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Chapter 6 Gases
6.3Pressure and Volume (Boyle’s Law)
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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.
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In Boyle’s Law, the product P x V is constant as long as T and n do not change.
P1V1 = 8.0 atm x 2.0 L = 16 atm L
P2V2 = 4.0 atm x 4.0 L = 16 atm L
P3V3 = 2.0 atm x 8.0 L = 16 atm L
Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)
PV Constant in Boyle’s Law
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Solving for a Gas Law Factor
The equation for Boyle’s Law can be rearranged tosolve for any factor.
P1V1 = P2V2 Boyle’s Law
To solve for V2 , divide both sides by P2.
P1V1 = P2V2
P2 P2
V1 x P1 = V2
P2
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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.
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
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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.
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Calculations with Boyle’s Law
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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 mm Hg after its pressure is changed to 2200 mm Hg at constant T?
1. Set up a data table:Conditions 1 Conditions 2 P1 = 550 mm Hg P2 = 2200 mm Hg
V1 = 8.0 L V2 =
Calculation with Boyle’s Law
?
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2. When pressure increases, volume decreases.
Solve Boyle’s Law for V2:
P1V1 = P2V2
V2 = V1 x P1
P2 V2 = 8.0 L x 550 mm Hg = 2.0 L
2200 mm Hg pressure ratio
decreases volume
Calculation with Boyle’s Law (Continued)
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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 decreases2) 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 B2) Pressure increases A
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Learning Check
If a sample of helium gas has a volume of 120 mLand a pressure of 850 mm Hg, what is the newvolume if the pressure is changed to 425 mm Hg ?
1) 60 mL 2) 120 mL 3) 240 mL
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3) 240 mL P1 = 850 mm Hg P2 = 425 mm Hg
V1 = 120 mL V2 = ??
V2 = V1 x P1 = 120 mL x 850 mm Hg = 240 mL P2 425 mm Hg
Pressure ratio
increases volume
Solution
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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 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.
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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
Learning Check
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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 atmVolume 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. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant).
1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg
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Solution
1) 200. mm Hg Data table Conditions 1 Conditions 2P1 = 600. mm Hg P2 = ???
V1 = 12.0 L V2 = 36.0 L
P2 = P1 x V1
V2
600. mm Hg x 12.0 L = 200. mm Hg 36.0 L
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