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 Cengage Learning. 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 Cengage Learning. All rights reserved. 5 3

Learning Objectives 1. Gas Pressure and Its Measurement a. Define pressure and its units. b. Convert units of pressure. 2. Empirical Gas Laws a. Express Boyle s law in words and as an equation. b. Use Boyle s law. c. Express Charles s law in words and as an equation. Copyright Cengage Learning. All rights reserved. 5 4

d. Use Charles s law. e. Express the combined gas law as an equation. f. State Avogadro s law. g. Define standard temperature and pressure (STP). 3. The Ideal Gas Law a. State what makes a gas an ideal gas. b. Learn the ideal gas law equation. c. Derive the empirical gas laws from the ideal gas law. Copyright Cengage Learning. All rights reserved. 5 5

d. Use the ideal gas law. e. Calculate gas density. f. Determine the molecular mass of a vapor. g. Use an equation to calculate gas density. 4. Stoichiometry Problems Involving Gas Volumes a. Solving stoichiometry problems involving gas volumes. Copyright Cengage Learning. All rights reserved. 5 6

5. Gas Mixtures; Law of Partial Pressures a. Learn the equation for Dalton s law of partial pressures. b. Define the mole fraction of a gas. c. Calculate the partial pressure and the mole fraction of a gas in a mixture. d. Describe how gases are collected over water and how to determine the vapor pressure of water. e. Calculate the amount of gas collected over water. Copyright Cengage Learning. All rights reserved. 5 7

6. Kinetic Theory of An Ideal Gas a. List the five postulates of the kinetic theory. b. Provide a qualitative description of the gas laws based on the kinetic theory. Copyright Cengage Learning. All rights reserved. 5 8

7. Molecular Speeds; Diffusion and Effusion a. Describe how the root-mean square (rms) molecular speed of gas molecules varies with temperature. b. Describe the molecular-speed distribution of gas molecules of different temperatures. c. Calculate the rms speed of a molecule. d. Define effusion and diffusion. e. Describe how individual gas molecules move while undergoing diffusion. f. Calculate the ratio of effusion rates of gases. Copyright Cengage Learning. All rights reserved. 5 9

Pressure, P The force exerted per unit area. It can be given by two equations: F P = P = dgh A The SI unit for pressure is the pascal, Pa. kg m 2 s kg = 2 2 m m s = Pa (pascal) kg m 3 m kg m = 2 2 s m s = Pa (pascal) Copyright Cengage Learning. All rights reserved. 5 12

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 Cengage Learning. All rights reserved. 5 14

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 gd h h = gd h d Hg Hg H2O H2O H O = h Hg 2 d Hg H 2 O Copyright Cengage Learning. All rights reserved. 5 15

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 Cengage Learning. All rights reserved. 5 16

Boyle s Law The volume of a sample of gas at constant temperature varies inversely with the applied pressure. The mathematical relationship: V 1 P In equation form: PV = constant PV i i = P V f f Copyright Cengage Learning. All rights reserved. 5 17

Figure A shows the plot of V versus P for 1.000 g O 2 at 0 C. This plot is nonlinear. Figure B shows the plot of ( 1 / V ) versus P for 1.000 g O 2 at 0 C. This plot is linear, illustrating the inverse relationship. Copyright Cengage Learning. All rights reserved. 5 18

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 Cengage Learning. All rights reserved. 5 19

When a 1.00-g sample of O 2 gas at 0 C is placed in a container at a pressure of 0.50 atm, it occupies a volume of 1.40 L. When the pressure on the O 2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume. Copyright Cengage Learning. All rights reserved. 5 20

? 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? V i = 38.7 ml P i = 751 mmhg T i = 21 C V f =? P f = 359 mmhg T f = 21 C V = f PV i P f i Copyright Cengage Learning. All rights reserved. 5 21

V i = 38.7 ml P i = 751 mmhg T i = 21 C V f =? P f = 359 mmhg T f = 21 C V = f PV i P f i V f = (38.7 ml)(751 mmhg) (359 mmhg) = 81.0 ml (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 22

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). Copyright Cengage Learning. All rights reserved. 5 24

Charles s Law The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K). The mathematical relationship: V T In equation form: V T V T i i = = constant V T f f Copyright Cengage Learning. All rights reserved. 5 25

A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume. As the air inside warms, the balloon expands to its orginial size. Copyright Cengage Learning. All rights reserved. 5 26

A 1.0-g sample of O 2 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 O 2 is doubled to 0.52 L. Copyright Cengage Learning. All rights reserved. 5 27

? You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO 3. According to your calculations, you should obtain 79.4 ml of CO 2 at 0 C and 760 mmhg. How many milliliters of gas would you obtain at 27 C? V i = 79.4 ml P i = 760 mmhg T i = 0 C = 273 K V f =? P f = 760 mmhg T f = 27 C = 300. K V = f T V f T i i Copyright Cengage Learning. All rights reserved. 5 28

V i = 79.4 ml P i = 760 mmhg T i = 0 C = 273 K V f = V = f (300. V f =? P f = 760 mmhg T f = 27 C = 300. K T V f T K)(79.4 ml) (273 K) i i = 87.3 ml (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 29

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. The mathematical relationship: In equation form: PV T PV i T i i = = constant P V f T f f V T P Copyright Cengage Learning. All rights reserved. 5 30

? Divers working from a North Sea drilling platform experience pressure of 5.0 10 1 atm at a depth of 5.0 10 2 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? V i = 5.0 L P i = 5.0 10 1 atm T i = 4 C = 277 K V f =? P f = 1.0 atm T f = 11 C = 284 K Copyright Cengage Learning. All rights reserved. 5 31

V i = 5.0 L P i = 5.0 10 1 atm T i = 4 C = 277 K V f =? P f = 1.0 atm T f = 11 C = 284 K V f = V = (284 K)(5.0 f T f PV i i T P (277 K)(1.0 atm) f x 10 i 1 atm)(5.0 L) = 2.6 x 10 2 L (2 significant figures) Copyright Cengage Learning. All rights reserved. 5 32

a. Decreasing the temperature at a constant pressure results in a decrease in volume. Subsequently increasing the volume at a constant temperature results in a decrease in pressure. b. Increasing the temperature at a constant pressure results in an increase in volume. Subsequently decreasing the volume at a constant temperature results in an increase in pressure. Copyright Cengage Learning. All rights reserved. 5 34

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, V m, 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 Cengage Learning. All rights reserved. 5 36

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 Cengage Learning. All rights reserved. 5 37

? You put varying amounts of a gas into a given container at a given temperature. Use the ideal gas law to show that the amount (moles) of gas is proportional to the pressure at constant temperature and volume. PV = nrt V n = P RT V where RT is constant. Copyright Cengage Learning. All rights reserved. 5 38

? A 50.0-L cylinder of nitrogen, N 2, 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 n = (17.1 atm)(50.0 L) n = L atm 0.08206 (296 K) mol K mass = 35.20 mol 28.02 mol g PV RT mass = 986 g (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 39

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 grams (per liter). To find molar mass, find the moles of gas, and then find the ratio of mass to moles. In equation form: PMm d = or Mm = RT drt P Copyright Cengage Learning. All rights reserved. 5 40

? What is the density of methane gas (natural gas), CH 4, at 125 C and 3.50 atm? M m = 16.04 g/mol P = 3.50 atm T = 125 C = 398 K d = M m RT P d = g (16.04 )(3.50 atm) mol L atm 0.08206 (398 K) mol K g d = 1.72 L (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 41

? 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 C 4 H 9.) What is the molecular formula of octane? Copyright Cengage Learning. All rights reserved. 5 42

d = 1.57 g/0.5000 L = 3.140 g/l P = 634 mmhg = 0.8342 atm T = 95.0 C = 368 K M = m drt P M m = 3.140 g L L atm 0.08206 mol K (0.8342 atm) g M m = 114 mol (3 significant figures) ( 368 K) Copyright Cengage Learning. All rights reserved. 5 43

Assume the flasks are closed. a. All flasks contain the same number of atoms. b. The gas with the highest molar mass, Xe, has the greatest density. c. The flask at the highest temperature (the one containing He) has the highest pressure. d. The number of atoms is unchanged. Copyright Cengage Learning. All rights reserved. 5 46

? When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO 3 was required to neutralize the spill. What volume of CO 2 was released by the neutralization at 735 mmhg and 20. C? Copyright Cengage Learning. All rights reserved. 5 48

First, write the balanced chemical equation: CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) Second, calculate the moles of CO 2 produced: Molar mass of CaCO 3 = 100.09 g/mol 1.2 x 10 3 g CaCO 3 1mol 100.09 CaCO 3 g CaCO 3 1mol 1mol CO 2 CaCO 3 Moles of CO 2 produced = 11.99 mol Copyright Cengage Learning. All rights reserved. 5 49

n = 11.99 mol P = 735 mmhg = 0.967 atm T = 20 C = 293 K V = nrt P V = L atm 11.99 0.08206 (293 K) mol K (0.967 atm) ( mol) = 2.98 10 2 L (2 significant figures) Copyright Cengage Learning. All rights reserved. 5 50

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 Cengage Learning. All rights reserved. 5 51

Originally (left), flask A contains He at 152 mmhg and flask B contains O 2 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 Cengage Learning. All rights reserved. 5 52

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 = P A + P B + P C +... Copyright Cengage Learning. All rights reserved. 5 53

? A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N 2, 0.0194 g O 2, 0.00640 g CO 2, 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 Cengage Learning. All rights reserved. 5 54

P N2 P O2 P CO2 P H2 O 1mol N2 L atm 0.0830 g N2 0.08206 ( 308 K) 28.01 g N2 mol K = = 0.749 atm 1L ( ) 100.0 ml 3 10 ml 1mol O2 L atm 0.0194 g O2 0.08206 ( 308 K) 32.00 g O2 mol K = = 0.153 atm 1L ( ) 100.0 ml 3 10 ml 1mol CO2 L atm 0.00640 g CO2 0.08206 ( 308 K) 44.01 g CO2 mol K = = 0.0368 atm 1L ( ) 100.0 ml 3 10 ml 1mol H2O L atm 0.00441 g H2O 0.08206 ( 308 K) 18.01 g H2O mol K = = 0.0619 atm 1L ( ) 100.0 ml 3 10 ml Copyright Cengage Learning. All rights reserved. 5 55

? 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? P N = 570.0 mmhg 2 P O = 103.0 mmhg 2 P CO = 40.0 mmhg 2 P H2 = O 47.0 mmhg Copyright Cengage Learning. All rights reserved. 5 57

Mole fraction of N 2 Mole fraction of O 2 570.0 mmhg 103.0 mmhg = = 760.0 mmhg 760.0 mmhg Mole fraction of CO 2 Mole fraction of H 2 O 40.0 mmhg 47.0 mmhg = = 760.0 mmhg 760.0 mmhg Mole fraction N 2 = 0.7500 Mole fraction O 2 = 0.1355 Mole fraction CO 2 = 0.0526 Mole fraction O 2 = 0.0618 Copyright Cengage Learning. All rights reserved. 5 59

a. Nothing happens to the pressure of H 2. b. The pressures are equal because the moles are equal. c. The total pressure is the sum of the pressures of the two gases. Because the pressures are equal, the total pressure is double the individual pressures. Copyright Cengage Learning. All rights reserved. 5 60

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 Cengage Learning. All rights reserved. 5 61

The reaction of Zn(s) with HCl(aq) produces hydrogen gas according to the following reaction: Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor. Copyright Cengage Learning. All rights reserved. 5 62

P = 769 mmhg At 19 C, P (See Table P P P H H = 2 2 P = = P H = 2 H 2 P + H 2 O = P P 5.6) H H 2 2 O O 16.5 mmhg 769 mmhg 16.5 mmhg 752.5 mmhg P H = 753 mmhg 2 (no decimal places) Copyright Cengage Learning. All rights reserved. 5 64

? You prepare nitrogen gas by heating ammonium nitrite: NH 4 NO 2 (s) N 2 (g) + 2H 2 O(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 NH 4 NO 2? P = 727 mmhg P vapor = 21.1 mmhg P gas = 706 mmhg T = 23 C = 296 K Molar mass NH 4 NO 2 = 64.06 g/mol Copyright Cengage Learning. All rights reserved. 5 65

P = 727 mmhg P vapor = 21.1 mmhg P gas = 706 mmhg T = 23 C = 296 K Molar mass NH 4 NO 2 = 64.06 g/mol V = nrt P 5.68 g NH4NO2 1mol NH4NO 2 64.04 g NH4NO2 1mol N 1mol 2 NH4NO2 = 0.886 mol N 2 gas Copyright Cengage Learning. All rights reserved. 5 66

P = 727 mmhg P vapor = 21.1 mmhg P gas = 706 mmhg T = 23 C = 296 K n = 0.0887 mol V = ( ) 0.0887 mol 706 mmhg V = nrt P L atm 0.08206 (296 K) mol K 1atm 760 mmhg = 2.32 L of N 2 (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 67

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 E 2 K = mv 2 m = mass v = velocity Copyright Cengage Learning. All rights reserved. 5 68

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 Cengage Learning. All rights reserved. 5 69

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 Cengage Learning. All rights reserved. 5 70

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. P u 1 N V mu Copyright Cengage Learning. All rights reserved. 5 72

Rearranging this relationship gives 2 PV Nmu The average kinetic energy of a molecule of mass m and average speed u is 1 / 2 mu 2. Thus PV is proportional to the average kinetic energy of the molecule. Copyright Cengage Learning. All rights reserved. 5 73

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 Cengage Learning. All rights reserved. 5 74

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 = 3RT M m Copyright Cengage Learning. All rights reserved. 5 75

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 Cengage Learning. All rights reserved. 5 80

Diffusion The process whereby a gas spreads out through another gas to occupy the space uniformly Below NH 3 diffuses through air. The indicator paper tracks its progress. Copyright Cengage Learning. All rights reserved. 5 82

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 effusion of molecules 1 M m Copyright Cengage Learning. All rights reserved. 5 84

? 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 H 2 2.016 = = Rate He 1 4.002 4.002 2.016 Hydrogen will diffuse more quickly by a factor of 1.4. Copyright Cengage Learning. All rights reserved. 5 85

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 Cengage Learning. All rights reserved. 5 87

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 + n 2 a/v 2 ). Values for a and b are found in Table 5.7 2 n a P + = 2 V ( V nb) nrt Copyright Cengage Learning. All rights reserved. 5 88

? Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO 2 that has a volume of 10.0 L at 25 C. Compare this with value with the pressure obtained from the ideal gas law. n = 2.00 mol V = 10.0 L T = 25 C = 298 K For CO 2 : a = 3.658 L 2 atm/mol 2 b = 0.04286 L/mol Copyright Cengage Learning. All rights reserved. 5 89

n = 2.00 mol V = 10.0 L T = 25 C = 298 K Ideal gas law: nrt P = V P = L atm 2.00 0.08206 (298 K) mol K ( mol) ( 10.0 L) = 4.89 atm (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 90

n = 2.00 mol V = 10.0 L T = 25 C = 298 K For CO 2 : a = 3.658 L 2 atm/mol 2 b = 0.04286 L/mol P = P = nrt n 2 ( V nb) 2 L atm 2 L atm 2.00 3.658 2 ( mol) 0.08206 ( 298 K) ( 2.00 mol) mol K 10.0 L 2.00 mol 0.04286 P = L mol V P actual = 4.79 atm (3 significant figures) 4.933 atm 2 a 0.146 atm ( 10.0 L) 2 mol Copyright Cengage Learning. All rights reserved. 5 91