Gases - General Chemistry - Slides | CHEM 142, Study notes of Chemistry

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Chapter 5: Gases 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5 Dalton’s Law of Partial Pressures 5.6 The Kinetic molecular Theory of Gases 5.7 Effusion and Diffusion 5.8 Collisions of Gas Particles with the Container Walls 5.9 Intermolecular Collisions 5.10 Real Gases 5.11 Chemistry in the Atmosphere 1 • Helium He 4.0 • Neon Ne 20.2 • Argon Ar 39.9 • Hydrogen H2 2.0 • Nitrogen N2 28.0 • Nitrogen Monoxide NO 30.0 • Oxygen O2 32.0 • Hydrogen Chloride HCL 36.5 • Ozone O3 48.0 • Ammonia NH3 17.0 • Methane CH4 16.0 Substances that are Gases under Normal Conditions Substance Formula M (g/mol) 2 Pressure of the Atmosphere • Called “Atmospheric pressure,” or the force exerted on earth’s surface by the gases in air • The force per unit area of these gases. • Can be measured using a barometer: Pressure = Force Area 5 Figure 5.1: A torricellian barometer. Density of Mercury = 13.6 g/cm3 760 mm column of 1 cm2 area weighs 76 cm x 1 cm2 x 13.6 g/cm3 = 1030 g = 1.03 kg = 2.28 lbs P = force / area = 2.28 pounds / cm2 = 14.7 pounds / in2 = 1.00 atm. 6 Common Units of Pressure Unit Atmospheric Pressure Scientific Field Used Pascal (Pa) = N/m2; 1.01325 x 105 Pa SI unit; physics, kilopascal (kPa) 101.325 kPa Chemistry bar 1.01325 bar Meteorology, Chemistry atmosphere (atm) 1 atm Chemistry millimeters of mercury 760 mmHg Chemistry, medicine, (mmHg), also called 760 torr biology “torr” pounds per square inch 14.7 lb/in2 Engineering (psi or lb/in2) 7 Figure 5.2: A simple manometer, a device for measuring the pressure of a gas in a container. Units of pressure here are mm of Hg or mm of water … 10 Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from the decomposition of Lime stone (CaCO3) in a closed-end manometer (i.e, vacuum is the reference pressure), the height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmHg, so we use the conversion factors from Table above to find the pressure in the other units. Solution: PCO2 (torr) = 341.6 mm Hg x = 341.6 torr 1 torr 1 mm Hg converting from mmHg to torr: converting from torr to atm: PCO2( atm) = 341.6 torr x = 0.4495 atm 1 atm 760 torr converting from atm to kPa: PCO2(kPa) = 0.4495 atm x = 45.54 kPa 101.325 kPa 1 atm 11 Boyle’s Law : P - V inverse proportional when n and T are constant in a gas sample • Pressure is inversely proportional to Volume • P = or V = or PV=k • Change of Conditions Problems if n and T are constant ! • P1V1 = k P2V2 = k • Then : P1V1 = P2V2 k V k P 12 Boyle’s Law : Balloon • A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant (which obviously is not true), what is the final volume of the balloon? • P1 = 1.0 atm P2 = 0.40 atm • V1 = 0.55 L V2 = ? • V2 = 15 Boyle’s Law : Balloon • A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant (which obviously is not true), what is the final volume of the balloon? • P1 = 1.0 atm P2 = 0.40 atm • V1 = 0.55 L V2 = ? • V2 = V1 x P1/P2 = (0.55 L) x (1.0 atm / 0.40 atm) • V2 = 1.4 L 16 Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm3, what will be the volume in L if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm3 to mL and then to liters, then we use the pressure ratio to obtain the final volume! Solution: V1 (cm3) V1 (ml) V1 (L) V2 (L) 1cm3 = 1 mL 1000mL = 1L x P1/P2 P1 = 1.23 atm P2 = 3.16 atm V1 = 15.8 cm3 V2 = unknown T and n remain constant V1 = 15.8 cm3 x x = L V2 = V1 x = P1 P2 17 Calculation Continued P2 =1 atm + 6 x 103 ft x x x P2 = 1 atm + _____________ mm Hg But 760 mm Hg = 1.00 atm, so V2 = = = V1 x P1 P2 P2 = 1 atm + 20 Calculation Continued P2 =1 atm + 6 x 103 ft x x x 0.3048 m 1 ft 1000 mm 1 m 1.026 g sH2O/cm3 13.6 g Hg/cm3 P2 = 1 atm + 138,000 mm Hg But 760 mm Hg = 1.00 atm, so P2 = 1 atm + 182 atm = 183 atm! V2 = = = 183 cm3 = 0.183 liters V1 x P1 P2 1.00 cm3 x 183 atm 1.00 atm P2 = 1 atm + (138,000 mm Hg) x (1.00 atm / 760 mm Hg) 21 Figure 5.7: Plots of V versus T (c) for several gases. 22 Charles Law Problem • A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. • V1 = 3.20 L T1 = 125oC = 398 K • V2 = 1.54 L T2 = ? • V1 / T1 = V2 / T2 So T2 = 25 Charles Law Problem • A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. • V1 = 3.20 L T1 = 125oC = 398 K • V2 = 1.54 L T2 = ? • T2 = T1 x ( V2 / V1) T2 = 398 K x = 192 K • T2 = 192 K oC = K - 273.15 = 192 - 273 oC = -81oC 1.54 L 3.20 L 26 Charles Law Problem - I • A balloon in Antarctica in a building is at room temperature ( 75o F ) and has a volume of 20.0 L . What will be its volume outside where the temperature is -70oF ? • V1 = 20.0 L V2 = ? • T1 = 75o F T2 = -70o F • o C = ( o F - 32 ) 5/9 • T1 = • T2 = 27 Antarctic Balloon Problem - II • V1 / T1 = V2 / T2 V2 = V1 x ( T2 / T1 ) • V2 = 20.0 L x • V2 = 14.6 L • The Balloon shrinks from 20 L to 15 L !!!!!!! • Just by going outside !!!!! 216.4 K 297.0 K 30 31 Martin S. Silberberg, Chemistry: The Molecular Nature of Mattar and Change. 24 Edition. Copyright © The McGraw-Hill Companies, Inc. All rights reserved. Breathing and the Gas Laws Rib cage expands Diaphragm contracts (moves down) Lungs fill with air Avogadro’s Law - Moles and Volume The volume of gas is directly proportional to the amount of gas (in moles), when measured at the same P and T: V α n or V = n x constant n1 = initial moles of gas V1 = initial volume of gas n2 = final moles of gas V2 = final volume of gas n1 V1 n2 V2 = or: n1 = n2 x V1 V2 32 Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles(2.298g H2) of Hydrogen and has a volume of 28.75 L. What mass of Hydrogen must be added to the balloon to increase it’s volume to 112.46 Liters. Assume T and P are constant. Plan: Volume and amount of gas are changing with T and P constant, so we will use Avogadro’s law, and the change of conditions form. Solution: n1 = 1.14 moles H2 n2 = 1.14 moles + ? moles V1 = 28.75 L V2 = 112.46 L T = constant P = constant n1 n2 V1 V2= n2 = n1 x = V2 V1 35 Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles(2.298g H2) of Hydrogen and has a volume of 28.75 L. What mass of Hydrogen must be added to the balloon to increase it’s volume to 112.46 Liters. Assume T and P are constant. Plan: Volume and amount of gas are changing with T and P constant, so we will use Avogadro’s law, and the change of conditions form. Solution: n1 = 1.14 moles H2 n2 = 1.14 moles + ? moles V1 = 28.75 L V2 = 112.46 L T = constant P = constant n1 n2 V1 V2= n2 = n1 x = 1.14 moles H2 x V2 V1 n2 = 4.4593 moles = 4.46 moles 112.46 L 28.75 L mass = moles x molecular mass mass = 4.46 moles x 2.016 g / mol mass = 8.99 g H2 gasadded mass = 8.99g - 2.30g = 6.69g 36 IDEAL GAS LAW • The ideal gas equation combines both Boyles Law and Charles Law into one easy-to-remember law: PV=nRT n = number of moles of gas in volume V • R = Ideal gas constant • R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1 • Later R = 8.314 J / (mol K) = 8.314 J mol-1 K-1 • An ideal gas is one for which both the volume of molecules and forces between the molecules are so small that they have insignificant effect on its P-V-T behavior. Independent of substance, in the limit that n/V →0, all gases behave ideally. Usually true below 2 atm. 37 Table 5.2 (P 150) Molar Volumes for Various Gases at 00 and 1 atm Gas Molar Volume (L) Oxygen (O2) 22.397 Nitrogen (N2) 22.402 Hydrogen (H2) 22.433 Helium (He) 22.434 Argon (Ar) 22.397 Carbon dioxide (CO2) 22.260 Ammonia (NH3) 22.079 40 4 Martin &. Silberberg, Chemistry: The Molecular Nature of Matter and Change. 2°4 Edition. Copyright © The McGraw-Hill Companies, Inc. All righte reserved. Standard Molar Volume n=1 mol P=1 atm (760 torr) n=1 mol P=1 atm (760 torr) T = 0°C (273 K) T = 0°C (273 K) T =0°C (273 K) V=22.4L V=22.4L V=22.4L Number of gas particles} Number of gas particles|Number of gas particles _= 6.022 x 10 = 6.022 x 107° = 6.022 x 10” n=1 mol P=1 atm (760 torr) Gas Law: Solving for Pressure Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC. Plan: Convert all information into the units required, and substitute into the Ideal Gas Equation ( PV=nRT ). Solution: nXe = = 38.37014471 mol Xe 5038 g Xe 131.3 g Xe / mol T = 18.8 oC + 273.15 K = 291.95 K 42 Ideal Gas Calculation - Nitrogen • Calculate the pressure in a container holding 375 g of Nitrogen gas. The volume of the container is 0.150 m3 and the temperature is 36.0 oC. • n = 375 g N2/ 28.0 g N2 / mol = 13.4 mol N2 • V = 0.150 m3 x 1000 L / m3 = 150 L • T = 36.0 oC + 273.15 = 309.2 K • PV=nRT so P= nRT/V • P = • • P = 2.27 atm ( 13.4 mol) ( 0.08206 L atm/mol K) ( 309.2 K) 150 L 45 Mass of Air in a Hot Air Balloon - Part I • Calculate the mass of air in a spherical hot air balloon that has a volume of 14,100 cubic feet when the temperature of the gas is 86 oF and the pressure is 748 mm Hg? (Avg. molar mass of gases in air is ~29 g/mol) • P = 748 mm Hg x 1atm / 760 mm Hg= 0.984 atm • V = 1.41 x 104 ft3x • T = 86 oF (convert to K) • n = PV/RT Then convert to grams (29 g/mol). 46 Mass of Air in a Hot Air Balloon - Part I • Calculate the mass of air in a spherical hot air balloon that has a volume of 14,100 cubic feet when the temperature of the gas is 86 oF and the pressure is 748 mm Hg? • P = 748 mm Hg x 1atm / 760 mm Hg= 0.984 atm • V = 1.41 x 104 ft3x (12 in/1 ft)3x(2.54 cm/1 in)3 x x (1ml/1 cm3) x ( 1L / 1000 mL) =3.99 x 105 L • T = oC =(86-32)5/9 = 30 oC 30 oC + 273 = 303 K 47 Ammonia Density Problem – by another method • Calculate the density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC. Density =d = mass per unit volume = m / L • P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm • T = 55 oC + 273 = 328 K n = mass / Molar mass = m / M PV = nRT so m/(MV) = n/V = P/(RT) so d = m/V = PM/(RT) • d = = • d = 0.626 g / L (more than 1000x less dense than water!) P x M R x T ( 0.989 atm) ( 17.03 g/mol) ( 0.08206 L atm/mol K) ( 328 K) 50 Applying the Temperature - Pressure Relationship (constant n and V) Problem: A copper tank is filled with compressed gas to a pressure of 4.28 atm at a temperature of 0.185 oF. What will be the pressure if the temperature is raised to 95.6 oC? Plan: The volume of the tank is not changed, and we only have to deal with the temperature change, so convert to SI units, and calculate the pressure ratio from the T ratio. Solution: T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC T1 = -17.68 oC + 273.15 K = 255.47 K T2 = 95.6 oC + 273.15 K = 368.8 K P1 P2 T1 T2 = = nR/V P2 = P1 x = ? T2 T1 P2 = 4.28 atm x = 6.18 atm 368.8 K 255.47 K 51 Change of Conditions :Problem -I • A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10 mL, what is the pressure? • P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm • P2 = ? • V1 = 45.9 L V2 = 3.10 ml = 0.00310 L • T1 = 25 oC + 273 = 298 K • T2 = 155 oC + 273 = 428 K 52 Change of 2 Conditions Problem II: continued • P1 x V1 P2 x V2 V2 = • V2 = = • V2 = 3,117.23 m3 = 3,120 m3 or 31 times the volume !!! T1 T2= P1V1T2 T1P2 ( 1.0 atm) ( 100 m3) (183 K) (298 K) (0.0197 atm) 55 Example Problem: Molar Mass of a Gas from its weight and P,V,T Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 mL flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan: Use the Ideal gas law to calculate n, then calculate the molar mass. Solution: P = 550.0 Torr x x = 0.724 atm1mm Hg 1 Torr 1.00 atm 760 mm Hg V = 250.0 ml x = 0.250 L1.00 L 1000 mL T = 25.0 oC + 273.15 K = 298.2 K n = P V R T n = M = mass / n = 56 Example Problem: Molar Mass of a Gas from its weight and P,V,T Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan: Use the Ideal gas law to calculate n, then calculate the molar mass. Solution: P = 550.0 Torr x x = 0.724 atm1mm Hg 1 Torr 1.00 atm 760 mm Hg V = 250.0 ml x = 0.250 L1.00 L 1000 ml T = 25.0 oC + 273.15 K = 298.2 K n = P V R T n = = 0.007393 mol (0.0821 L atm/mol K) (298.2 K) (0.724 atm)(0.250 L) M = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol 57 Chemical Equation Calc - III Reactants Products Molecules Moles Mass Molecular Weight g/mol Atoms (Molecules) Avogadro’s Number 6.02 x 1023 Solutions Molarity moles / liter Gases PV = nRT 60 Sodium Azide Decomposition-I • Sodium Azide (NaN3) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 . • 2 NaN3 (s) 2 Na (s) + 3 N2 (g) • mol NaN3 = • mol N2= mol NaN3 x 3 mol N2/2 mol NaN3 • V = nRT/P 61 Sodium Azide Decomposition-I • Sodium Azide (NaN3) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 . • 2 NaN3 (s) 2 Na (s) + 3 N2 (g) • mol NaN3 = 60.0 g NaN3 / (65.02 g NaN3 / mol NaN3) = = 0.9228 mol NaN3 • mol N2= 0.9228 mol NaN3 x (3 mol N2/2 mol NaN3) = 1.38 mol N2 62 Like Example 5.6 (P 151) Problem: Calculate the volume of carbon dioxide produced by the thermal decomposition of 36.8 g of calcium carbonate at 715mm Hg, and 370 C according to the reaction below: CaCO3(s) CaO(s) + CO2 (g) Solution: 36.8g x = 0.368 mol CaCO3 1 mol CaCO3 100.1g CaCO3 Moles of CO2 = mol CaCO3 x (1 mol CaCO3 / 1 mol CO2) V = =nRTP (0.368 mol CO2)(0.08206 L atm/mol K)(310.15K) (715mmHg/(760mm Hg/atm) V = 9.96 L 65 Figure 5.9: Partial pressure of each gas in a mixture of gases depends on the number of moles of that gas (ni). Defined as Pi = (ni/ntotal) Ptotal 66 Ideal Gas Mixtures • Ideal gas equation applies to the mixture as a whole and to each gas species (i) individually: PtotalV = ntotalRT Ptotal is what is measured by pressure guages. Σ ni = n1+n2+n3…= ntotal PiV = niRT for all species i. Pi harder to measure. Note: Σ Pi = P1+P2+P3…= Ptotal i i 67 Partial Pressures Problem - #1 cont. • nHe = 0.10 g He / 4.003 g He / mol He • = 0.025 mol He • PHe = nHeRT/V • PHe = • PHe = 0.30 atm • PTotal = PCO2 + PHe = 0.812 atm + 0.30 atm • PTotal = 1.11 atm (0.025 mol) ( 0.08206 L atm / mol K) ( 290 K ) ( 2.00 L ) 70 Mole fraction of species i: xi = ni/ntotal • PiV = niRT for all i, and Σ Pi = P1+P2+P3…= Ptotal • Σ ni = n1+n2+n3…= ntotal • Σ xi = x1+x2+x3…= 1.00000 • Pi = niRT/V = (ni/ntotal)ntotalRT/V = xi . Ptotal 71 Example Problem using mole fractions • A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ? • Total # moles = ntotal = • xNe = nNe / ntotal = • PNe = xNe PTotal = • xAr = • PAr = • xXe = • PXe = 72 Figure 5.10: Production of O2(g) by thermal decomposition of KCIO3: It is mixed with vapor pressure of water. 75 Collection of Hydrogen gas over Water - Vapor pressure - I • 2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g) • Calculate the mass of Hydrogen gas collected over water if 156 mL of gas is collected at 20oC and 769 mm Hg. • PTotal = PH2 + PH2O PH2 = PTotal - PH2O FROM TABLE PH2 = • T = 20oC + 273 = 293 K • PH2 = • V = 0.156 L • nH2 = PH2V/RT = • Convert to mass 76 Collection of Hydrogen gas over Water - Vapor pressure - I • 2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g) • Calculate the mass of Hydrogen gas collected over water if 156 mL of gas is collected at 20oC and 769 mm Hg. • PTotal = PH2 + PH2O PH2 = PTotal - PH2O FROM TABLE PH2 = 769 mm Hg - 17.5 mm Hg = 752 mm Hg • T = 20oC + 273 = 293 K • PH2 = 752 mm Hg /760 mm Hg /1 atm = 0.989 atm • V = 0.156 L 77 Gas Law Stoichiometry - I - NH3 + HCl Problem: A slide separating two containers is removed, and the gases are allowed to mix and react. The first container with a volume of 2.79 L contains Ammonia gas at a pressure of 0.776 atm and temperature = 18.7 oC. The second with a volume of 1.16 L contains HCl gas at a pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid ammonium chloride will be formed, and what will be remaining in the container, and what is the pressure? Plan: This is a limiting reactant problem, so we must calculate the moles of each reactant using the gas law to determine the limiting reagent. Then we can calculate the mass of product, and determine what is left in the combined volume of the container, and the conditions. Solution: Equation: NH3 (g) + HCl(g) NH4Cl(s) TNH3 = 18.7 oC + 273.15 = 291.9 K NH3 HCl 80 Gas Law Stoichiometry - II - NH3 + HCl ni = PiV RT nNH3 = = 0.0903 mol NH3 ÷ 1 (0.776 atm) (2.79 L) (0.0821 L atm/mol K) (291.9 K) nHCl = = 0.0451 mol HCl ÷ 1 (0.932 atm) (1.16 L) (0.0821 L atm/mol K) (291.9 K) limiting reactant Therefore the product will be 0.0451 mol NH4Cl or 2.28 g NH4Cl Ammonia remaining = 0.0903 mol - 0.0451 mol = 0.0452 mol NH3 V = 1.16 L + 2.79 L = 3.95 L PNH3= = = 0.274 atm nNH3RT V (0.0452 mol) (0.0821 L atm/mol K) (291.9 K) (3.95 L) 81 Postulates of Kinetic Molecular Theory The kinetic energy associated with the molecules’ velocity: KE = (1/2) mass x u2 The temperature of matter is a measure of its thermal energy, such that T is proportional to the average KE. so that their velocity (u) is proportional to √T. Gas molecules make collisions with the walls, and on average simply reverse their momentum in the direction perpendicular to the wall. This gives rise to the pressure (force per unit area) on the wall. 82 Figure 5.13: (a) In collision: exact reversal of the x component of the velocity, so that momentum change on impact is 2 x mass x ux. (b) The # impacts on the shaded walls per unit time can be calculated from the velocity and the number density. Proportional to both. (c) Force on wall = the average momentum change per unit time = (momentum change per impact) x (# impacts per unit time) 85 Velocity and Energy • Kinetic Energy = KE = ½ mu2 for one molecule • Average Kinetic Energy of a mole of gas (KEavg) = 3/2 RT independent of gas identity (use R in J/mol K) • Average Kinetic Energy of one molecule (KEavg) = 3/2 RT / NA independent of mass, identity = ½ mu2 (mean value of u2) ½ mu2 = 3/2 RT/NA u2 = 3RT/(mNA) = 3RT/M since m = M/NA √(u2) = √(3RT/M) “Root mean square velocity” Must use R in J/mol K, and remember that 1 J = 1 kg m2/s2 86 Figure 5.16: A plot of the relative number of N2 molecules that have a given velocity at three temperatures. Also: At same T, heavier gases go slower, as 1/√M: Average u2/u1 = (M1/M2)1/2 Kinetic energy per mole = ½ Mu2 = 3/2 RT, independent of M HEAT! 87 Molecular Mass and Molecular Speeds - II For methane , CH4: M = 16.04 g/mole = 16.04 x 10 - 3 kg/mole √(u2) = √(3RT/M) R = 8.314 J/mol K = 8.314 kg m2/(s2 mol K) √(u2) = 90 Molecular Mass and Molecular Speeds - II For methane , CH4: M = 16.04 g/mole = 16.04 x 10 - 3 kg/mole √(u2) = √(3RT/M) R = 8.314 J/mol K = 8.314 kg m2/(s2 mol K) √(u2) = √ [(3x 8.314 kg m2/(s2 mol K) x 300 K)/ 16.04 x 10 - 3 kg/mole] Note: 1 J = 1 kg m2/s2 √(u2) = 6.838 x 102 m/s = 683.8 m/s Root mean square velocity 91 Molecular Mass and Molecular Speeds - III for Carbon dioxide CO2 = 44.01 g/mole u = 4.124 x 102 (J/kg)1/2 = 412.4 m/s 6.213 x 10 - 21 J/molecule = 3.654 x 10 - 26 kg/molecule (u2) m = = 7.308 x 10 - 26 kg/molecule44.01 x 10 - 3 kg/mole 6.022 x 1023 molecules/mole EK = 6.213 x 10 - 21 J/molecule = 1/2mu2 EK = x T = 1.5 x x 300K = 3 R 2 NA 8.314 J/mol K 6.022 x 1023 molecules/mole 92 Figure 5.19: (a) demonstration of the relative diffusion rates of NH3 and HCL molecules through air. 95 Diffusion Rates: proportional to average velocity or to 1/√M • Rate1/Rate2 = u1/u2 = (M2/M1)1/2 • HCl = 36.46 g/mol NH3 = 17.03 g/mol • RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2 • RateNH3 = RateHCl x 1.463 96 Figure 5.18: Effusion rate of a gas into an evacuated chamber = collision rate with imaginary surface. 97