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Chapter 7 Gases

Chapter 7 Gases. Kinetic Theory of Gases. Particles of a gas Move rapidly in straight lines and are in constant motion. Have kinetic energy that increases with an increase in temperature. Are very far apart. Have essentially no attractive (or repulsive) forces.

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Chapter 7 Gases

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  1. Chapter 7 Gases

  2. Kinetic Theory of Gases Particles of a gas • Move rapidly in straight lines and are in constant motion. • Have kinetic energy that increases with an increase in temperature. • Are very far apart. • Have essentially no attractive (or repulsive) forces. • Have very small volumes compared to the volume of the container they occupy.

  3. Properties of Gases • Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

  4. Barometer • A barometer measures the pressure exerted by the gases in the atmosphere. • The atmospheric pressure is measured as the height in mm of the mercury column.

  5. Learning Check A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense 2) H2O is heavier 3) air is more dense than H2O

  6. Solution A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense

  7. Pressure • A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area • One atmosphere (1 atm) is 760 mm Hg. • 1 mm Hg = 1 torr1.00 atm = 760 mm Hg = 760 torr

  8. Units of Pressure • In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).

  9. Learning Check A. What is 475 mm Hg expressed in atm?1) 475 atm2) 0.638 atm3) 3.61 x 105 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg?1) 2.00 mm Hg2) 1520 mm Hg3) 22,300 mm Hg

  10. Solution A. What is 475 mm Hg expressed in atm? 2) 0.638 atm 485 mm Hg x 1 atm = 0.638 atm 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) 1520 mm Hg 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm

  11. Boyle’s Law • The pressure of a gas is inversely related to its volume when T and n are constant. • If volume decreases, the pressure increases.

  12. PV Constant in Boyle’s Law • The product P x V remains constant as long as T and n do not change.P1V1 = 8.0 atm x 2.0 L = 16 atm LP2V2 = 4.0 atm x 4.0 L = 16 atm LP3V3 = 2.0 atm x 8.0 L = 16 atm L • Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)

  13. Solving for a Gas Law Factor • The equation for Boyle’s Law can be rearranged to solve for any factor. • To solve for V2, divide both sides by P2. P1V1 = P2V2Boyle’s Law P2 P2 P1V1 = V2 P2 • Solving for P2 P2 = P1 V1 V2

  14. PV in Breathing Mechanics • When the lungs expand, the pressure in the lungs decreases. • Inhalation occurs as air flows towards the lower pressure in the lungs.

  15. PV in Breathing Mechanics • When the lung volume decreases, pressure within the lungs increases. • Exhalation occurs as air flows from the higher pressure in the lungs to the outside.

  16. Calculation with Boyle’s Law Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T? 1. Set up a data table Conditions 1 Conditions 2 P1 = 50 mm Hg P2 = 200 mm Hg V1 = 8 L V2 = ?

  17. Calculation with Boyle’s Law (continued) 2.When pressure increases, volume decreases. Solve Boyle’s Law for V2: P1V1 = P2V2 V2 = V1P1 P2 V2 = 8 L x 50 mm Hg = 2 L 200 mm Hg pressure ratio decreases volume

  18. Learning Check The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume? At a new pressure of 420 mm Hg, what is the new volume? 1) 60 mL 2) 120 mL 3) 240 mL

  19. Solution The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume? B) If P decreases, V increases. At a new pressure of 420 mm Hg, what is the new volume of the cylinder? 3) 240 mL

  20. Learning Check A sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm. What is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L B) 6.4 L C) 12.8 L

  21. Solution A) 3.2 L Solve for V2: P1V1 = P2V2 V2 = V1P1 P2 V2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (Temperature is constant).

  22. Learning Check A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.) 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg

  23. Solution 1) 200. mm Hg Data table Conditions 1 Conditions 2P1 = 600. mm Hg P2 = ???V1 = 12.0 L V2 = 36.0 L P2 = P1 V1 V2 600. mm Hg x 12.0 L = 200. mm Hg 36.0 L

  24. Charles’ Law • The Kelvin temperature of a gas is directly related to the volume (P and n are constant). • When the temperature of a gas increases, its volume increases.

  25. Charles’ Law V and T • For two conditions, Charles’ Law is written V1 = V2 (P and n constant) T1 T2 • Rearranging Charles’ Law to solve for V2 V2 = V1T2 T1

  26. Learning Check Solve Charles’ Law expression for T2. V1 = V2 T1 T2

  27. Solution V1 = V2 T1 T2 Cross multiply to give V1T2 = V2T1 Isolate T2 by dividing through by V1 V1T2 = V2T1 V1 V1 T2 = V2T1 V1

  28. Calculations Using Charles’ Law A balloon has a volume of 785 mL at 21°C. If The temperature drops to 0°C, what is the new volume of the balloon (P constant)? 1. Set up data table: Conditions 1 Conditions 2 V1 = 785 mL V2 = ? T1 = 21°C = 294 K T2 = 0°C = 273 K Be sure that you always use the Kelvin (K) temperature in gas calculations.

  29. Calculations Using Charles’ Law (continued) 2. Solve Charles’ law for V2 V1 = V2 T1 T2 V2 = V1 T2 T1 V2 = 785 mL x 273 K = 729 mL 294 K

  30. Learning Check A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443°C 2) 170°C 3) – 82°C

  31. Solution 170°C T2 = T1V2 V1 T2 = 291 K x 640 mL = 443 K 420 mL = 443 K – 273 K = 170°C

  32. Gay-Lussac’s Law: P and T • The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n. P1 = P2 T1 T2

  33. Calculation with Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (V and n constant) • Set up a data table. Conditions 1 Conditions 2 P1 = 2.0 atm P2 = T1 = 18°C + 273 T2 = 62°C + 273 = 291 K = 335 K ?

  34. Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P2 P1 = P2 T1 T2 P2 = P1 T2 T1 P2 = 2.0 atm x 335 K = 2.3 atm 291 K

  35. Learning Check Use the gas laws to complete with • Increases 2) Decreases A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12.0 L to 24.0 L. D. Volume _______when T changes from 15.0 °C to 45.0°C.

  36. Solution Use the gas laws to complete with 1) Increases 2) Decreases A. Pressure 1) Increases, when V decreases. B. When T decreases, V 2) Decreases. C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C

  37. Next Time • We complete Chapter 7

  38. Combined Gas Law • The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P1 V1 = P2 V2 T1 T2

  39. Combined Gas Law Calculation A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? 1. Set up Data Table Conditions 1 Conditions 2 P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L (180 mL) V2 = 90.0 mL T1 = 29°C + 273 = 302 K T2 = ??

  40. Combined Gas Law Calculation (continued) 2. Solve for T2 P1 V1 = P2 V2 T1 T2 T2 = T1 P2V2 P1V1 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K – 273 = 331 °C

  41. Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)?

  42. Solution Data Table T1 = 308 K T2 = -95°C + 273 = 178K V1 = 675 mL V2 = ??? P1 = 646 mm Hg P2 = 802 mm Hg Solve for T2 V2 = V1 P1 T2 P2T1 V2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K

  43. Avogadro's Law: Volume and Moles • The volume of a gas is directly related to the number of moles of gas when T and P are constant.V1 = V2n1 n2

  44. Learning Check If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L

  45. Solution 3) 2.4 L Conditions 1 Conditions 2 V1 = 1.5 L V2 = ??? n1 = 0.75 mole He n2 = 1.2 moles He V2 = V1n2n1V2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

  46. STP • The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg)

  47. Molar Volume • At STP, 1 mole of a gas occupies a volume of 22.4 L. • The volume of one mole of a gas is called the molar volume.

  48. Molar Volume as a Conversion Factor • The molar volume at STP can be used to form conversion factors.22.4 Land1 mole 1 mole 22.4 L

  49. Learning Check A. What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g

  50. Solution • 1) 5.60 L 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He

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