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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