1. 5-1 Introduction to Gases

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 The Distinction of Gases from Liquids and Solids 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 densites under normal conditions. 5. Gases are miscible.

4. 5-4 Figure 5.1 The three states of matter.

5. 5-5 Kinetic Theory of Gases Kinetic Theory Postulates: Gas particles are sizeless relative to the volume of the gas Gas particles are in constant rapid motion Gas particles have elastic collisions; means no kinetic energy is lost on impact. The absolute temperature is directly proportional to the kinetic energy of a gas. Gas particles have no attraction to each other; i.e. no inter particle froces.

6. 5-6 6 Properties of Gases Expand to completely fill their container Take the Shape of their container Low Density –much less than solid or liquid state Compressible Mixtures of gases are always homogeneous Fluid

7. 5-7 Atmospheric Pressure

8. 5-8 8 Air Pressure Constantly present when air present Decreases with altitude –less air Varies with weather conditions Measured using a barometer –Column of mercury supported by air pressure –Longer mercury column supported = higher pressure –Force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury

9. 5-9 Measuring Gas Pressure Barometer - A device to measure atmospheric pressure. Pressure is defined as force divided by area. The force is the force of gravity acting on the air molecules. Manometer - A device to measure gas pressure in a closed container.

10. 5-10 Standard Atmospheric Pressure pressure of the atmosphere is balanced by pressure exerted by mercury 760 mm at 273 K at sea level 1 atm = 760 mm Hg = 760 torr barometer Torricelli

11. 5-11 Figure 5.3 A mercury barometer.

12. 5-12 12 Figure 12.2: Mercury Barometer

13. 5-13 Pressure Measurement Open Tube Manometer gas Now what is pushing harder, the gas or the atomosphere?

14. 5-14 Pressure Measurement Open Tube Manometer gas Now what is pushing harder, the gas or the atmosphere? Neither, both the same.

15. 5-15 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? = 15 mm

16. 5-16 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? Gas in the canister If the atmospheric pressure is 766 mm, then what is the pressure of the canister? = 15 mm

17. 5-17 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? Gas in the canister If the atmospheric pressure is 766 mm, then what is the pressure of the canister? P = 766 + 15 = 781 mm (torr) = 15 mm gas

18. 5-18 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? gas

19. 5-19 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? The atmosphere What is the pressure of the gas if the atmosphere is 766 mm? = 13 mm gas

20. 5-20 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? The atmosphere What is the pressure of the gas if the atmosphere is 766 mm? 753 mm = 13 mm gas

21. 5-21 Figure 5.4 Two types of manometer

22. 5-22 Dr. Wolf’s CHM 101 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 mmHgchemistry, medicine, biology torr760 torrchemistry pounds per square inch (psi or lb/in 2 ) 14.7lb/in 2 engineering bar1.01325 barmeteorology, chemistry, physics

23. 5-23 Units for Expressing Pressure UnitValue Atmosphere1 atm Pascal (Pa)1 atm = 1.01325 x 10 5 Pa Kilopascal (kPa)1 atm = 101.325 kPa mmHg1 atm = 760 mmHg Torr1 atm = 760 torr Bar1 atm = 1.01325 bar mbar1 atm = 1013.25 mbar psi1 atm = 14.7 psi

24. 5-24 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 = 291.4mmHg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. SOLUTION: PLAN: Construct conversion factors to find the other units of pressure. 291.4mmHg1torr 1mmHg = 291.4torr 291.4torr 1atm 760torr = 0.3834atm 0.3834atm 101.325kPa 1atm = 38.85kPa

25. 5-25 Parameters Affecting Gases Pressure (P); atm, mmHg, torr, lbs/in 2 Volume (V); L, mL Temperature (T); K (only) Number of Moles (n)

26. 5-26 Dr. Wolf’s CHM 101 Boyle’s Law - The relationship between volume and the pressure of a gas. (Temperature is kept constant.)

27. 5-27 Dr. Wolf’s CHM 101 Boyle’s Law V  n and T are fixed 1 P PV = constantV = constant / P

28. 5-28 Dr. Wolf’s CHM 101 Charles’s Law - The relationship between volume and the temperature of a gas. (Pressure is kept constant.)

29. 5-29 Dr. Wolf’s CHM 101 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 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 When the amount of gas, n, is constant

30. 5-30 Dr. Wolf’s CHM 101 An experiment to study the relationship between the volume and amount of a gas. Avogadro’s Law - The volume of a gas is directly proportionate to the amount of gas. (Pressure and temperature kept constant.) V = constant x n V n = constant V  n Twice the amount gives twice the volume

31. 5-31 Dr. Wolf’s CHM 101 If the constant pressure is 1 atm and the constant temperature is 0 o C, 1 mole of any gas has a volume of 22.4 L. This is known as the Standard Molar Volume

32. 5-32 Dr. Wolf’s CHM 101 IDEAL GAS LAW Boyle’s Law = constant V P V = Charles’s Law constant X T Avogadro’s Law constant X n fixed n and Tfixed n and Pfixed P and T THE IDEAL GAS LAW PV = nRT R = PV nT = 1atm x 22.414L 1mol x 273.15K = 0.0821atm-L Mol-K

33. 5-33 Dr. Wolf’s CHM 101 Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8cm 3 at 1.12atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64atm. Assuming constant temperature, what is the new volume of air (inL)? 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 = 1.12atmP 2 = 2.64atm V 1 = 24.8cm 3 V 2 = unknown n and T are constant 24.8cm 3 1mL 1cm 3 L 10 3 mL = 0.0248L = V 1 P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P 1 V 1 = P 2 V 2 P1V1P1V1 P2P2 V 2 = =0.0248L 1.12atm 2.46atm =0.0105L 1cm 3 =1mL 10 3 mL=1L xP 1 /P 2 = R =

34. 5-34 Dr. Wolf’s CHM 101 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 1.00x10 3 torr. It is filled with helium at 23 0 C and 0.991atm 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 o C P 2 (torr) 1atm=760torr x T 2 /T 1 K= 0 C+273.15 P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P1P1 T1T1 P2P2 T2T2 = 0.991atm 1atm 760 torr = 753 torr P 2 =P1P1 T2T2 T1T1 = 753 torr 373K 296K = 949 torr

35. 5-35 Dr. Wolf’s CHM 101 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 55dm 3. When 1.10mol of He is added to the blimp, the volume is 26.2dm 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 = 1.10moln 2 = unknown V 1 = 26.2dm 3 V 2 = 55.0dm 3 P and T are constant P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = V1V1 n1n1 V2V2 n2n2 = n 2 = n 1 V2V2 V1V1 n 2 = 1.10mol 55.0dm 3 26.2dm 3 = 2.31mol 4.003g He mol He = 4.84g Hen 2 - n 1 = 2.31 -1.10 = 1.21 mol He

36. 5-36 Dr. Wolf’s CHM 101 Sample Problem 5.5Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: A steel tank has a volume of 438L and is filled with 0.885kg 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 = 438LT = 21 0 C (convert to K) n = 0.885kg (convert to mol)P = unknown 21 0 C + 273.15 = 294K 0.885kg 10 3 g kg mol O 2 32.00g O 2 = 27.7mol O 2 P = nRT V = 27.7mol 294K atm*L mol*K 0.0821x x 438L = 1.53atm

37. 5-37 Dr. Wolf’s CHM 101 Sample Problem 5.6 Calculating Gas Density PROBLEM: Calculate the density (in g/L) of carbon dioxide and the number of molecules per liter (a) at STP (0 0 C and 1 atm) and (b) at ordinary room conditions (20. 0 C and 1.00atm). 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 Avogardro’s number. d = mass/ VPV = nRT n / V = P / RT d= RT M x P d = 44.01g/molx 1atm atm*L mol*K 0.0821x 273K = 1.96g/L 1.96g L mol CO 2 44.01g CO 2 6.022x10 23 molecules mol = 2.68x10 22 molecules CO 2 /L (a) Mass = n M mult. both sides by M

38. 5-38 Dr. Wolf’s CHM 101 Sample Problem 5.6 Calculating Gas Density continued (b)= 1.83g/L d = 44.01g/molx 1atm 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

39. 5-39 Dr. Wolf’s CHM 101 Calculating the Molar Mass, M, of a Gas Since PV = nRT Then n = PV / RT And n = mass / M So m / M = PV / RT And M = mRT / PV Or M = d RT / P

40. 5-40 Dr. Wolf’s CHM 101 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 = 213mL Mass of flask + gas = 78.416g T = 100.0 0 C Mass of flask = 77.834g P = 754 torr Is the calculated molar mass consistent with the liquid being cyclohexane? m = (78.416 - 77.834)g= 0.582g M = m RT P V 0.582g atm*L mol*K 0.0821373K x 0.213Lx0.992atm =84.4g/mol M of C 6 H 12 is 84.16g/mol and the calculated value is within experimental error. x =

41. 5-41 Dr. Wolf’s CHM 101 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 Partial Pressure of a Gas in a Mixture of Gases Gases mix homogeneously. Each gas in a mixture behaves as if it is the only gas present, e.g. its pressure is calculated from PV = nRT with n equal to the number of moles of that particular gas......called the Partial Pressure. The gas pressure in a container is the sum of the partial pressures of all of the gases present.

42. 5-42 Dr. Wolf’s CHM 101 Often in gas experiments the gas is collected “over water.” So the gases in the container includes water vapor as a gas. The water vapor’s partial pressure contributes to the total pressure in the container. Collecting Gas over Water

43. 5-43 Dr. Wolf’s CHM 101 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 0.75atm 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  18 O 2 partial pressure P 18 O 2 divide by 100 multiply by P total  18 O 2 = 4.0mol% 18 O 2 100 = 0.040 = 0.030atm P =  x P total = 0.040 x 0.75atm 18 O 2

44. 5-44 Dr. Wolf’s CHM 101 Sample Problem 5.9 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 738torr and the volume is 523mL. At the temperature of the gas (23 0 C), the vapor pressure of water is 21torr. 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 717torr atm 760torr = 0.943atm

45. 5-45 Dr. Wolf’s CHM 101 Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water continued 0.943atm0.523Lx n C2H2C2H2 = atm*L mol*K 0.0821 x296K = 0.203mol 0.203mol 26.04g C 2 H 2 mol C 2 H 2 = 0.529 g C 2 H 2

46. 5-46 Dr. Wolf’s CHM 101 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 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).

47. 5-47 Dr. Wolf’s CHM 101 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 765torr and 225 0 C is needed to form 35.5g of Cu from 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 ) 35.5g Cu mol Cu 63.55g Cu 1mol H 2 1 mol Cu = 0.559mol H 2 0.559mol H 2 x498K atm*L mol*K 0.0821x 1.01atm = 22.6L

48. 5-48 Dr. Wolf’s CHM 101 Sample Problem 5.11Using 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 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g 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 = 5.25L T = 293Kn = unknown P = 0.950atm n = PV RT Cl 2 x5.25L = 0.950atm atm*L mol*K 0.0821x293K =0.207mol 17.0g 39.10g K mol K =0.435mol K 0.207mol Cl 2 2mol KCl 1mol Cl 2 0.435mol K 2mol KCl 2mol K Cl 2 is the limiting reactant. 0.414mol KCl 74.55g KCl mol KCl = 30.9 g KCl = 0.414mol KCl formed = 0.435mol KCl formed

49. 5-49 Dr. Wolf’s CHM 101 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(K k ) of the particles is constant. Postulate 1: Particle Volume Postulate 2: Particle Motion Postulate 3: Particle Collisions

50. 5-50 Dr. Wolf’s CHM 101 A molecular description of Boyle’s Law Boyle’s Law V  n and T are fixed 1 P

51. 5-51 Dr. Wolf’s CHM 101 A molecular description of Charles’s Law n and P are fixedCharles’s Law V  T

52. 5-52 Dr. Wolf’s CHM 101 A molecular description of Dalton’s law of partial pressures. Dalton’s Law of Partial Pressures P total = P 1 + P 2 + P 3 +...

53. 5-53 Dr. Wolf’s CHM 101 A molecular description of Avogadro’s Law Avogadro’s Law V  n P and T are fixed

54. 5-54 Dr. Wolf’s CHM 101 Kinetic-Molecular Theory Gas particles are in motion and have a molecular speed, . But they are moving at various speeds, some very slow, some very fast, but most near the average speed of all of the particles,   (avg). Since kinetic energy is defined as ½ mass x (speed) 2, we can define the average kinetic energy, E k(avg) = ½m   (avg) Since energy is a function of temperature, the average energy and, hence, average molecular speed will increase with temperature.

55. 5-55 Dr. Wolf’s CHM 101 The relationship between average kinetic energy and temperature is given as, E k(avg) = 3/2 (R/N A ) x T (where R is the gas constant in energy units, 8.314 J/mol-K N A is Avogadro’s number, and T temperature in K.) An increase in temperature results in an increase in average molecular kinetic energy. To have the same average kinetic energy, heavier atoms must have smaller speeds. The root-mean-square speed,   (rms), is the speed where a molecule has the average kinetic energy. The relationship between   (rms) and molar mass is:   (rms) = (3RT/ M ) ½ So the speed (or rate of movement) is: rate  ( M ) ½

56. 5-56 Dr. Wolf’s CHM 101 Relationship between molar mass and molecular speed.

57. 5-57 Dr. Wolf’s CHM 101 Graham’s Law of Effusion The rate of effusion of a gas is inversely related to the square root of its molar mass. Effusion is the process by which a gas in a closed container moves through a pin-hole into an evacuated space. This rate is proportional to the speed of a molecule so..... Rate of effusion  ( M ) ½ So doing two identical effusion experiments measuring the rates of two gases, one known, one unknown, allows the molecular mass of the unknown to be determined. = rate A rate B ( M B / M A ) ½

58. 5-58 Dr. Wolf’s CHM 101 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 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 = 2.002( 16.04/ 4.003 ) ½ =

59. 5-59 Dr. Wolf’s CHM 101 End of Chapter 5