Kinetic Molecular Theory

1. Kinetic Molecular Theory Postulates of the Kinetic Molecular Theory of Gases • Gases consist of particles (atoms or molecules) in constant, straight-line motion. • 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). • 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

2. 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

3. Collisions of Gas ParticlesPressure = collisions on container walls

4. Changing the Size of the Container • 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.

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

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

7. K = ºC + 273 Temperature Always use temperature in Kelvin when working with gases. Std temperature = 273 K ºF -459 32 212 ºC -273 0 100 K 0 273 373 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

8. Standard Temperature & Pressure 0°C 273 K 1 atm 101.325 kPa - OR - STP STP Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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

10. Boyle’s Law Illustrated Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 404

11. Boyle’s Law • The pressure and volume of a gas are inversely related • at constant mass & temp PV = k P1 x V1 = P2 x V2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

12. Boyle’s Law example A quantity of gas under a pressure of 106.6 kPa has a volume of 380 cm3. What is the volume of the gas at standard pressure, if the temperature is held constant? P1 x V1 = P2 x V2 (106.6 kPa) x (380 cm3) = (101.3 kPa) x (V2) V2 = 400 cm3

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

14. 300 K • 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

15. 600 K 300 K • Either the volume will increase to 2 liters at 1 atm

16. 600 K 300 K Or the pressure will increase to 2 atm.

17. Charles’ Law • The volume and absolute temperature (K) of a gas are directly related • at constant mass & pressure V1 / T1 = V2 / T2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

18. Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume P1 / T1 = P2 / T2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

19. P1V1 T1 P2V2 T2 = Combined Gas Law P T V T PV T PV = k P1V1T2 =P2V2T1 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

20. The Combined Gas Law A quantity of gas has a volume of 400 cm3 at STP. What volume will it occupy at 35oC and 83.3 kPa? (101.325 kPa) x (400 cm3) = (83.3 kPa) x (V2) 273 K 308 K P1 = 101.325 kPa T1 = 273 K V1 = 400 cm3 P2 = 83.3 kPa T2 = 35oC + 273 = 308 K V2 = ? cm3 V2 = 548.9 cm3

21. The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 cm3. What volume will it occupy at 20oC and 93.3 kPa? (101.325 kPa) x (500 cm3) = (93.3 kPa) x (V2) 273 K 293 K P1 = 101.325 kPa T1 = 273 K V1 = 500 cm3 P2 = 93.3 kPa T2 = 20oC + 273 = 293 K V2 = ? cm3 V2 = 582.8 cm3

22. Molar Volume (Avogadro) 1 mol of all gases @ STP have a volume of 22.4 L Avogadro’s Law V1/n1 = V2/n2 Timberlake, Chemistry 7th Edition, page 268

23. 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.

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

25. 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.

26. Gas Problems 1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.00C and 3.25 atm? Use proper sig figs. g/mol = (0.010g/ml) x (.082atm L/mol K) x (262 K) x (1/3.25 atm) x (1000ml/1 L) Molar mass = ? g/mol D = 0.010 g/ml T = 262 K P = 3.25 atm R = .082 (L atm)/(mol K) P = 66 g/mol

27. Gas Problems 2. What is the volume of 3.35 mol of gas which has a measured temperature of 47.00C and a pressure of 185 kPa? Use proper sig figs. (185 kPa) x (V) = (3.35 mol) x (8.314 L kPa/mol K) x (320 K) PV = nRT V = ? L n = 3.35 mol T = 320 K P = 185 kPa R = 8.314 (L kPa)/(mol K) V = 48.2 L

28. Ptotal = P1 + P2 + ... Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. 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 gasesdetermines each gas’s partial pressure. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

29. Gas Mixtures and Dalton’s Law

30. Gas Collected Over Water When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

31. 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. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.6 kPa WORK: Ptotal = PH2+ PH2O 94.4 kPa = PH2 + 2.6 kPa PH2 = 91.8 kPa Look up water-vapor pressure on p.10 for 22°C. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

32. 3 mol He 7 mol gas 4 mol Ne 7 mol gas 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. 41.7 kPa PHe = (97.4 kPa) = 55.7 kPa (97.4 kPa) PNe = =

33. 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? PKr = 380 mm Hg Ptotal = ? PAr = 2 atm Pxe 607.8 kPa V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters

34. …just add them up PKr = 380 mm Hg Ptotal = ? PAr = 2 atm Pxe 607.8 kPa 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” PT = PAr + PKr + PXe + … P1 x V1 = P2 x V2 P1 x V1 = P2 x V2 (6 atm) (0.5 L) = (X atm) (2L) (0.5 atm) (3L) = (X atm) (2L) PKr = 0.75 atm Pxe = 1.5 atm PT = 1 atm + 0.75 atm + 1.5 atm PT = 3.25 atm

35. Partial Pressure A gas is collected over water at 649 torr and 26.00C. If its volume when collected is 2.99 L, what is its volume at STP? Use proper sig figs. (83.1 x 2.99) / 299 = (101.325 x V2) / 273 P1V1/T1 = P2V2/T2 PT = PG + Pw V2 = ? L V1 = 2.99 L T1 = 299 K T2 = 273 K PT = 649 torr P1 = 86.5 kPa – 3.4 kPa = 83.1 kPa P2 = 101.325 kPa V2 = 2.24 L

36. Gas Stoichiometry Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric acid. Pressure =107.3 kPa; temperature = 88oC. Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

37. Gas Stoichiometry Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric acid. Pressure =107.3 kPa; temperature = 88oC. Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g) 38.2 g excess V = ? L H2 P = 107.3 kPa T = 88oC (361 K) At STP, we’d use 22.4 L per mol, but we aren’t at STP.

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

39. A B C Balloon Riddle 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?