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Chapter 5 The Gaseous State Contents and Concepts Gas Laws We will investigate the quantitative relationships that describe the behavior of gases. 1. Gas Pressure and Its Measurement 2. Empirical Gas Laws 3. The Ideal Gas Law 4. Stoichiometry Problems Involving Gas Volumes 5. Gas Mixtures; Law of Partial Pressures Copyright © Houghton Mifflin Company. All rights reserved. 5|2 Kinetic-Molecular Theory This section will develop a model of gases as molecules in constant random motion. 6. Kinetic Theory of Gases 7. Molecular Speeds; Diffusion and Effusion 8. Real Gases Copyright © Houghton Mifflin Company. All rights reserved. 5|3 Gases are compressible. Gas volume depends on temperature and pressure. Pressure, volume, temperature, and amount of a gas are related by the ideal gas law: PV = nRT Copyright © Houghton Mifflin Company. All rights reserved. 5|4 Pressure, P The force exerted per unit area It can be given by two equations: F P A P dgh The SI unit for pressure is the pascal, Pa. kg m s 2 kg m2 m s2 Pa (pascal) Copyright © Houghton Mifflin Company. All rights reserved. kg m kg 2 m 3 m s m s2 Pa (pascal) 5|5 Pressure exerted by a column of fluid Volume: V hA Mass: m Vd (h=height, A=area) (d =density) Force: F ma Vdg hAdg (g =acceleration due to gravity) Pressure: P F / A (hAdg ) / A hdg Copyright © Houghton Mifflin Company. All rights reserved. 5|6 Copyright © Houghton Mifflin Company. All rights reserved. 5|7 A barometer is a device for measuring the pressure of the atmosphere. A manometer is a device for measuring the pressure of a gas or liquid in a vessel. Copyright © Houghton Mifflin Company. All rights reserved. 5|8 The water column would be higher because its density is less by a factor equal to the density of mercury to the density of water. P gdh gdHg hHg gdH2O hH2O hH2O hHg d Hg d H2O Copyright © Houghton Mifflin Company. All rights reserved. 5|9 Empirical Gas Laws All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties. The studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 10 Boyle’s Law The volume of a sample of gas at constant temperature varies inversely with the applied pressure. 1 The mathematical relationship: V P In equation form: PV constant PiVi PfVf Copyright © Houghton Mifflin Company. All rights reserved. 5 | 11 Figure A shows the plot of V versus P for 1.000 g O2 at 0°C. This plot is nonlinear. Copyright © Houghton Mifflin Company. All rights reserved. Figure B shows the plot of (1/V) versus P for 1.000 g O2 at 0°C. This plot is linear, illustrating the inverse relationship. 5 | 12 At one atmosphere the volume of the gas is 100 mL. When pressure is doubled, the volume is halved to 50 mL. When pressure is tripled, the volume decreases to onethird, 33 mL. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 13 When a 1.00-g sample of O2 gas at 0C is placed in a containerat a pressure of 0.50 atm, it occupies a volume of 1.40 L. When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 14 ? A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant? Vi = 38.7 mL Pi = 751 mmHg Ti = 21°C PV i i = PV f f Copyright © Houghton Mifflin Company. All rights reserved. Vf = ? Pf = 359 mmHg Tf = 21°C PV Vf i i Pf 5 | 15 Vi = 38.7 mL Pi = 751 mmHg Ti = 21°C Vf = ? Pf = 359 mmHg Tf = 21°C PiVi Vf Pf (38.7 mL)(751mmHg) Vf (359 mmHg) = 81.0 mL (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 16 A graph of V versus T is linear. Note that all lines cross zero volume at the same temperature, -273.15°C. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 17 The temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero. This is the basis of the absolute temperature scale, the Kelvin scale (K). When working with gas laws (and almost anything related to thermochemistry), the first thing you should do is convert the temperature to Kelvin Copyright © Houghton Mifflin Company. All rights reserved. 5 | 18 Charles’s Law The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K). The mathematical relationship: In equation form: V T V constant T Vi Vf Ti Tf Copyright © Houghton Mifflin Company. All rights reserved. 5 | 19 A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume. Copyright © Houghton Mifflin Company. All rights reserved. As the air inside warms, the balloon expands to its orginial size. 5 | 20 A 1.0-g sample of O2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L. When the absolute temperature of the sample is raised to 200 K, the volume of the O2 is doubled to 0.52 L. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 21 ? You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C? Vi = 79.4 mL Pi = 760 mmHg Ti = 0°C = 273 K Vi Vf = Ti Tf Copyright © Houghton Mifflin Company. All rights reserved. Vf = ? Pf = 760 mmHg Tf = 27°C = 300. K TfVi Vf Ti 5 | 22 Vi = 79.4 mL Pi = 760 mmHg Ti = 0°C = 273 K Vf = ? Pf = 760 mmHg Tf = 27°C = 300. K TfVi Vf Ti (300. K)(79.4 mL) Vf (273 K) = 87.3 mL (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 23 Combined Gas Law The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature. T The mathematical relationship: V P PV In equation form: constant T PiVi PfVf Ti Tf Copyright © Houghton Mifflin Company. All rights reserved. 5 | 24 ? Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C? Vi = 5.0 L Pi = 5.0 × 101 atm Ti = 4°C = 277 K Copyright © Houghton Mifflin Company. All rights reserved. Vf = ? Pf = 1.0 atm Tf = 11°C = 284 K 5 | 25 Vi = 5.0 L Pi = 5.0 × 101 atm Ti = 4°C = 277 K Vf = ? Pf = 1.0 atm Tf = 11°C = 284. K Tf PiVi Vf Ti Pf (284 K)(5.0 x 101 atm)(5.0L) Vf (277 K)(1.0 atm) = 2.6 x 102 L (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 26 Avogadro’s Law Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 27 Standard Temperature and Pressure (STP) The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure. The molar volume, Vm, of a gas at STP is 22.4 L/mol. The volume of the yellow box is 22.4 L. To its left is a basketball. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 28 Ideal Gas Law P1V1=P2V2 or PV=constant (at constant n and T) V1/T1=V2/T2 or V/T=constant (at constant n and P) Avogadro’s Law V1/V2=n1/n2 or V/n=constant (ant constant P and T) PV/nT=constant=R PV=nRT Copyright © Houghton Mifflin Company. All rights reserved. 5 | 29 Ideal Gas Law The ideal gas law is given by the equation PV=nRT The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 30 ? A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder? V = 50.0 L P = 17.1 atm T = 23°C = 296 K PV n RT (17.1atm )(50.0L) n L atm 0.08206 (296 K) m ol K 28.01g mass 35.20mol mol Copyright © Houghton Mifflin Company. All rights reserved. mass = 986 g (3 significant figures) 5 | 31 Gas Density and Molar Mass Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to density (grams per liter). To find molar mass, find the moles of gas, and then find the ratio of mass to moles. n P PV = nRT or V RT m nMm PMm d= = V V RT Copyright © Houghton Mifflin Company. All rights reserved. dRT or Mm P 5 | 32 ? What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm? Mm = 16.04 g/mol P = 3.50 atm T = 125°C = 398 K Mm P d RT g (16.04 )(3.50 atm) mol d L atm 0.08206 (398 K) mol K Copyright © Houghton Mifflin Company. All rights reserved. g d 1.72 L (3 significant figures) 5 | 33 ? A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of octane? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 34 d = 1.57 g/0.5000 L = 3.140 g/L P = 634 mmHg = 0.8342 atm dRT Mm P T = 95.0°C = 368.2 K Mm g L atm 3.140 0.08206 368.2K L mol K (0.8342atm) g M m 114 mol (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 35 Molar mass = 114 g/mol Empirical formula: C4H9 Empirical formula molar mass = 57 g/mol g 114 mol 2 n g 57 mol Molecular formula: C8H18 Copyright © Houghton Mifflin Company. All rights reserved. 5 | 36 Stoichiometry and Gas Volumes Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 37 ? When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20.°C? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 38 First, write the balanced chemical equation: CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) Second, calculate the moles of CO2 produced: Molar mass of CaCO3 = 100.09 g/mol 1 mol CaCO3 1 mol CO 2 1.2 x 10 g CaCO3 100.09g CaCO3 1 mol CaCO3 3 Moles of CO2 produced = 11.99 mol Copyright © Houghton Mifflin Company. All rights reserved. 5 | 39 n = 12.0 mol P = 735 mmHg = 0.967 atm T = 20°C = 293 K nRT V P L atm 12.0 mol 0.08206 (293 K) mol K V (0.967 atm) = 3.0 × 102 L (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 40 Gas Mixtures Dalton found that in a mixture of unreactive gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 41 Originally (left), flask A contains He at 152 mmHg and flask B contains O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg Copyright © Houghton Mifflin Company. All rights reserved. 5 | 42 Partial Pressure The pressure exerted by a particular gas in a mixture Dalton’s Law of Partial Pressures The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture: P = PA + PB + PC + . . . Copyright © Houghton Mifflin Company. All rights reserved. 5 | 43 ? A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 44 1 mol N2 0.0830g N2 28.01g N2 L atm 0.08206 308 K mol K 0.749 atm PN2 100.0mL 13L 10 mL 1 mol O 2 L atm 0.0194g O 2 0.08206 308 K 32.00 g O 2 mol K 0.153 atm PO2 100.0mL 13L 10 mL 1 mol CO 2 L atm 0.00640g CO 2 0.08206 308 K 44.01g CO 2 mol K 0.0368atm PCO 2 100.0mL 13L 10 mL 1 mol H2 O L atm 0.00441g H2 O 0.08206 308 K 18.01 g H O mol K 2 0.0619atm PH2O 1L 100.0mL 3 10 mL Copyright © Houghton Mifflin Company. All rights reserved. 5 | 45 PN2 0.749atm PO2 0.153atm PCO 2 0.0368atm PH2O 0.0619atm P PN2 PO2 PCO 2 PH2O P = 1.00 atm Copyright © Houghton Mifflin Company. All rights reserved. 5 | 46 ? The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air? PN2 570.0mmHg PO2 103.0Hg PCO 2 40.0 mmHg PH2O 47.0 mmHg Copyright © Houghton Mifflin Company. All rights reserved. 5 | 47 P PN2 PO2 PCO 2 PH2O 570.0 mmHg 103.0 mmHg 40.0 mmHg 47.0 mmHg P = 760.0 mmHg Copyright © Houghton Mifflin Company. All rights reserved. 5 | 48 Mole fraction of N2 570.0 mmHg 760.0 mmHg Mole fraction of O2 103.0 mmHg 760.0 mmHg Mole fraction of CO2 40.0 mmHg 760.0 mmHg Mole fraction of H2O 47.0 mmHg 760.0 mmHg Mole fraction N2 = 0.7500 Mole fraction O2 = 0.1355 Mole fraction CO2 = 0.0526 Mole fraction O2 = 0.0618 Copyright © Houghton Mifflin Company. All rights reserved. 5 | 49 Collecting Gas Over Water Gases are often collected over water. The result is a mixture of the gas and water vapor. The total pressure is equal to the sum of the gas pressure and the vapor pressure of water. The partial pressure of water depends only on temperature and is known (Table 5.6). The pressure of the gas can then be found using Dalton’s law of partial pressures Copyright © Houghton Mifflin Company. All rights reserved. 5 | 50 The reaction of Zn(s) with HCl(aq) produces hydrogen gas according to the following reaction: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 51 Copyright © Houghton Mifflin Company. All rights reserved. 5 | 52 P 769 mmHg At 19C, PH2O 16.5 mmHg (See T able5.6) P PH2 PH2O PH2 P PH2O PH2 769 mmHg 16.5 mmHg PH2 752.5mmHg PH2 753 mmHg (no decimalplaces) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 53 ? You prepare nitrogen gas by heating ammonium nitrite: NH4NO2(s) N2(g) + 2H2O(l) If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2? P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K Copyright © Houghton Mifflin Company. All rights reserved. Molar mass NH4NO2 = 64.05 g/mol 5 | 54 P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K Molar mass NH4NO2 = 64.04 g/mol nRT V P 1 mol CaCo3 1 mol CO 2 5.68 g CaCO3 64.04 g CaCO3 1 mol CaCO3 = 0.8869 mol CO2 gas Copyright © Houghton Mifflin Company. All rights reserved. 5 | 55 P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K n = 0.8869 mol nRT V P L atm 0.0887mol 0.08206 (296 K) mol K V 1 atm 706 mmHg 760 mmHg = 2.32 L of CO2 (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 56 Kinetic-Molecular Theory (Kinetic Theory) A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion. Kinetic energy is related to the mass and velocity: 1 EK mv 2 2 m = mass v = velocity Program demonstrating kinetic-molecular theory: http://people.chem.byu.edu/rbshirts/research/boltzmann_3d Copyright © Houghton Mifflin Company. All rights reserved. 5 | 57 Postulates of the Kinetic Theory 1. Gases are composed of molecules whose sizes are negligible. 2. Molecules move randomly in straight lines in all directions and at various speeds. 3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide. 4. When molecules collide with each other, the collisions are elastic. 5. The average kinetic energy of a molecule is proportional to the absolute temperature. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 58 An elastic collision is one in which no kinetic energy is lost. The collision on the left causes the ball on the right to swing the same height as the ball on the left had initially, with essentially no loss of kinetic energy. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 59 Each of the gas laws can be derived from the postulates. For the ideal gas law: P frequency of collision x average force Copyright © Houghton Mifflin Company. All rights reserved. 5 | 60 The average force depends on the mass of the molecules, m, and its average speed, u; it depends on momentum, mu. The frequency of collision is proportional to the average speed, u, and the number of molecules, N, and inversely proportional to the volume, V. 1 P u N mu V Copyright © Houghton Mifflin Company. All rights reserved. 5 | 61 Rearranging this relationship gives PV Nmu2 The average kinetic energy of a molecule of mass m and average speed u is 1/2mu2. Thus PV is proportional to the average kinetic energy of the molecule. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 62 However, the average kinetic energy is also proportional to the absolute temperature and the number of molecules, N, is proportional to moles of molecules. We now have PV nT Inserting the proportionality constant, R, gives PV nRT Copyright © Houghton Mifflin Company. All rights reserved. 5 | 63 Molecular Speeds According to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases. Root-Mean Square (rms) Molecular Speed, u A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy u Copyright © Houghton Mifflin Company. All rights reserved. 3RT Mm 5 | 64 When using the equation R = 8.3145 J/(mol K) T must be in kelvins Mm must be in kg/mol Copyright © Houghton Mifflin Company. All rights reserved. 5 | 65 ? What is the rms speed of carbon dioxide molecules in a container at 23°C? T = 23°C = 296 K CO2 molar mass = 0.04401 kg/mol Copyright © Houghton Mifflin Company. All rights reserved. u rms 3RT Mm 5 | 66 Recall kg m 2 J s2 u rms kg m 2 2 296 K 3 8.3145 s mol K kg 0.04401 mol urms 2 m 1.68x105 2 s urms Copyright © Houghton Mifflin Company. All rights reserved. m 4.10 x10 s 2 5 | 67 ? Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor? 1 Rate H2 Rate He 2.016 1 4.002 2.016 4.002 Hydrogen will diffuse more quickly by a factor of 1.4. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 68 Copyright © Houghton Mifflin Company. All rights reserved. 5 | 69 a. He will reach the end first because it has a smaller molar mass. b. Open the valves at two different times, allowing Ar more time by a factor equal to the square root of the ratio of molar masses of Ar to He, or approximately 3.16 times longer. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 70 Maxwell predicted the distributions of molecular speeds at various temperatures. The graph shows 0°C and 500°C. 3/ 2 M f (u) 4 2 RT u 2 exp(Mu 2 / 2RT ) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 71 Diffusion The process whereby a gas spreads out through another gas to occupy the space uniformly Below NH3 diffuses through air. The indicator paper tracks its progress. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 72 Effusion The process by which a gas flows through a small hole in a container. A pinprick in a balloon is one example of effusion. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 73 Graham’s Law of Effusion At constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas. rate of effusionof molecules Copyright © Houghton Mifflin Company. All rights reserved. 1 Mm 5 | 74 Real Gases At high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 75 At high pressure, some of the assumptions of the kinetic theory no longer hold true: 1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible. 2. At high pressure, the intermolecular forces (Postulate 3) are not negligible. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 76 Van der Waals Equation An equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior The term V becomes (V – nb). The term P becomes (P + n2a/V2). Values for a and b are found in Table 5.7 n 2a P V nb nRT 2 V Copyright © Houghton Mifflin Company. All rights reserved. 5 | 77 Other Resources Visit the student website at http://www.college.hmco.com/pic/ebbing9e Copyright © Houghton Mifflin Company. All rights reserved. 5 | 78