1. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 1 of 46 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 6: Gases

2. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 2 of 46 Contents 6-1Properties of Gases: Gas Pressure 6-2The Simple Gas Laws 6-3Combining the Gas Laws: The Ideal Gas Equation and The General Gas Equation 6-4Applications of the Ideal Gas Equation 6-5Gases in Chemical Reactions 6-6Mixtures of Gases

3. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 3 of 46 Contents 6-6 Mixtures of Gases 6-7Kinetic—Molecular Theory of Gases 6-8Gas Properties Relating to the Kinetic—Molecular Theory 6-9Nonideal (Real) Gases  Focus On Earth’s Atmosphere

4. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 4 of 46  The gaseous states of three halogens.  Most common gases are colorless  H 2, O 2, N 2, CO and CO 2 6-1 Properties of Gases: Gas Pressure

5. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 5 of 46 The Concept of Pressure  The pressure exerted by a solid.  Both cylinders have the same mass  They have different areas of contact P (Pa) = Area (m 2 ) Force (N)

6. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 6 of 46 Liquid Pressure  The pressure exerted by a liquid depends on:  The height of the column of liquid.  The density of the column of liquid. P = g ·h ·d

7. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 7 of 46 Barometric Pressure Standard Atmospheric Pressure 1.00 atm, 760 mm Hg, 760 torr, 101.325 kPa, 1.01325 bar

8. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 8 of 46 Manometers

9. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 9 of 46 6-2 Simple Gas Laws  Boyle 1662 P  1 V PV = constant

10. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 10 of 46 Relating Gas Volume and Pressure – Boyle’s Law. The volume of a large irregularly shaped, closed tank can be determined. The tank is first evacuated and then connected to a 50.0 L cylinder of compressed nitrogen gas. The gas pressure in the cylinder, originally at 21.5 atm, falls to 1.55 atm after it is connected to the evacuated tank. What is the volume of the tank? EXAMPLE 6-4

11. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 11 of 46 EXAMPLE 6-4 P 1 V 1 = P 2 V 2 V2 =V2 = P1V1P1V1 P2P2 = 694 L V tank = 644 L

12. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 12 of 46 Charles’s Law Charles 1787 Gay-Lussac 1802 V  T V = b T

13. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 13 of 46 Standard Temperature and Pressure  Gas properties depend on conditions.  Define standard conditions of temperature and pressure (STP). P = 1 atm = 760 mm Hg T = 0°C = 273.15 K

14. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 14 of 46 Avogadro’s Law  Gay-Lussac 1808  Small volumes of gases react in the ratio of small whole numbers.  Avogadro 1811  Equal volumes of gases have equal numbers of molecules and  Gas molecules may break up when they react.

15. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 15 of 46 Formation of Water

16. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 16 of 46 Avogadro’s Law V  n or V = c n At STP 1 mol gas = 22.4 L gas At an a fixed temperature and pressure:

17. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 17 of 46 6-3 Combining the Gas Laws: The Ideal Gas Equation and the General Gas Equation  Boyle’s lawV  1/P  Charles’s lawV  T  Avogadro’s law V  n PV = nRT V  nT P

18. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 18 of 46 The Gas Constant R =R = PV nT = 0.082057 L atm mol -1 K -1 = 8.3145 m 3 Pa mol -1 K -1 PV = nRT = 8.3145 J mol -1 K -1 = 8.3145 m 3 Pa mol -1 K -1

19. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 19 of 46 Using the Gas Law

20. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 20 of 46 The General Gas Equation R =R = = P2V2P2V2 n2T2n2T2 P1V1P1V1 n1T1n1T1 = P2P2 T2T2 P1P1 T1T1 If we hold the amount and volume constant:

21. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 21 of 46 Using the Gas Laws

22. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 22 of 46 6-4 Applications of the Ideal Gas Equation PV = nRT and n = m M PV = m M RT M = m PV RT Molar Mass Determination

23. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 23 of 46 Determining a Molar Mass with the Ideal Gas Equation. Polypropylene is an important commercial chemical. It is used in the synthesis of other organic chemicals and in plastics production. A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (δ water = 0.9970 g cm -3 ) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene? Strategy: Determine V flask. Determine m gas. Use the Gas Equation. EXAMPLE 6-10

24. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 24 of 46 Determine V flask : V flask = m H 2 O  d H 2 O = (138.2410 g – 40.1305 g)  (0.9970 g cm -3 ) Determine m gas : = 0.1654 g m gas = m filled - m empty = (40.2959 g – 40.1305 g) = 98.41 cm 3 = 0.09841 L EXAMPLE 6-10

25. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 25 of 46 Use the Gas Equation: PV = nRT PV = m M RT M = m PV RT M = (0.9741 atm)(0.09841 L) (0.6145 g)(0.08206 L atm mol -1 K -1 )(297.2 K) M = 42.08 g/mol EXAMPLE 5-6

26. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 26 of 46 Gas Densities PV = nRT and d = m V PV =PV = m M RT MP RTV m = d =, n = m M

27. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 27 of 46 6-5 Gases in Chemical Reactions  Stoichiometric factors relate gas quantities to quantities of other reactants or products.  Ideal gas equation relates the amount of a gas to volume, temperature and pressure.  Law of combining volumes can be developed using the gas law.

28. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 28 of 46 Using the Ideal gas Equation in Reaction Stoichiometry Calculations. The decomposition of sodium azide, NaN 3, at high temperatures produces N 2 (g). Together with the necessary devices to initiate the reaction and trap the sodium metal formed, this reaction is used in air-bag safety systems. What volume of N 2 (g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN 3 is decomposed? 2 NaN 3 (s) → 2 Na(l) + 3 N 2 (g) EXAMPLE 6-12

29. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 29 of 46 Determine moles of N 2 : Determine volume of N 2 : n N 2 = 70 g NaN 3  1 mol NaN 3 65.01 g NaN 3  3 mol N 2 2 mol NaN 3 = 1.62 mol N 2 = 41.1 L P nRT V = = EXAMPLE 6-12 (735 mm Hg) (1.62 mol)(0.08206 L atm mol -1 K -1 )(299 K) 1.00 atm  760 mm Hg

30. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 30 of 46 6-6 Mixtures of Gases  Partial pressure  Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone.  Gas laws apply to mixtures of gases.  Simplest approach is to use n total, but....

31. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 31 of 46 Dalton’s Law of Partial Pressure The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture.

32. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 32 of 46 Partial Pressure P tot = P a + P b +… V a = n a RT/P tot and V tot = V a + V b +… VaVa V tot n a RT/P tot n tot RT/P tot = = nana n tot PaPa P tot n a RT/V tot n tot RT/V tot = = nana n tot nana = aa Recall

33. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 33 of 46 Pneumatic Trough P tot = P bar = P gas + P H 2 O

34. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 34 of 46 6-7 Kinetic Molecular Theory  Particles are point masses in constant, random, straight line motion.  Particles are separated by great distances.  Collisions are rapid and elastic.  No force between particles.  Total energy remains constant.

35. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 35 of 46 Pressure – Assessing Collision Forces  Translational kinetic energy,  Frequency of collisions,  Impulse or momentum transfer,  Pressure proportional to impulse times frequency

36. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 36 of 46 Pressure and Molecular Speed  Three dimensional systems lead to: u m is the modal speed u av is the simple average u rms

37. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 37 of 46 Pressure Assume one mole: PV=RT so: N A m = M: Rearrange:

38. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 38 of 46 Distribution of Molecular Speeds

39. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 39 of 46 Determining Molecular Speed

40. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 40 of 46 Temperature Modify: PV=RT so: Solve for e k : Average kinetic energy is directly proportional to temperature!

41. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 41 of 46 6-8 Gas Properties Relating to the Kinetic-Molecular Theory  Diffusion  Net rate is proportional to molecular speed.  Effusion  A related phenomenon.

42. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 42 of 46 Graham’s Law  Only for gases at low pressure (natural escape, not a jet).  Tiny orifice (no collisions)  Does not apply to diffusion.  Ratio used can be:  Rate of effusion (as above)  Molecular speeds  Effusion times  Distances traveled by molecules  Amounts of gas effused.

43. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 43 of 46 6-9 Nonideal (Real) Gases  Compressibility factor PV/nRT = 1  Deviations occur for real gases.  PV/nRT > 1 - molecular volume is significant.  PV/nRT < 1 – intermolecular forces of attraction.

44. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 44 of 46 Real Gases

45. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 45 of 46 van der Waals Equation P + n2an2a V2V2 V – nb = nRT

46. Prentice-Hall © 2007 General Chemistry: Chapter 6 Slide 46 of 46 End of Chapter Questions  A problem is like a knot in a ball of wool:  If you pull hard on any loop: ◦The knot will only tighten. ◦The solution (undoing the knot) will not be achieved.  If you pull lightly on one loop and then another: ◦You gradually loosen the knot. ◦As more loops are loosened it becomes easier to undo the subsequent ones.  Don’t pull too hard on any one piece of information in your problem, it tightens.