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

2. 5-2 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. 5-3 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. 5-4 Figure 5.1 States of Matter

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

6. 5-6 Figure 5.4 Two Types of Manometer

7. 5-7 Table 5.2 Common Units of Pressure Atmospheric PressureUnitScientific Field chemistryatmosphere (atm)1 atm pascal (Pa) kilopascal (kPa) 1.01325 x 10 5 Pa; 101.325 kPa SI unit; physics, chemistry; (1 Pa = 1 N/m 2 ) millimeters of mercury (Hg) 760 mm Hgchemistry, medicine, biology torr760 torrchemistry pounds per square inch (psi or lb/in 2 ) bar0.01325 barmeteorology, chemistry, physics engineering 14.7 lb/in 2

8. 5-8 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 mm Hg x 1 torr 1 mm Hg = 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

9. 5-9 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. 5-10 Figure 5.5 Relationship between volume and pressure of a gas Boyle’s Law

11. 5-11 Boyle’s Law V  n and T are fixed 1 P PV = constantV = constant / P P = pressure V = volume n = number of moles of gas T = temperature (volume is inversely proportional to pressure) or

12. 5-12 Figure 5.6 Relationship between volume and temperature of a gas Charles’s Law

13. 5-13 Boyle’s Law: n and T are fixed V  1 P Charles’s Law: V  T P and n are fixed V T = constant V = constant x T Amonton’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 (Boyle’s + Charles’s)

14. 5-14 Figure 5.7 An experiment to study the relationship between volume and amount of a gas Avogadro’s Law V  n (P and T fixed) At fixed T and P, equal volumes of any ideal gas contain equal numbers of particles (or moles).

15. 5-15 Figure 5.8 Standard Molar Volume STP 0 o C (273.15 K) 1 atm (760 torr) Standard Molar Volume: 22.4141 L or 22.4 L

16. 5-16 THE IDEAL GAS LAW PV = nRT nRT P PV = nRT or V = Boyle’s Law V = constant P R = PV nTnT = 1 atm x 22.414 L 1 mol x 273.15 K = 0.0821atm L mol K V = Charles’s Law constant x T Avogadro’s Law constant x n At fixed n and T:At fixed n and P:At fixed P and T:

17. 5-17 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 P 1 = 1.12 atmP 2 = 2.64 atm V 1 = 24.8 cm 3 V 2 = unknown n and T are constant 24.8 cm 3 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 1.12 atm 2.64 atm =0.0105 L 1 cm 3 = 1 mL 10 3 mL = 1 L x P 1 /P 2 or xx x

18. 5-18 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 10 3 torr. It is filled with helium at 23 o C and 0.991 atm and placed in boiling water at exactly 100 o C. Will the safety valve open? PLAN: SOLUTION: P 1 (atm)T 1 and T 2 ( o C) P 1 (torr)T 1 and T 2 (K) P 1 = 0.991 atmP 2 = unknown T 1 = 23 o CT 2 = 100 o C P 2 (torr) 1 atm = 760 torr x T 2 /T 1 K = o C + 273.15 P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P1P1 T1T1 P2P2 T2T2 = 0.991 atm x 1 atm 760 torr = 753 torr P 2 =P1P1 T2T2 T1T1 = 753 torr x 373 K 296 K = 949 torr n and V are constant or (valve will not open)

19. 5-19 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 (V 2 ). When 1.10 mol of He (n 1 ) are added to the blimp, the volume is 26.2 dm 3 (V 1 ). How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: SOLUTION: We are given the initial n 1 and V 1 and the final V 2. We need 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 He to add 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 = n 1 V2V2 V1V1 n 2 = 1.10 mol 55.0 dm 3 26.2 dm 3 = 2.31 mol 4.003 g He mol He = 4.84 g He or x 2.31 mol - 1.10 mol = 1.21 mol x

20. 5-20 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 o C. PLAN: SOLUTION: V, T and mass, which can be converted to moles (n), are given. Use the ideal gas law to find P. V = 438 LT = 21 o C (convert to K) n = 0.885 kg (convert to mol)P = unknown 21 o C + 273.15 = 294 K 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 K atm L mol K 0.0821x x 438 L = 1.53 atm x

21. 5-21 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. 5-22 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 (0 o C and 1 atm) and (b) at ordinary room conditions (20. o 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, the molar mass can be determined. Convert mass/L to molecules/L using Avogadro’s number. d = mass/volumePV = nRTV = nRT/Pd= RTRT M x P d = 44.01 g/molx 1 atm atm L mol K 0.0821x 273 K = 1.96 g/L 1.96 g L mol CO 2 44.01 g CO 2 6.022 x 10 23 molecules mol = 2.68 x 10 22 molecules CO 2 /L (a) xx

23. 5-23 Sample Problem 5.6(continued) (b) = 1.83 g/L d = 44.01 g/molx 1 atm x 293 K atm L mol K 0.0821 1.83 g L mol CO 2 44.01 g CO 2 6.022 x 10 23 molecules mol = 2.50 x 10 22 molecules CO 2 /Lx x

24. 5-24 The Molar Mass of a Gas n =n = mass M = PV RT M = d RT P mRT PV d = m V

25. 5-25 Figure 5.11 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

26. 5-26 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 (C 6 H 12 ). She uses the Dumas method and obtains the following data to determine its molar mass: 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 o C Mass of flask = 77.834 g P = 754 torr Is the calculated molar mass consistent with the liquid being cyclohexane? m = (78.416 - 77.834)g= 0.582 g of gas M = m RT VP = 0.582 g atm L mol K 0.0821373 K x x 0.213 Lx0.992 atm =84.4 g/mol M of C 6 H 12 is 84.16 g/mol - the calculated value is within experimental error RTRT M x P d = mRT VPVP M =

27. 5-27 Dalton’s Law of Partial Pressures P total = P 1 + P 2 + P 3 +... P 1 =  1 x P total and  1 is the mole fraction  1 = n1n1 n 1 + n 2 + n 3 +... = n1n1 n total In a mixture of unreacting gases, the total pressure is equal to the sum of the partial pressures of the individual gases. where

28. 5-28 Sample Problem 5.8 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 0.75 atm 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  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

29. 5-29 Figure 5.12 Collecting a water-insoluble gaseous reaction product and determining its pressure

30. 5-30 Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water PLAN: SOLUTION: The difference in pressures will give P for C 2 H 2. The ideal gas law allows a determination of 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 ) reacts 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 o C), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? P total P C2H2C2H2 n C2H2C2H2 -P H2OH2O n = PV RT x M P C2H2C2H2 = (738 - 21) torr = 717 torr 717 torr x atm 760 torr = 0.943 atm g C2H2C2H2

31. 5-31 Sample Problem 5.9(continued) 0.943 atm0.523 Lx n C2H2C2H2 = atm L mol K 0.0821 x296 K = 0.0203 mol C 2 H 2 0.0203 mol x 26.04 g C 2 H 2 mol C 2 H 2 = 0.529 g C 2 H 2

32. 5-32 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 molar ratio from balanced equation Figure 15.13 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)

33. 5-33 Sample Problem 5.10Using Gas Variables to Find Amount of Reactants and Products PROBLEM: A laboratory-scale method for reducing a metal oxide is to heat it with H 2. The pure metal and H 2 O are products. What volume of H 2 at 765 torr and 225 o C is needed to form 35.5 g of Cu from copper(II) oxide? SOLUTION: 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 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 x498 K atm L mol K 0.0821x 1.01 atm = 22.6 L x

34. 5-34 Sample Problem 5.11 Using the Ideal Gas Law in a Limiting Reactant Problem 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? SOLUTION: 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. 2K( s ) + Cl 2 ( g ) 2KCl( s ) V = 5.25 L T = 293 Kn = unknown P = 0.950 atm n = PV RT Cl 2 x5.25 L = 0.950 atm atm L mol K 0.0821x293 K 0.207 mol Cl 2 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 0.435 mol K x 2 mol KCl 2 mol K = 0.435 mol KCl formed Cl 2 is the limiting reactant 0.414 mol KCl x 74.55 g KCl mol KCl = 30.9 g KCl =

35. 5-35 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

36. 5-36 Figure 5.14 Distribution of molecular speeds at three temperatures The most probable speed increases as the temperature increases The average kinetic energy, E k, is proportional to the absolute temperature For N 2 gas

37. 5-37 Figure 5.15 A molecular description of Boyle’s Law

38. 5-38 Figure 5.16 A molecular description of Dalton’s law of partial pressures

39. 5-39 Figure 5.17 A molecular description of Charles’s Law

40. 5-40 Figure 5.18 A molecular description of Avogadro’s Law

41. 5-41 E k = 1/2 mass x speed 2 E k = 1/2 massx u 2 u 2 is the average of the squares of the molecular speeds; its square root equals u rms u rms = √ 3RT3RT M R = 8.314 joule/mol K Why do equal numbers of molecules of two different gases, such as O 2 and H 2, occupy the same volume (c.f. standard molar volume)? = root-mean-square speed; a molecule moving at this speed has the average kinetic energy At constant T, two gases possess the same kinetic energy; thus, the heavier gas must be moving more slowly. u rms  1 √M√M

42. 5-42 Figure 5.19 Relationship between molar mass and molecular speed At a given temperature, gases with lower molar masses have higher most probable speeds

43. 5-43 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 EFFUSION: the process by which a gas escapes from its container through a tiny hole into an evacuated space (related to the rms speed) rate A /rate B = M B 1/2 /M A 1/2 The same relationships pertain to gaseous diffusion rates!

44. 5-44 Sample Problem 5.12 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 of each gas. Find the molar mass of both gases and find the inverse square root of their masses. M of CH 4 = 16.04 g/mol M of He = 4.003 g/mol CH 4 He rate = √ 16.04 4.003 = 2.002

45. 5-45 Diffusion of a gas particle through a space filled with other particles Figure 5.20 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)

46. 5-46 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. 5-47 The behavior of several real gases with increasing external pressure Figure 5.21 At moderately high P: intermolecular attractions dominate At very high P: molecular volume effects dominate

48. 5-48 The effect of intermolecular attractions on measured gas pressure Figure 5.22

49. 5-49 The effect of molecular volume on measured gas volume Figure 5.23

50. 5-50 The van der Waals equation for n moles of a real gas (P + n 2 a/V 2 ) (V - nb) = nRT a and b are the van der Waals constants (adjusts P up) (adjusts V down)