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H 2. Gases. Gas Bag. Paris 1783. N 2. Chapter 10. Why gases are studied separately. Many common compounds exist as gases Gases transport matter and energy across the globe (i.e., weather) Compared with those of liquids and solids, the behavior of gases is easiest to model.

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  1. H2 Gases Gas Bag Paris 1783 N2 Chapter 10

  2. Why gases are studied separately • Many common compounds exist as gases • Gases transport matter and energy across the globe (i.e., weather) • Compared with those of liquids and solids, the behavior of gases is easiest to model.

  3. Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. (No solubility rules!) • Gases have much lower densities than liquids and solids. • Density of a gas given in g/L (vice g/mL for liquids) NO2

  4. Force Area mass Fig. 10.2 760 mm Hg Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 101,325 Pa = 101.325 kPa 1 atm = 760 mm Hg = 760 torr Height of column ∝1/density

  5. Using a Manometer to Measure Gas Pressure Fig. 10.3 For P ≈ 1 atm

  6. Pressure-Volume Relationship: Boyle’s Law P1V1 = P2V2 • Temperature-Volume Relationship: Charles’s Law V ∝ T Volume-Amount Relationship: Avogadro’s Law V ∝ n The Ideal Gas Law: PV = nRT

  7. Add Hg Add Hg Add Hg V decreases As P (h) increases Fig. 10.6

  8. Boyle’s Law Fig 10.7 Constant temperature Constant amount of gas P·V = constant P1·V1 = P2·V2

  9. (626 mm Hg) (500. mL) P1·V1 = 355 mm Hg P2 A sample of argon gas occupies a volume of 500. mL at a pressure of 626 mm Hg. What is the volume of the gas (in mm Hg) if the pressure is reduced at constant temperature to 355 mm Hg? P1 = 626 mm Hg P2 = 356 mm Hg V1 = 500. mL V2 = ? P1·V1 = P2·V2 V2 = = 882 mL

  10. As T increases V increases

  11. Variation of gas volume with temperature at constant pressure: Charles’ Law Fig 10.8 Temperature must be in Kelvin T (K) = t (°C) + 273.15 V∝T V = constant ·T Constant pressure Constant amount of gas

  12. 6.60 L · 296.0 K V2·T1 = 13.20 L V1 A sample of hydrogen gas occupies 13.20 L at 22.8 °C. At what temperature will the gas occupy half that volume if the pressure remains constant? V1 = 13.20 L V2 = 1.60 L T1 = 296.0 K T2 = ? T1 = 22.8 (°C) + 273.15 (K) = 296.0 K T2 = = 148 K 148 K - 273.15 = -125 °C

  13. Avogadro’s Law Fig 10.10 V∝ number of moles (n) V = constant ·n Constant pressure Constant temperature

  14. Avogadro’s Hypothesis Fig 10.11 Molar Volume

  15. C3H8 + 5O2 3CO2 + 4H2O 1 mole C3H8 3 mole CO2 1 volume C3H8 3 volumes CO2 Propane burns in oxygen to form carbon dioxide and water vapor. How many volumes of carbon dioxide are obtained from one volume of propane at the same temperature and pressure? At constant T and P

  16. PV = nRT The conditions 0 °C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L:

  17. Fig 10.13 Comparison of Molar Volumes at STP • One mole of an ideal gas occupies 22.41 L STP • One mole of various real gases at STP occupy:

  18. (1 mol HCl) V = n = (49.8 g) = 1.37 mol (36.45 g HCl) (1.37 mol)(0.08216 ) (273.15 K) V = 1 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 °C = 273.15 K P = 1 atm PV = nRT V = 30.6 L

  19. PV = nRT useful when P, V, n, and T do not change Modify equation when P, V, and/or T change: • Initial state (1) of gas: Combined Gas Law • Final state (2) of gas: Eqn [10.8]

  20. m V P(MM) = RT dRT P Density (d) Calculations m is the mass of the gas in g d = MM is the molar mass of the gas Molar Mass (MM) of a Gaseous Substance d is the density of the gas in g/L MM = • What happens to the density of a gas if: • it is heated at constant volume? • It is compressed at constant temperature? • Additional gas is added at constant volume?

  21. What is the volume of CO2 produced at 37.0°C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 6 mol CO2 g C6H12O6 mol C6H12O6 mol CO2V CO2 x 1 mol C6H12O6 1 mol C6H12O6 x 180 g C6H12O6 L•atm mol•K nRT (0.1867 mol) (0.0821 ) (310.15 K) = P 1.00 atm Volumes of Gases in Chemical Reations = 0.1867 mol CO2 5.60 g C6H12O6 V = = 4.75 L

  22. Gas Mixtures and Partial Pressures V and T are constant P1 P2 Ptotal= P1 + P2 Dalton’s Law of Partial Pressures

  23. PA = nART nBRT V V PB = nA nB nA + nB nA + nB XB = XA = ni mole fraction (Xi) = nT Consider a case in which two gases, A and B, are in a container of volume V. nA is the number of moles of A nB is the number of moles of B PT = PA + PB PA = XAPT PB = XBPT Pi = Xi PT Dalton’s Law of Partial Pressures

  24. 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2 2 Fig. 10.16 Collecting a water-insoluble gas over water Bottle full of oxygen gas and water vapor

  25. 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2 2 Sample Exercise 10.12 p 413 A 0.811 g sample of KClO3 is partially decomposed to produce oxygen gas over water. The volume of gas collected is 0.250 L at 26 °C and 765 torr total pressure. How many grams of O2 are collected? PT PO = - PH O 2 2 PV = nRT

  26. Pressure of water vapor vs temperature App B p 1111

  27. Kinetic Molecular Theory of Gases • A gas is composed of widely-separated molecules. The molecules can be considered to be points; that is, they possess mass but have negligible volume. • Gas molecules are in constant random motion. • Collisions among molecules are perfectly elastic. • The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. • KE ∝ T

  28. Fig 10.18 The Effect of Temperature on Molecular Speeds urms≡ root-mean-square speed Hot molecules are fast, cold molecules are slow.

  29. The distribution of speeds of three different gases at the same temperature 3RT urms = (MM) Fig 10.19 The Effect of Molecular Mass on Molecular Speeds R = 8.314 J/(mol K) Heavy molecules are slow, light molecules are fast.

  30. An “Ideal Gas” • An ideal gas “obeys” PV = nRT • i.e., calculated value ≈ experimental value Assumptions: • Gas molecules do not exert any force (attractive or repulsive) on each other • i.e., collisions are perfectly elastic • Volume of molecules themselves is negligible compared to volume of container • i.e., the molecules are considered to be points

  31. Deviations from Ideal Behavior • Assumptions made in the kinetic-molecular model: • negligible volume of gas molecules themselves • no attractive forces between gas molecules These breakdown at high pressure and/or low temperature.

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