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1 Chapter 6Gases 6.3 Pressure and Volume (Boyle’s Law)

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1 1 Chapter 6Gases 6.3 Pressure and Volume (Boyle’s Law)

2 2 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.

3 3 In Boyle’s 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 Boyle’s Law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyle’s Law

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

5 5 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

6 6 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

7 7 Calculations with Boyle’s Law

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 = Calculation with Boyle’s Law ?

9 9 2. When pressure increases, volume decreases. Solve Boyle’s Law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 x 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 Boyle’s Law (Continued)

10 10 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

11 11 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

12 12 Learning Check If a sample of helium gas 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 ? 1) 60 mL 2) 120 mL3) 240 mL

13 13 3) 240 mL P 1 = 850 mm HgP 2 = 425 mm Hg V 1 = 120 mLV 2 = ?? V 2 = V 1 x P 1 = 120 mL x 850 mm Hg = 240 mL P 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?

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.

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 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 V 2 = V 1 x 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.)

18 18 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

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


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