
to a vacuum pump.
If you turn on the vacuum pump, so that air is removed from inside the
bottle, the rubber stretched over the bottom begins to extend inside the
bottle. You can assume the temperature of the air inside the bottle
remains constant. A student in a previous class was asked to explain this
phenomena using ideas of the SPM mechanisms for pressure and
Newton’s Laws. You are to evaluate her explanation according to the
criteria of accuracy, completeness, and logical reasoning and clarity. If
you find the explanation to be poor, indicate what is wrong and make
the corrections. (This student did not fill out a Macro/Micro table.)
(1) When the vacuum pump is turned on, air is removed from inside the
bottle to the outside of the bottle, so the total # of air particles inside
the bottle decreases. (2) This causes the #collisions/sec of the
particles with the walls to decrease. (3) Because the air is being
evacuated from the bottle, the average speed of the particles increases,
and this causes the oomph/collision to increase. (4) According to the SPM
mechanism for pressure, pressure = #collisions/sec x oomph/collision.
(5) Since the #collisions/sec decreases, and the oomph/collision
increases, these two factors cancel each other out and the air pressure
inside the bottle, pbot, remains the same. (6) Because air was moved from
inside the bottle to outside the bottle, air pressure outside the bottle,
pout, increased.
(7) Since pout is greater than pbot, there is an unbalanced force on the
rubber bottom in a direction towards the inside of the bottle. (8)
Therefore, the rubber bottom moves inside the bottle.
This explanation is poor because it is inaccurate. To identify the problematic parts of
the explanation, I’ve numbered each of her sentences. (1) and (2) are okay. (3) is
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Activity 3: Effects of Pressure Differences
wrong! Removing air from the bottle does not cause the average speed of the particles
to increase. Only an increase in temperature can cause the average speed of the
particles to increase. However, we are told the temperature remains constant, so that
means that the average speed also remains constant, and the oomph/collision also
remains constant. (4) is an okay statement of the SPM mechanism for pressure. (5) is
wrong! The # collisions/sec does decrease, but the oomph/collision remains constant,
so the air pressure inside the bottle decreases. (6) is mainly wrong! Even though air
was moved from inside the bottle to outside the bottle, we are assuming the air
pressure outside the bottle remains essentially constant (or increases by a very small
amount)—because there is so much air outside. (7) and (8) are okay.
S2: A partially inflated balloon is placed inside a large glass container. A
vacuum pump is connected to the container, as shown in the following
picture.
(a) When the vacuum pump is turned on, what do you predict will
happen to the balloon?
I predict the balloon will expand.
(b) Justify your prediction by writing an explanation. Let pcont be the
pressure inside the large glass container, and let pbal be the pressure
inside the balloon.
Fill out a Macro/Micro table for the air in the large glass container (but
not inside the balloon). Use ⇑, ⇓, ⇔.
Temp.
⇔
Mass
⇓
Volume
⇔
Avg.
#
#
Oomph/
Pressure
Speed
particles
collisions/sec
collision
pcont
⇔
⇓
⇓
⇔
⇓
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Cycle 4
Write the explanation narrative. The narrative should consist of two
parts. In the first part, use the SPM mechanisms for pressure to explain
what happens to the pressure of air inside the large glass container. (Use
the Macro/Micro table as your guide.) In the second part, compare pbal
and pcon and use Newton’s Laws to explain what happens to the balloon.
During this process we can assume the temperature of the air in the container
remains constant, which means the average speed of the air particles and the
oomph/collision also remain constant. After the pump is turned on, air is removed
from the container; thus the mass of the air and the total # particles of air in the
container will decrease. This means the # collisions/sec will decrease. According to
the SPM mechanism for pressure, pressure ~ # collisions/sec x oomph/collision.
Thus, the pressure of the air inside the glass container will decrease.
Since the air inside the balloon is not connected to the vacuum pump, the air pressure
inside the balloon remains constant during this process. Since pcon < pbal, the force
the container air exerts on the balloon surface is less than the force the balloon air
exerts balloon surface. There is thus an unbalanced force pushing outward on the
balloon surface. By Newton’s law, this means the balloon surface will begin moving
outward, causing the balloon to expand.
Although not requested, a force diagram, like the one below, is helpful:
Participate in a whole class discussion to review the answers to the
Summarizing questions.
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