1 / 50

The Chemistry of Gases

The Chemistry of Gases. A Gas. Uniformly fills any container. Mixes completely with any other gas Exerts pressure on its surroundings. Pressure. is equal to force/unit area SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa 1 standard atmosphere = 1 atm =

Download Presentation

The Chemistry of Gases

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. The Chemistry of Gases

  2. A Gas • Uniformly fills any container. • Mixes completely with any other gas • Exerts pressure on its surroundings.

  3. Pressure • is equal to force/unit area • SI units = Newton/meter2 = 1 Pascal (Pa) • 1 standard atmosphere = 101,325 Pa • 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

  4. Effect of Atmospheric Pressure

  5. Standard Pressure Units • 1 atmosphere = • 760 mm Hg = 760 torr • 1.013 x 105 Pa = 101.3 kPa • 14.7 psi (pounds per square inch)

  6. Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this as atmospheres and as Pascals. 49 torr| 1 atm = 6.4 x 10-2 atm | 760 torr 49 torr | 1.013 x 105 Pa = 6.5 x 103 Pa | 760 torr

  7. Boyle’s Law* • Pressure  Volume = Constant (T = constant) • P1V1 = P2V2 (T = constant) • V 1/P (T = constant) • (*Holds precisely only at very low pressures.)

  8. A gas that strictly obeys Boyle’s Law is called an ideal gas.

  9. 1.53 L of sulfur dioxide gas at 5600 Pa is changed to 15000 Pa at constant temperature. What is its new volume? P1V1 =P2V2 5600 x 1.53 = 15000 x V2 V2 = 0.57 L 1.0 L of hydrogen gas at 345 torr is connected to an empty 2.5 L flask and allowed to combine. What is its new pressure? 345 x 1.0 = P2 x 3.5 P2 = 98.6 torr

  10. Charles’s Law • The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. • V = bT (P = constant) b = a proportionality constant

  11. Balloon with Liquid N2

  12. Charles’s Law

  13. Always Change to KELVIN! K = C + 273

  14. A sample of gas at 15o C and 1 atm has a volume of 2.58 L. What is its new volume at 38o C and 1 atm pressure? Pressure remains constant at 1 atm T1 = 15 + 273 = 288 K T2 = 38 + 273 = 311 K V1 = V2 or V1T2 = V2T1 T1 T2 2.58 x 311 = V2 x 288 V2 = 2.79 L

  15. Gay-Lussac’s Law • Pressure is directly proportional to the Kelvin Temperature • P = k T • or P1 = P2 • T1 T2

  16. Standard Temperature and Pressure • “STP” • P = 1 atmosphere • T = C • The molar volume of an ideal gas is 22.42 liters at STP

  17. Combined Gas Law • When all three change use: • P1 V1 = P2 V2 • T1 T2 • orP1V1 T2 = P2V2 T1

  18. A cylinder contains 255 mL of CO2 at 27o C and 812 torr. What volume would this occupy at STP? P1= 812 torr T1= 27o C + 273= 300 K V1= 255 mL P2= 1 atm= 760 torr T2 = 0o C = 273 K P1V1T2=P2V2T1 (812)(255)(273)=V2(760)(300) V2= 247.9 mL

  19. Avogadro’s Law • For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). • V = an • a = proportionality constant • V = volume of the gas • n = number of moles of gas

  20. Ideal Gas Law • An equation of state for a gas. • “state” is the condition of the gas at a given time. • PV = nRT

  21. Ideal Gas Law • PV = nRT • R = proportionality constant = 0.08206 L atm  mol • P = pressure in atm • V = volume in liters • n = moles • T = temperature in Kelvins • Holds closely at P < 1 atm

  22. Ideal Gas Laws • Can be used to calculate molar masses by gas density • density= mass = molar mass • volume molar volume • At STP, molar volume is 22.4 L

  23. Suppose you have a sample of ammonia gas with a volume of 856 mL at a pressure of 932 torr and 28o C. What is the mass of the ammonia present? P= 932 torr / 760 torr = 1.23 atm V = 0.856 L T = 28 + 273= 301 K n = PV = (1.23)(.856) = 0.0426 moles RT (.0821)(301) mass = moles x MM= 0.0426 x 17= 0.724g

  24. Dalton’s Law of Partial Pressures • For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . Used primarily when collecting gases over water.

  25. Mixtures of helium and oxygen are used for deep scuba diving. If 46 L oxygen and 12 L He at 1.0 atm are put into a 5.0 L tank at 25oC, calculate the partial pressures of each gas and the total pressure in the tank. n = PV = (1)(46) and (1)(12) RT (.0821)(298) (.0821)(298) nO2= 1.9 mol nHe = 0.49 mol

  26. In the tank, V= 5.0 L PO2 =nRT / V = (1.9)(.0821)(298) / 5.0 = 9.3 atm Phe = (0.49)(.0821)(298)/ 5.0 = 2.4 atm Ptotal = 9.3 + 2.4 = 11.7 atm

  27. Homework • #1 p223ff 8, 12, 15, 17, 18, 30

  28. Stoichiometry and Gases A sample of solid potassium chlorate decomposes by the equation 2 KClO3 (s) -----> 2 KCl (s) + 3 O2 (g) If 650 mL of oxygen at 22oC and 754 torr is collected over water, how much KClO3 decomposed? PH2O = 21 torr (from table of vapor pressure) PO2 = 754 - 21 = 733 torr / 760 = 0.964 atm nO2 = PV /RT = (.964)(.650) / (.0821)(295) = 0.0259 0.0259 mol O2 | 2 mol KClO3 | 122.6 g KClO3 = 2.11 g | 3 mol O2 | 1 mol KClO3 KClO3

  29. Homework • #2 p226ff 32, 35, 38, 39, 41, 42, 44

  30. Kinetic Molecular Theory • 1. Volume of individual particles is  zero. • 2. Collisions of particles with container walls cause pressure exerted by gas. • 3. Particles exert no forces on each other. • 4. Average kinetic energy  Kelvin temperature of a gas.

  31. The Meaning of Temperature • Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

  32. Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. • Effusion: describes the passage of gas into an evacuated chamber.

  33. Diffusion Rates

  34. Effusion: Diffusion:

  35. Find the ratio of the effusion rates of UF6 versus H2. Rate H2 = MMUF6 = 352 =13.2 Rate UF6 MMH2 2

  36. Homework • #3 p228ff 53, 54, 68, 70, 72, 76 • #4 p229ff 77, 89, 91, 92 • Super XCR #97

  37. Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

  38. Real Gases   corrected pressure corrected volume Pideal Videal

More Related