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Chapter 10 Gases

Chapter 10 Gases. Characteristics of Gases. Unlike liquids and solids, gases expand to fill their containers; are highly compressible; have extremely low densities. F A. P =. Pressure. Pressure is the amount of force applied to an area.

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Chapter 10 Gases

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  1. Chapter 10Gases

  2. Characteristics of Gases • Unlike liquids and solids, gases • expand to fill their containers; • are highly compressible; • have extremely low densities.

  3. F A P = Pressure • Pressure is the amount of force applied to an area. • Atmospheric pressure is the weight of air per unit of area.

  4. Units of Pressure • Pascals • 1 Pa = 1 N/m2 • Bar • 1 bar = 105 Pa = 100 kPa

  5. Units of Pressure • mm Hg or torr • These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. • Atmosphere • 1.00 atm = 760 torr = 101.3 kPa

  6. Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

  7. Standard Pressure • Normal atmospheric pressure at sea level is referred to as standard pressure. • It is equal to • 1.00 atm • 760 torr (760 mm Hg) • 101.325 kPa

  8. Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

  9. PV = k • Since • V = k (1/P) • This means a plot of V versus 1/P will be a straight line. As P and V areinversely proportional A plot of V versus P results in a curve.

  10. Examples • What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant?

  11. V T = k • i.e., Charles’s Law • The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line.

  12. Examples • At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 L at constant pressure?

  13. V = kn • Mathematically, this means Avogadro’s Law • The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

  14. Examples • A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC? • What volume of gas could the can release at 22ºC and 743 torr?

  15. Combining these, we get nT P V= Ideal-Gas Equation • So far we’ve seen that V=1/P (Boyle’s law) V=T (Charles’s law) V=n (Avogadro’s law)

  16. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

  17. nT P nT P V= V= R Ideal-Gas Equation The relationship then becomes or PV = nRT

  18. Examples • Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr? • A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?

  19. Examples • Mercury can be produced by the following decomposition: • HgO • What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at 400.ºC and 740 torr?

  20. n V P RT = Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get

  21. m V P RT = Densities of Gases • We know that • moles  molecular mass = mass n  = m • So multiplying both sides by the molecular mass ( ) gives

  22. m V P RT d = = Densities of Gases • Mass  volume = density • So, Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.

  23. P RT dRT P d =  = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes

  24. Examples • Complete and balance the following equation: • NaHCO3 + HCl • calculate the mass of sodium hydrogen carbonate necessary to produce 2.87-L of carbon dioxide at 25ºC and 2.00 atm. • If 27 L of gas are produced at 26ºC and 745 torr when 2.6-L of HCl are added what is the concentration of HCl?

  25. Examples • Consider the following reaction • NH3 + O2 NO2 + H2O (balance first!) • What volume of NO2at 1.0 atm and 1000.ºC can be produced from 10.0 L of NH3 and excess O2 at the same temperature and pressure? • What volume of O2 measured at STP will be consumed when 10.0 kg NH3 is reacted?

  26. Dalton’s Law ofPartial Pressures • The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. • In other words, • Ptotal = P1 + P2 + P3 + …

  27. Partial Pressures • When one collects a gas over water, there is water vapor mixed in with the gas. • To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

  28. Examples • The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen? • What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

  29. Examples 4.00 L CH4 1.50 L N2 3.50 L O2 • When these valves are opened, what is each partial pressure and the total pressure? 0.752 atm 2.70 atm 4.58 atm

  30. Example • N2O can be produced by the following reaction NH4NO3(s)  N2O(g) + 2 H2O(l) • What volume of N2O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH4NO3? (the vapor pressure of water at 22ºC is 21 torr)

  31. Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

  32. Main Tenets of Kinetic-Molecular Theory Gases consist of large numbers of molecules that are in continuous, random motion.

  33. Main Tenets of Kinetic-Molecular Theory The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.

  34. Main Tenets of Kinetic-Molecular Theory Attractive and repulsive forces between gas molecules are negligible.

  35. Main Tenets of 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.

  36. Main Tenets of Kinetic-Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature.

  37. Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

  38. Effusion The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.

  39. Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.

  40. (KE)avg = NA(1/2 mu2) • [m is mass of one gas particle in kg] • (KE)avg = 3/2 RT so….. urms = (3RT/Mkg)1/2 • Where Mkgis the molar mass in kilograms, and R has the units 8.314 kg-m2 / s2-mol-K or J/K-mol. • The velocity will be in m/s

  41. KE1 = KE2 1/2 m1v12 = 1/2 m2v22 m1 v22 = m2 v12 v2 v22 m1m2 = v1 v12 = Graham's Law 

  42. Example • Calculate the root mean square velocity of carbon dioxide at 25ºC.

  43. Examples • A compound effuses through a porous cylinder 1.60 times faster than chlorine. What is it’s molar mass?

  44. If 0.00251 mol of NH3 effuses through a hole in 2.47 min, how much HCl would effuse in the same time?

  45. A sample of N2 effuses through a hole in 38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions?

  46. Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

  47. Real Gases Even the same gas will show wildly different behavior under high pressure at different temperatures.

  48. Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.

  49. Corrections for Nonideal Behavior • The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. • The corrected ideal-gas equation is known as the van der Waals equation.

  50. (P + ) (V−nb) = nRT n2a V2 The van der Waals Equation

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