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Chapter 10 Gases

Chapter 10 Gases. Characteristics of Gases. Unlike liquids and solids, gases Expand to fill their containers. Are highly compressible. Have extremely low densities. F A. P =. Pressure. Pressure is the amount of force applied to an area:.

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Chapter 10 Gases

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  1. Chapter 10Gases

  2. Characteristics of Gases • Unlike liquids and solids, gases • Expand to fill their containers. • Are highly compressible. • Have extremely low densities.

  3. F A P = Pressure • Pressure is the amount of force applied to an area: • Atmospheric pressure is the weight of air per unit of area.

  4. Units of Pressure • Pascals • 1 Pa = 1 N/m2 • Bar • 1 bar = 105 Pa = 100 kPa

  5. 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. • Atmosphere • 1.00 atm = 760 torr

  6. Manometer The manometer is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

  7. Standard Pressure • Normal atmospheric pressure at sea level is referred to as standard pressure. • It is equal to • 1.00 atm • 760 torr (760 mmHg) • 101.325 kPa

  8. Sample Exercise 10.2Using 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 ( FIGURE 10.3), 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? • (a) The pressure of the gas equals the atmospheric pressure plus h: • (b) To calculate the pressure in kPa, we employ the conversion factor • Between atmospheres and kPa:

  9. Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

  10. PV = k • Since • V = k (1/P ) • This means a plot of V versus 1/P will be a straight line. P andV are Inversely Proportional A plot of V versus P results in a curve.

  11. Charles’s Law • The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

  12. V T = k • So, • A plot of V versus T will be a straight line. Charles’s Law

  13. V = kn • Mathematically, this means Avogadro’s Law • The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

  14. nT P V Ideal-Gas Equation • So far we’ve seen that V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law) • Combining these, we get

  15. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

  16. nT P V nT P V= R Ideal-Gas Equation The relationship then becomes or PV = nRT

  17. Sample Exercise 10.4Using the Ideal-Gas Equation Calcium carbonate, CaCO3(s), the principal compound in limestone, decomposes upon heating to CaO(s) and CO2(g). A sample of CaCO3 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 CO2 gas were generated?

  18. Sample Exercise 10.5Calculating the Effect of Temperature Changes on Pressure 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?

  19. Sample Exercise 10.6Calculating the Effect of Changing P and T on Gas Volume 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.

  20. n V P RT = Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get

  21. m V P RT = Densities of Gases • We know that • Moles  molecular mass = mass n  = m • So multiplying both sides by the molecular mass () gives

  22. m V P RT d = = 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.

  23. P RT d = dRT P  = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: becomes

  24. Sample Exercise 10.7Calculating Gas Density What is the density of carbon tetrachloride vapor at 714 torr and 125 C?

  25. Sample Exercise 10.8Calculating the Molar Mass of a Gas 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 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. 137.456 g  134.567 g = 2.889 g 1067.9 g  134.567 g = 933.3 g d = 2.889 g/0.936 L = 3.09 g/L

  26. Sample Exercise 10.9Relating a Gas Volume to the Amount of Another Substance in a Reaction Automobile air bags are inflated by nitrogen gas generated by the rapid decomposition of sodium azide, NaN3: 2 NaN3(s)  2 Na(s) + 3 N2(g) If an air bag has a volume of 36 L and is to be filled with nitrogen gas at 1.15 atm and 26.0 C, how many grams of NaN3 must be decomposed?

  27. 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, • Ptotal = P1 + P2 + P3 + … P1 = X1Pt Where X1 is the mole fraction (n1/nt).

  28. Sample Exercise 10.10Applying Dalton’s Law of Partial Pressures A mixture of 6.00 g O2(g) and 9.00 g CH4(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 = PO2 + PCH4 = 0.281 atm + 0.841 atm = 1.122 atm

  29. Sample Exercise 10.11Relating Mole Fractions and PartialPressures A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO2, 18.0 mol percent O2, and 80.5 mol percent Ar. (a) Calculate the partial pressure of O2 in the mixture if the total pressure of the atmosphere is to be 745 torr. (b) If this atmosphere is to be held in a 121-L space at 295 K, how many moles of O2 are needed? PO2 = (0.180)(745 torr) = 134 torr

  30. Continued

  31. 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. Ptotal = Pgas + Pwater

  32. Sample Exercise 10.12Calculating the Amount of Gas Collected over Water When a sample of KClO3 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 O2 are collected? (b) How many grams of KClO3 were decomposed? PO2 = 765 torr  25 torr = 740 torr

  33. Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

  34. Main Tenets of Kinetic-Molecular Theory Gases consist of large numbers of molecules that are in continuous, random motion. The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.

  35. Main Tenets of Kinetic-Molecular Theory Attractive and repulsive forces between gas molecules are negligible.

  36. Main Tenets of Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the averagekinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

  37. Main Tenets of Kinetic-Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature.

  38. Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

  39. Effusion The difference in the rates of effusion for helium and argon, for example, explains why a helium balloon would deflate faster.

  40. Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.

  41. Molecular Effusion and Diffusion • The average kinetic energy (e) of a gas is related to its mass: e = ½ mu 2 • Consider two gases at the same temperature: the lighter gas has a higher rms(Root-Mean-Square) speed than the heavier gas. • Mathematically: • The lower the molar mass, M, the higher the rms speed for that gas at a constant temperature. rms speed: is the speed of a gas molecule having average kinetic energy.

  42. Sample Exercise 10.14Calculating a Root-Mean-Square Speed Calculate the rms speed of the molecules in a sample of N2 gas at 25 C.

  43. Graham's Law of Effusion Consider two gases with molar masses M1 and M2, with effusion rates, r1 and r2, respectively: • The relative rate of effusion is given by Graham’s law: Only those molecules that hit the small hole will escape through it. • Therefore, the higher the rms speed, the more likely that a gas molecule will hit the hole. • We can show that:

  44. Sample Exercise 10.15Applying Graham’s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is 0.355 times the rate at which O2 gas effuses at the same temperature. Calculate the molar mass of the unknown and identify it.

  45. Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

  46. Real Gases Even the same gas will show wildly different behavior under high pressure at different temperatures.

  47. 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.

  48. 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.

  49. n 2a V 2 (P + ) (V−nb) = nRT The van der Waals Equation

  50. Sample Exercise 10.16Using the van der Waals Equation 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 Cl2(g) in 22.41 L at 0.0 C.

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