1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.3Pressure and Volume (Boyle’s Law)

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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 the pressure (P) increases, then the volume decreases

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Boyle’s law if 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 can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyle’s Law

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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 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. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 P and V in Inhalation During inhalation, the lungs expand the pressure in the lungs decreases air flows toward the lower pressure in the lungs

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 P and V in Exhalation During exhalation, lung volume decreases pressure within the lungs increases air flows from the higher pressure in the lungs to the outside

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Using the Gas Laws

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of an 8.0-L sample of Freon gas initially at 550 mmHg after its pressure is changed to 2200 mmHg at constant T? STEP 1 Organize the data in a table of initial and final conditions. Conditions 1Conditions 2Know Predict P 1 = 550 mmHgP 2 = 2200 mmHg P increases V 1 = 8.0 LV 2 = ? V decreases Example of Using Boyle’s Law

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 STEP 2 Rearrange the gas law for the unknown. Solve Boyle’s law for V 2. When pressure increases, volume decreases. P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 STEP 3 Substitute values into the gas law to solve for the unknown. V 2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume Example of Using Boyle’s Law (continued)

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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 cylinder B 2) Pressure increases cylinder A

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 Learning Check If the helium in a cylinder has a volume of 120 mL and a pressure of 850 mmHg, what is the new volume if the pressure changes to 425 mmHg inside the cylinder? 1) 60 mL 2) 120 mL3) 240 mL

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 STEP 1 Organize the data in a table of initial and final conditions. Conditions 1Conditions 2 Know Predict P 1 = 850 mmHgP 2 = 425 mmHg P decreases V 1 = 120 mLV 2 = ? V increases STEP 2 Rearrange the gas law for the unknown. Solve Boyle’s law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 Solution

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 STEP 3 Substitute values into the gas law to solve for the unknown. V 2 = P 1 V 1 = 120 mL x 850 mmHg = 240 mL (3) P 2 425 mmHg Pressure ratio increases volume Solution (continued)

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 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 and n constant), is the new volume represented by A, B, or C?

16. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 16 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 and n constant), the new volume is represented by the smaller balloon A.

17. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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 and n constant)? 1) 3.2 L2) 6.4 L3) 12.8 L Learning Check

18. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 18 STEP 1 Organize the data in a table of initial and final conditions. Conditions 1Conditions 2 Know Predict P 1 = 0.70 atmP 2 = 1.40 atm P increases V 1 = 6.4 LV 2 = ? V decreases STEP 2 Rearrange the gas law for the unknown. Solve Boyle’s law for V 2. P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 Solution

19. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 19 STEP 3 Substitute values into the gas law to solve for the unknown. V 2 = 6.4 L x 0.70 atm = 3.2 L (1) 1.40 atm Volume decreases when there is an increase in the pressure (at constant T and n). Solution (continued)

20. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 20 A sample of oxygen gas has a volume of 12.0 L at 600. mmHg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg Learning Check A gauge indicates the pressure of a gas in a tank.

21. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 21 STEP 1 Organize the data in a table of initial and final conditions. Conditions 1Conditions 2 Know Predict P 1 = 600 mmHgP 2 = ? P decreases V 1 = 12.0 LV 2 = 36.0 L V increases STEP 2 Rearrange the gas law for the unknown. P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 Solution

22. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 22 STEP 3 Substitute values into the gas law to solve for the unknown. 600. mmHg x 12.0 L = 200. mmHg (1) 36.0 L Solution (continued)