1. 1 Chapter 7 Gases 7.1 Properties of Gases 7.2 Gas Pressure

2. 2 Kinetic Theory of Gases A gas consists of small particles that move rapidly in straight lines have essentially no attractive (or repulsive) forces are very far apart have very small volumes compared to the volumes of the containers they occupy have kinetic energies that increase with an increase in temperature

3. 3 Properties of Gases Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

4. 4 Gas pressure is the force acting on a specific area Pressure (P) = force area has units of atm, mmHg, torr, lb/in. 2 and kilopascals(kPa). 1 atm = 760 mmHg (exact) 1 atm = 760 torr 1 atm = 14.7 lb/in. 2 1 atm = 101.325 kPa Gas Pressure

5. 5 Atmospheric Pressure Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth

6. 6 Atmospheric Pressure (continued) Atmospheric pressure is about 1 atmosphere at sea level depends on the altitude and the weather is lower at high altitudes where the density of air is less is higher on a rainy day than on a sunny day

7. Chapter 7 Gases 7.3 Pressure and Volume (Boyle’s Law) 7

8. 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 (V) decreases 8

9. 9 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

10. 10 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

11. 11 Boyle’s Law and Breathing: Inhalation During inhalation, the lungs expand the pressure in the lungs decreases air flows towards the lower pressure in the lungs

12. 12 Boyle’s Law and Breathing: Exhalation During exhalation, lung volume decreases pressure within the lungs increases air flows from the higher pressure in the lungs to the outside

13. 13 Guide to Calculations with Gas Laws

14. 14 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 after its pressure is changed from 550 mmHg to 2200 mmHg at constant T? STEP 1 Set up a data table: Conditions 1Conditions 2 Know Predict P 1 = 550 mmHgP 2 = 2200 mmHg P increases V 1 = 8.0 LV 2 = ? V decreases Calculation with Boyle’s Law

15. 15 STEP 2 Solve Boyle’s law for V 2. When pressure increases, volume decreases. P 1 V 1 = P 2 V 2 V 2 = V 1 x P 1 P 2 STEP 3 Set up problem V 2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume Calculation with Boyle’s Law (continued)

16. 16 Chapter 7 Gases 7.4 Temperature and Volume (Charles’s Law)

17. 17 Charles’s Law In Charles’s law, the Kelvin temperature of a gas is directly related to the volume P and n are constant when the temperature of a gas increases, its volume increases

18. 18 For two conditions, Charles’s law is written V 1 = V 2 (P and n constant) T 1 T 2 Rearranging Charles’s law to solve for V 2 gives T 2 x V 1 = V 2 x T 2 T 1 T 2 V 2 = V 1 x T 2 T 1 Charles’s Law: V and T

19. 19 A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? STEP 1 Set up data table: Conditions 1 Conditions 2 Know Predict V 1 = 785 mL V 2 = ? V decreases T 1 = 21 °C T 2 = 0 °C = 294 K = 273 K T decreases Be sure to use the Kelvin (K) temperature in gas calculations. Calculations Using Charles’s Law

20. 20 Calculations Using Charles’s Law (continued) STEP 2 Solve Charles’s law for V 2 : V 1 = V 2 T 1 T 2 V 2 = V 1 x T 2 T 1 Temperature factor decreases T STEP 3 Set up calculation with data: V 2 = 785 mL x 273 K = 729 mL 294 K

21. 21 Chapter 7 Gases 7.5 Temperature and Pressure (Gay-Lussac’s Law)

22. 22 Gay-Lussac’s Law: P and T In Gay-Lussac’s law, the pressure exerted by a gas is directly related to the Kelvin temperature V and n are constant P 1 = P 2 T 1 T 2

23. 23 Chapter 7 Gases 7.6 The Combined Gas Law

24. Summary of Gas Laws The gas laws can be summarized as follows: 24

25. 25 The combined gas law uses Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law

26. 26 A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? Step 1 Set up data table: Conditions 1Conditions 2 P 1 = 0.800 atm P 2 = 3.20 atm V 1 = 0.180 L (180 mL) V 2 = 90.0 mL T 1 = 29 °C + 273 = 302 KT 2 = ? Combined Gas Law Calculation

27. 27 STEP 2 Solve for T 2 P 1 V 1 =P 2 V 2 T 1 T 2 T 2 = T 1 x P 2 x V 2 P 1 V 1 STEP 3 Substitute values to solve for unknown. T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180. mL T 2 = 604 K  273 = 331 °C Combined Gas Law Calculation (continued)

28. 28 Chapter 7 Gases 7.7 Volume and Moles (Avogadro’s Law)

29. 29 Avogadro's Law: Volume and Moles Avogadro’s law states that the volume of a gas is directly related to the number of moles (n) of gas T and P are constant V 1 = V 2 n 1 n 2

30. 30 A gas has a pressure at 2.0 atm at 18 °C. What is the new pressure when the temperature is 62 °C? (V and n constant) STEP 1 Set up a data table: Conditions 1 Conditions 2 Know Predict P 1 = 2.0 atm P 2 = ? P increases T 1 = 18 °C + 273 T 2 = 62 °C + 273 T increases = 291 K = 335 K Calculation with Gay-Lussac’s Law

31. 31 Calculation with Gay-Lussac’s Law (continued) STEP 2 Solve Gay-Lussac’s Law for P 2 : P 1 = P 2 T 1 T 2 P 2 = P 1 x T 2 T 1 STEP 3 Substitute values to solve for unknown: P 2 = 2.0 atm x 335 K = 2.3 atm 291 K Temperature ratio increases pressure

32. 32 The volumes of gases can be compared at STP (Standard Temperature and Pressure) when they have the same temperature Standard temperature (T) = 0 °C or 273 K the same pressure Standard pressure (P) = 1 atm (760 mmHg) STP

33. 33 Molar Volume The molar volume of a gas is measured at STP (standard temperature and pressure) is 22.4 L for 1 mole of any gas

34. 34 Molar Volume as a Conversion Factor The molar volume at STP has about the same volume as 3 basketballs can be used to form 2 conversion factors: 22.4 L and 1 mole 1 mole 22.4 L

35. 35 Guide to Using Molar Volume

36. 36 Using Molar Volume What is the volume occupied by 2.75 moles of N 2 gas at STP? STEP 1 Given: 2.75 moles of N 2 Need: Liters of N 2 STEP 2 Write a plan: Use the molar volume to convert moles to liters.

37. 37 Using Molar Volume (continued) STEP 3 Write equalities and conversion factors: 1 mole of gas = 22.4 L 1 mole gas and 22.4 L 22.4 L 1 mole gas STEP 4 Substitute data and solve: 2.75 moles N 2 x 22.4 L = 61.6 L of N 2 1 mole N 2

38. 38 Chapter 7 Gases 7.8 The Ideal Gas Law

39. 39 The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT Rearranging this expression gives the expression called the ideal gas law. PV = nRT Ideal Gas Law

40. 40 The universal gas constant, R, can be calculated using the molar volume at STP when calculated at STP, uses a temperature of 273 K, a pressure of 1.00 atm, a quantity of 1.00 mole of a gas, and a molar volume of 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1.00 mole)(273K) n T = 0.0821 L  atm mole  K Universal Gas Constant, R

41. Summary of Units for Ideal Gas Constants 41

42. Guide to Using the Ideal Gas Law 42

43. 43 Chapter 7 Gases 7.9 Partial Pressure (Dalton’s Law)

44. 44 The partial pressure of a gas is the pressure of each gas in a mixture is the pressure that gas would exert if it were by itself in the container Partial Pressure

45. 45 Dalton’s Law of Partial Pressures indicates that pressure depends on the total number of gas particles, not on the types of particles the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases P T = P 1 + P 2 + P 3 +..... Dalton’s Law of Partial Pressures

46. 46 Dalton’s Law of Partial Pressures (continued)

47. 47 For example, at STP, one mole of a pure gas in a volume of 22.4 L will exert the same pressure as one mole of a gas mixture in 22.4 L. V = 22.4 L Gas mixtures Total Pressure 0.5 mole O 2 0.3 mole He 0.2 mole Ar 1.0 mole 1.0 mole N 2 0.4 mole O 2 0.6 mole He 1.0 mole 1.0 atm

48. 48 Scuba Diving When a scuba diver is below the ocean surface, the increased pressure causes more N 2 (g) to dissolve in the blood. If a diver rises too fast, the dissolved N 2 gas can form bubbles in the blood, a dangerous and painful condition called “the bends.” For deep descents, helium, which does not dissolve in the blood, is added to O 2.

49. Guide to Solving for Partial Pressure 49

50. 50 Gases We Breathe The air we breathe is a gas mixture contains mostly N 2 and O 2 and small amounts of other gases

51. 51 Blood Gases In the lungs, O 2 enters the blood, while CO 2 from the blood is released. In the tissues, O 2 enters the cells, which releases CO 2 into the blood.

52. 52 Blood Gases (continued) In the body, O 2 flows into the tissues because the partial pressure of O 2 is higher in blood and lower in the tissues. CO 2 flows out of the tissues because the partial pressure of CO 2 is higher in the tissues and lower in the blood. Partial Pressures (mmHg) in Blood and Tissue Gas Oxygenated Blood Deoxygenated Blood Tissues O 2 100 mmHg 40 mmHg 30 mmHg or less CO 2 40 mmHg 46 mmHg 50 mmHg or greater

53. 53 Changes in Partial Pressures of Blood Gases During Breathing