1. 5-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 5 Gases and the Kinetic-Molecular Theory แก๊สกับทฤษฎีจลน์โมเลกุล

2. 5-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. 5-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 5.1 Some Important Industrial Gases Methane (CH 4 ) Ammonia (NH 3 ) Chlorine (Cl 2 ) Oxygen (O 2 ) Ethylene (C 2 H 4 ) natural deposits; domestic fuel from N 2 +H 2 ; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Name (Formula)Origin and Use Atmosphere-Biosphere Redox Interconnections

4. 5-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. 5-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.1 The three states of matter.

6. 5-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.2 Effect of atmospheric pressure on objects at the Earth’s surface.

7. 5-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.3 A mercury barometer.

8. 5-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.4 Two types of manometer closed-end open-end

9. 5-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 5.2 Common Units of Pressure Atmospheric PressureUnitScientific Field chemistryatmosphere(atm)1 atm* pascal(Pa); kilopascal(kPa) 1.01325x10 5 Pa; 101.325 kPa SI unit; physics, chemistry millimeters of mercury(Hg) 760 mm Hg*chemistry, medicine, biology torr760 torr*chemistry pounds per square inch (psi or lb/in 2 ) 14.7lb/in 2 engineering bar1.01325 barmeteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary.

10. 5-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.1 Converting Units of Pressure PROBLEM:A geochemist heats a limestone (CaCO 3 ) sample and collects the CO 2 released in an evacuated flask attached to a closed- end manometer. After the system comes to room temperature,  h = 291.4 mm Hg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. SOLUTION: PLAN: Construct conversion factors to find the other units of pressure. 291.4 mmHg x 1torr 1 mmHg = 291.4 torr 291.4 torr x 1 atm 760 torr = 0.3834 atm 0.3834 atm x 101.325 kPa 1 atm = 38.85 kPa

11. 5-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.5 The relationship between the volume and pressure of a gas. Boyle’s Law

12. 5-12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.6 The relationship between the volume and temperature of a gas. Charles’s Law

13. 5-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Boyle’s Lawn and T are fixed V  1 P Charles’ s Law V  T P and n are fixed V T = constant V = constant x T Amontons’s Law P  T V and n are fixed P T = constant P = constant x T combined gas law V  T P V = constant x T P PV T = constant V x P= constantV = constant / P

14. 5-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.7 An experiment to study the relationship between the volume and amount of a gas.

15. 5-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.8 Standard molar volume.

16. 5-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.9 The volume of 1 mol of an ideal gas compared with some familiar objects.

17. 5-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. THE IDEAL GAS LAW PV = nRT IDEAL GAS LAW nRT P PV = nRT or V = Boyle’s Law V = constant P R = PV nT = 1atm x 22.414L 1mol x 273.15K = 0.0821atm*L mol*K V = Charles’s Law constant x T Avogadro’s Law constant x n fixed n and T fixed n and P fixed P and T Figure 5.10 R = universal gas constant 3 significant figures

18. 5-18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM:Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm 3 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)? PLAN: SOLUTION: V 1 in cm 3 V 1 in mL V 1 in L V 2 in L unit conversion gas law calculation P 1 = 1.12 atmP 2 = 2.64 atm V 1 = 24.8 cm 3 V 2 = unknown P and T are constant 24.8 cm 3 x 1 mL 1 cm 3 L 10 3 mL = 0.0248 L P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P 1 V 1 = P 2 V 2 P1V1P1V1 P2P2 V 2 = =0.0248 L x 1.12 atm 2.46 atm = 0.0105 L 1cm 3 =1mL 10 3 mL=1L xP 1 /P 2 x

19. 5-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.3 Applying the Temperature-Pressure Relationship PROBLEM:A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x10 3 torr. It is filled with methane at 23 0 C and 0.991 atm and placed in boiling water at exactly 100 0 C. Will the safety valve open? PLAN:SOLUTION: P 1 (atm)T 1 and T 2 ( 0 C) P 1 (torr)T 1 and T 2 (K) P 1 = 0.991atmP 2 = unknown T 1 = 23 0 CT 2 = 100 0 C P 2 (torr) 1atm=760torr x T 2 /T 1 K= 0 C+273.15 P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P1P1 T1T1 P2P2 T2T2 = 0.991 atm x 1 atm 760 torr = 753 torr P 2 = P1T2P1T2 T1T1 = 753 torr x 373K 296K = 949 torr

20. 5-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.4 Applying the Volume-Amount Relationship PROBLEM:A scale model of a blimp rises when it is filled with helium to a volume of 55 dm 3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: SOLUTION: We are given initial n 1 and V 1 as well as the final V 2. We have to find n 2 and convert it from moles to grams. n 1 (mol) of He n 2 (mol) of He mol to be added g to be added x V 2 /V 1 x M subtract n 1 n 1 = 1.10 moln 2 = unknown V 1 = 26.2 dm 3 V 2 = 55.0 dm 3 P and T are constant P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = V1V1 n1n1 V2V2 n2n2 = n 2 = V2n1V2n1 V1V1 n 2 = 1.10 mol x 55.0 dm 3 26.2 dm 3 = 2.31 mol x 4.003 g He mol He = 9.24 g He

21. 5-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM:A steel tank has a volume of 438 L and is filled with 0.885 kg of O 2. Calculate the pressure of O 2 at 21 0 C. PLAN: SOLUTION: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. V = 438 LT = 21 0 C (convert to K) n = 0.885 kg (convert to mol)P = unknown 21 0 C + 273.15 = 294.15K 0.885 kg x 10 3 g kg mol O 2 32.00 g O 2 = 27.7 mol O 2 P = nRT V = 27.7 mol 294.15K atm*L mol*K 0.0821x x 438 L = 1.53 atm x

22. 5-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.6Using 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. PLAN: SOLUTION: 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. Which of the following balanced equations describes the reaction? (1) A 2 + B 2 2AB(2) 2AB + B 2 2AB 2 (4) 2AB 2 A 2 + 2B 2 (3) A + B 2 AB 2 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. 5-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. density = m/V n = m/ M The Density of a Gas PV = nRTPV = (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. 5-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.7 Calculating Gas Density PROBLEM:To apply a green chemistry approach, a chemical engineer uses waste CO 2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO 2 and the number of molecules (a) at STP (0 0 C and 1 atm) and (b) at room conditions (20. 0 C and 1.00 atm). PLAN: SOLUTION: 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/volumePV = nRTV = nRT/P d= RT M x P d = 44.01 g/molx 1atm atm*L mol*K 0.0821x 273.15K = 1.96 g/L 1.96 g L mol CO 2 44.01 g CO 2 6.022x10 23 molecules mol = 2.68x10 22 molecules CO 2 /L (a) x xx

25. 5-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.6 continued Calculating Gas Density (b) = 1.83 g/L d = 44.01 g/molx 1 atm x 293K atm*L mol*K 0.0821 1.83g L mol CO 2 44.01g CO 2 6.022x10 23 molecules mol = 2.50x10 22 molecules CO 2 /L xx

26. 5-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.11 Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)

27. 5-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.8 Finding the Molar Mass of a Volatile Liquid PROBLEM:An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data: PLAN: SOLUTION: Use unit conversions, mass of gas, and density- M relationship. Volume of flask = 213 mL Mass of flask + gas = 78.416 g T = 100.0 0 C Mass of flask = 77.834 g P = 754 torr Calculate the molar mass of the liquid. m = (78.416 - 77.834) g= 0.582 g M = m RT VP = 0.582 g atm*L mol*K 0.0821373K x x 0.213 Lx0.992 atm = 84.4 g/mol

28. 5-28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Dalton’s Law of Partial Pressures P total = P 1 + P 2 + P 3 +... P 1 =  1 x P total where  1 is the mole fraction  1 = n1n1 n 1 + n 2 + n 3 +... = n1n1 n total Mixtures of Gases Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present.

29. 5-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.9 Applying Dalton’s Law of Partial Pressures PROBLEM:In a study of O 2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N 2, 17 mol% 16 O 2, and 4.0 mol% 18 O 2. (The isotope 18 O will be measured to determine the O 2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18 O 2 in the mixture. PLAN: SOLUTION: Find the  and P from P total and mol% 18 O 2. 18 O 2 mol% 18 O 2  partial pressure P 18 O 2 divide by 100 multiply by P total  18 O 2 = 4.0 mol% 18 O 2 100 = 0.040 = 0.030 atm P =  x P total = 0.040 x 0.75 atm 18 O 2

30. 5-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Molar Mass of a Gas n = mass M = PV RT M = d RT P m RT VP d = m V

31. 5-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 5.3 Vapor Pressure of Water (P ) at Different T H2OH2O T( 0 C)P (torr)T( 0 C)P (torr) 0 5 10 11 12 13 14 15 16 18 20 22 24 26 28 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 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

32. 5-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.12 Collecting a water-insoluble gaseous reaction product and determining its pressure.

33. 5-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water PLAN: SOLUTION: The difference in pressures will give us the P for the C 2 H 2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. PROBLEM: Acetylene (C 2 H 2 ), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC 2 ) reaction with water: CaC 2 ( s ) + 2H 2 O( l ) C 2 H 2 ( 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 (23 0 C), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? P total P C2H2C2H2 n C2H2C2H2 g C2H2C2H2 P H2OH2O n = PV RT x M P C2H2C2H2 = (738-21)torr = 717torr 717 torr atm 760torr = 0.943 atm x

34. 5-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.9 continued Calculating the Amount of Gas Collected Over Water 0.943 atm0.523Lx n C2H2C2H2 = atm*L mol*K 0.0821 x 296K = 0.0203 mol 0.0203 mol x 26.04 g C 2 H 2 mol C 2 H 2 = 0.529 g C 2 H 2 n = PV RT

35. 5-35 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ideal gas law molar ratio from balanced equation 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

36. 5-36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.11 Using Gas Variables to Find Amount of Reactants and Products 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 H 2 reduces the metal oxide, forming the pure metal and H 2 O. On a laboratory scale, what volume of H 2 at 765 torr and 225 0 C is needed to reduce 35.5 g of copper(II) oxide? SOLUTION: 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 H 2 gas. mass (g) of Cu mol of Cu mol of H 2 L of H 2 divide by M molar ratio use known P and T to find V CuO( s ) + H 2 ( g ) Cu( s ) + H 2 O( g ) 35.5 g Cu x mol Cu 63.55 g Cu 1 mol H 2 1 mol Cu = 0.559 mol H 2 0.559 mol H 2 x 498K atm*L mol*K 0.0821x 1.01 atm = 22.6 L x

37. 5-37 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.12 Using the Ideal Gas Law in a Limiting-Reactant Problem 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 L of chlorine gas at 0.950 atm and 293K reacts with 17.0 g of potassium? SOLUTION: 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. 2K( s ) + Cl 2 ( g ) 2KCl( s ) V = 5.25 L T = 293Kn = unknown P = 0.950 atm n = PV RT Cl 2 x5.25 L = 0.950 atm atm*L mol*K 0.0821x293K = 0.207 mol 17.0 g x 39.10 g K mol K = 0.435 mol K 0.207 mol Cl 2 x 2 mol KCl 1 mol Cl 2 = 0.414 mol KCl formed Cl 2 is the limiting reactant. 0.414 mol KCl x 74.55 g KCl mol KCl = 30.9 g KCl

38. 5-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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(E k ) of the particles is constant. Postulate 1: Particle Volume Postulate 2: Particle Motion Postulate 3: Particle Collisions

39. 5-39 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.14 Distribution of molecular speeds at three temperatures.

40. 5-40 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.15 A molecular description of Boyle’s Law.

41. 5-41 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.16 A molecular description of Dalton’s law of partial pressures.

42. 5-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.17 A molecular description of Charles’s Law.

43. 5-43 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Avogadro’s Law V  n E k = 1/2 mass x speed 2 E k = 1/2 mass x u  2 u 2 is the root-mean-square speed u rms = √ 3RT M 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. rate of effusion  1 √M√M

44. 5-44 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.18 A molecular description of Avogadro’s Law.

45. 5-45 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.19 Relationship between molar mass and molecular speed. E k = 3/2 (R/N A ) T

46. 5-46 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.13 Applying Graham’s Law of Effusion PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH 4 ). SOLUTION: 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. M of CH 4 = 16.04g/mol M of He = 4.003g/mol CH 4 He rate = √ 16.04 4.003 = 2.002

47. 5-47 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. 5-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B5.1 Variations in pressure, temperature, and composition of the Earth’s atmosphere.

49. 5-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

50. 5-50 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

51. 5-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 5.4 Molar Volume of Some Common Gases at STP (0 0 C and 1 atm) Gas Molar Volume (L/mol) Condensation Point ( 0 C) He H 2 Ne Ideal gas Ar N 2 O 2 CO Cl 2 NH 3 22.435 22.432 22.422 22.414 22.397 22.396 22.390 22.388 22.184 22.079 -268.9 -252.8 -246.1 --- -185.9 -195.8 -183.0 -191.5 -34.0 -33.4

52. 5-52 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.21 The behavior of several real gases with increasing external pressure.

53. 5-53 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.22 The effect of intermolecular attractions on measured gas pressure.

54. 5-54 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.23 The effect of molecular volume on measured gas volume.

55. 5-55 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 5.5 Van der Waals Constants for Some Common Gases 0.034 0.211 1.35 2.32 4.19 0.244 1.39 1.36 6.49 3.59 2.25 4.17 5.46 He Ne Ar Kr Xe H 2 N 2 O 2 Cl 2 CO 2 CH 4 NH 3 H 2 O 0.0237 0.0171 0.0322 0.0398 0.0511 0.0266 0.0391 0.0318 0.0562 0.0427 0.0428 0.0371 0.0305 Gas a atm*L 2 mol 2 b L mol Van der Waals equation for n moles of a real gas adjusts P upadjusts V down