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Chapter 5

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

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Chapter 5

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  1. Chapter 5 Gases and the Kinetic-Molecular Theory

  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. 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.

  4. The three states of matter. Figure 5.1

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

  6. Figure 5.3 A mercury barometer.

  7. 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 5.1 Common Units of Pressure Unit Atmospheric Pressure Scientific Field *This is an exact quantity; in calculations, we use as many significant figures as necessary.

  8. 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 Sample Problem 5.1 Converting Units of Pressure SOLUTION: 291.4 mmHg = 291.4 torr 291.4 torr = 0.3834 atm 0.3834 atm = 38.85 kPa

  9. Figure 5.4 The relationship between the volume and pressure of a gas. Boyle’s Law

  10. Figure 5.5 The relationship between the volume and temperature of a gas. Charles’s Law

  11. 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’sLaw n and T are fixed V = constant / P Charles’sLaw V a T P and n are fixed V = constant x T Amontons’sLaw P a T V and n are fixed P = constant x T Combined gas law

  12. Figure 5.6 An experiment to study the relationship between the volume and amount of a gas.

  13. Standard molar volume. Figure 5.7

  14. Figure 5.8 The volume of 1 mol of an ideal gas compared with some familiar objects.

  15. 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 5.9 PV = nRT 3 significant figures R = = = R is the universal gas constant fixed n and T fixed n and P fixed P and T Boyle’s Law Charles’s Law Avogadro’s Law V = constant X n V = V = constant X T

  16. 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 V2 = = 0.0105 L = 0.0248 L 2.46 atm P2 Sample Problem 5.2 Applying the Volume-Pressure Relationship 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 = 0.0248 L V1 in L xP1/P2 P1V1 = P2V2 = V2 in L

  17. P1 P2 = = T1 T2 760 torr P2V2 0.991 atm P1V1 1 atm n2T2 n1T1 T2 373K P2 = P1 = 753 torr T1 296K Sample Problem 5.3 Applying the Pressure-Temperature Relationship 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

  18. = = n2 = n1 V1 V2 V2 P2V2 P1V1 n1 V1 n2 n2T2 n1T1 55.0 dm3 4.003 g He n2 = 1.10 mol 26.2 dm3 mol He 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 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 g to be added = 9.24 g He = 2.31 mol

  19. 103 g mol O2 0.885kg kg 32.00 g O2 atm*L nRT 24.7 mol x 0.0821 294.15K x mol*K V P = 438 L 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 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 = 27.7 mol O2 = 1.53 atm =

  20. (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 below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K. New figures go here. 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).

  21. The Density of a Gas density = m/V n = m/M PV = nRT PV = (m/M)RT 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.

  22. 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) = 2.68x1022 molecules CO2/L

  23. 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 Calculating Gas Density continued (b) = 1.83 g/L = 2.50x1022 molecules CO2/L

  24. 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 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: 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 = =

  25. 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

  26. 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 Sample Problem 5.9 Applying Dalton’s Law of Partial Pressures PLAN: mol% 18O2 SOLUTION: = 0.040 divide by 100 c 18O2 = 0.030 atm multiply by Ptotal

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

  28. Table 5.2 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

  29. Figure 5.10 Collecting a water-insoluble gaseous reaction product and determining its pressure.

  30. 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 738torr and the volume is 523mL. At the temperature of the gas (230C), the vapor pressure of water is 21torr. How many grams of acetylene are collected? P P g C2H2 C2H2 C2H2 PV atm P n = RT 760torr n C2H2 Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water 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 = 0.943atm 717torr H2O x M

  31. n = C2H2 26.04g C2H2 0.0203mol mol C2H2 atm*L 0.0821 mol*K Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water continued 0.943atm x 0.523L = 0.0203mol x 296K = 0.529 g C2H2

  32. Figure 15.11 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). amount (mol) of gas B amount (mol) of gas A P,V,T of gas A P,V,T of gas B ideal gas law ideal gas law molar ratio from balanced equation

  33. 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 torr and 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 x mol*K x 498K 1.01 atm Sample Problem 5.11 Using Gas Variables to Find Amounts of Reactants and Products mass (g) of Cu SOLUTION: divide by M 35.5 g Cu = 0.559 mol H2 mol of Cu molar ratio 0.559 mol H2 = 22.6 L mol of H2 use known P and T to find V L of H2

  34. 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? 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) 0.950 atm T = 293K n = unknown = = 0.207 mol PV x 293K RT 2 mol KCl 2 mol KCl atm*L 0.207 mol Cl2 17.0g = 0.435 mol K 0.0821 2 mol K 1 mol Cl2 mol*K mol K 39.10 g K 74.55 g KCl mol KCl Sample Problem 5.12 Using the Ideal Gas Law in a Limiting-Reactant Problem SOLUTION: x 5.25L n = Cl2 = 0.414 mol KCl formed Cl2 is the limiting reactant. 0.435 mol K = 0.435 mol KCl formed 0.414 mol KCl = 30.9 g KCl

  35. 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.

  36. Distribution of molecular speeds at three temperatures. Figure 5.12

  37. Figure 5.13 A molecular description of Boyle’s Law.

  38. Figure 5.14 A molecular description of Dalton’s law of partial pressures.

  39. Figure 5.15 A molecular description of Charles’s Law.

  40. 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 = 8.314Joule/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.

  41. A molecular description of Avogadro’s Law. Figure 5.16

  42. Figure 5.17 Relationship between molar mass and molecular speed. Ek = 3/2 (R/NA) T

  43. 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 Sample Problem 5.13 Applying Graham’s Law of Effusion SOLUTION: M of CH4 = 16.04g/mol M of He = 4.003g/mol = 2.002

  44. Figure 5.18 Diffusion of a gas particle through a space filled with other particles. distribution of molecular speeds mean free path collision frequency

  45. Table 5.3 Molar Volume of Some Common Gases at STP (00C and 1 atm) Molar Volume (L/mol) Condensation Point (0C) Gas 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 He H2 Ne Ideal gas Ar N2 O2 CO Cl2 NH3

  46. Figure 5.19 The behavior of several real gases with increasing external pressure.

  47. Figure 5.20 The effect of intermolecular attractions on measured gas pressure.

  48. Figure 5.21 The effect of molecular volume on measured gas volume.

  49. Van der Waals equation for n moles of a real gas atm*L2 L a b mol2 mol Gas He Ne Ar Kr Xe H2 N2 O2 Cl2 CO2 CH4 NH3 H2O 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 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 Table 5.54 Van der Waals Constants for Some Common Gases adjusts P up adjusts V down

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