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Chapter 13

Chapter 13. Gases and the Gas Laws Mythbuster – cansI Intro to gas laws. Gas Laws in Action: workers steam cleaned this tanker car and then sealed up the container, they came back the following morning to this disaster. ( video ). Kinetic Theory Assumptions for an Ideal Gas.

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Chapter 13

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  1. Chapter 13 Gases and the Gas Laws Mythbuster – cansI Intro to gas laws

  2. Gas Laws in Action: workers steam cleaned this tanker car and then sealed up the container, they came back the following morning to this disaster. (video)

  3. Kinetic Theory Assumptions for an Ideal Gas • Gas particles are in constant, random motion • Gas particles themselves have virtually no volume • Gas particles do not attract nor repel each other • No kinetic energy is lost when particles collide • If gases are at the same temp. they have the same KE

  4. NOTE: • Real gases (actual gases) do NOT obey all the assumptions made by the kinetic theory (only ideal gases behave this way- we will get exceptions later)

  5. Factors (variables) that Affect Gases • Number of gas particles present • Temperature • Pressure • Volume of the sample Animation (explanation) Animation changing each of the variables (graph)

  6. In a tied off balloon the pressure from the outside = pressure from the inside (in this chapter we will look at how changing the factors, changes these values) =

  7. STP • Standard temperature and standard pressure • Standard temperature = 0° C (273 K) • Standard pressure = 101.3 kPa (1 atm or 760 mm Hg)

  8. Boyles Law • States that the volume of a gas is inversely proportional to its pressure if the temperature remains constant • As pressure goes up, volume goes down and vice versa if temperature is constant • PV = k

  9. PVP x V P1=1 atm V1= 800 ml 800 P2=2 atm V2= 400 ml 800 P3=3 atm V3= 267 ml 800 P1V1 = 800 P2V2 = 800 So: P1V1 = P2V2 Boyles Law

  10. If .600 L (V1)of a gas at 100.0 kPa (P1) changes to 62.4 kPa.(P2) What is the new volume if temperature remains constant? P1V1 = P2V2 (100.0 kPa) (.600L) = (62.4 kPa) (V2) .96153 L = V2 .962 L = V2

  11. Note: you do not need to convert units as long as they match on both sides of the equation A 185 ml sample has a pressure of 4.2 atm. What is it’s pressure when the volume is 250 ml if temperature remains constant? P1V1 = P2V2 (4.2 atm)(185 ml) = P2 (250 ml) 3.1 atm = P2

  12. Charles Law • Jacque Charles investigated the property of changing temperature on the volume of a gas (saw w/ each ° C change the volume changed by 1/273rd) • Charles Law – volume of a fixed mass of gas is directly proportional to its kelvin temperature if the pressure is constant • Ex. Helium balloon deflates when walking outside on a cold day

  13. Charles Law: V1 = V2 T1 T2 V1T2 = V2T1 or *** “T” must be in Kelvin ( K = C +273)

  14. A balloon inflated in an air conditioned room at 28° C (T1) has a volume of 4.0 L (V1). If it is heated to a temperature of 58 °C ( T2), what is the new volume (V2) of the balloon if pressure remains constant? V1 = V2 T1 T2 T1 = 28 + 273 = 301K T2 = 58 + 273 = 331 K

  15. V1T2 = V2 T1 (4.0 L ) (331 K) = V2 (301 K) 4.4 L = V2

  16. Adjust the volume of 609 ml of a gas at 83°C to standard temperature.

  17. Eggs and Gas Laws

  18. Gay-Lussac’s Law • States that the pressure of a gas is directly proportional to the Kelvin temperature if volume is kept constant • Ex. Spray paint can (rigid container) in a bonfire P1 = P2 T1 T2 P1T2 = P2T1 or “T” must be in Kelvin

  19. As temperature increases, pressure has to increase proportionately to keep same volume

  20. The pressure of a gas in a tank is 3.20 atm (P1) at 22.0 °C (T1). If the temperature is raised to 60.0 °C (T2), what is the new pressure (P2) if volume is held constant? • T1 = 22.0 + 273 = 295 K • T2 = 60.0 + 273 = 333 K

  21. P1 = P2 T1 T2 P1T2 = P2 T1 (3.20 atm) (333K) = P2 295 K 3.61 atm = P2

  22. The Combined Gas Law • Many times it is hard to keep a variable constant (and only deal with changing 2 variables at a time), so we have to use all the laws together • Combined Gas Law: all the variables of pressure, temperature and volume change (only thing that is constant is the number of particles)

  23. P1V1 = P2V2 T1 T2 or P1V1T2 = P2V2T1

  24. Find the volume of a gas at STP if it measures 806 ml at 26.0° C and 103.0 kPa P1V1 = P2V2 T1 T2 P1 = 103.0 kPa V1 = 806 ml T1 = 26.0 + 273 = 299 K P2 = 101.3 kPa (standard pressure) V2 = ? T2 = 273 K (standard temperature)

  25. (103.0 kPa) (806 ml) = (101.3 kPa) (V2) 299 K 273 K (103.0 kPa) (806ml) (273 K) = V2 (299 K) (101.3 kPa) 748 ml = V2

  26. Gases and the MOLE Rock me Avogadro • Avogadro’s Principle: at equal temperatures and equal pressures, equal volumes of gases contain the same number of molecules • Molar Volume: volume occupied by 1 mole of any gas under STP (0 °C, 101.3 kPa) is 22.4 L (conversion factor= 22.4 L/1 mole)

  27. O2 He 1 mole He at STP 6.02 x 10 23 atoms He 4.00 g 22.4 L 1 mole O2 at STP 6.02 x 10 23 molecules of O2 32.0 g 22.4 L

  28. What is the volume of 8.6 g of Cl2 at STP? • Convert g  moles (molar mass) • Convert moles  volume (22.4 L/ mole 1) .12 moles 8.6 g Cl2 1 mole Cl2 22.4 L 2) 71.0 g Cl2 1 mole .12 moles 2.7 L

  29. Ideal Gas • Combines Avogadro’s principle, Boyles, Charles and Gay-Lussac’s Law into a statement w/ P, V, T and # moles • Changing one variable will affect the other 3 variables • Ideal Gas Equation: PV = nRT

  30. PV = nRt • n = # of mole • R = Ideal Gas Constant ( experimentally determined constant based on Avogadro’s # and STP  dependent on unit used for pressure)

  31. Pressure in: • atm use: R= .0821 L· atm/ mole ·K • kPa use : R = 8.314 L ·kPa/ mole· K • mm Hg use: R = 62.4 L· mm Hg/ mole ·K

  32. Calculate the number of moles of gas contained in a 3.0 L vessel at 30Ō K and a pressure of 1.50 atm. • PV = nRT • PV = n R= .0821 RT (1.50 atm) (3.0 L) = n (.0821 · 30ŌK) n = .18 moles

  33. Applying the Ideal Gas Law • Calculate molar mass: n (# of moles) = mass of gas = m Molar mass M So: PV = nRT PV = mRT or M M = mRT PV

  34. Calculate density: • D= m/V M = mRT PV (substitute D for m/v in this equation M = DRT P Or D = MP RT

  35. Deviations from Ideal Behavior • Ideal Gases: have no attractive forces and do not take up space (volume) • Real Gases: • Occupy definite volume (take up space)- but volume is small • Under normal conditions real gases behave like ideal gases (follow all gas laws)

  36. Under high pressures: particles are forced close together and can’t compress any further, also attractive forces take over • So: real gases will liquefy instead of disappearing like Boyle predicted • Same is true under really low temperatures

  37. Gas Laws Test • Formulas, R values and periodic table will be given to you • > 40 questions • 12 multiple choice • 13 fill in the blank (need to know who did what law/PTV card/variables) • 7 calculations (1 for each formula, 1 using 22.4 L= 1 mole, 1 PV= nRt)

  38. Know: • Who did each law • What each law stands for • Scenerios with each law • Absolute zero • STP • Molar Volume /avogadros principle • Variables on a gas • Real gas vs ideal gas

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