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Gases: Their Properties & Behavior

9. Gases: Their Properties & Behavior. Gas Pressure 01. Gas Pressure 02. Units of pressure: atmosphere (atm) Pa (N/m 2 , 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar (1.01325 bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in 2 (14.696 lb/in 2 = 1 atm)

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Gases: Their Properties & Behavior

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  1. 9 Gases: Their Properties & Behavior Chapter 09

  2. Gas Pressure 01 Chapter 09

  3. Gas Pressure 02 • Units of pressure: atmosphere (atm) Pa (N/m2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar (1.01325 bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in2 (14.696 lb/in2 = 1 atm) in Hg (29.921 in Hg = 1 atm) Chapter 09

  4. Pressure–Volume Law (Boyle’s Law): Boyle’s Law 01 Chapter 09

  5. Pressure–Volume Law (Boyle’s Law): Boyle’s Law 01 Chapter 09

  6. Boyle’s Law 02 • Pressure–Volume Law (Boyle’s Law): • The volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure. P1 x V1 = K1 P2 x V2 = K1 P1 x V1 = P2 x V2 Chapter 09

  7. 726 mmHg x 946 mL P1 x V1 = 154 mL V2 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg 5.3 Chapter 09

  8. As T increases V increases Chapter 09

  9. T(K)=Temperature in Kelvin T (K) = t (0C) + 273.15 Charles’ Law 01 • Temperature–Volume Law (Charles’ Law): Chapter 09

  10. V T µ V 1 = k T 1 1 Charles’ Law 01 • Temperature–Volume Law (Charles’ Law): • The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin temperature of the gas. Chapter 09

  11. Temperature must be in Kelvin Variation of gas volume with temperature at constant pressure. P1<P2<P3<P4 Third Law of Thermodynamics states that it's impossible to reach absolute zero. VaT V = constant x T T (K) = t (0C) + 273.15 V1/T1 = V2/T2 Chapter 09

  12. 1.54 L x 398.15 K V2 x T1 = 3.20 L V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T2 = = 192 K Chapter 09

  13. Avogadro’s Law 01 • The Volume–Amount Law (Avogadro’s Law): Chapter 09

  14. Avogadro’s Law 01 • The Volume–Amount Law (Avogadro’s Law): • At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. Chapter 09

  15. 4NH3 + 5O2 4NO + 6H2O 1 volume NH3 1 volume NO 1 mole NH3 1 mole NO Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? At constant T and P Chapter 09

  16. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT Chapter 09

  17. The conditions 0 0C 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. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) PV = nRT R = 0.082057 L • atm / (mol • K) Chapter 09

  18. The Ideal Gas Law 01 • Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. • R =The gas constant • R = 0.08206 L·atm·K–1·mol–1 Chapter 09

  19. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 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 0C = 273.15 K P = 1 atm PV = nRT V = 30.6 L Chapter 09

  20. P1 = 1.20 atm P2 = ? T1 = 291 K T2 = 358 K nR P = = T V P2 P1 T2 358 K T1 T2 T1 291 K = 1.20 atm x P2 = P1 x Argon is an inert gas used in lightbulbs to retard the oxidation and vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? n, V and Rare constant PV = nRT = constant = 1.48 atm Chapter 09

  21. Calculation of Molar Mass USING GAS DENSITY The density of air at 15 oC and 1.00 atm is 1.23 g/L. What is the molar mass of air? Molar Mass = Mass / Number of moles 1. Calc. moles of air. V = 1.00 L, m = 1.23 g, P = 1.00 atm, T = (273.15+15 )= 288 K n = PV/RT = 0.0423 mol, m = 1.23 g • Calc. molar mass mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol Chapter 09

  22. Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal= P1 + P2 Chapter 09

  23. PA = nART nBRT V V PB = 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 Chapter 09

  24. PT = PA = + nBRT nBRT nART nART RT V V V V V PB = PA / PT (nA + nB) PA / PT = XA nB nA nA nA + nB nA + nB nA + nB XA = XB = = = PA = XAPT PB = XBPT Pi = XiPT PT = PA + PB Chapter 09

  25. 0.116 8.24 + 0.421 + 0.116 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = XiPT PT = 1.37 atm = 0.0132 Xpropane = Ppropane = 0.0132 x 1.37 atm = 0.0181 atm Chapter 09

  26. Gas Stoichiometry 01 • In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles. • Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10): 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l) Chapter 09

  27. What is the volume of CO2 produced at 370 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 0.187 mol x 0.0821 x 310.15 K = P 1.00 atm Gas Stoichiometry 5.60 g C6H12O6 = 0.187 mol CO2 V = = 4.76 L Chapter 09

  28. Gases: P, V, T & n Chapter 09

  29. Kinetic Molecular Theory • This theory presents physical properties of gases in terms of the motion of individual molecules. • Average Kinetic Energy  Kelvin Temperature • Gas molecules are points separated by a great distance • Particle volume is negligible compared to gas volume • Gas molecules are in rapid random motion • Gas collisions are perfectly elastic ( No Change of Total Kinetic Energy) • Gas molecules experience no attraction or repulsion Chapter 09

  30. Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. Chapter 09

  31. Kinetic Molecular Theory • Average Kinetic Energy (KE) is given by: U: The Root–Mean–Square Speed Chapter 09

  32. Kinetic Molecular Theory 03 • Average Kinetic Energy (KE) is given by: (1000 miles/hr) U: The Root–Mean–Square Speed rms speed at 25 °C Chapter 09

  33. Kinetic Molecular Theory 05 • Maxwell speed distribution curves. Chapter 09

  34. Kinetic Molecular Theory 04 • The Root–Mean–Square Speed: is a measure of the average molecular speed. • R=8.314 J/K.mol 1J = 1Kg.m2/s2 • Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in miles/hr at 25°C. Taking square root of both sides gives the equation He: miles/hr Chapter 09

  35. Diffusion of Gases Diffusion: The mixing of different gases by molecular motion with frequent molecular collisions.

  36. Graham’s Law 02 • Effusion is when gas molecules escape , through a tiny hole into a vacuum. Chapter 09

  37. Graham’s Law 03 • Graham’s Law: Rate of effusion is proportional to its rms speed, urms. • For two gases at same temperature and pressure: Chapter 09

  38. A Problem to Consider • How much faster would H2 gas effuse through an opening than methane, CH4? So hydrogen effuses 2.8 times faster than CH4 Chapter 09

  39. Behavior of Real Gases 01 • Deviations result from assumptions about ideal gases. • Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. • Volume of the molecules is negligibly small compared with that of the container. Chapter 09

  40. Behavior of Real Gases 02 • At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures. As a result, the pressure of real gases will be smaller than the ideal value Chapter 09

  41. Behavior of Real Gases 03 • The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value. Chapter 09

  42. Behavior of Real Gases 05 • Corrections for non-ideality require van der Waals equation. Excluded Volume IntermolecularAttractions Chapter 09

  43. A Problem to Consider • If sulfur dioxide were an “ideal” gas, the pressure at 0 oC exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure. Use the following values for SO2 a = 6.865 L2.atm/mol2 b = 0.05679 L/mol Chapter 09

  44. R= 0.0821 L. atm/mol. K T = 273.2 K V = 22.41 L a = 6.865 L2.atm/mol2 b = 0.05679 L/mol What is the Real pressure of one mole of a gas, with a Volume of 22.4 L at STP? • First, let’s rearrange the van der Waals equation to solve for pressure. Chapter 09

  45. A Problem to Consider • The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure. Chapter 09

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