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Chapter 5 The Gaseous State

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1 Chapter 5 The Gaseous State

2 Contents and Concepts Gas Laws
We will investigate the quantitative relationships that describe the behavior of gases. Gas Pressure and Its Measurement Empirical Gas Laws The Ideal Gas Law Stoichiometry Problems Involving Gas Volumes Gas Mixtures; Law of Partial Pressures Copyright © Cengage Learning. All rights reserved.

3 Kinetic-Molecular Theory
This section will develop a model of gases as molecules in constant random motion. Kinetic Theory of Gases Molecular Speeds; Diffusion and Effusion Real Gases Copyright © Cengage Learning. All rights reserved.

4 Learning Objectives Gas Pressure and Its Measurement
a. Define pressure and its units. b. Convert units of pressure. 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 e. Express the combined gas law as an equation.
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.

6 e. Calculate gas density. f. Determine the molecular mass of a vapor.
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.

7 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.

8 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.

9 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.

10 a. Explain how and why a real gas is different from an ideal gas.
8. Real Gases a. Explain how and why a real gas is different from an ideal gas. b. Use the van der Waals equation. Copyright © Cengage Learning. All rights reserved.

11 Gases differ from liquids and solids: They are compressible.
Pressure, volume, temperature, and amount are related. Copyright © Cengage Learning. All rights reserved.

12 Pressure, P The force exerted per unit area.
It can be given by two equations: The SI unit for pressure is the pascal, Pa. Copyright © Cengage Learning. All rights reserved.

13 Other Units atmosphere, atm mmHg torr bar
Copyright © Cengage Learning. All rights reserved.

14 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.

15 Suppose that you set up two barometers like the one shown in Figure 5
Suppose that you set up two barometers like the one shown in Figure 5.2. In one of the barometers you use mercury, and in the other you use water. Which of the barometers would have a higher column of liquid, the one with Hg or H2O? Explain your answer. 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. Change to Figure 5.2 with no caption Copyright © Cengage Learning. All rights reserved.

16 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.

17 Boyle’s Law The volume of a sample of gas at constant temperature varies inversely with the applied pressure. The mathematical relationship: In equation form: Copyright © Cengage Learning. All rights reserved.

18 Figure A shows the plot of V versus P for 1. 000 g O2 at 0°C
Figure A shows the plot of V versus P for g O2 at 0°C. This plot is nonlinear. Figure B shows the plot of (1/V) versus P for g O2 at 0°C. This plot is linear, illustrating the inverse relationship. Copyright © Cengage Learning. All rights reserved.

19 At one atmosphere the volume of the gas is 100 mL
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 one-third, 33 mL. Change to Figure 5.5 with no caption Copyright © Cengage Learning. All rights reserved.

20 When a 1.00-g sample of O2 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 O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume. Change to Figure 5.6 with no captions Copyright © Cengage Learning. All rights reserved.

21 A volume of oxygen gas occupies 38. 7 mL at 751 mmHg and 21°C
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 Vf = ? Pf = 359 mmHg Tf = 21°C Copyright © Cengage Learning. All rights reserved.

22 (3 significant figures)
Vi = 38.7 mL Pi = 751 mmHg Ti = 21°C Vf = ? Pf = 359 mmHg Tf = 21°C = 81.0 mL (3 significant figures) Copyright © Cengage Learning. All rights reserved.

23 Change to Figure 5.9 with no caption
A graph of V versus T is linear. Note that all lines cross zero volume at the same temperature, −273.15°C. Copyright © Cengage Learning. All rights reserved.

24 The temperature -273. 15°C is called absolute zero
The temperature °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.

25 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: Copyright © Cengage Learning. All rights reserved.

26 As the air inside warms, the balloon expands to its orginial size.
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.

27 A 1. 0-g sample of O2 at a temperature of 100 K and a pressure of 1
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. Change to Figure 5.10 with no caption Copyright © Cengage Learning. All rights reserved.

28 You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3
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 Vf = ? Pf = 760 mmHg Tf = 27°C = 300. K Copyright © Cengage Learning. All rights reserved.

29 (3 significant figures)
Vi = 79.4 mL Pi = 760 mmHg Ti = 0°C = 273 K Vf = ? Pf = 760 mmHg Tf = 27°C = 300. K = 87.3 mL (3 significant figures) Copyright © Cengage Learning. All rights reserved.

30 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: Copyright © Cengage Learning. All rights reserved.

31 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 Vf = ? Pf = 1.0 atm Tf = 11°C = 284 K Copyright © Cengage Learning. All rights reserved.

32 (2 significant figures)
Vi = 5.0 L Pi = 5.0 × 101 atm Ti = 4°C = 277 K Vf = ? Pf = 1.0 atm Tf = 11°C = 284 K = 2.6  102 L (2 significant figures) Copyright © Cengage Learning. All rights reserved.

33 Concept Check 5.2 To conduct some experiments, a 10.0-L fl ask equipped with a movable plunger, as illustrated here, is filled with enough H2 gas to come to a pressure of 20 atm. a. In the first experiment, we decrease the temperature in the flask by 10°C and then increase the volume. Predict how the pressure in the flask changes during each of these events and, if possible, how the final pressure compares to your starting pressure. b. Once again we start with the pressure in the fl ask at 20 atm. The flask is then heated 10°C, followed by a volume decrease. Predict how the pressure in the flask changes during each of these events and, if possible, how the final pressure compares to your starting pressure. Copyright © Cengage Learning. All rights reserved.

34 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.

35 Avogadro’s Law Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules. Copyright © Cengage Learning. All rights reserved.

36 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 © Cengage Learning. All rights reserved.

37 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.

38 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. Copyright © Cengage Learning. All rights reserved.

39 (3 significant figures)
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 mass = 986 g (3 significant figures) Copyright © Cengage Learning. All rights reserved.

40 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: Copyright © Cengage Learning. All rights reserved.

41 What is the density of methane gas (natural gas), CH4, at 125°C and 3
What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm? Mm = g/mol P = 3.50 atm T = 125°C = 398 K Copyright © Cengage Learning. All rights reserved.

42 A 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 © Cengage Learning. All rights reserved.

43 d = 1.57 g/ L = g/L P = 634 mmHg = atm T = 95.0°C = 368 K Copyright © Cengage Learning. All rights reserved.

44 Empirical formula molar mass = 57 g/mol
Empirical formula: C4H9 Empirical formula molar mass = 57 g/mol Molecular formula: C8H18 Copyright © Cengage Learning. All rights reserved.

45 Concept Check 5.3 Three 3.0-L fl asks, each at a pressure of 878 mmHg, are in a room. The fl asks contain He, Ar, and Xe, respectively. a. Which of the fl asks contains the most atoms of gas? b. Which of the fl asks has the greatest density of gas? c. If the He fl ask were heated and the Ar flask cooled, which of the three flasks would be at the highest pressure? d. Say the temperature of the He was lowered while that of the Xe was raised. Which of the three flasks would have the greatest number of moles of gas? Copyright © Cengage Learning. All rights reserved.

46 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.

47 Stoichiometry and Gas Volumes
Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa. Copyright © Cengage Learning. All rights reserved.

48 When a 2. 0-L bottle of concentrated HCl was spilled, 1
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 © Cengage Learning. All rights reserved.

49 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 = g/mol Moles of CO2 produced = mol Copyright © Cengage Learning. All rights reserved.

50 (3 significant figures)
n = mol P = 735 mmHg = atm T = 20°C = 293 K = 2.98 × 102 L (3 significant figures) Copyright © Cengage Learning. All rights reserved.

51 Gas Mixtures Dalton found that in a mixture of unreactive gases, each gas acts as if it were the only gas in the mixture as far as pressure is concerned. Copyright © Cengage Learning. All rights reserved.

52 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 © Cengage Learning. All rights reserved.

53 Dalton’s Law of Partial Pressures
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 © Cengage Learning. All rights reserved.

54 A mL sample of air exhaled from the lungs is analyzed and found to contain g N2, g O2, g CO2, and 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.

55 Copyright © Cengage Learning. All rights reserved.

56 P = 1.00 atm Copyright © Cengage Learning. All rights reserved.

57 The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, mmHg; oxygen, mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air? Copyright © Cengage Learning. All rights reserved.

58 570.0 mmHg 103.0 mmHg 40.0 mmHg 47.0 mmHg P = 760.0 mmHg
Copyright © Cengage Learning. All rights reserved.

59 Mole fraction of N2 Mole fraction of O2 Mole fraction of CO2
Mole fraction of H2O Mole fraction N2 = Mole fraction O2 = Mole fraction CO2 = Mole fraction O2 = Copyright © Cengage Learning. All rights reserved.

60 Concept Check 5.4 A flask equipped with a valve contains 3.0 mol of H2 gas. You introduce 3.0 mol of Ar gas into the fl ask via the valve and then seal the fl ask. a. What happens to the pressure of just the H2 gas in the flask after the introduction of the Ar? If it changes, by what factor does it do so? b. How do the pressures of the Ar and the H2 in the fl ask compare? c. How does the total pressure in the flask relate to the pressures of the two gases? Nothing happens to the pressure of H2. The pressures are equal because the moles are equal. 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.

61 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.

62 Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
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 © Cengage Learning. All rights reserved.

63 Copyright © Cengage Learning. All rights reserved.

64 Copyright © Cengage Learning. All rights reserved.

65 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 Molar mass NH4NO2 = g/mol Copyright © Cengage Learning. All rights reserved.

66 P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K
Molar mass NH4NO2 = g/mol = mol N2 gas Copyright © Cengage Learning. All rights reserved.

67 (3 significant figures)
P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K n = mol = 2.32 L of N2 (3 significant figures) Copyright © Cengage Learning. All rights reserved.

68 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: m = mass v = velocity Copyright © Cengage Learning. All rights reserved.

69 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.

70 An elastic collision is one in which no kinetic energy is lost
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.

71 frequency of collision x average force
Each of the gas laws can be derived from the postulates. For the ideal gas law: frequency of collision x average force Copyright © Cengage Learning. All rights reserved.

72 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. Copyright © Cengage Learning. All rights reserved.

73 Rearranging this relationship gives
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 © Cengage Learning. All rights reserved.

74 Inserting the proportionality constant, R, gives
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 Inserting the proportionality constant, R, gives Copyright © Cengage Learning. All rights reserved.

75 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 Copyright © Cengage Learning. All rights reserved.

76 When using the equation R = 8.3145 J/(mol · K). T must be in Kelvins
Mm must be in kg/mol Copyright © Cengage Learning. All rights reserved.

77 What is the rms speed of carbon dioxide molecules in a container at 23°C?
T = 23°C = 296 K CO2 molar mass = kg/mol Copyright © Cengage Learning. All rights reserved.

78 Copyright © Cengage Learning. All rights reserved.

79 Concept Check 5.6 Consider the experimental
apparatus shown. In this setup, each round flask contains a gas, and the long tube contains no gas (that is, it is a vacuum). a. We use 1.0 mol of He for experiment X and 1.0 mol of Ar for experiment Y. If both valves are opened at the same time, which gas would you expect to reach the end of the long tube first? b. If you wanted the Ar to reach the end of the long tube at the same time as the He, what experimental condition (that is, you cannot change the equipment) could you change to make this happen? Copyright © Cengage Learning. All rights reserved.

80 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.

81 Maxwell predicted the distributions of molecular speeds at various temperatures. The graph shows 0°C and 500°C. Copyright © Cengage Learning. All rights reserved.

82 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 © Cengage Learning. All rights reserved.

83 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 © Cengage Learning. All rights reserved.

84 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. Copyright © Cengage Learning. All rights reserved.

85 Both hydrogen and helium have been used as the buoyant gas in blimps
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? Hydrogen will diffuse more quickly by a factor of 1.4. Copyright © Cengage Learning. All rights reserved.

86 Real Gases At high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below. Change to Figure 5.30 with no caption Copyright © Cengage Learning. All rights reserved.

87 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.

88 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 Copyright © Cengage Learning. All rights reserved.

89 Use the van der Waals equation to calculate the pressure exerted by 2
Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO2 that has a volume of 10.0 L at 25°C. Compare this with value with the pressure obtained from the ideal gas law. For CO2: a = L2 atm/mol2 b = L/mol n = 2.00 mol V = 10.0 L T = 25°C = 298 K Copyright © Cengage Learning. All rights reserved.

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

91 (3 significant figures)
n = 2.00 mol V = 10.0 L T = 25°C = 298 K For CO2: a = L2 atm/mol2 b = L/mol Pactual = 4.79 atm (3 significant figures) Copyright © Cengage Learning. All rights reserved.


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