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Chapter 5

Chapter 5. Gases and the Kinetic-Molecular Theory แก๊สกับทฤษฎีจลน์โมเลกุล. Gases and the Kinetic Molecular Theory. 5.1 An Overview of the Physical States of Matter. 5.2 Gas Pressure and Its Measurement. 5.3 The Gas Laws and Their Experimental Foundations.

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Chapter 5

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  1. Chapter 5 Gases and the Kinetic-Molecular Theory แก๊สกับทฤษฎีจลน์โมเลกุล

  2. Gases and the Kinetic Molecular Theory 5.1 An Overview of the Physical States of Matter 5.2 Gas Pressure and Its Measurement 5.3 The Gas Laws and Their Experimental Foundations 5.4 Further Applications of the Ideal Gas Law 5.5 The Ideal Gas Law and Reaction Stoichiometry 5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior 5.7 Real Gases: Deviations from Ideal Behavior

  3. Name (Formula) Origin and Use Methane (CH4) Ammonia (NH3) Chlorine (Cl2) Oxygen (O2) Ethylene (C2H4) natural deposits; domestic fuel from N2+H2; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Table 5.1 Some Important Industrial Gases Atmosphere-Biosphere Redox Interconnections

  4. An Overview of the Physical States of Matter The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure. 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity. 4. Most gases have relatively low densities under normal conditions. 5. Gases are miscible.

  5. The three states of matter. Figure 5.1

  6. Figure 5.2 Effect of atmospheric pressure on objects at the Earth’s surface.

  7. Figure 5.3 A mercury barometer.

  8. closed-end open-end Figure 5.4 Two types of manometer

  9. pascal(Pa); kilopascal(kPa) 1.01325x105Pa; 101.325 kPa SI unit; physics, chemistry atmosphere(atm) 1 atm* chemistry millimeters of mercury(Hg) 760 mm Hg* chemistry, medicine, biology torr 760 torr* chemistry pounds per square inch (psi or lb/in2) 14.7lb/in2 engineering bar 1.01325 bar meteorology, chemistry, physics Table 5.2 Common Units of Pressure Unit Atmospheric Pressure Scientific Field *This is an exact quantity; in calculations, we use as many significant figures as necessary.

  10. PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Dh = 291.4 mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. 1torr 1 mmHg 1 atm 760 torr 101.325 kPa 1 atm Converting Units of Pressure Sample Problem 5.1 SOLUTION: 291.4 mmHg x = 291.4 torr 291.4 torr x = 0.3834 atm 0.3834 atm x = 38.85 kPa

  11. Figure 5.5 The relationship between the volume and pressure of a gas. Boyle’s Law

  12. Figure 5.6 The relationship between the volume and temperature of a gas. Charles’s Law

  13. 1 T T V a P P P V x P = constant = constant P V T T = constant V a V = constant x = constant PV T Boyle’s Law n and T are fixed V = constant / P V a T P and n are fixed Charles’ s Law V = constant x T Amontons’s Law P a T V and n are fixed P = constant x T combined gas law

  14. Figure 5.7 An experiment to study the relationship between the volume and amount of a gas.

  15. Standard molar volume. Figure 5.8

  16. Figure 5.9 The volume of 1 mol of an ideal gas compared with some familiar objects.

  17. PV 1atm x 22.414L 0.0821atm*L nT 1mol x 273.15K mol*K IDEAL GAS LAW PV = nRT or V = nRT P constant P THE IDEAL GAS LAW Figure 5.10 R = universal gas constant 3 significant figures PV = nRT = R = = fixed P and T fixed n and P fixed n and T Boyle’s Law Avogadro’s Law Charles’s Law V = V = constant x T V = constant x n

  18. PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? unit conversion 1 mL L 1 cm3 103 mL gas law calculation P2V2 P1V1 n2T2 n1T1 P1V1 1.12 atm 0.0105 L V2 = = = 0.0248 L x 2.46 atm P2 Applying the Volume-Pressure Relationship Sample Problem 5.2 PLAN: SOLUTION: P and T are constant V1 in cm3 P1 = 1.12 atm P2 = 2.64 atm 1cm3=1mL V1 = 24.8 cm3 V2 = unknown V1 in mL 103 mL=1L 24.8 cm3 x x = 0.0248 L V1 in L P1V1 = P2V2 = xP1/P2 V2 in L

  19. P1 P2 = = T1 T2 760 torr P2V2 P1V1 0.991 atm x 1 atm n2T2 n1T1 P1T2 373K P2 = = 753 torr x T1 296K Applying the Temperature-Pressure Relationship Sample Problem 5.3 PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and 0.991 atm and placed in boiling water at exactly 1000C. Will the safety valve open? PLAN: SOLUTION: P1(atm) T1 and T2(0C) P1 = 0.991atm P2 = unknown 1atm=760torr K=0C+273.15 T1 = 230C T2 = 1000C P1(torr) T1 and T2(K) x T2/T1 P2(torr) = 753 torr = 949 torr

  20. = = n2 = V1 V2 V2n1 P2V2 P1V1 n1 V1 n2 n2T2 n1T1 55.0 dm3 4.003 g He n2 = 1.10 mol x 26.2 dm3 mol He Applying the Volume-Amount Relationship Sample Problem 5.4 PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams. n1(mol) of He SOLUTION: P and T are constant x V2/V1 n1 = 1.10 mol n2 = unknown n2(mol) of He V1 = 26.2 dm3 V2 = 55.0 dm3 subtract n1 mol to be added x M = 2.31 mol x g to be added = 9.24 g He

  21. 103 g mol O2 0.885 kg x kg 32.00 g O2 atm*L nRT x 294.15K 27.7 mol x 0.0821 mol*K V P = 438 L Solving for an Unknown Gas Variable at Fixed Conditions Sample Problem 5.5 PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 210C. PLAN: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. SOLUTION: V = 438 L T = 210C (convert to K) n = 0.885 kg (convert to mol) P = unknown 210C + 273.15 = 294.15K x = 27.7 mol O2 = 1.53 atm =

  22. (1) A2 + B2 2AB (2) 2AB + B2 2AB2 (3) A + B2 AB2 (4) 2AB2 A2 + 2B2 Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K and the volume does not change. Which of the following balanced equations describes the reaction? PLAN: We know P, T, and V, initial and final, from the pictures. Note that the volume doesn’t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume. SOLUTION: Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).

  23. The Density of a Gas density = m/V n = m/M PV = nRT PV = (m/M)RT Density = m/V = M x P/ RT • The density of a gas is directly proportional to its molar mass. • The density of a gas is inversely proportional to the temperature.

  24. M x P RT d = 44.01 g/mol atm*L x 1atm 0.0821 mol*K 1.96 g mol CO2 6.022x1023 molecules x 273.15K L 44.01 g CO2 mol Sample Problem 5.7 Calculating Gas Density PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (00C and 1 atm) and (b) at room conditions (20.0C and 1.00 atm). PLAN: Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number. d = mass/volume PV = nRT V = nRT/P SOLUTION: = 1.96 g/L d = (a) x x x = 2.68x1022 molecules CO2/L

  25. 44.01 g/mol x 1 atm d = x 293K 6.022x1023 molecules mol atm*L 0.0821 mol*K 1.83g mol CO2 L 44.01g CO2 Sample Problem 5.6 continued Calculating Gas Density (b) = 1.83 g/L x x = 2.50x1022 molecules CO2/L

  26. Figure 5.11 Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)

  27. Volume of flask = 213 mL T = 100.00C P = 754 torr Mass of flask + gas = 78.416 g Mass of flask = 77.834 g atm*L 0.0821 m RT 0.582 g 373K x mol*K = 84.4 g/mol VP 0.213 L x 0.992 atm Finding the Molar Mass of a Volatile Liquid Sample Problem 5.8 PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data: Calculate the molar mass of the liquid. PLAN: Use unit conversions, mass of gas, and density-M relationship. SOLUTION: m = (78.416 - 77.834) g = 0.582 g x M = =

  28. n1 n1 = ntotal n1 + n2 + n3 +... c1 = Mixtures of Gases • Gases mix homogeneously in any proportions. • Each gas in a mixture behaves as if it were the only gas present. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + ... P1 = c1 x Ptotal where c1 is the mole fraction

  29. PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2, and 4.0 mol% 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture. Find the c and P from Ptotal and mol% 18O2. 18O2 18O2 4.0 mol% 18O2 = c 18O2 100 P = c x Ptotal = 0.040 x 0.75 atm 18O2 18O2 partial pressure P 18O2 Applying Dalton’s Law of Partial Pressures Sample Problem 5.9 PLAN: mol% 18O2 SOLUTION: = 0.040 divide by 100 c = 0.030 atm 18O2 multiply by Ptotal

  30. mass PV = M RT m RT m d = VP V d RT M = P The Molar Mass of a Gas n = M =

  31. Table 5.3 Vapor Pressure of Water (P ) at Different T H2O T(0C) P (torr) T(0C) P (torr) 0 5 10 11 12 13 14 15 16 18 20 22 24 26 28 4.6 6.5 9.2 9.8 10.5 11.2 12.0 12.8 13.6 15.5 17.5 19.8 22.4 25.2 28.3 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 31.8 42.2 55.3 71.9 92.5 118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760.0

  32. Figure 5.12 Collecting a water-insoluble gaseous reaction product and determining its pressure.

  33. PROBLEM: Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reaction with water: CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq) For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is 738 torr and the volume is 523 mL. At the temperature of the gas (230C), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? P g P C2H2 C2H2 C2H2 PV atm P n = RT 760torr n C2H2 Calculating the Amount of Gas Collected Over Water Sample Problem 5.10 PLAN: The difference in pressures will give us the P for the C2H2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. SOLUTION: = (738-21)torr = 717torr Ptotal = 0.943 atm 717 torr x H2O x M

  34. n = C2H2 PV = RT 26.04 g C2H2 0.0203 mol x mol C2H2 atm*L 0.0821 mol*K Calculating the Amount of Gas Collected Over Water Sample Problem 5.9 continued 0.943 atm x 0.523L = 0.0203 mol n 296K x = 0.529 g C2H2

  35. Figure 15.13 Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or product and the gas variables pressure (P), volume (V), and temperature (T). P,V,T of gas A amount (mol) of gas A amount (mol) of gas B P,V,T of gas B ideal gas law ideal gas law molar ratio from balanced equation

  36. PROBLEM: Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H2 reduces the metal oxide, forming the pure metal and H2O. On a laboratory scale, what volume of H2 at 765 torrand 2250C is needed to reduce 35.5 g of copper(II) oxide? PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas. CuO(s) + H2(g) Cu(s) + H2O(g) mol Cu 1 mol H2 63.55 g Cu 1 mol Cu atm*L 0.0821 mol*K x 498K 1.01 atm Sample Problem 5.11 Using Gas Variables to Find Amount of Reactants and Products SOLUTION: mass (g) of Cu divide by M 35.5 g Cu x x = 0.559 mol H2 mol of Cu molar ratio mol of H2 0.559 mol H2 x = 22.6 L use known P and T to find V L of H2

  37. PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25 Lof chlorine gas at 0.950 atmand 293Kreacts with 17.0 g of potassium? PLAN: After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product. P = 0.950 atm V = 5.25 L 2K(s) + Cl2(g) 2KCl(s) T = 293K n = unknown 0.950 atm 0.207 mol = = PV x 293K RT 2 mol KCl atm*L 0.207 mol Cl2 x 17.0 g x = 0.435 mol K 0.0821 1 mol Cl2 mol*K mol K 39.10 g K 74.55 g KCl mol KCl Using the Ideal Gas Law in a Limiting-Reactant Problem Sample Problem 5.12 SOLUTION: x 5.25 L n = Cl2 Cl2 is the limiting reactant. = 0.414 mol KCl formed 0.414 mol KCl x = 30.9 g KCl

  38. Postulate 1: Particle Volume Postulate 2: Particle Motion Postulate 3: Particle Collisions Postulates of the Kinetic-Molecular Theory Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume. Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. Collisions are elastic therefore the total kinetic energy(Ek) of the particles is constant.

  39. Figure 5.14 Distribution of molecular speeds at three temperatures.

  40. Figure 5.15 A molecular description of Boyle’s Law.

  41. Figure 5.16 A molecular description of Dalton’s law of partial pressures.

  42. Figure 5.17 A molecular description of Charles’s Law.

  43. Ek = 1/2 mass x u2 u 2 is the root-mean-square speed urms = √ 3RT M 1 rate of effusion a √M Avogadro’s Law V a n Ek = 1/2 mass x speed2 R = 8.314 Joule/mol*K Graham’s Law of Effusion The rate of effusion of a gas is inversely related to the square root of its molar mass.

  44. A molecular description of Avogadro’s Law. Figure 5.18

  45. Figure 5.19 Relationship between molar mass and molecular speed. Ek = 3/2 (R/NA) T

  46. PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4). PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses. rate √ He 16.04 = rate 4.003 CH4 Applying Graham’s Law of Effusion Sample Problem 5.13 SOLUTION: M of CH4 = 16.04g/mol M of He = 4.003g/mol = 2.002

  47. Figure 5.20 Diffusion of a gas particle through a space filled with other particles. distribution of molecular speeds mean free path collision frequency

  48. Variations in pressure, temperature, and composition of the Earth’s atmosphere. Figure B5.1

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