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The Gas Laws

The Gas Laws. Chemistry Dr. May. Gaseous Matter. Indefinite volume and no fixed shape Particles move independently of each other Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids. Avogadro’s Number.

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The Gas Laws

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  1. The Gas Laws Chemistry Dr. May

  2. Gaseous Matter • Indefinite volume and no fixed shape • Particles move independently of each other • Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids

  3. Avogadro’s Number • One mole of a gas contains Avogadro’s number of molecules • Avogadro’s number is 6.02 x 1023 or 602,000,000,000,000,000,000,000

  4. Gas Hydrogen (H2) Nitrogen (N2) Oxygen (O2) Fluorine (F2) Chlorine (Cl2) Molar Mass 2 grams/mole 28 grams/mole 32 grams/mole 38 grams/mole 70 grams/mole Diatomic Gas Elements

  5. Gas Helium Neon Argon Krypton Xenon Radon Molar Mass 4 grams/mole 20 grams/mole 40 grams/mole 84 grams/mole 131 grams/mole 222 grams/mole Inert Gas Elements

  6. Gas Carbon Dioxide Carbon Monoxide Sulfur Dioxide Methane Ethane Freon 14 Formula Molar Mass CO2 44 g/mole CO 28 g/mole SO2 64 g/mole CH4 16 g/mole CH3CH3 30 g/mole CF4 88 g/mole Other Important Gases

  7. One Mole of Oxygen Gas (O2) • Has a mass of 32 grams • Occupies 22.4 liters at STP • 273 Kelvins (0oC) • One atmosphere (101.32 kPa)(760 mm) • Contains 6.02 x 1023 molecules (Avogadro’s Number)

  8. Mole of Carbon Dioxide (CO2) • Has a mass of 44 grams • Occupies 22.4 liters at STP • Contains 6.02 x 1023 molecules

  9. One Mole of Nitrogen Gas (N2) • Has a mass of 28 grams • Occupies 22.4 liters at STP • Contains 6.02 x 1023 molecules

  10. Mass Volume at STP Molecules 2.0 grams 22.4 liters 6.02 x 1023 Mole of Hydrogen Gas (H2)

  11. Molar Volume Standard Temperature Standard Pressure 22.4 liters/mole 0 oC 273 Kelvins 1 atmosphere 101.32 kilopascals 760 mm Hg Standard Conditions (STP)

  12. liters  milliliters milliliters  liters o C  Kelvins Kelvins  o C mm  atm atm  mm atm  kPa kPa  atm Multiply by 1000 Divide by 1000 Add 273 Subtract 273 Divide by 760 Multiply by 760 Multiply by 101.32 Divide by 101.32 Gas Law Unit Conversions

  13. Charles’ Law • At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins V1 = V2 T1 T2 As the temperature goes up, the volume goes up

  14. Boyle’s Law • At constant temperature, the volume of a gas is inversely proportional to the pressure. P1V1 = P2V2 As the pressure goes up, the volume goes down

  15. Combined Gas Law P1V1 = P2V2 T1 T2 • Standard Pressure (P) = 101.32 kPa, 1 atm, or 760 mm Hg • Standard Temperature (T) is 273 K • Volume (V) is in liters, ml or cm3

  16. Charles’ Law Problem A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume? 1. Convert oC to Kelvins 25 + 273 = 298 K 38 + 273 = 311 K 2. Insert into formula

  17. Charles’ Law Solution V1 = V2 T1 T2 V1 = 2 liters V2 = Unknown T1 = 298 K T2 = 311 K 2 = V2 298 311

  18. Charles’ Law Solution 2 = V2 298 311 298 V2 = (2) 311 V2 = 622 298 V2 = 2.09 liters

  19. Charles’ Law Problem Answer A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume? V2 = 2.09 liters

  20. Boyle’s Law Problem A balloon has a volume of 2.0 liters at 743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume? 1. Convert pressure to the same units 743  760 = .98 atm 2. Insert into formula

  21. Boyle’s Law Solution P1V1 = P2V2 P1 = 0.98 atm P2 = 2.5 atm V1 = 2.0 liters V2 = unknown 0.98 (2.0) = 2.5 V2

  22. Boyle’s Law Solution P1V1 = P2V2 0.98 (2.0) = 2.5 V2 V2= 0.98 (2.0) 2.5 V2 = 0.78 liters

  23. Boyle’s Law Problem Answer A balloon has a volume of 2.0 liters at 743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume? V2 = 0.78 liters

  24. Combined Gas Law Problem A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions? 1. Convert 25 oC to Kelvins 25 + 273 = 298 K 2. Standard pressure is 101.32 kPa 3. Standard temperature is 273 K 4. Insert into formula

  25. Combined Gas Law Solution P1V1 = P2V2 T1 T2 P1 = 98 kPa P2 = 101.32 kPa V1 = 2.0 liters V2 = unknown T1 = 298 K T2 = 273 K

  26. Combined Gas Law Solution P1V1 = P2V2 T1 T2 98 (2.0) = 101.32 V2 298 273 (298) (101.32) V2 = (273) (98) (2.0)

  27. Combined Gas Law Solution P1V1 = P2V2 T1 T2 (298) (101.32) V2 = (273) (98) (2.0) V2= 273 (98) (2.0) (298) (101.32)

  28. Combined Gas Law Solution P1V1 = P2V2 T1 T2 V2= 273 (98) (2.0) (298) (101.32) V2=53508=1.77 liters 30193

  29. Combined Gas Law Problem Answer A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions? V2 = 1.77 liters

  30. Combined Gas Law – V2 P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 P1V1T2= V2 P2T1

  31. The End • This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology • Please send suggestions and comments to rmay@nccvt.k12.de.us

  32. The Ideal Gas Law Chemistry Dr. May

  33. Kinetic Molecular Theory Molecules of an ideal gas • Are dimensionless points • Are in constant, straight-line motion • Have kinetic energy proportional to their absolute temperature • Have elastic collisions • Exert no attractive or repulsive forces on each other

  34. Ideal Gas Law PV = nRT P = pressure in kilopascals (kPa) or atmospheres (atm) V = volume in liters n = moles T = temperature in Kelvins R = universal gas constant

  35. Pressure (P) Volume (V) Moles (n) Temperature (T) The universal gas constant (R) Atm or kPa Always liters Moles Kelvins 0.0821 ( P in atm) or 8.3 (P in kPa) Ideal Gas Law: PV = nRT

  36. Universal Gas Constant R = 0.0821 if P = atmospheres R = 8.3 if P = kilopascals R = PV nT

  37. Deriving R for P in Atmospheres R = PV nT Assume n = 1 mole of gas Standard P = 1atmosphere Standard V = molar volume = 22.4 liters Standard T = 273 Kelvins

  38. R Value When P Is In Atmospheres R = PV nT R = (1) (22.4) (1) 273 R = 0.0821atm Liters mole Kelvins

  39. Deriving R For P In Kilopascals R = PV nT Assume n = 1 mole of gas Standard P = 101.32 kilopascals Standard V = molar volume = 22.4 liters Standard T = 273 Kelvins

  40. R Value When P Is In Kilopascals R = PV nT R = (101.32) (22.4) (1) 273 R = 8.3kPa Liters mole Kelvins

  41. Ideal Gas Law - Pressure PV = nRT P = nRT V Solves for pressure when moles, temperature, and volume are known

  42. Ideal Gas Law - Volume PV = nRT V = nRT P Solves for volume when moles, temperature, and pressure are known

  43. Ideal Gas Law - Temperature PV = nRT T = PV nR Solves for temperature when moles, pressure, and volume are known

  44. Ideal Gas Law - Moles PV = nRT n = PV RT Solves for moles when pressure, temperature, and volume are known

  45. Ideal Gas Law Problem What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ? • V = 2.3 liters • P = 1.2 atmospheres • T = 25 oC = 298 Kelvins • R = 0.0821 since P is in atms. • Find moles (n), then grams

  46. Ideal Gas Law Solution (moles) PV = nRT 1.2 (2.3) = n (0.0821) (298) n = 1.2 ( 2.3) = 0.11 moles (0.0821) (298)

  47. Ideal Gas Law Solution (Grams) Grams = moles x molecular weight (MW) • Moles = 0.11 • Molecular Weight of N2 = 28 g/mole • Grams = 0.11 x 28 = 3.1 grams

  48. Ideal Gas Law Answer What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ? The answer is 0.11 moles and 3.1 grams

  49. The End • This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology • Please send suggestions and comments to rmay@nccvt.k12.de.us

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