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

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

  4. States of Matter Figure 5.1

  5. A Mercury Barometer Pressure = force/area A device used to measure atmospheric pressure Figure 5.3

  6. Two Types of Manometer Figure 5.4

  7. pascal (Pa) kilopascal (kPa) 1.01325 x 105Pa; 101.325 kPa SI unit; physics, chemistry; (1 Pa = 1 N/m2) 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) bar 0.01325 bar meteorology, chemistry, physics Table 5.2 Common Units of Pressure Unit Atmospheric Pressure Scientific Field engineering 14.7 lb/in2

  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. 1 torr 1 mm Hg 1 atm 760 torr 101.325 kPa 1 atm Sample Problem 5.1 Converting Units of Pressure SOLUTION: 291.4 mm Hg x = 291.4 torr 291.4 torr x = 0.3834 atm 0.3834 atm x = 38.85 kPa

  9. Three laws (Boyle’s, Charles’s and Avogadro’s) are combined to describe a universal relationship among the key gas variables (volume, pressure, temperature, amount). This universal relationship is known as the Ideal Gas Law. Let’s examine the three individual laws first, and then see how they are combined to generate the Ideal Gas Law.

  10. Relationship between volume and pressure of a gas Boyle’s Law Figure 5.5

  11. 1 P Boyle’sLaw V a n and T are fixed (volume is inversely proportional to pressure) PV = constant or V = constant / P P = pressure V = volume n = number of moles of gas T = temperature

  12. Relationship between volume and temperature of a gas Charles’s Law Figure 5.6

  13. 1 T T V a P P P = constant P V T T = constant V a V = constant x = constant PV T Boyle’sLaw: n and T are fixed Charles’sLaw: V a T P and n are fixed V = constant x T Amonton’sLaw: P a T V and n are fixed P = constant x T Combined Gas Law: (Boyle’s + Charles’s)

  14. An experiment to study the relationship between volume and amount of a gas Avogadro’s Law Figure 5.7 V a n (P and T fixed) At fixed T and P, equal volumes of any ideal gas contain equal numbers of particles (or moles).

  15. Standard Molar Volume STP 0 oC (273.15 K) 1 atm (760 torr) Standard Molar Volume: 22.4141 L or 22.4 L Figure 5.8

  16. PV 1 atm x 22.414 L 0.0821atm L nT 1 mol x 273.15 K mol K PV = nRT or V = nRT P constant P THE IDEAL GAS LAW PV = nRT R = = = At fixed n and T: At fixed n and P: At fixed P and T: Boyle’s Law Charles’s Law Avogadro’s Law V = constant x n V = V = constant x T

  17. 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)? 1 mL L 1 cm3 103mL P2V2 P1V1 n2T2 n1T1 1.12 atm V2 = = 0.0105 L = 0.0248 L 2.64 atm Sample Problem 5.2 Applying the Volume-Pressure Relationship PLAN: SOLUTION: n and T are constant V1 in cm3 P1 = 1.12 atm P2 = 2.64 atm 1 cm3= 1 mL V1 = 24.8 cm3 V2 = unknown V1 in mL 103 mL = 1 L 24.8 cm3 x x = 0.0248 L V1 in L = P1V1 = P2V2 or x P1/P2 V2 in L P1V1 x P2

  18. P1 P2 = = T1 T2 760 torr P2V2 0.991 atm x P1V1 1 atm n2T2 n1T1 T2 373 K P2 = P1 = 753 torr x T1 296 K Sample Problem 5.3 Applying the Temperature-Pressure Relationship PROBLEM: A 1 L steel tank is fitted with a safety valve that opens if the internal pressure exceeds 1.00 x 103 torr. It is filled with helium at 23 oC and 0.991 atm and placed in boiling water at exactly 100 oC. Will the safety valve open? PLAN: SOLUTION: n and V are constant P1(atm) T1 and T2(oC) P1 = 0.991 atm P2 = unknown 1 atm = 760 torr K = oC + 273.15 T1 = 23 oC T2 = 100 oC P1(torr) T1 and T2(K) or x T2/T1 P2(torr) = 753 torr = 949 torr (valve will not open)

  19. = = 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 (V2). When 1.10 mol of He (n1) are added to the blimp, the volume is 26.2 dm3 (V1). How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: We are given the initial n1 and V1 and the final V2. We need 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 or mol to be added x M = 2.31 mol x g He to add = 4.84 g He 2.31 mol - 1.10 mol = 1.21 mol x

  20. 103g mol O2 0.885 kg x kg 32.00 g O2 atm L nRT 27.7 mol x 0.0821 x 294 K 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 21oC. PLAN: V, T and mass, which can be converted to moles (n), are given. Use the ideal gas law to find P. SOLUTION: V = 438 L T = 21oC (convert to K) n = 0.885 kg (convert to mol) P = unknown = 27.7 mol O2 21oC + 273.15 = 294 K x = 1.53 atm

  21. The Density of a Gas PV = nRT or PV = m/M x RT where m = mass and M = molar mass m/V = d = (M x P)/RT where d = density

  22. M x P RT 44.01 g/mol x 1 atm x 273 K atm L 0.0821 mol K 1.96 g mol CO2 6.022 x 1023 molecules L 44.01 g CO2 mol Sample Problem 5.6 Calculating the Density of a Gas PROBLEM: Calculate the density (in g/L) of carbon dioxide and the number of molecules per liter (a) at STP (0oC and 1 atm) and (b) at ordinary room conditions (20.oC 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, the molar mass can be determined. Convert mass/L to molecules/L using Avogadro’s number. d = mass/volume PV = nRT V = nRT/P d = SOLUTION: d = (a) = 1.96 g/L x x = 2.68 x 1022 molecules CO2/L

  23. 44.01 g/mol x 1 atm d = x 293 K atm L 0.0821 mol K 1.83 g mol CO2 6.022 x 1023 molecules L 44.01 g CO2 mol Sample Problem 5.6 (continued) (b) = 1.83 g/L x = 2.50 x 1022molecules CO2/L x

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

  25. Determining the molar mass of an unknown volatile liquid [based on the method of J.B.A. Dumas (1800 -1884)] M = mRT/PV or M = dRT/P Figure 5.11

  26. Volume of flask = 213 mL T = 100.0 oC P = 754 torr Mass of flask + gas = 78.416 g Mass of flask = 77.834 g M x P RT atm L 0.0821 mRT m RT 0.582 g 373 K x mol K = 84.4 g/mol VP VP 0.213 L x 0.992 atm Sample Problem 5.7 Finding the Molar Mass of a Volatile Liquid PROBLEM: An organic chemist isolates from a petroleum sample a colorless liquid with the properties of cyclohexane (C6H12). She uses the Dumas method and obtains the following data to determine its molar mass: Is the calculated molar mass consistent with the liquid being cyclohexane? PLAN: Use unit conversions, mass of gas and density-M relationship. d = SOLUTION: m = (78.416 - 77.834)g = 0.582 g of gas M = x M = = M of C6H12 is 84.16 g/mol - the calculated value is within experimental error

  27. n1 n1 = ntotal n1+ n2 + n3 +... c1 = Dalton’s Law of Partial Pressures In a mixture of unreacting gases, the total pressure is equal to the sum of the partial pressures of the individual gases. Ptotal = P1 + P2 + P3 + ... where P1 = c1 x Ptotal and c1is the mole fraction

  28. 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 0.75 atm 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 partial pressure P 18O2 Sample Problem 5.8 Applying Dalton’s Law of Partial Pressures PLAN: mol% 18O2 = 0.040 SOLUTION: divide by 100 c18O2 = 0.030 atm 18O2 multiply by Ptotal

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

  30. PROBLEM: Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts 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 (23 oC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? P P C2H2 C2H2 PV atm -P n = RT 760 torr g n C2H2 C2H2 Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water PLAN: The difference in pressures will give P for C2H2. The ideal gas law allows a determination of n. Converting n to grams requires the molar mass, M. SOLUTION: = (738 - 21) torr = 717 torr Ptotal 717 torr x H2O = 0.943 atm x M

  31. n = C2H2 26.04 g C2H2 0.0203 mol x mol C2H2 atm L 0.0821 mol K Sample Problem 5.9 (continued) 0.943 atm x 0.523 L = 0.0203 mol C2H2 x 296 K = 0.529 g C2H2

  32. Summary of the stoichiometric relationships between 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 Figure 15.13

  33. PROBLEM: A laboratory-scale method for reducing a metal oxide is to heat it with H2. The pure metal and H2O are products. What volume of H2 at 765 torr and 225 oC is needed to form 35.5 g of Cu from copper(II) oxide? PLAN: This problem requires stoichiometry and the gas laws; write a balanced equation and use the moles of Cu to calculate moles 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 498 K 1.01 atm Sample Problem 5.10 Using Gas Variables to Find Amount of Reactants and Products mass (g) of Cu SOLUTION: divide by M 35.5 g Cu x x = 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) react with the halogens (Group 7A) to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium? PLAN: Write a balanced equation, and use the ideal gas law to find the number of moles of reactants, the limiting reactant and the moles of product. P = 0.950 atm V = 5.25 L 2K(s) + Cl2(g) 2KCl(s) 0.950 atm T = 293 K n = unknown = 0.207 mol Cl2 PV x 293 K RT 2 mol KCl 2 mol KCl atm L 0.207 mol Cl2x 17.0 g x = 0.435 mol K 0.0821 1 mol Cl2 2 mol K 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 5.11 SOLUTION: x 5.25 L n = = Cl2 = 0.414 mol KCl formed 0.435 mol K x = 0.435 mol KCl formed 0.414 mol KCl x = 30.9 g KCl Cl2 is the limiting reactant

  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 The most probable speed increases as the temperature increases For N2 gas The average kinetic energy, Ek, is proportional to the absolute temperature Figure 5.14

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

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

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

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

  41. Ek = 1/2 massx u2 u2 is the average of the squares of the molecular speeds; its square root equals urms urms= √ 3RT M 1 urms a √M Why do equal numbers of molecules of two different gases, such as O2 and H2, occupy the same volume (c.f. standard molar volume)? At constant T, two gases possess the same kinetic energy; thus, the heavier gas must be moving more slowly. Ek= 1/2 mass x speed2 root-mean-square speed; a molecule moving at this speed has the average kinetic energy = R = 8.314 joule/mol K

  42. Relationship between molar mass and molecular speed Figure 5.19 At a given temperature, gases with lower molar masses have higher most probable speeds

  43. 1 rate of effusion a √M EFFUSION: the process by which a gas escapes from its container through a tiny hole into an evacuated space Graham’s Law of Effusion The rate of effusion of a gas is inversely related to the square root of its molar mass. (related to the rms speed) rateA/rateB = MB1/2/MA1/2 The same relationships pertain to gaseous diffusion rates!

  44. 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 of 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 5.12 Applying Graham’s Law of Effusion SOLUTION: M of CH4= 16.04 g/mol M of He = 4.003 g/mol = 2.002

  45. Diffusion of a gas particle through a space filled with other particles Distribution of Molecular Speeds Mean Free Path: the average distance a molecule travels between collisions at a given T and P Collision Frequency: the average number of collisions per second (has implications for chemical reaction rates) Figure 5.20

  46. Real Gases Molecules are not points of mass. There are attractive and repulsive forces between molecules. Real gases approach ideal behavior at high T and low P.

  47. Figure 5.21 The behavior of several real gases with increasing external pressure At moderately high P: intermolecular attractions dominate At very high P: molecular volume effects dominate

  48. The effect of intermolecular attractions on measured gas pressure Figure 5.22

  49. The effect of molecular volume on measured gas volume Figure 5.23

  50. The van der Waals equation for n moles of a real gas (adjusts V down) (P + n2a/V2) (V - nb) = nRT (adjusts P up) a and b are the van der Waals constants

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