Chapter 8 Student Notes PHW

# Chapter 8 Student Notes PHW - Chapter 8 Objectives 8.1...

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Unformatted text preview: Chapter 8 Objectives 8.1. Explain how gas molecules differ from liquid and solid molecules. molecules. 8.2. Outline the basic relationships between pressure, volume, temperature, and the number of moles observed with gases. gases. 8.3. Include a discussion of how the relationships in 8.2 above can be can applied experimentally to identify unknown gases using density and molecular mass. 8.4. Provide an overview of the Kinetic Theory of Gases and include a brief discussion of average kinetic energy and molecular speed. molecular 8.5. Differentiate between the processes of diffusion and effusion. 8.5. effusion. Introduce Graham’s Law. Graham’ Law. 8/12/2009 Zumdahl Chapter 5 1 The Person Behind the Science Evangelista Torricelli (1608-1647) Highlights – In 1641, moved to Florence to assist the astronomer Galileo – Designed first barometer Barometer P = – It was Galileo who suggested that Evangelista Torricelli use mercury in his vacuum experiments – Torricelli filled a four-foot long glass fourtube with mercury and inverted the tube into a dish Moments in a Life – Succeeded Galileo as professor of mathematics at the University of Pisa – Asteroid (7437) Torricelli named in his honor 8/12/2009 Zumdahl Chapter 5 2 1 Units of Pressure Pascal (Pa) 8/12/2009 Zumdahl Chapter 5 Boyle’s Law 1662 Charles’ Law 1787 V∝P -1 V∝T 3 P1V1 = P2V2 at a fixed T V1 / V2 = T1 / T2 at a fixed P Avogadro V∝n 1811 n = no. of moles V ∝ n T P-1 at a fixed T & P Proportionality constant found to be independent of gas identity Zumdahl Chapter 5 4 The above relationships work well up to pressures above 1 atm 8/12/2009 2 Boyle’s Experiments The product of the pressure and volume, PV, of a sample of gas is a constant at a PV, constant temperature: at fixed T and n 8/12/2009 Zumdahl Chapter 5 5 Sample Problem Using Boyle’s Law The long cylinder of a bicycle pump has a volume of 1131 cm3 and is filled with air at a pressure of 1.02 atm. The outlet valve is sealed shut, and the pump handle is pushed down until the volume of the air is 517 is 3. The temperature of the air trapped inside does not change. cm Compute the pressure inside the pump. (at fixed T & n) (1.02 atm)(1131 cm3) = P2(517 cm3) 8/12/2009 Zumdahl Chapter 5 6 3 Charles’ Law: The Effect of Temperature on Gas Volume Charles’ For all gases, Charles’ data extrapolate gases, Charles’ to −273.15 oC in the limit where V → 0 (Fundamental limit for T) T) (a) (b) 8/12/2009 Zumdahl Chapter 5 Pa = Pb na > n b 7 The Absolute (Kelvin) Temperature Scale According to Charles’ Data: Charles’ where α = (273.15 oC)-1 independent of gas identity = coefficient of thermal expansion and V0 ≡ volume when t = 0. The Kelvin (K) temperature scale: T(K) ≡ 273.15 + t(C) Gas volume is proportional to T(K): (fixed P & n) 8/12/2009 Zumdahl Chapter 5 8 4 The Ideal Gas Equation of State From the work of Boyle, Charles, and Avogadro, we know that V is proportional to nT/P and that the proportionality constant is nT/ universal, i.e., the same for all gases: universal, PV = nRT (R is the universal ideal gas constant) R = 8/12/2009 Zumdahl Chapter 5 9 Sample Problem At one point during its ascent, a weather balloon filled with helium at a volume of 1.0 × 104 L at 1.00 atm and 30oC reaches an altitude z at which the temperature is −10 oC, yet the balloon volume is unchanged. What is the pressure at altitude z? n1 = n2 PV1 P2V2 1 = n1T1 n2T2 1.00 atm 30 ° C = V1 = V2 x atm − 10 ° C ▬► 1.00 atm ▬► 303.15K = x atm 263.15K x = 0.868 atm 8/12/2009 Zumdahl Chapter 5 10 5 What is the Value for R? One mole of an ideal gas at T = 0 oC and P = 1atm is found to occupy a volume of 22.414 L. R= (1atm)(22.414L) PV = = 0.08206 L atm mol-1 K-1 (1.00 mol)(273.15 K) nT (101.325x 103 N m-2 ) (22.414 x 10-3 m3 ) R= = 8.3145 J mol-1 K-1 (1.00 mol) (273.15K) 8/12/2009 Zumdahl Chapter 5 11 Physical Properties of an Ideal Gas In an ideal gas, the molecules are far enough apart (on average) that 1. The potential energy of interaction between gas molecules is negligibly small. 2. The volume occupied by the gas molecules is a negligible fraction of the total volume. 8/12/2009 Zumdahl Chapter 5 12 6 Sample Problem What mass of hydrogen gas is needed to fill a weather balloon to a volume of 10,000. L at 1.00 atm and 30.0 °C? Strategy 1. Use PV = nRT 2. Find the number of moles 3. Convert moles to mass using moles = mass divided by molar mass 8/12/2009 Zumdahl Chapter 5 13 Sample Problem (con’t) What mass of hydrogen gas is needed to fill a weather balloon to a volume of 10,000. L at 1.00 atm and 30.0 °C? n= PV RT (1.00 atm) (1x10 4 L) (0.08206 Latmmol -1 K -1 )( 3 03.15K) n= m ⇒ m = nM M 8/12/2009 n = 402 moles m grams H2 = 402 mol x 2.015 g H2 1 mol H2 = 810 g H2 Zumdahl Chapter 5 14 7 Deviations from Ideal Gas Behavior One mole of an ideal gas at T = 0 oC and P = 1atm is found to occupy a volume of 22.414 L. 8/12/2009 Zumdahl Chapter 5 15 Gas Density and Molar Mass Rearrange PV = nRT PV = 8/12/2009 m RT M → Zumdahl Chapter 5 m P M = V RT m P =d= M V RT 16 8 Sample Problem Calculate the density of gaseous hydrogen at a pressure of 1.32 atm and a temperature of -45oC. 1.32 atm 2 g H2 x −1 −1 (0.082 L atm mol K )(233.15K) 1mol H 2 8/12/2009 = d = 0.142 g/L Zumdahl Chapter 5 17 Sample Problem Fluorocarbons are compounds of fluorine and carbon. A 45.60 g sample of a gaseous fluorocarbon contains 7.94 g of carbon and 37.66 g of fluorine and fluorine occupies 7.40 L at STP (P = 1.00 atm and T = 273.15 K). Determine the (P approximate molar mass of the fluorocarbon and give its molecular formula. molecular formula. ⎛ 1mol C ⎞ ⎜ ⎟ n C = 7.94 g C x ⎜ ⎟ = 0.661mol C ⎝ 12 g C ⎠ ⎛ 1mol F ⎞ n F = 37.66 g F x ⎜ ⎜ 19 g F ⎟ = 1.982 mol F ⎟ ⎝ ⎠ M =d } Empirical Formula is m RT where d= P V −1 −1 ⎛ 45.60g ⎞⎛ 0.082 L atm mol K x 273K ⎞ ~ ⎜ ⎟ ⎟⎜ M =⎜ ⎟ 1atm ⎝ 7.40L ⎠⎝ ⎠ 140 g/mol ~ 2 MCF3 Therefore, molecular formula is 8/12/2009 Zumdahl Chapter 5 18 9 Mixtures of Gases Dalton’s Law of Partial Pressures The pressure exerted by a mixture of ideal gases is the sum of the the pressures that each one would exert if it occupied the container alone. P1 = n RT n RT n1RT , P3 = 3 , P2 = 2 V V V Ptotal = P1 + P2 + P3 = (n1 + n 2 + n 3 ) RT V ⎛ RT ⎞ Ptotal = n Total ⎜ ⎟ ⎝ V ⎠ 8/12/2009 Zumdahl Chapter 5 19 Mole Fractions and Partial Pressures The mole fraction of a component in a mixture is defined as the number of moles of the component divided by the total number of moles present. Mole Fraction of A = X A XA = nA nA = n tot n A + n B + ... + n N For ideal gases: PA V = n A RT Ptot V = n tot RT divide equations PA V n A RT P n n = or A = A or PA = A Ptot Ptot V n tot RT Ptot n tot n tot 8/12/2009 Zumdahl Chapter 5 → 20 10 Sample Problem A solid hydrocarbon is burned in air in a closed container, producing a mixture producing of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to mixture contain 0.340 g of water vapor, 0.792 g of carbon dioxide, 0.288 g of oxygen, 3.790 g of nitrogen, and no other gases. Calculate the mole fraction and partial fraction pressure of carbon dioxide in this mixture. n tot = n H 2 O + n CO 2 + n O 2 + n N 2 n tot = X CO2 0.34 0.792 0.288 3.790 + + + = 0.1809 18 44 32 28 n CO 2 0.018 = = = 0.0995 n tot 0.1809 PCO 2 = X CO 2 Ptot = 0.0995 x 3.34 atm = 0.332 atm 8/12/2009 Zumdahl Chapter 5 21 Mixtures of Gases Dalton’s Law of Partial Pressures The pressure exerted by a mixture of ideal gases is the sum of the the pressures that each one would exert if it occupied the container alone. P1 = n RT n RT n1RT , P3 = 3 , P2 = 2 V V V Ptotal = P1 + P2 + P3 = (n1 + n 2 + n 3 ) RT V ⎛ RT ⎞ Ptotal = n Total ⎜ ⎟ ⎝ V ⎠ 8/12/2009 Zumdahl Chapter 5 22 11 Mole Fractions and Partial Pressures The mole fraction of a component in a mixture is defined as the number of moles of the component divided by the total number of moles present. Mole Fraction of A = X A XA = nA nA = n tot n A + n B + ... + n N For ideal gases: PA V = n A RT Ptot V = n tot RT divide equations PA V n A RT P n n = or A = A or PA = A Ptot Ptot V n tot RT Ptot n tot n tot 8/12/2009 → Zumdahl Chapter 5 23 Sample Problem A solid hydrocarbon is burned in air in a closed container, producing a mixture producing of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to mixture contain 0.340 g of water vapor, 0.792 g of carbon dioxide, 0.288 g of oxygen, 3.790 g of nitrogen, and no other gases. Calculate the mole fraction and partial fraction pressure of carbon dioxide in this mixture. n tot = n H 2 O + n CO 2 + n O 2 + n N 2 n tot = X CO2 0.34 0.792 0.288 3.790 + + + = 0.1809 18 44 32 28 n CO 2 0.018 = = = 0.0995 n tot 0.1809 PCO 2 = X CO 2 Ptot = 0.0995 x 3.34 atm = 0.332 atm 8/12/2009 Zumdahl Chapter 5 24 12 The Kinetic Molecular Theory of Gases The Ideal Gas Law is an empirical relationship based on experimental observations (Boyle, Charles, & Avogadro). The Kinetic Molecular Theory is a simple theoretical model that attempts to explain the behavior of gases. 8/12/2009 Zumdahl Chapter 5 25 The Kinetic Molecular Theory of Gases 1. The molecules of an ideal gas are constantly moving in random directions with a distribution of speeds. Collisions of the particles with the walls of the speeds. container are the cause of the pressure exerted by the gas. The gas molecules occupy a negligibly small fraction of the total volume. 2. The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant collisions, velocities. The gases are assumed to neither attract or repel each other. The velocities. The collisions of the molecules with each other and with the walls of the container of are elastic, i.e., no energy is lost during a collision. Essentially, the gas elastic, molecules behave like tiny billiard balls undergoing ceaseless random motion. random motion. 3. The average kinetic energy of a collection of gas particles is found to be directly proportional to the temperature of the gas in degrees Kelvin. Kelvin. See pages 351 − 357 in text for derivation of key results 8/12/2009 Zumdahl Chapter 5 26 13 Pressure ∝ (impulse per collision) x (rate of collisions with walls) impulse per collision ∝ rate of collisions with walls ∝ rate of collisions with walls ∝ => 8/12/2009 Zumdahl Chapter 5 P ∝ (m × u) × (N/V) × u 27 => But, the molecules of a gas are But, constantly moving in random directions with a distribution of speeds. speeds. u2 ≡ 8/12/2009 Zumdahl Chapter 5 Mean-square speed Meanof all molecules 28 14 Temperature and Molecular Motion As shown in the text derivation, the proportionality constant ― ― ― relating PV to Nmu2 is 1/3, i.e., PV = Nmu2/3 = nNAmu2/3 Therefore, PV/n = RT = = ideal gas law molar kinetic energy 8/12/2009 Zumdahl Chapter 5 29 The Meaning of Temperature u 2 = 3RT M The temperature of a gas in degrees K is a measure of the random motions of the gas atoms. Higher T => greater random motion 8/12/2009 Zumdahl Chapter 5 30 15 Maxwell-Boltzmann Distribution of Speeds N2 f(u) = 4π (m/2πkBT)3/2 u2 exp(−mu2/2kBT) f(u) 4π (m/2π exp(− kB ≡ R / NA = 1.38 x 10−23 J K−1 molecule−1 Temperature is a measure of the average kinetic energy of molecules when their speeds have a Maxwell-Boltzmann Maxwelldistribution. i.e., when the molecules are at thermal equilibrium. distribution. equilibrium. 8/12/2009 Zumdahl Chapter 5 31 Distributions of Molecular Speeds u2 = uavg = 3RT M 8RT π M ump : uavg : urms = 1.000 : 1.128: 1.225 8/12/2009 Zumdahl Chapter 5 32 16 Sample Problem At a certain temperature, the root-mean-square-speed of the root- mean- squaremolecules of hydrogen in a sample of gas is 1055 m s-1. Compute the root-mean square speed of molecules of oxygen at the same roottemperature. temperature. Strategy 1. Find T for the H2 gas with a urms = 1055 m s-1 2. Find urms of O2 at the same temperature 8/12/2009 Zumdahl Chapter 5 33 Strategy 1. Find T for the H2 gas with a urms = 1055 m s-1 u rm s (u T H 2 rm s = = u ) (u 2 = H 2 rm s 3 R T M = 2 3 R T M ) 2 M H 2 3 R = (1055 m s-1)2 (0.002016 kg mol-1) / 3 (8.315 J K-1 mol-1) 8/12/2009 Zumdahl Chapter 5 34 17 Strategy 2. Find urms of O2 at the same temperature At a constant temperature, urms is proportional to M−1/2. Therefore, urms for O2 = = 1055 m s-1 x urms for O2 = 264.8 m s-1 8/12/2009 Zumdahl Chapter 5 35 Gaseous Diffusion and Effusion Diffusion: Mixing of gases via their ceaseless random motions Distance diffused by NH uavg(NH3) ____________3 ~ _____ Distance diffused by HCl = MHCl ____ 1/2 (M ) NH3 = 1.5 uavg(HCl) (HCl) approximately correct Effusion: Rate of passage of a gas through a tiny orifice in a chamber Rate of effusion of gas 1 uavg(1) __________________ = ______ = Rate of effusion of gas 2 uavg(2) (M ) M2 ___ 1/2 1 Graham’s Law of Effusion Graham’ 8/12/2009 Zumdahl Chapter 5 36 18 Collisions of Gas Molecules with Container Walls See Text, pages 361-363 for details ZA ≡ no. of collisions per unit time of gas molecules with an area A of container wall 1.00 atm of O2 at T = 27 oC makes 2.72 x 1023 collisions per second per cm2 of container surface 8/12/2009 Zumdahl Chapter 5 37 Intermolecular Collisions See Text, pages 363−365 for details 363− Z ≡ no. of collisions per unit time between gas molecules Z = (d ≡ molecular diameter) Z−1 = mean time between collisions λ ≡ = mean distance traveled between collisions λ = uavg Z−1 = For O2 at T = 27 oC and P = 1 atm: atm: Z 8/12/2009x 109 collisions/s = 4.0 Z −1 Chapter 5 Zumdahl = 2.5 x 10−10 s 7 λ ~ 1 x 10−38m 19 Real Gases Ideal Gas behavior is generally observed under conditions of low pressure and high temperature. For an ideal gas, PV/nRT = 1 at all pressures. PV/nRT For real gases, attractive forces dominate at intermediate pressures pressures 8/12/2009 Zumdahl Chapter 5 39 and repulsive forces dominate at high pressures Van der Waals Equation of State Johannes van der Waals (1873) – Correction to ideal gas equation of state for attractive forces in gases (and liquids) – Correction to ideal gas equation of state to account for volume occupied by the molecules 8/12/2009 Zumdahl Chapter 5 40 20 The Person Behind the Science Johannes van der Waals (1837-1923) Highlights – 1873 first to realize the necessity of taking into account the volumes of molecules and – intermolecular forces (now generally called "van der Waals forces") in establishing the relationship between the pressure, volume and temperature of gases and liquids. Moments in a Life – 1910 awarded Nobel Prize in Physics 8/12/2009 Zumdahl Chapter 5 41 Chapter 5: Gases 5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.7. 5.8. 5.9. 5.10. 5.11. 8/12/2009 Early Experiments The Gas Laws of Boyle, Charles and Avogadro The Ideal Gas Law Gas Stoichiometry Dalton’s Laws of Partial Pressure Dalton’ The Kinetic Molecular Theory of Gases Effusion and Diffusion Collisions of Gas Particles with the Container Walls Intermolecular Collisions Real Gases Chemistry in the Atmosphere Zumdahl Chapter 5 42 21 ...
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