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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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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

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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

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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

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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

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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

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

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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 ?

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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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Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.7 Volume and Moles (Avogadro’s Law) Balloons rise in the air because helium.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.7 Volume and Moles (Avogadro’s Law) Balloons rise in the air because helium.

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