2. Kinetic Molecular Theory Postulates of the Kinetic Molecular Theory of Gases 1.Gases consist of particles (atoms or molecules) in constant, straight-line motion. 2.Gas particles do not attract or repel each other (no interactions). Particles collide with each other and surfaces elastically. Collisions with walls of container define pressure (P = F/A). 3.Gas particles are small, compared with the distances between them. Hence, the volume (size) of the individual particles can be assumed to be negligible (zero). 4. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas

3. Properties of Gases Gases expand to fill any container. –random motion, no attraction Gases are fluids (like liquids). –particles flow easily Gases have very low densities. –lots of empty space; particles spaced far apart Gases are easily compressible. –empty space reduced to smaller volume Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

4. Collisions of Gas Particles Pressure = collisions on container walls

5. In a smaller container - particles have less room to move. Particles hit the sides of the container more often. This causes an increase in pressure. As volume decreases: pressure increases. Changing the Size of the Container

6. Pressure = Force/Area KEY UNITS AT SEA LEVEL (also known as standard pressure) 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Sea level

7. Barometers Mount Everest Sea level On top of Mount Everest Sea level

8. Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 Always use temperature in Kelvin when working with gases. Std temperature = 273 K Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

9. STP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa - OR - STP Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

10. Boyle’s Law 1 atm 4 Liters As the pressure on a gas increases 2 atm 2 Liters As the pressure on a gas increases - the volume decreases Pressure and volume are inversely related

11. Boyle’s Law Illustrated Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 404

12. b The pressure and volume of a gas are inversely related at constant mass & temp Boyle’s Law PV = k Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Volume (mL) Pressure (torr) P. V (mL. torr) 10.0 20.0 30.0 40.0 760.0 379.6 253.2 191.0 7.60 x 10 3 7.59 x 10 3 7.60 x 10 3 7.64 x 10 3 P 1 x V 1 = P 2 x V 2

13. Boyle’s Law example A quantity of gas under a pressure of 106.6 kPa has a volume of 380 cm 3. What is the volume of the gas at standard pressure, if the temperature is held constant? P 1 x V 1 = P 2 x V 2 (106.6 kPa) x (380 cm 3 ) = (101.3 kPa) x (V 2 ) V 2 = 400 cm 3

14. Charles’s Law Timberlake, Chemistry 7 th Edition, page 259

15. If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to 600 K one of 2 things happens 300 K

16. Either the volume will increase to 2 liters at 1 atm 300 K 600 K

17. 300 K 600 K Or the pressure will increase to 2 atm.

18. The volume and absolute temperature (K) of a gas are directly related –at constant mass & pressure Charles’ Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Volume (mL) Temperature (K) V / T (mL / K) 40.0 44.0 47.7 51.3 273.2 298.2 323.2 348.2 0.146 0.148 0.147 V 1 / T 1 = V 2 / T 2

19. The pressure and absolute temperature (K) of a gas are directly related –at constant mass & volume Gay-Lussac’s Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Temperature (K) Pressure (torr) P/T (torr/K) 248691.62.79 273760.02.78 298828.42.78 3731,041.22.79 P 1 / T 1 = P 2 / T 2

20. = kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

21. A quantity of gas has a volume of 400 cm 3 at STP. What volume will it occupy at 35 o C and 83.3 kPa? P 1 = 101.325 kPa T 1 = 273 K V 1 = 400 cm 3 P 2 = 83.3 kPa T 2 = 35 o C + 273 = 308 K V 2 = ? cm 3 (101.325 kPa) x (400 cm 3 ) = (83.3 kPa) x (V 2 ) 273 K 308 K V 2 = 548.9 cm 3 The Combined Gas Law

22. When measured at STP, a quantity of gas has a volume of 500 cm 3. What volume will it occupy at 20 o C and 93.3 kPa? P 1 = 101.325 kPa T 1 = 273 K V 1 = 500 cm 3 P 2 = 93.3 kPa T 2 = 20 o C + 273 = 293 K V 2 = ? cm 3 (101.325 kPa) x (500 cm 3 ) = (93.3 kPa) x (V 2 ) 273 K293 K V 2 = 582.8 cm 3

23. Molar Volume (Avogadro) Timberlake, Chemistry 7 th Edition, page 268 1 mol of all gases @ STP have a volume of 22.4 L Avogadro’s Law V 1 /n 1 = V 2 /n 2

24. Ideal Gas Law PV = nRT Brings together all gas properties. P = pressure V = volume (must be in liters) n = moles R = universal gas constant (0.082 or 8.314) T = temperature (must be in Kelvin) Can be derived from experiment and theory.

25. Ideal Gas Law What is the pressure of 0.18 mol of a gas in a 1.2 L flask at 298 K? P = ? atm n = 0.18 mol T = 298 K V = 1.2 L R =.082 (L x atm)/(mol x K) P x (1.2 L) = (0.18 mol) x (.082) x (298 K) P = 3.7 atm PV = nRT

26. Gas Density D = (MM)P/RT Larger particles are more dense. Gases are more dense at higher pressures and lower temperatures D = density P = pressure MM = molar mass R = universal gas constant T = temperature (must be in Kelvin) Can be derived from experiment and theory.

27. Gas Problems 1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.0 0 C and 3.25 atm? Use proper sig figs. 2. What is the volume of 3.35 mol of gas which has a measured temperature of 47.0 0 C and a pressure of 185 kPa? Use proper sig figs.

28. Gas Problems 1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.0 0 C and 3.25 atm? Use proper sig figs. Molar mass = ? g/mol D = 0.010 g/ml T = 262 K P = 3.25 atm R =.082 (L atm)/(mol K) g/mol = (0.010g/ml) x (.082atm L/mol K) x (262 K) x (1/3.25 atm) x (1000ml/1 L) P = 66 g/mol

29. Gas Problems 2. What is the volume of 3.35 mol of gas which has a measured temperature of 47.0 0 C and a pressure of 185 kPa? Use proper sig figs. PV = nRT V = ? L n = 3.35 mol T = 320 K P = 185 kPa R = 8.314 (L kPa)/(mol K) (185 kPa) x (V) = (3.35 mol) x (8.314 L kPa/mol K) x (320 K) V = 48.2 L

30. Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem In a gaseous mixture, a gas’s partial pressure is the one the gas would exert if it were by itself in the container. The mole ratio in a mixture of gases determines each gas’s partial pressure.

31. Gas Mixtures and Dalton’s Law

32. Gas Collected Over Water When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

33. GIVEN: P H 2 = ? P total = 94.4 kPa P H 2 O = 2.6 kPa WORK: P total = P H 2 + P H 2 O 94.4 kPa = P H 2 + 2.6 kPa P H 2 = 91.8 kPa Dalton’s Law Hydrogen gas is collected over water at 22°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure on p.10 for 22°C. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

34. 41.7 kPa Dalton’s Law The total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa. What is the partial pressure of each gas. P He = P Ne = 3 mol He 7 mol gas (97.4 kPa) = 55.7 kPa 4 mol Ne 7 mol gas (97.4 kPa) = ? ?

35. Dalton’s Law Suppose you are given four containers – three filled with noble gases. The first 1 L container is filled with argon and exerts a pressure of 2 atm. The second 3 liter container is filled with krypton and has a pressure of 380 mm Hg. The third 0.5 L container is filled with xenon and has a pressure of 607.8 kPa. If all these gases were transferred into an empty 2 L container…what would be the pressure in the “new” container? P Ar = 2 atm P xe 607.8 kPa P Kr = 380 mm Hg P total = ? V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters

36. …just add them up P Ar = 2 atm P xe 607.8 kPa P Kr = 380 mm Hg P total = ? V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters Dalton’s Law of Partial Pressures “Total Pressure = Sum of the Partial Pressures” P T = P Ar + P Kr + P Xe + … P T = 1 atm + 0.75 atm + 1.5 atm P T = 3.25 atm P 1 x V 1 = P 2 x V 2 (0.5 atm) (3L) = (X atm) (2L) (6 atm) (0.5 L) = (X atm) (2L) P Kr = 0.75 atm P xe = 1.5 atm

37. Partial Pressure A gas is collected over water at 649 torr and 26.0 0 C. If its volume when collected is 2.99 L, what is its volume at STP? Use proper sig figs. P 1 V 1 /T 1 = P 2 V 2 /T 2 P T = P G + P w V 2 = ? L V 1 = 2.99 L T 1 = 299 K T 2 = 273 K P T = 649 torr P 1 = 86.5 kPa – 3.4 kPa = 83.1 kPa P 2 = 101.325 kPa (83.1 x 2.99) / 299 = (101.325 x V 2 ) / 273 V 2 = 2.24 L

38. Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric acid. Pressure =107.3 kPa; temperature = 88 o C. Zn (s) + 2 HCl (aq)  ZnCl 2 (aq) + H 2 (g) Gas Stoichiometry

39. Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric acid. Pressure =107.3 kPa; temperature = 88 o C. Zn (s) + 2 HCl (aq)  ZnCl 2 (aq) + H 2 (g) 38.2 g excessV = ? L H 2 P = 107.3 kPa T = 88 o C (361 K) Gas Stoichiometry At STP, we’d use 22.4 L per mol, but we aren’t at STP.

40. Pressure and Balloons A B = pressure exerted ON balloon A = pressure exerted BY balloon B When balloon is being filled: P A > P B When balloon is filled and tied: P A = P B When balloon deflates: P A < P B

41. When the balloons are untied, will the large balloon (A) inflate the small balloon (B); will they end up the same size or will the small balloon inflate the large balloon? Why? Balloon Riddle A B C