1. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 8 Gases 8.1 Gases and Kinetic Theory 8.2 Gas Pressure 8.8 Ideal Gas Law * You do not need to know Boyle’s (8.3), Charles’ (8.4), Gay-Lussac’s (8.5), Avogadro’s (8.7) or the Combined gas (8.6) laws. They are all contained within the ideal gas law. We will not cover Dalton’s law (8.9).

2. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 2 Particles of a gas  Move rapidly in straight lines.  Have kinetic energy that increases with an increase in temperature.  Are very far apart.  Have essentially no attractive (or repulsive) forces.  Have very small volumes compared to the volume of the container they occupy. 8.1 Kinetic Theory of Gases

3. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 3 Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). Properties of Gases

4. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 4 A barometer measures the pressure exerted by the gases in the atmosphere. The atmospheric pressure is measured as the height in mm of the mercury column. 8.2 Barometer

5. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 5 A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense 2) H 2 O is heavier 3) air is more dense than H 2 O Learning Check

6. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 6 A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense Solution

7. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 7  A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area  One atmosphere (1 atm) is 760 mm Hg.  1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr Pressure

8. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 8 In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa). Units of Pressure

9. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 9 A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.638 atm 3) 3.61 x 10 5 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3)22,300 mm Hg Learning Check

10. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 10 A. What is 475 mm Hg expressed in atm? 2) 0.638 atm 485 mm Hg x 1 atm = 0.638 atm 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) 1520 mm Hg 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm Solution

11. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 11 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 8.8 Ideal Gas Law

12. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 12 The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 273 K (0°C ) Standard pressure (P) 1 atm (760 mm Hg) STP

13. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 13 At STP, 1 mole of a gas occupies a volume of 22.4 L. The volume of one mole of a gas is called the molar volume. Molar Volume

14. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 14 The molar volume at STP can be used to form conversion factors. 22.4 L and 1 mole 1 mole 22.4 L Molar Volume as a Conversion Factor

15. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 15 A. What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g2) 0.357 g3) 1.43 g Learning Check

16. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 16 A. 1) 5.60 L 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He Solution

17. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 17 The universal gas constant, R, can be calculated using the molar volume of a gas at STP. At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T = 0.0821 L atm mole K Note there are four units associated with R. Universal Gas Constant, R

18. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 18 Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? Learning Check

19. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 19 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1 mole) (273K) = 62.4 L mm Hg mole K Solution

20. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 20 Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N 2 O at 23°C, what is the pressure (mm Hg) in the tank? Learning Check

21. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 21 1. Adjust the units of the given properties to match the units of R. V = 20.0 L, T = 296 K, n = 2.8 moles, P = ? 2. Rearrange the ideal gas law for P. P = nRT V P = (2.8 moles)(62.4 L mm Hg)(296 K) (20.0 L) (mole K) = 2.6 x 10 3 mm Hg Solution

22. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 22 A cylinder contains 5.0 L of O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Learning Check

23. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 23 1. Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O 2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = 0. 18 mole O 2 x 32.0 g O 2 = 5.8 g O 2 1 mole O 2 Solution

24. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 24 What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) RT (0.0821 L atm/mole K)(303K) = 0.00703 mole 2. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mole mole 0.00703 mole Molar Mass of a Gas

25. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 25 Gases in Equations The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors. Problem: What volume (L) of Cl 2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s)

26. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 26 Gases in Equations (continued) 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1.5 g ? L 1.2 atm, 300K 1. Calculate the moles of Cl 2 needed. 1.5 g Al x 1 mole Al x 3 moles Cl 2 = 0.083 mole Cl 2 27.0 g Al 2 moles Al 2. Place the moles Cl 2 in the ideal gas equation. V = nRT = (0.083 mole Cl 2 )(0.0821 Latm/moleK)(300K) P 1.2 atm = 1.7 L Cl 2

27. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 27 What volume (L) of O 2 at 24°C and 0.950 atm are needed to react with 28.0 g NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

28. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 28 1. Calculate the moles of O 2 needed. 28.0 g NH 3 x 1 mole NH 3 x 5 mole O 2 17.0 g NH 3 4 mole NH 3 = 2.06 mole O 2 2. Place the moles O 2 in the ideal gas equation. V = nRT = (2.06 moles)(0.0821 L atm/moleK)(297K) P 0.950 atm = 52.9 L O 2 Solution

29. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 29 Solution V 1 = V 2 T 1 T 2 Cross multiply to give V 1 T 2 =V 2 T 1 Isolate T 2 by dividing through by V 1 V 1 T 2 =V 2 T 1 T 2 =V 2 T 1 V 1 V 1 V 1