Transcript (Boyle`s Law).

Chapter 6
Gases
6.3
Pressure and Volume (Boyle’s Law)
Copyright © 2009 by Pearson Education, Inc.
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Boyle’s Law
Boyle’s law states that
• the pressure of a gas
is inversely related to
its volume when T and
n are constant.
• if volume decreases,
the pressure
increases.
Copyright © 2009 by Pearson Education, Inc.
2
PV Constant in Boyle’s Law
In Boyle’s law, the product
as T and n do not change.
P1V1 = 8.0 atm x
P2V2 = 4.0 atm x
P3V3 = 2.0 atm x
P x V is constant as long
2.0 L = 16 atm L
4.0 L = 16 atm L
8.0 L = 16 atm L
Boyle’s law can be stated as
P1V1 = P2V2 (T, n constant)
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Solving for a Gas Law Factor
The equation for Boyle’s law can be rearranged to
solve for any factor.
P1V1 = P2V2
Boyle’s law
To solve for V2 , divide both sides by P2.
P1V1
P2
V1 x
P1
P2
=
P2V2
P2
=
V2
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Boyles’ Law and Breathing
During an inhalation,
• the lungs expand.
• the pressure in the
lungs decreases.
• air flows towards
the lower pressure
in the lungs.
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Boyles’ Law and Breathing
During an exhalation,
• lung volume
decreases.
• pressure within the
lungs increases.
• air flows from the
higher pressure in
the lungs to the
outside.
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6
Calculations with Boyle’s Law
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Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems.
What is the new volume (L) of a 8.0 L sample of Freon
gas initially at 550 mmHg after its pressure is changed
to 2200 mmHg at constant T and n?
1. Set up a data table:
Conditions 1
P1 = 550 mmHg
V1 = 8.0 L
Conditions 2
P2 = 2200 mmHg
V2 = ?
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Calculation with Boyle’s Law
(continued)
2. When pressure increases, volume decreases.
Solve Boyle’s law for V2:
P1V1 = P2V2
V2
V2
= V1 x P1
P2
= 8.0 L x 550 mmHg =
2200 mmHg
2.0 L
pressure ratio
decreases volume
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Learning Check
For a cylinder containing helium gas, indicate if
cylinder A or cylinder B represents the new volume for
the following changes (n and T are constant).
1) pressure decreases
2) pressure increases
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Solution
For a cylinder containing helium gas, indicate if
cylinder A or cylinder B represents the new volume for
the following changes (n and T are constant).
1) pressure decreases B
2) pressure increases A
Copyright © 2009 by Pearson Education, Inc.
11
Learning Check
If a sample of helium gas has a volume of 120 mL
and a pressure of 850 mmHg, what is the new
volume if the pressure is changed to 425 mmHg?
1) 60 mL
2) 120 mL
3) 240 mL
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Solution
3) 240 mL
P1 = 850 mmHg
V1 = 120 mL
V2 = V1 x P1 =
P2
P2 = 425 mmHg
V2 = ??
120 mL x 850 mmHg = 240 mL
425 mmHg
Pressure ratio
increases volume
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Learning Check
A sample of helium gas in a balloon has a volume of
6.4 L at a pressure of 0.70 atm. At 1.40 atm (T is
constant), is the new volume represented by A, B, or
C?
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Solution
A sample of helium gas in a balloon has a volume of
6.4 L at a pressure of 0.70 atm. At a higher pressure
(T constant), the new volume is represented by the
smaller balloon A.
Copyright © 2009 by Pearson Education, Inc.
15
Learning Check
If the sample of helium gas has a volume of 6.4 L
at a pressure of 0.70 atm, what is the new
volume when the pressure is increased to 1.40
atm (T constant)?
A) 3.2 L
B) 6.4 L
C) 12.8 L
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Solution
If the sample of helium gas has a volume of 6.4 L
at a pressure of 0.70 atm, what is the new
volume when the pressure is increased to 1.40
atm (T constant)?
A) 3.2 L
V2 =
V1 x P1
P2
V2 =
6.4 L x 0.70 atm = 3.2 L
1.40 atm
Volume decreases when there is an increase in
the pressure (temperature is constant.)
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Learning Check
A sample of oxygen gas has a
volume of 12.0 L at 600. mmHg.
What is the new pressure when
the volume changes to 36.0 L?
(T and n constant).
1) 200. mmHg
2) 400. mmHg
3) 1200 mmHg
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Solution
1) 200. mmHg
Data Table
Conditions 1
P1 = 600. mmHg
V1 = 12.0 L
Conditions 2
P2
= ???
V2
= 36.0 L
P2 = P1 x
V1
V2
600. mmHg x 12.0 L = 200. mmHg
36.0 L
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Learning Check
If the sample of nitrogen (gas) has a volume of
360 mL at a pressure of 720 mmHg, what is the new
volume when the pressure is increased to 1.20 atm
(T constant)?
A) 284 mL
B) 456 mL
C) 2160 mL
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Solution
We need to make the units for initial and final
pressure the same:
1.20 atm x 760 mmHg
1 atm
=
912 mmHg
V2 =
V1 x P1
P2
V2 =
360 mL x 720 mmHg = 284 mL (A)
912 mmHg
Volume decreases when there is an increase in
the pressure (temperature is constant.)
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