Chapter%205%20Student%20Notes%20Part%201%20PHW - Chapter 5:...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 5: Gases 5.1. Explain how gas molecules differ from liquid and solid molecules. molecules. 5.2. Outline the basic relationships between pressure, volume, temperature, and the number of moles observed with gases. gases. 5.3. Include a discussion of how the relationships in 5.2 above can be can applied experimentally to identify unknown gases using density and molecular mass. 5.4. Provide an overview of the Kinetic Theory of Gases and include a brief discussion of average kinetic energy and molecular speed. molecular 5.5. Differentiate between the processes of diffusion and effusion. effusion. Introduce Graham's Law. Graham' Law. 9/23/2008 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 It was Galileo who suggested that Evangelista Torricelli use mercury in his vacuum experiments Torricelli filled a four-foot long glass tube with mercury and inverted the tube into a dish Moments in a Life Succeeded Galileo as professor of mathematics at the University of Pisa Barometer P = Asteroid (7437) Torricelli named in his honor 9/23/2008 Zumdahl Chapter 5 2 1 Units of Pressure Pascal (Pa) 9/23/2008 Zumdahl Chapter 5 3 Boyle's Law 1662 VP VT Vn -1 P1V1 = P2V2 at a fixed T Charles' Law 1787 V1 / V2 = T1 / T2 at a fixed P at a fixed T & P Avogadro 1811 n = no. of moles V n T P-1 9/23/2008 Proportionality constant found to be independent of gas identity Zumdahl Chapter 5 4 The above relationships work well up to pressures above 1 atm 2 Boyle's Experiments The product of the pressure and volume, PV, of a sample of gas is a constant at a constant temperature: at fixed T and n 9/23/2008 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) 9/23/2008 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) Pa = Pb na > n b 9/23/2008 Zumdahl Chapter 5 7 The Absolute (Kelvin) Temperature Scale According to Charles' Data: Charles' where = (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) 9/23/2008 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 = (R is the universal ideal gas constant) 9/23/2008 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? PV1 P2V2 1 = n1T1 n2T2 1.00 atm 30 C = x atm - 10 C n1 = n2 V1 = V2 303.15K 1.00 atm = x atm 263.15K x = 0.868 atm 9/23/2008 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 R= (101.325x 103 N m-2 ) (22.414 x 10-3 m3 ) = 8.3145 J mol-1 K-1 (1.00 mol) (273.15K) 9/23/2008 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. 9/23/2008 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 9/23/2008 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 = 402 moles 2.015 g H2 1 mol H2 n= m m = nM M m grams H2 = 402 mol x = 810 g H2 Zumdahl Chapter 5 9/23/2008 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. 9/23/2008 Zumdahl Chapter 5 15 Gas Density and Molar Mass PV = nRT PV = m RT M Rearrange m P M = V RT m P =d= M V RT 9/23/2008 Zumdahl Chapter 5 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 = d = 0.142 g/L 9/23/2008 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 } Empirical Formula is M =d 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 9/23/2008 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 RT V Ptotal = P1 + P2 + P3 = (n1 + n 2 + n 3 ) RT Ptotal = n Total V 9/23/2008 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 9/23/2008 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 9/23/2008 Zumdahl Chapter 5 21 11 ...
View Full Document

This note was uploaded on 09/04/2009 for the course CHEM 1310 taught by Professor Cox during the Spring '08 term at Georgia Institute of Technology.

Ask a homework question - tutors are online