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Chemistry-140 Lecture 30

Chemistry-140 Lecture 30. Chapter Highlights pressure measurements concept of STP gas laws (Charle’s, Boyle’s, Avogadro’s) ideal gas law applications of ideal gas law partial pressures kinetic theory of gases diffusion & effusion (Graham’s law) van der Waal’s equation.

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Chemistry-140 Lecture 30

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  1. Chemistry-140 Lecture 30 • Chapter Highlights • pressure measurements • concept of STP • gas laws (Charle’s, Boyle’s, Avogadro’s) • ideal gas law • applications of ideal gas law • partial pressures • kinetic theory of gases • diffusion & effusion (Graham’s law) • van der Waal’s equation Chapter 12: Gases

  2. Chemistry-140 Lecture 30 • Gas: a substance that expands to fill its container and attains the container's shape • is highly compressible • usually nonmetallic • simple molecular formula • low molar mass Definition of a Gas

  3. Chemistry-140 Lecture 30 • Only substances that are gaseous under normal conditions of temperature and pressure are called gases • A substance that is normally a liquid or solid is called a vapour in the gas state. example water vapour • Gases form homogeneous mixtures regardless of the amounts and characteristics of the components. Characteristics of a Gas

  4. Chemistry-140 Lecture 30 • Pressure: the force a gas exerts on the walls of its vessel per unit area: P = F/Area . • Newtons are the SI units of force. (1 N = 1 kg-m/s2) • Pascals are the SI units of pressure. (1 Pa = 1 N/m2) • Atmospheric pressure: the gas pressure most commonly measured. The mass of a column of atmosphere 1 m2 in cross-sectional area and extending to the top of the atmosphere exerts a force of 1.01 x 105 N Pressure

  5. Chemistry-140 Lecture 30 Barometric Pressure • Atmospheric pressure is measured with a barometer and is called barometric pressure • The standard pressure, (1 atmosphere, atm), is the pressure required to support a mercury column to a height of 760 mm = 760 Torr • = 1.01325 x 105 Pa • = 101.325 kPa

  6. Chemistry-140 Lecture 30 • Four variables can adequately describe a gas sample: • T = Temperature (generally expressed in Kelvin) • n = amount of material (generally in moles) • P = pressure (atmospheres is most common) • V = volume (litres is most common) Gas Laws

  7. V2 = Chemistry-140 Lecture 30 • The volume of a fixed amount of gas (n) at constant temperature (T) is inversely proportional to the pressure of the gas; PV = constant (at constant T and n). • Comparing the gas at different pressures: Boyle’s Law: The Pressure-Volume Relationship P1V1 = P2V2 = P3V3….

  8. = Chemistry-140 Lecture 30 • The volume of a fixed amount of gas (n) at constant pressure (P) is directly proportional to the temperature of the gas; = constant • Comparing the gas at different pressures: Charle’s Law: The Temperature-Volume Relationship V2 =

  9. V2 = = Chemistry-140 Lecture 30 • The volume of a gas at constant pressure and temperature is directly proportional to the amount of gas present, expressed in moles; = constant • Comparing two gas samples; Avogadro’s Law: The Quantity-Volume Relationship

  10. Chemistry-140 Lecture 30 • Boyle's law: VR (@ constant n, T) • Charles's law: VRT (@ constant n, P) • Avogadro's law VRn (@ constant P, T) • This can combine to give a more general law: VR The Ideal Gas Equation or PV = nRT where R = the gas constant

  11. Chemistry-140 Lecture 30 • An ideal gas: one that can be described by the ideal-gas equation; PV = nRT • The usual value of the ideal gas constant is • R = 0.08206 L-atm/K-mol • Temperature must be expressed in Kelvin and the units of volume and pressure must match the units of R • (also note that R can have the value = 8.3145 J/K-mol) An Ideal Gas

  12. Chemistry-140 Lecture 30 STP: Standard Temperature & Pressure • STP: The conditions 0.00 °C (273.15 K) and 1 atm are referred to as standard temperature and pressure, STP. At STP, the volume of 1 mol of an ideal gas is 22.41 L. this known as the molar volume of a gas at STP

  13. Chemistry-140 Lecture 30 Question (similar to example 12.6): Calcium carbonate, CaCO3(s), decomposes to CaO(s) and CO2(g). A sample of CaCO3(s) is decomposed and the CO2(g) collected in a 250 mL flask. After decomposition, the gas has a pressure of 1.3 atm at a temperature of 31oC. How many moles of CO2(g) were generated? Applications of the Ideal Gas Equation

  14. Chemistry-140 Lecture 30 Answer: Step 1: Identify the unknown quantity and known quantities. Use units consistent with R. n = ? R = 0.0821 L-atm/mol-K P = 1.3 atm V = 250 mL = 0.250 L T = 31 oC = (31 + 273) K = 304 K Applications of the Ideal Gas Equation

  15. Chemistry-140 Lecture 30 • Answer: • Step 2: Rearrange the ideal gas equation and • solve for n. • n = • = • = Applications of the Ideal Gas Equation 0.013 mol CO2(g)

  16. Chemistry-140 Lecture 30 Question (similar to example 12.4): A sample of argon gas is confined to a 1.00 L tank at 27.0oC. The pressure in the tank is 4.15 atm. The gas is allowed to expand into a larger vessel. Upon expansion, the temperature drops to 15.0oC and the pressure drops to 655 Torr. What is the final volume of the gas? Applications of the Ideal Gas Equation

  17. Chemistry-140 Lecture 30 Answer: Step 1: Identify the unknown quantity and tabulate the known quantities in units consistent with those in R. Notice that we are missing both V and n in the final state!! Applications of the Ideal Gas Equation

  18. Chemistry-140 Lecture 30 Answer: Step 2: Since the number of moles of gas does not change we can calculate the number of moles initially present and know how many were present in the final state n = = = Applications of the Ideal Gas Equation 0.168 mol Ar(g)

  19. Chemistry-140 Lecture 30 Answer: Step 3: We can then use this to calculate the final volume. V2 = = = Applications of the Ideal Gas Equation 4.62 L Ar(g)

  20. Chemistry-140 Lecture 30 Question: The gas pressure in an aerosol can is 1.5 atm at 25oC. Assuming that the gas obeys the ideal-gas equation what would the pressure be if the can was heated to 450oC? Applications of the Ideal Gas Equation

  21. Chemistry-140 Lecture 30 Answer: Step 1: Identify the unknown quantity and tabulate the known quantities in units consistent with those in R. Applications of the Ideal Gas Equation

  22. = and = then = Chemistry-140 Lecture 30 Answer: Step 2: Since the can is a closed container, the volume and # of moles cannot change; V1 = V2 and n1 = n2. Therefore: Applications of the Ideal Gas Equation

  23. Chemistry-140 Lecture 30 Answer: Step 3: Rearrange and calculate P2 P2 = = = = Applications of the Ideal Gas Equation 3.6 atm

  24. Chemistry-140 Lecture 31 • Chapter Highlights • pressure measurements • concept of STP • gas laws (Charle’s, Boyle’s, Avogadro’s) • ideal gas law • applications of ideal gas law • partial pressures • kinetic theory of gases • diffusion & effusion (Graham’s law) • van der Waal’s equation Chapter 12: Gases

  25. Chemistry-140 Lecture 31 • Recall that: • d (density) = m(mass)/V(volume) • n(moles) = m(mass)/M(molar mass) • So if we write the ideal-gas equation as = • substitute for n: = or = • thus: d = and M = Molar Mass & Gas Density

  26. Chemistry-140 Lecture 31 Question (similar to excercise 12.7): What is the density of carbon tetrachloride, CCl4, vapour at 714 Torr and 125oC? Molar Mass & Gas Density

  27. Chemistry-140 Lecture 31 Answer: Step 1: Recall that d = We need the molar mass of CCl4 M(CCl4) = (12.0 + 4(35.5)) = 154 g/mol d = = Molar Mass & Gas Density 4.43 g/L

  28. Chemistry-140 Lecture 31 Question (no corresponding example): A flask is evacuated and found to weigh 134.567 g. It is filled to a pressure of 735 Torr at 31OC with a gas of unknown molar mass and then reweighed; 137.456 g. The flask is then filled with water and weighed again; 1067.9 g. What is the molar mass of the unknown gas? (density of water at 31oC is 0.997 g/cm3) Molar Mass & Gas Density

  29. Chemistry-140 Lecture 31 Answer: Step 1: We need to know the volume of the flask. We are given the mass of water when the flask is filled, so we can use the density of water to calculate the volume of the flask. V = = 936 cm3 = Molar Mass & Gas Density 0.936 L

  30. Chemistry-140 Lecture 31 Answer: Step 2: Since we now know the volume of the flask and the mass of gas is easily calculated, we can obtain the gas density and use this to get the molar mass. m(gas) = (137.456 g - 134.567 g) = d = = M = = Molar Mass & Gas Density 2.889 g 3.09 g/L 79.7 g/mol

  31. Partial pressure: The pressure of each component gas in a mixture. The total pressure is the sum of the partial pressure. • Pt = P1 + P2 + P3 ... • Each partial pressure is then given by Pi = • and Pt = (n1 + n2 + n3 ...) Chemistry-140 Lecture 31 Gas Mixtures & Partial Pressure

  32. The partial pressures of the gases present in a mixture are given by the mole fraction, Xi of the gases in the mixture; = = = Xi Chemistry-140 Lecture 31 Gas Mixtures & Partial Pressure Pi = XiPt

  33. Chemistry-140 Lecture 31 Question (similar to example 12.1): A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 is placed in a 15.0 L vessel at 0 oC. What is the partial pressure of each gas, and what is the total pressure in the vessel? Partial Pressures

  34. Chemistry-140 Lecture 31 Answer: Step 1: Since each gas behaves independently, we can calculate the pressure that each would exert if the other were not present. Convert masses to moles n(O2) = (6.00 g O2) = n(CH4) = (9.00 g CH4) = Partial Pressures 0.188 mol O2 0.563 mol CH4

  35. Answer: Step 2: Use the ideal-gas equation to calculate the partial pressure of each gas. P(O2) = = = P(CH4) = = = Chemistry-140 Lecture 31 Partial Pressures 0.281 atm 0.841 atm

  36. Chemistry-140 Lecture 31 Answer: Step 3: We can now calculate the total pressure from Pt = P1 + P2 + P3 ... Pt = (0.281 atm) + (0.841 atm) Pt = Partial Pressures 1.12 atm

  37. Chemistry-140 Lecture 31 Partial Pressures • Gases are often generated in the laboratory. Either as a product of a reaction or as a gaseous reactant to be used in a chemical reaction.

  38. Chemistry-140 Lecture 31 Question: The industrial synthesis of nitric acid involves the reaction of nitrogen dioxide gas with water. 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) How many moles of HNO3 can be prepared using 450 L of NO2(g) at a pressure of 5.00 atm and a temperature of 295 K? Partial Pressures

  39. Chemistry-140 Lecture 31 Answer: Step 1: Use the ideal-gas equation to determine the moles of NO2. n = = = Partial Pressures 92.9 mol

  40. Answer: Step 2: Use the stoichiometry of the equation to calculate the moles of nitric acid produced. (92.9 mol NO2) = 61.9 mol HNO3 Chemistry-140 Lecture 31 Partial Pressures

  41. Chemistry-140 Lecture 31 Trapped Gases • A common way to trap and measure the gas formed is a technique called displacement. The gas that is collected is saturated with water vapour. Total pressure inside the jar is then: Ptotal = Pgas + Pwater

  42. Chemistry-140 Lecture 31 Question (similar to example 12.12): A sample of KClO3 is partially decomposed producing O2 that is collected over water. The volume collected is 0.250 L at 26 oC and 765 Torr total pressure. 2 KClO3(s) 2 KCl(s) + 3 O2(g) Knowing the vapour pressure of water is 25 Torr at 26 oC, calculate how many grams of KClO3 decomposed. Trapped Gases

  43. Chemistry-140 Lecture 31 Answer: Step 1: We know V and T but not P(O2). This can be determined from: P(O2) = (765 Torr) - (25 Torr) = n(O2) = = = Trapped Gases 740 Torr 9.92 x 10-3 mol O2

  44. Answer: Step 2: We can use the stoichiometry of the equation to calculate the moles of KClO3 and then convert to grams. (9.92 x 10-3 mol O2) = 0.811 g KClO3 Chemistry-140 Lecture 31 Trapped Gases

  45. Chemistry-140 Lecture 32 • Chapter Highlights • pressure measurements • concept of STP • gas laws (Charle’s, Boyle’s, Avogadro’s) • ideal gas law • applications of ideal gas law • partial pressures • kinetic theory of gases • diffusion & effusion (Graham’s law) • van der Waals equation Chapter 12: Gases

  46. Chemistry-140 Lecture 32 • The ideal-gas equation explains how gases behave. • Kinetic-molecular theory: • Explains why ideal gases behave as they do. Kinetic Molecular Theory

  47. Chemistry-140 Lecture 32 • Gases consist of large numbers of molecules in continuous, random motion. • The volume of all the molecules is negligible compared to the total volume in which the gas is contained. • Attractive and repulsive interactions among gas molecules are negligible. Kinetic Molecular Theory

  48. Chemistry-140 Lecture 32 • The collisions are elastic. Energy is transferred between molecules during collisions. • The average kinetic energy is proportional to the absolute temperature. Kinetic Molecular Theory

  49. KE = 1/2mu2 KE = the average kinetic energy u = root mean squared speed m = mass of the molecule Chemistry-140 Lecture 32 • Although kinetic energy at a given temperature is the same for all gases, the molecular speeds are different. At a constant kinetic energy, as the molecular mass increases, the molecular speed decreases. Molecular Speeds

  50. Chemistry-140 Lecture 32 • Pressure is caused by gas molecules bombarding the container walls. The total force of these collisions depends on the number of collisions & the average force per collision. Kinetic Molecular Theory Explains...

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