2. GASES “ UP, UP AND AWWWAAAAYYYYYYYY”

3. KINETIC THEORY OF GASES 1.A gas consists of small particles that move rapidly in straight lines. 2.have essentially no attractive (or repulsive) forces. 3.are very far apart. 4.have very small volumes compared to the volume of the container they occupy. 5.have kinetic energies that increase with an increase in temperature. 6.Gas Pressure is the collisions between the gas particles and the walls of the container and each other. 2 Copyright © 2009 by Pearson Education, Inc.

4. PROPERTIES THAT DESCRIBE A GAS Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). 3 Copyright © 2009 by Pearson Education, Inc.

5. FACTORS AFFECTING GAS PRESSURE Amount of Gas- low pressure = fewer particles high pressure = more particles Volume – decrease the volume = increase in pressure Temperature- increase temperature = increase pressure

6. GAS PRESSURE Gas pressure is a force acting on a specific area. Pressure (P) = force area has units of atm, mmHg, torr, lb/in. 2, and kilopascals(kPa). 1 atm = 760 mm Hg (exact) 1 atm = 760 torr 1 atm = 14.7 lb/in. 2 1 atm = 101 325 Pa 1 atm = 101.325 kPa 5

7. BAROMETER A barometer measures the pressure exerted by the gases in the atmosphere. It indicates atmospheric pressure as the height in mm of the mercury column. 6 Copyright © 2009 by Pearson Education, Inc.

8. Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm Copyright © 2009 by Pearson Education, Inc. Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. Atmospheric Pressure

9. LEARNING CHECK 1. What is 475 mmHg expressed in atm? A) 475 atm B) 0.625 atm C) 3.61 x 10 5 atm 2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg? A) 2.00 mmHg B) 1520 mmHg C)22 300 mmHg 8

10. SOLUTION 1. What is 475 mmHg expressed in atm? B) 0.625 atm 475 mmHg x 1 atm = 0.625 atm 760 mmHg 2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg? B) 1520 mmHg 2.00 atm x 760 mmHg = 1520 mmHg 1 atm 9

11. ALTITUDE AND ATMOSPHERIC PRESSURE Atmospheric pressure is about 1 atmosphere at sea level. It depends on the altitude and the weather. It is lower at higher altitudes, where the density of air is less. It is higher on a rainy day than on a sunny day. 10 Copyright © 2009 by Pearson Education, Inc.

12. LEARNING CHECK 1. The downward pressure of the Hg in a barometer is _____ than (as) the pressure of the atmosphere. A) greater B) less C) the same 2. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because A) H 2 O is less dense than mercury. B) H 2 O is heavier than mercury. C) air is more dense than H 2 O. 11

13. SOLUTION 1.The downward pressure of the Hg in a barometer is C) the same (as) the pressure of the atmosphere. 2. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because A) H 2 O is less dense than mercury. 12

14. TEMPERATURE CONVERSION EQUATIONS 2 equations to use: K = o C + 273 o C = K – 273

15. PRACTICE PROBLEM #1 240 o C = ____K K = o C + 273 K = 240 + 273 K = 513

16. PRACTICE PROBLEM #2 350K = ____ o C o C = K – 273 o C = 350 – 273 o C = 77

17. PRACTICE PROBLEM #3 510K = ____ o C o C = K – 273  o C = 510 – 273  o C = 237

18. PRACTICE PROBLEM #4 20 o C = ____K K = o C + 273 K = 20 + 273 K = 293

19. BOYLE’S LAW Boyle’s law For a given mass of a gas at a constant temperature… The volume of a gas varies inversely with the pressure. P 1 V 1 =P 2 V 2

20. CHARLES’ LAW Charles’ law For a given mass of a gas at a constant pressure… The volume of a gas varies directly proportional with the temperature. V 1 /T 1 =V 2 /T 2

21. GUY LUSSAC’S LAW Guy Lussac’s law For a given mass of a gas at a constant volume… The pressure of a gas varies directly proportional with the temperature. P 1 / T 1 =P 2 /T 2

22. COMBINED GAS LAWS If all gas laws are combined ( nothing is held constant) Then the formula becomes P 1 V 1 = P 2 V 2 T 1 T 2

23. DALTON’S LAW OF PARTIAL PRESSURE The individual pressure of each gas in a mixture is called partial pressure of a gas. John Dalton suggested that the total pressure of a mixture of gases is equal to the sum of the partial pressure of the individual gases in the mixture.(constant volume and temperature. P total = P 1 + P 2 + P 3 +…..

24. AVOGADRO LAW Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules". For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

25. CALCULATIONS USING AVOGADRO LAW If the volume of 2.67 g of SF 6 gas at 1.143 atm and 28.5 º C in 2.93m 3, what is the mass of SF 6 gas in a container with a volume of 543.9 m 3 at the same pressure and temperature?

26. AVOGARDRO’S PRINCIPLE Equal volumes of gases at the same temperature and pressure contain the same number of particles. Gas is mostly empty space and can be compressed.. 1mole=6.02 x 10 23 particles =22.4dm 3 Standard temperature = 0ºC= 273 K Standard pressure = 1 atm = 101.3 kPa = 760 mm Hg

27. IDEAL GAS An ideal gas is a gas composed of molecules with mass but no attractive forces and no volume to each particle. Follows all gas laws at all conditions of temperature and pressure. The volume of a gas at a specific temperature and pressure is directly proportional to the number of particles in the gas. The number of moles, n, is also directly proportional to the number of particles.

28. IDEAL GAS Thus, the number of moles is directly proportional to the volume of the gas P 1 V 1 = P 2 V 2 T 1 n T 2 n Constant for any ideal gas At STP, 1 mole = 22.4 dm 3, P = 101.3 kPa and T = 273 K P 1 V 1 = (101.3)(22.4) T 1 n (273) (1) = 8.31kPa dm 3 /mole K = R

29. IDEAL GAS Drop the subscripts PV = R T n PV=n RT

30. IDEAL GAS Molecular mass can be determined from the ideal gas law if you remember that n = mass/ atomic mass PV=n RT PV=mass RT atomic mass atomic mass = mass RT PV

31. DIFFUSION AND GRAHAM’S LAW The rate varies directly as the velocity of the molecules. At the same temperature, molecules of smaller mass diffuse faster than larger molecules because smaller molecules travel faster. Diffusion is the spontaneous spreading of particle throughout a given volume until they are uniformly distributed. All gases do not diffuse at the same rate.

32. DIFFUSION AND GRAHAM’S LAW Thomas Graham formulated this principle. Grahams law states that the relative rates at which two gases under identical conditions of temperature and pressure will diffuse very inversely as the square root of the molecular masses of the gases. At the same temperature, the average kinetic energy must be the same, so V 1 = V 2 m 2 m 1

33. GAS STOICHIOMETRY I KNOW YOU ARE SO HAPPY YOU GET TO DO STOICHIOMETRY AGAIN!!!!!!

34. TWO TYPES OF GAS STOICHIOMETRY PROBLEMS 1.STP gases 2.Non-STP gases

35. Gas Stoichiometry We will use the ideal gas law to establish the number of moles of a gas in a given volume at any temperature and pressure. This allows us to determine quantities used and produced in chemical reactions. We can solve at STP or not at STP. Remember!! STP = standard temperature & pressure 1atm and 273 K

36. Gas Stoichiometry Steps for Solving Gas Stoichiometry problems 1. Write a balanced equation for the reaction 2. Convert all amounts to moles 3. Compare molar amounts using stoichiometry ratios from the balanced equation 4. Convert the moles into the units required, using the ideal gas law Practice @ STP A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O 2 are present?

37. PRACTICE A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O 2 are present? Gas Stoichiometry @ STP

38. EXAMPLE #1 What volume of hydrogen gas will form at STP when 30.0 g of sodium react with an excess of hydrochloric acid, HCl?

39. EXAMPLE #2 What volume of hydrogen at STP will form when 2.00 g of calcium react with an excess of water?

40. Gas Stoichiometry e.g. What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18 o C and 100.3 kPa?

41. Gas Stoichiometry e.g. What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18 o C and 100.3 kPa? G:H 2 SO 4(aq) + Fe (s)  H 2(g) + FeSO 4(s) mass Fe = 40.0g temperature = 18 o C = 291 K pressure = 100.3 kPa

42. Gas Stoichiometry What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18 o C and 100.3 kPa? M.M. of Fe =58g/mol U: volume of H 2 M.R. = 1 to 1 1. mass Fe  moles Fe  moles H 2 2. PV = nRT V = nRT P

43. Gas Stoichiometry e.g. What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18 o C and 100.3 kPa? 1. moles H 2 = 40.0g Fe = 0.7168 mol H 2

44. Gas Stoichiometry e.g. What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18 o C and 100.3 kPa? 1. moles H 2 = 40.0g Fe = 0.7168 mol H 2 2. V = (0.7168)(8.314)(291.15) 100.3 = 17 L H 2

45. GAS STOICHIOMETRY Moles  Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas lawNon-STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conversion Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

46. 1 mol CaCO 3 100.09g CaCO 3 GAS STOICHIOMETRY PROBLEM What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 1.26 mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

47. WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 GAS STOICHIOMETRY PROBLEM What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm 3  kPa/mol  K Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

48. WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = 0.597 mol O 2 GAS STOICHIOMETRY PROBLEM How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 dm 3  kPa/mol  K 4 Al + 3 O 2  2 Al 2 O 3 15.0 L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT  Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

49. 2 mol Al 2 O 3 3 mol O 2 GAS STOICHIOMETRY PROBLEM How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

50. GAS LAWS RE-EXPLAINED Boyles Law: initial pressure equals final pressure times final volume  P 1 V 1  P 2 V 2 Charles Law: the ratio of volume to temperature of a given gas at fixed pressure is constant  V 1 /T 1 = V 2 /T 2 Gay-Lussac’s Law: the ratio of pressure to temperature of a given gas at fixed volume is constant  P 1 /T 1 = P 2 / T 2 Graham’s law : the diffusion or effusion rate of a gas is inversely proportional to its molar mass.

51. Combined Gas Law: the combined law incorporates all of the 3 main laws: Boyles, Charles, and Gay Lussac’s  P 1 V 1 /T 1 = P 2 V 2 /T 2 The Ideal Gas Law: relates the amount of gas produced in a reaction  PV=nRT where n= moles, R= This is the proportionality constant Dalton’s Law of Partial Pressures: the total pressure of a mixture of gases equals the sum of the pressures  P total = P 1 +P 2 etc. Gas Stoichiometry - use the molar volume of a gas (22.4 L) if at STP If not at stp, use The Ideal Gas law and stoichiometry

52. THE END