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1 Chapter 11Gases 11.3Pressure and Volume (Boyles Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Presentation on theme: "1 Chapter 11Gases 11.3Pressure and Volume (Boyles Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings."— Presentation transcript:

1 1 Chapter 11Gases 11.3Pressure and Volume (Boyles Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 2 Boyles Law Boyles 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 © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

3 3 In Boyles Law The product P x V is constant as long as T and n do not change. P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L Boyles Law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyles Law

4 4 Solving for a Gas Law Factor The equation for Boyles Law can be rearranged to solve for any factor. P 1 V 1 = P 2 V 2 Boyles Law To obtain V 2, divide both sides by P 2. P 1 V 1 = P 2 V 2 P 2 P 2 P 1 V 1 = V 2 P 2

5 5 PV in Breathing In inhalation, The lungs expand. The pressure in the lungs decreases. Air flows towards the lower pressure in the lungs. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

6 6 PV in Breathing In exhalation Lung volume decreases. Pressure within the lungs increases. Air flows from the higher pressure in the lungs to the outside. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

7 7 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

8 8 Freon-12, CCl 2 F 2, 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 1Conditions 2 P 1 = 550 mm HgP 2 = 2200 mm Hg V 1 = 8.0 LV 2 = (predict smaller V 2 ) Calculation with Boyles Law ?

9 9 2. Because pressure increases, we predict that the volume will decrease. Solve Boyles Law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 V 2 = 8.0 L x 550 mm Hg = 2.0 L 2200 mm Hg pressure ratio decreases volume Calculation with Boyles Law (Continued)

10 10 Learning Check For a cylinder containing helium 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 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

11 11 Solution For a cylinder containing helium 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

12 12 Learning Check If the helium in a cylinder has a volume of 120 mL and a pressure of 850 mm Hg, what is the new volume if the pressure is changed to 425 mm Hg inside the cylinder? 1) 60 mL 2) 120 mL3) 240 mL

13 13 3) 240 mL P 1 = 850 mm Hg P 2 = 425 mm Hg V 1 = 120 mL V 2 = ?? V 2 = P 1 V 1 = 120 mL x 850 mm Hg = 240 mL P 2 425 mm Hg Pressure ratio increases volume Solution

14 14 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? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

15 15 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 © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

16 16 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 LB) 6.4 LC) 12.8 L Learning Check

17 17 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 V 2 = V 1 P 1 P 2 V 2 = 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.) Solution

18 18 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 Learning Check

19 19 1) 200. mm Hg Data table Conditions 1Conditions 2 P 1 = 600. mm HgP 2 = ??? (lower) V 1 = 12.0 LV 2 = 36.0 L P 2 = P 1 V 1 V 2 600. mm Hg x 12.0 L = 200. mm Hg 36.0 L Solution


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