1. © 2012 Pearson Education, Inc. Chapter 10 Gases Dr. Subhash C. Goel South Georgia College Douglas, GA Lecture Presentation

2. Gases © 2012 Pearson Education, Inc. Characteristics of Gases Unlike liquids and solids, gases –Expand to fill their containers. –Are highly compressible. –Have extremely low densities. –Examples: He, Ne, Ar, Kr, Xe H 2, N 2, O 2, Cl 2 CO 2, NH 3, SO 2, CH 4

3. Tro, Principles of Chemistry: A Molecular Approach3 Gases Pushing Gas molecules are constantly in motion. As they move and strike a surface, they push on that surface. push = force If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting. pressure = force per unit area

4. Tro, Principles of Chemistry: A Molecular Approach4 The Effect of Gas Pressure The pressure exerted by a gas can cause some amazing and startling effects. Whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure. The bigger the difference in pressure, the stronger the flow of the gas. If there is something in the gas’s path, the gas will try to push it along as the gas flows.

5. Gases © 2012 Pearson Education, Inc. Pressure is the amount of force applied to an area: Pressure Atmospheric pressure is the weight of air per unit of area. P = FAFA

6. Gases © 2012 Pearson Education, Inc. Units of Pressure Pascals –1 Pa = 1 N/m 2 [1 N = 1 kg-m/s 2 ] Bar –1 bar = 10 5 Pa = 100 kPa

7. Gases © 2012 Pearson Education, Inc. Units of Pressure mmHg or torr – These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury.

8. Gases © 2012 Pearson Education, Inc. Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to –1.00 atm –14.7 psi –760 torr (760 mmHg) –101.325 kPa

9. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 10.1 Converting Pressure Units Solution Analyze In each case we are given the pressure in one unit and asked to convert it to another unit. Our task, therefore, is to choose the appropriate conversion factors. Plan We can use dimensional analysis to perform the desired conversions. Solve (a)To convert atmospheres to torr, we use the relationship 760 torr = 1 atm: Note that the units cancel in the required manner. (b)We use the same relationship as in part (a). To get the appropriate units to cancel, we must use the conversion factor as follows: (c)The relationship 760 torr = 101.325 kPa allows us to write an appropriate conversion factor for this problem: 760 torr = 101.325 (a) Convert 0.357 atm to torr. (b) Convert 6.6  10  2 torr to atmospheres. (c) Convert 147.2 kPa to torr.

10. Gases © 2012 Pearson Education, Inc. Manometer The manometer is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

11. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 10.2 Using a Manometer to Measure Gas Pressure On a certain day a laboratory barometer indicates that the atmospheric pressure is 764.7 torr. A sample of gas is placed in a flask attached to an open-end mercury manometer and a meter stick is used to measure the height of the mercury in the two arms of the U tube. The height of the mercury in the open-end arm is 136.4 mm, and the height in the arm in contact with the gas in the flask is 103.8 mm. What is the pressure of the gas in the flask (a) in atmospheres, (b) in kilopascals?

12. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Continued Solve (a)The pressure of the gas equals the atmospheric pressure plus h: We convert the pressure of the gas to atmospheres: (b)To calculate the pressure in kPa, we employ the conversion factor Between atmospheres and kPa: Sample Exercise 10.2 Using a Manometer to Measure Gas Pressure

13. Tro, Principles of Chemistry: A Molecular Approach13 Boyle’s Law Pressure of a gas is inversely proportional to its volume. constant T and amount of gas graph P vs. V is curve graph P vs. 1/V is straight line As P increases, V decreases by the same factor. P × V = constant P 1 × V 1 = P 2 × V 2 Robert Boyle (1627–1691)

14. Gases © 2012 Pearson Education, Inc. P and V are Inversely Proportional

15. Tro, Principles of Chemistry: A Molecular Approach15 Boyle’s Law: A Molecular View Pressure is caused by the molecules striking the sides of the container. When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant. This results in increasing the pressure.

16. Tro, Principles of Chemistry: A Molecular Approach16 Charles’s Law Volume is directly proportional to temperature. constant P and amount of gas graph of V vs. T is straight line As T increases, V also increases. Kelvin T = Celsius T + 273 V = constant × T if T measured in Kelvin Jacques Charles (1746–1823)

17. Tro, Principles of Chemistry: A Molecular Approach17 If the lines are extrapolated back to a volume of “0,” they all show the same temperature, –273.15 °C, called absolute zero. If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line.

18. Tro, Principles of Chemistry: A Molecular Approach18 Example 3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L? Since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does. V 1 = 2.57 L, V 2 = 2.80 L, t 1 = 0.00 °C t 2, K and °C Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, V 2, T 2 T1T1 T(K) = t(°C) + 273.15,

19. Tro, Principles of Chemistry: A Molecular Approach19 Avogadro’s Law volume directly proportional to the number of gas molecules V = constant × n while P and T remain constant more gas molecules = larger volume count number of gas molecules by moles Amedeo Avogadro (1776–1856)

20. Gases © 2012 Pearson Education, Inc. Avogadro’s Law Equal volume of gases at constant P and T have same number of moles of gas

21. Tro, Principles of Chemistry: A Molecular Approach21 Exercise 4: A 0.225-mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? Since n and V are directly proportional, when the volume increases, the moles should increase, and it does. V 1 = 4.65 L, V 2 = 6.48 L, n 1 = 0.225 mol n 2, and added moles Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, V 2, n 1 n2n2 mol added = n 2 – n 1,

22. Gases © 2012 Pearson Education, Inc. Ideal-Gas Equation V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law) So far we’ve seen that Combining these, we get V V  nT P

23. Gases © 2012 Pearson Education, Inc. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

24. Gases © 2012 Pearson Education, Inc. Ideal-Gas Equation The relationship then becomes nT P V V  nT P V = R or PV = nRT

25. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 5 Using the Ideal-Gas Equation Solution Analyze We are given the volume (250 mL), pressure (1.3 atm), and temperature (31  C) of a sample of CO 2 gas and asked to calculate the number of moles of CO 2 in the sample. Solve In analyzing and solving gas law problems, it is helpful to tabulate the information given in the problems and then to convert the values to units that are consistent with those for R (0.08206 L-atm/mol-K). In this case the given values are Remember: Absolute temperature must always be used when the ideal-gas equation is solved. We now rearrange the ideal-gas equation (Equation 10.5) to solve for n Calcium carbonate, CaCO 3 (s), the principal compound in limestone, decomposes upon heating to CaO(s) and CO 2 (g). A sample of CaCO 3 is decomposed, and the carbon dioxide is collected in a 250-mL flask. After decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31  C. How many moles of CO 2 gas were generated? Plan Because we are given V, P, and T, we can solve the ideal-gas equation for the unknown quantity, n.

26. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 6 Calculating the Effect of Temperature Changes on Pressure Solution Analyze We are given the initial pressure (1.5 atm) and temperature (25  C) of the gas and asked for the pressure at a higher temperature (450  C). Plan The volume and number of moles of gas do not change, so we must use a relationship connecting pressure and temperature. Converting temperature to the Kelvin scale and tabulating the given information, we have Solve To determine how P and T are related, we start with the ideal-gas equation and isolate the quantities that do not change (n, V, and R) on one side and the Variables (P and T) on the other side. Because the quotient P/T is a constant, we can write (where the subscripts 1 and 2 represent the initial and final states, respectively). Rearranging to solve for P 2 and substituting the given data give The gas pressure in an aerosol can is 1.5 atm at 25  C. Assuming that the gas obeys the ideal-gas equation, what is the pressure when the can is heated to 450  C?

27. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 7 Calculating the Effect of Changing P and T on Gas Volume Solution Analyze We need to determine a new volume for a gas sample when both pressure and temperature change. Plan Let’s again proceed by converting temperatures to kelvins and tabulating our information. Because n is constant, we can use Equation 10.8. Solve Rearranging Equation 10.8 to solve for V 2 gives An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend until the pressure is 0.45 atm. During ascent, the temperature of the gas falls from 22  C to  21  C. Calculate the volume of the balloon at its final altitude.

28. Tro, Principles of Chemistry: A Molecular Approach28 Density at Standard Conditions Density is the ratio of mass to volume. Density of a gas is generally given in g/L. mass of 1 mole = molar mass volume of 1 mole at STP = 22.4 L

29. Tro, Principles of Chemistry: A Molecular Approach29 Practice 8: Calculate the density of N 2 (g) at STP. The units and magnitude are reasonable. N2N2 d N2, g/L Check: Solution: Conceptual Plan: Relationships: Given: Find: MMd

30. Gases © 2012 Pearson Education, Inc. Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get nVnV P RT =

31. Gases © 2012 Pearson Education, Inc. We know that –Moles  molecular mass = mass Densities of Gases So multiplying both sides by the molecular mass (  ) gives n   = m P  RT mVmV =

32. Gases © 2012 Pearson Education, Inc. Densities of Gases Mass  volume = density So, Note: One needs to know only the molecular mass, the pressure, and the temperature to calculate the density of a gas. P  RT mVmV = d =

33. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 9 Calculating Gas Density Solution Analyze We are asked to calculate the density of a gas given its name, its pressure, and its temperature. From the name we can write the chemical formula of the substance and determine its molar mass. Plan We can use Equation 10.10 to calculate the density. Before we can do that, however, we must convert the given quantities to the appropriate units, degrees Celsius to kelvins and pressure to atmospheres. We must also calculate the molar mass of CCl 4. Solve The absolute temperature is 125 + 273 = 398 K. The pressure is (714 torr) (1 atm/760 torr) = 0.939. The molar mass of CCl 4 is 12.01 + (4) (35.45) = 153.8 g/mol. Therefore, Check If we divide molar mass (g/mol) by density (g/L), we end up with L/mol. The numerical value is roughly 154/4.4 = 35. That is in the right ballpark for the molar volume of a gas heated to 125  C at near atmospheric pressure, so our answer is reasonable. What is the density of carbon tetrachloride vapor at 714 torr and 125  C?

34. Gases © 2012 Pearson Education, Inc. Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: becomes P  RT d = dRT P  =

35. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 10 Calculating the Molar Mass of a Gas Solution Analyze We are given the temperature (31  C) and pressure (735 torr) for a gas, together with information to determine its volume and mass, and we are asked to calculate its molar mass. Solve The gas mass is the difference between the mass of the flask filled with gas and the mass of the evacuated flask: The gas volume equals the volume of water the flask can hold, calculated from the mass and density of the water. The mass of the water is the difference between the masses of the full and evacuated flask: Rearranging the equation for density (d = m/V), we have A large evacuated flask initially has a mass of 134.567 g. When the flask is filled with a gas of unknown molar mass to a pressure of 735 torr at 31  C, its mass is 137.456 g. When the flask is evacuated again 137.456 g  134.567 g = 2.889 g 1067.9 g  134.567 g = 933.3 g and then filled with water at 31  C, its mass is 1067.9 g. (The density of water at this temperature is 0.997 g/mL.) Assuming the ideal-gas equation applies, calculate the molar mass of the gas. Plan We need to use the mass information given to calculate the volume of the container and the mass of the gas in it. From this we calculate the gas density and then apply Equation 10.11 to calculate the molar mass of the gas.

36. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 10 Calculating the Molar Mass of a Gas Continued Knowing the mass of the gas (2.889 g) and its volume (0.936 L), we can calculate the density of the gas: After converting pressure to atmospheres and temperature to kelvins, we can use Equation 10.11 to calculate the molar mass: 2.889 g/0.936 L = 3.09 g/L

37. Gases © 2012 Pearson Education, Inc. Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, P total = P 1 + P 2 + P 3 + …

38. Tro, Principles of Chemistry: A Molecular Approach38 The partial pressure of each gas in a mixture can be calculated using the ideal gas law.

39. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 11: Applying Dalton’s Law of Partial Pressures Solution Analyze We need to calculate the pressure for two gases in the same volume and at the same temperature. Solve We first convert the mass of each gas to moles: We use the ideal-gas equation to calculate the partial Pressure of each gas: According to Dalton’s law of partial pressures (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures: A mixture of 6.00 g O 2 (g) and 9.00 g CH 4 (g) is placed in a 15.0-L vessel at 0  C. What is the partial pressure of each gas, and what is the total pressure in the vessel? Pt = P O 2 + P CH 4 = 0.281 atm + 0.841 atm = 1.122 atm Plan Because each gas behaves independently, we can use the ideal gas equation to calculate the pressure each would exert if the other were not present. The total pressure is the sum of these two partial pressures.

40. Gases © 2012 Pearson Education, Inc. Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure. P total = P gas + P water

41. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 12 Calculating the Amount of Gas Collected over Water Solution (a) Analyze We need to calculate the number of moles of O 2 gas in a container that also contains water vapor. Plan We are given values for V and T. To use the ideal-gas equation to calculate the unknown, n O 2, we must know the partial pressure of O 2 in the system. We can calculate this partial pressure from the total pressure (765 torr) and the vapor pressure of water. Solve The partial pressure of the O 2 gas is the difference between the total pressure and the pressure of the water vapor at 26  C, 25 torr (Appendix B): P O 2 = 765 torr  25 torr = 740 torr We use the ideal-gas equation to calculate the number of moles of O 2 : When a sample of KClO 3 is partially decomposed in the setup shown in Figure 10.15, the volume of gas collected is 0.250 L at 26  C and 765 torr total pressure. (a) How many moles of O 2 are collected? (b) How many grams of KClO 3 were decomposed?

42. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Continued (b) Analyze We need to calculate the number of moles of reactant KClO 3 decomposed. Plan We can use the number of moles of O 2 formed and the balanced chemical equation to determine the number of moles of KClO 3 decomposed, which we can then convert to grams of KClO 3. Solve From Equation 10.17, we have 2 mol. The molar mass of KClO 3 is 122.6 g/mol. Thus, we can convert the number of moles of O 2 from part (a) to moles of KClO 3 and then to grams of KClO 3 : Sample Exercise 10.12 Calculating the Amount of Gas Collected over Water

43. Gases © 2012 Pearson Education, Inc. Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

44. Gases © 2012 Pearson Education, Inc. Main Tenets of Kinetic-Molecular Theory Gases consist of large numbers of molecules that are in continuous, random motion.

45. Gases © 2012 Pearson Education, Inc. Main Tenets of Kinetic-Molecular Theory The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.

46. Gases © 2012 Pearson Education, Inc. Main Tenets of Kinetic-Molecular Theory Attractive and repulsive forces between gas molecules are negligible.

47. Gases © 2012 Pearson Education, Inc. Main Tenets of Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = ½ mu 2

48. Gases © 2012 Pearson Education, Inc. 48 Gas Laws Explained— Boyle’s Law Boyle’s law states that the volume of a gas is inversely proportional to the pressure. Decreasing the volume forces the molecules into a smaller space. More molecules will collide with the container at any one instant, increasing the pressure.

49. Gases © 2012 Pearson Education, Inc. 49 Gas Laws Explained— Charles’s Law Charles’s law states that the volume of a gas is directly proportional to the absolute temperature. Increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently. –on average In order to keep the pressure constant, the volume must then increase.

50. Gases © 2012 Pearson Education, Inc. 50 Gas Laws Explained— Avogadro’s Law Avogadro’s law states that the volume of a gas is directly proportional to the number of gas molecules. Increasing the number of gas molecules causes more of them to hit the wall at the same time. In order to keep the pressure constant, the volume must then increase.

51. Gases © 2012 Pearson Education, Inc. 51 Gas Laws Explained— Dalton’s Law of Partial Pressures Dalton’s law states that the total pressure of a mixture of gases is the sum of the partial pressures. Kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact. Therefore, the molecules behave independently of each other, each gas contributing its own collisions to the container with the same average kinetic energy. Since the average kinetic energy is the same, the total pressure of the collisions is the same.

52. Gases © 2012 Pearson Education, Inc. 52 Dalton’s Law and Pressure Since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side.

53. Gases © 2012 Pearson Education, Inc. Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

54. Gases © 2012 Pearson Education, Inc. Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.

55. Copyright © Cengage Learning. All rights reserved.5 | 55 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

56. Copyright © Cengage Learning. All rights reserved.5 | 56 When using the equation R = 8.3145 J/(mol · K). T must be in Kelvins M m must be in kg/mol

57. Copyright © Cengage Learning. All rights reserved.5 | 57 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.

58. Gases © 2012 Pearson Education, Inc. Graham's Law  M 2  M 1 = r1r1 r2r2 M 1 and M 2 are the molar masses of gases. This indicate that lighter gas has higher effusion rate.

59. Gases © 2012 Pearson Education, Inc. Sample Exercise 13 Calculating a Root-Mean-Square Speed Solution Solve We must convert each quantity in our equation to SI units. We will also use R in units of J/mol-K (Table 10.2) to make the units cancel correctly. Calculate the rms speed of the molecules in a sample of N 2 gas at 25  C.

60. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 14 Applying Graham’s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is 0.355 times the rate at which O 2 gas effuses at the same temperature. Calculate the molar mass of the unknown and identify it. Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I 2.

61. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 61 Practice 15—Calculate the ratio of the rate of effusion for oxygen to hydrogen. O 2, 32.00 g/mol; H 2 2.016 g/mol Solution: Conceptual Plan: Relationships: Given: Find: MM O2, MM H2 rateA/rateB This means that, on average, the O 2 molecules are traveling at ¼ the speed of H 2 molecules.

62. Gases © 2012 Pearson Education, Inc. Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

63. Gases © 2012 Pearson Education, Inc. Real Gases Even the same gas will show wildly different behavior under high pressure at different temperatures.

64. Gases © 2012 Pearson Education, Inc. Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.

65. Gases © 2012 Pearson Education, Inc. Corrections for Nonideal Behavior The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. The corrected ideal-gas equation is known as the van der Waals equation.

66. Gases © 2012 Pearson Education, Inc. The van der Waals Equation ) (V − nb) = nRT n2aV2n2aV2 (P +

67. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 16 Using the van der Waals Equation Solution Solve Substituting n = 1.000 mol, R = 0.08206 L-atm/mol-K, T = 273.2 K, V = 22.41 L, a = 6.49 L 2 -atm/mol 2, and b = 0.0562 L/mol: If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0  C, it would exert a pressure of 1.000 atm. Use the van der Waals equation and Table 10.3 to estimate the pressure exerted by 1.000 mol of Cl 2 (g) in 22.41 L at 0.0  C.