Properties of Gases

1. 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

2. Kinetic Molecular Theory Postulates of the Kinetic Molecular Theory of Gases • Gases consist of tiny particles (atoms or molecules) • These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero). Gases have low densities. 3. The particles are in constant random straight-line motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas. 4. The particles are assumed not to attract or to repel each other. 5. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas

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. 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

7. 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?

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

9. 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

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

11. 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

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

13. 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

14. 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

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

16. 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

17. 600 K 300 K • Either the volume will increase to 2 liters at 1 atm

18. 600 K 300 K Or the pressure will increase to 2 atm.

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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.

26. 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

27. 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.

28. Gas Problems 1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.00C 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 47.00C and a pressure of 185 kPa? Use proper sig figs.

29. 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

30. 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) P = 48.2 L

31. 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

32. Gas Mixtures and Dalton’s Law

33. 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.

34. 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

35. 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 = =

36. 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

37. …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

38. 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

39. 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)

40. 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.