1 / 11

Chapter 5

Chapter 5. Gases. Volume liter milliliter cubic centimeters cubic meters Temperature Kelvin Celsius K = o C + 273.15 O C = K – 273.15. Units. Units. Pressure mm Hg atmosphere psi kilopascals 1 atm = 760 mm Hg = 760 torr = 101.3 kPa = 14.7 psi. PV = nRT P = pressure

curt
Download Presentation

Chapter 5

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. Chapter 5 Gases

  2. Volume liter milliliter cubic centimeters cubic meters Temperature Kelvin Celsius K = oC + 273.15 OC = K – 273.15 Units

  3. Units • Pressure • mm Hg • atmosphere • psi • kilopascals • 1 atm = 760 mm Hg = 760 torr = 101.3 kPa = 14.7 psi

  4. PV = nRT P = pressure V = volume n = number of moles R = constant T = temperature (ALWAYS IN KELVIN) Ideal Gas Law

  5. R values 0.0821 atm*l moles*K 62.4 mm Hg*l moles*K 8.314 __J__ moles*K 8.314 l*kPa moles*K

  6. FormulasMolar Mass and Density MM= mRT PV D = MMP RT

  7. Density • Pressure • Compressing a gas increases its density. • Temperature • Increasing the temperature of a gas decreases its density.

  8. Gas Stoichiometry Law of Combining Volumes The volume ratio of any two gases in a reaction at constant temperature and pressure is the same as the reacting mole ratio.

  9. Combined Gas Law P1V1 = P2V2 T1 T2

  10. 2H2(g) + O2(g) 2H2O(l) What volume of H2(g) at 25oC and 1.00 atm is required to react with 1.00 L of O2(g) at the same temperature and pressure? What mass of H2O(l) is formed in a., assuming a yield of 85.2% a. 1.00 l O2 * 2 l H2 = 2.00 l H2 1 l O2 n = PV nO2 = (1.00atm)(1.00 l) = .0409 moles O2 RT (0.0821 atm*l )(298 K) moles*K .0409 moles O2 * 2 moles H2O = .0918 moles H2O 1 moles O2 .0918 moles H2O * 18.02 g H2O = 1.62 g H2O 1 mole H2O 1.62 g H2O * 85.2% = 1.26 g H2O

  11. Chapter 5 Assignment • Vocabulary p. 125 • Summary Problem p. 126 • Chapter 5 Questions and Problems beginning p. 126 • 4, 6, 12, 14, 15, 18 20, 22, 26, 32, 34, 36, 40

More Related