1 / 44

Chemistry I Honors

Chemistry I Honors. Unit 9: Gases. Objectives # 1-3: Introduction to the Kinetic Theory of Gases. The Kinetic Theory Assumptions of the Kinetic Theory Gases are composed of tiny particles that are arranged far apart from each other

sarila
Download Presentation

Chemistry I Honors

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemistry I Honors Unit 9: Gases

  2. Objectives #1-3: Introduction to the Kinetic Theory of Gases • The Kinetic Theory • Assumptions of the Kinetic Theory • Gases are composed of tiny particlesthat are arranged far apart from each other • Gases are composed of individual atoms such as in the element neon or molecules such as in the element oxygen

  3. Objectives #1-3: Introduction to the Kinetic Theory of Gases • Collisions that may occur between gas particles are elastic with no net loss of energy • Gas particles are in constant, random motion • No attractiveforces exist between gas particles • The temperatureof a gas depends on the average kinetic energy of the particles • (video clip Ch. 3 “Common Gas Properties”)

  4. Objectives #1-3: Introduction to the Kinetic Theory of Gases • The Kinetic Theory and its Implications on the Properties of Gases • The temperature of a gas is directlyrelated to its kinetic energy • The temperature at which no kinetic energy is present is called absolute zero; this temperature is 0 on the Kelvin scale and -273oon the Celsius scale • Gases are able to expand freely due to their randommotion and the lack of attractive forces between the particles of a gas • Gas particles are tiny and can move past each other with ease due to the lack of attractive forces between gas particles

  5. Objectives #1-3: Introduction to the Kinetic Theory of Gases • The density of gases is generally less than other substances due to the ability of gases to movefreely • Gases can be compressed due to the great distances between gas particles • The ability of gases to spread out spontaneously or diffuseis due to the rapid motion of gas particles; particles with less mass and faster velocities spread out at a faster rate than heavier, slower gas particles (This is why you smell cookies baking! )

  6. Objectives #1-3: Introduction to the Kinetic Theory of Gases • Due to their small size, gas particles can be made to pass through smallopenings; this characteristic of effusion depends on the velocity and molar mass of the gas particles present; the rate of effusion is directly proportional to the velocity of the gas particles and inversely proportional to the molar mass of the gas particles • When gas particles collide against a surface they exhibit pressure; which is defined as the force per unit area

  7. Objectives #4-9 : Properties of Gases • Common Units of Pressure:

  8. Objectives #4-9: Properties of Gases • Pressure Conversion Examples: • Convert 2.5 atm to mmHg 2.5 atm x 760 mmHg = 1900 mmHg 1 atm • Convert 300. Pa to atm 300. Pa x 1 atm_____ = .00296 atm 101325 Pa

  9. Objectives #4-9: Properties of Gases Dalton’s Law of Partial Pressure • The total pressureof a mixture of gases is equal to the sum of the partial pressuresof each of the individual gases in the mixture PT = P1 + P2 + P3 + …. • This concept must be kept in mind when gases are collected over water in the laboratory; the vapor pressure of water must be subtracted from the measured pressure of the gas in order to obtain the true pressure of the gas being collected Pgas = Patm - Pwater

  10. Objectives #4-9: Properties of Gases • The vapor pressure due to water increases with increasing temperature (see attached chart) Molar Volume • at standard temperature and pressure (STP), which are defined as O o C and 1atm, 1 mole of any gas occupies 22.4 L; this is referred to as molar volume

  11. *Interpreting Vapor Pressure Charts

  12. Examples: • Which substance has the highest vapor pressure? • Which substance has the lowest vapor pressure? • Boiling occurs when the vapor pressure of a substance equals the external atmospheric pressure. At what external pressure will ethanol boil at 40oC? • A substance is volatile when it readily evaporates. Which substance is the most volatile? 5. Which substance has the strongest intermolecular forces? 6. Which substance has the weakest intermolecular forces?

  13. Objectives #4-9: Properties of Gases • Relationships Among Gas Characteristics • Amount of Gas vs. Pressure (assumes temperature and volume are held constant) • What is happening at the particle level:

  14. Objectives #4-9: Properties of Gases • As the amount of gas in a container increases, the pressure increases • This illustrates a directrelationship between the amt. of the gas and the pressure of the gas

  15. Objectives #4-9: Properties of Gases • Pressure vs. Volume of Gases (assumes constant temperature) • What is happening at the particle level:

  16. Objectives #4-9: Properties of Gases • As the pressureof a fixed amount of gas increases, its volume decreases • This illustrates an inverse relationship

  17. Objectives #4-9: Properties of Gases • Temperature vs. Volume (assumes constant pressure) • What is happening at the particle level:

  18. Objectives #4-9: Properties of Gases • As the Kelvin temperature of a fixed amount of gas increases, its volume increases • This illustrates a direct relationship:

  19. Objectives #4-9: Properties of Gases • Temperature vs. Pressure (assuming constant volume) • What is happening at the particle level:

  20. Objectives #4-9: Properties of Gases • As the Kelvin temperatureof a fixed amount of gas increases, its pressure increases • This illustrates a direct relationship

  21. Objectives #4-9: Properties of Gases Ideal vs. Real Gases • Ideal gases always follow the kinetictheoryunder any conditions • Real gas particles do have attractiveforces among each other • Real gases no longer act as ideal gases under conditions of highpressure and extremely lowtemperature

  22. Objective #10 Solving problems Involving the Gas Laws *the Gas Laws *the Gas Laws are mathematical formulas based on the relationships discussed in the previous section of notes P1V1 / T1 = P2V2 / T2 (Combined Gas Law) PV = nRT (Ideal Gas Law) (video clip “Ch. 4 Ideal Gas Law) • Combined Gas Law *examples:

  23. To what pressure must a gas be compressed to get it into a 9.00 L tank if it occupies 90.0 L at 1.00 atm? Can a variable be eliminated? P1V1 = P2V2 P1V1 / V2 = P2 (1.00 atm) (90.0 L) / 9.00 L = P2 10.0 atm = P2

  24. A container with a movable piston contains .89 L of methane gas at 100.50C. If the temperature of the gas drops to 11.3oC, what is the new volume of the gas? K = C + 273 = 100.5oC + 273 = 374 K

  25. K = C + 273 = 11.3oC + 273 = 284 K Can a variable be eliminated? V1 / T1 = V2 / T2 V1T2 = T1V2 V1T2 / T1 = V2 (.89 L) (284 K) / 374 K = V2 .676 L = V2

  26. 3. A sample of gas occupies a volume of 5.0 L at a pressure of 650. torr and a temperature of 24oC . We want to put the gas in a 100. ml container that can only withstand a pressure of 3.0 atm. What temperature must be maintained so that the container does not explode?

  27. 5.0 L = 5000 ml 650. torr = .855 atm 24oC = 297 K P1V1 / T1 = P2V2 / T2 Can a variable be eliminated? P1V1T2 = T1P2V2 T2 = T1P2V2 / P1V1

  28. T2 = T1P2V2 / P1V1 T2 = (297 K) (3.0 atm) (100. ml) / = (.855 atm) (5000 ml) = 21 K 4. A sample of gas occupies 2.00 L at STP. What volume will it occupy at 27oC and 200. mmHg?

  29. P1V1 / T1 = P2V2 / T2 Can a variable be eliminated? P1V1T2 = T1P2V2 P1V1T2 / T1P2 = V2 (760 mm Hg) (2.00 L) (300 K) / (273 K) (200. mm Hg) = V2 8.35 L = V2

  30. Additional Problems *Recall that at STP conditions, 1 mole of any gas = 22.4 L *examples: • Calculate the volume of .55000 moles of gas at STP. .55000 moles X 22.4 L / 1 mole = 12.320 L

  31. Calculate the moles of gas contained in 350 L of gas at STP. 350 L X 1 mole / 22.4 L = 16 moles • Calculate the mass in grams of 3.50 L of chlorine gas. 3.50 L X 71.0 g / 22.4 L = 11.1 g

  32. Objective #11 The Ideal Gas Law and Gas Stoichiometry PV=nRT *ideal gas constant: R = PV / nt R = ( 1 atm) (22.4 L) / (1 mole) (273K) = .0821 L . atm. / mole.K *examples: • What is the temperature of a .65 L sample of fluorine gas at 620. torr which contains 1.3 mol fluorine? T = PV / nR

  33. *examples: • What is the temperature of a .65 L sample of fluorine gas at 620. torr which contains 1.3 mol fluorine? T = PV / nR = (620 torr / 760 torr) (.65 L) / (1.3 mol) (.0821 L.atm / mol.K) = 5.0 K

  34. A 25.0 gram sample of argon gas is placed inside a container with a volume of 10.0 L at a temperature of 65oC. What is the pressure of argon at this temperature? PV=nRT P = nRT / V = (25.0 g / 39.9 g) (.0821) (338 K) / 10.0 L = 1.74 atm.

  35. *there are two types of gas stochiometry problems: at STP non STP *examples:

  36. Calculate the volume of hydrogen gas that can be produced from the reaction of 5.00 g of zinc reacted in an excess of hydrochloric acid. Assume STP conditions. Zn + 2HCl --› H2 + ZnCl2 5.00 g Zn X 1 mole Zn / 65.4 g Zn X 1 mole H2 / 1 mole Zn X 22.4 L / 1 mole H2 = 1.71 L

  37. Calculate the volume in liters of oxygen gas that can be produced from the decomposition of 3.50 X 1024 formula units of potassium chlorate. Assume STP conditions. 2KClO3 --› 2KCl + 3O2 3.50 X 1024 formula units KClO3 X 1 mole KClO3 / 6.02 X 1023 f. units X 3 mole O2 / 2 mole KClO3 X 22.4 L / 1 mole = 195 L

  38. Calculate the volume of hydrogen produced at 1.50 atm and 19oC by the reaction of 26.5 g of calcium metal with excess water. The vapor pressure of water is 16.5 mmHg. Ca + 2H2O --› H2 + Ca(OH)2

  39. *use amount of given reactant and stoichiometry to determine moles of gas desired in problem: 26.5 g Ca X 1 mole Ca / 40.1 g Ca X 1 mole H2 / 1 mole Ca = .661 mole H2

  40. *determine net pressure of gas: 16.5 mmHg X 1atm/760 mmHg = .0217 atm 1.50 atm - .0217 atm = 1.48 atm *use moles of gas found in ideal gas law to calculate volume of gas: PV=nRT V = nRT / P = (.661 moles) (.0821) (292 K) / 1.48 atm = 10.7 L

  41. Calculate the volume of chlorine gas produced at 1.25 atm at 25oC from the reaction of 5.00 g of sodium chloride and an excess of fluorine. 2NaCl + F2 --› 2NaF + Cl2 5.00 g NaCl X 1 mole NaCl / 58.5 g NaCl X 1 mole Cl2 / 2 mole NaCl = .0427 mole Cl2

  42. PV = nRT V = nRT / P = (.0427 mole) (.0821) (298 K) /1.25 atm = .836 L

More Related