1. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 1 of 41 Chapter 6: Gases Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition

2. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 2 of 41 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 © 2002General Chemistry: Chapter 6Slide 3 of 41 Contents 6-6Mixtures of Gases 6-7Kinetic—Molecular Theory of Gases 6-8Gas Properties Relating to the Kinetic—Molecular Theory 6-9Non-ideal (real) Gases Focus on The Chemistry of Air-Bag Systems

4. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 4 of 41 Barometric Pressure Standard Atmospheric Pressure 1.00 atm 760 mm Hg, 760 torr 101.325 kPa 1.01325 bar 1013.25 mbar

5. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 5 of 41 Manometers

6. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 6 of 41 6-2 Simple Gas Laws Boyle 1662 P  1 V PV = constant

7. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 7 of 41 Example 5-6 Relating Gas Volume and Pressure – Boyle’s Law. P 1 V 1 = P 2 V 2 V2 =V2 = P1V1P1V1 P2P2 = 694 L V tank = 644 L

8. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 8 of 41 Charles’s Law Charles 1787 Gay-Lussac 1802 V  T V = b T

9. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 9 of 41 STP 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

10. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 10 of 41 Avogadro’s Law Avogadro 1811 –Equal volumes of gases have equal numbers of molecules and –Gas molecules may break up when they react.

11. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 11 of 41 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:

12. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 12 of 41 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

13. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 13 of 41 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

14. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 14 of 41 The General Gas Equation R =R = = P2V2P2V2 n2T2n2T2 P1V1P1V1 n1T1n1T1 = P2P2 T2T2 P1P1 T1T1 If we hold the amount and volume constant:

15. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 15 of 41 6-4 Applications of the Ideal Gas Equation

16. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 16 of 41 Molar Mass Determination PV = nRT and n = m M PV = m M RT M = m PV RT

17. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 17 of 41 Example 6-10 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=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.

18. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 18 of 41 Example 5-6 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

19. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 19 of 41 Example 5-6 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

20. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 20 of 41 Gas Densities PV = nRT and d = m V PV =PV = m M RT MP RTV m = d =, n = m M

21. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 21 of 41 6-5 Gases in Chemical Reactions Stoichiometric factors relate gas quantities to quantities of other reactants or products. Ideal gas equation used to relate the amount of a gas to volume, temperature and pressure. Law of combining volumes can be developed using the gas law.

22. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 22 of 41 Example 6-10 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)

23. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 23 of 41 Example 6-10 Determine moles of N 2 : Determine volume of N 2 : n N 2 = 70 g N 3  1 mol NaN 3 65.01 g N 3 /mol N 3  3 mol N 2 2 mol NaN 3 = 1.62 mol N 2 = 41.1 L P nRT V = = (735 mm Hg) (1.62 mol)(0.08206 L atm mol -1 K -1 )(299 K) 760 mm Hg 1.00 atm

24. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 24 of 41 Dalton’s Law of Partial Pressure

25. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 25 of 41 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

26. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 26 of 41 Pressure Assume one mole: PV=RT so: N A m = M: Rearrange:

27. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 27 of 41 6-8 Gas Properties Relating to the Kinetic-Molecular Theory Diffusion –Net rate is proportional to molecular speed. Effusion –A related phenomenon.

28. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 28 of 41 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.

29. Prentice-Hall © 2002General Chemistry: Chapter 6Slide 29 of 41 Chapter 6 Questions 9, 13, 18, 31, 45, 49, 61, 63,