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Chapter 13 Gases

Chapter 13 Gases. Some Gases in Our Lives. Air: oxygen O 2 nitrogen N 2 ozone O 3 argon Ar carbon dioxide CO 2 water H 2 O Noble gases : helium He neon Ne krypton Kr xenon Xe Other gases: fluorine F 2 chlorine Cl 2 ammonia NH 3

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Chapter 13 Gases

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  1. Chapter 13Gases

  2. Some Gases in Our Lives Air: oxygen O2 nitrogen N2 ozone O3 argon Ar carbon dioxide CO2 water H2O Noble gases: helium He neon Ne krypton Kr xenon Xe Other gases: fluorine F2 chlorine Cl2 ammonia NH3 methane CH4 carbon monoxide CO nitrogen dioxide NO2 sulfur dioxide SO2

  3. 13.1 The Nature of Gases • Gases are compressible Why can you put more air in a tire, but can’t add more water to a glass full of water? • Gases have low densities Dsolid or liquid = 2 g/mL Dgas 2 g/L

  4. Nature of Gases 1. Why does a round balloon become spherical when filled with air? 2. Suppose we filled this room halfway with water. Where would pressure be exerted?

  5. Nature of Gases • Gases fill a container completely and uniformly • Gases exert a uniform pressure on all inner surfaces of their containers

  6. Barometers 760 mmHg atm pressure

  7. Unit of Pressure One atmosphere (1 atm) • Is the average pressure of the atmosphere at sea level • Is the standard of pressure • P = Force Area 1.00 atm = 760 mm Hg = 760 torr

  8. Types of Pressure Units Pressure Used in 760 mm Hg or 760 torr Chemistry 14.7 lb/in.2 U.S. pressure gauges 29.9 in. Hg U.S. weather reports 101.3 kPa (kilopascals) Weather in all countries except U.S. 1.013 bars Physics and astronomy

  9. Check Yourself 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 has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense 2) Hg is heavier 3) air is more dense than H2O

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

  11. Check Yourself A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.625 atm 3) 3.61 x 105 atm B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3) 22,300 mm Hg

  12. Solutions A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm (B) 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg 14.7 psi 1.00 atm (B)

  13. 13.2 P and V Changes – Boyles Law P1 P2 V1 V2

  14. Boyle's Law • The pressure of a gas is inversely related to the volume when T does not change • Then the PV product remains constant P1V1 = P2V2 P1V1= 8.0 atm x 2.0 L = 16 atm L P2V2= 4.0 atm x 4.0 L = 16 atm L

  15. PV Problem Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?

  16. PV Calculation Prepare a list of given information: Initial conditions Final conditions P1 = 50 mm Hg P2 = 200 mm Hg V1 = 1.6 L V2 = ? ?

  17. Find New Volume (V2) Solve for V2: P1V2 = P2V2 V2 = V1 x P1 /P2 V2 = 1.6 L x 50 mm Hg = 0.4 L 200 mm Hg

  18. Check Yourself A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain. 1) 3.2 L 2) 6.4 L 3) 12.8 L

  19. Solution A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) 6.4 L x 0.70 atm = 3.2 L (1) 1.40 atm Volume must decrease to cause an increase in the pressure

  20. Check Yourself A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg

  21. Solution A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 600. mm Hg x 12.0 L = 200. mmHg (1) 36.0 L Pressure decrease when volume increases.

  22. 13.3 Charles’ Law – Volume & Temp. V = 125 mL V = 250 mL T = 273 K T = 546 K Observe the V and T of the balloons. How does volume change with temperature?

  23. Charles’ Law: V and T At constant pressure, the volume of a gas is directly related to its absolute (K) temperature V1 = V2 T1 T2

  24. Check Yourself Use Charles’ Law to complete the statements below: 1. If final T is higher than initial T, final V is (greater, or less) than the initial V. 2. If final V is less than initial V, final T is (higher, or lower) than the initial T.

  25. Solution V1 = V2 T1 T2 1. If final T is higher than initial T, final V is (greater) than the initial V. 2. If final V is less than initial V, final T is (lower) than the initial T.

  26. V and T Problem A balloon has a volume of 785 mL on a Fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

  27. VT Calculation Complete the following setup: Initial conditions Final conditions V1 = 785 mL V2 = ? T1 = 21°C = 294 K T2 = 0°C = 273 K V2 = _______ mL x __ K = _______ mL V1 K Check your answer: If temperature decreases, V should decrease.

  28. Check Yourself A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 1) 443°C 2) 170°C 3) - 82°C

  29. Solution A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 2) 170°C T2 = 291 K x 640 mL = 443 K 420 mL = 443 K - 273 K = 170°C

  30. Gay-Lussac’s Law: P and T The pressure exerted by a confined gas is directly related to the temperature (Kelvin) at constant volume. P (mm Hg) T (°C) 936 100 761 25 691 0

  31. Check Yourself Use Gay-Lussac’s law to complete the statements below: 1. When temperature decreases, the pressure of a gas (decreases or increases). 2. When temperature increases, the pressure of a gas (decreases or increases).

  32. Solution 1. When temperature decreases, the pressure of a gas (decreases). 2. When temperature increases, the pressure of a gas (increases).

  33. PT Problem A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant) T = 18°C T = 62°C

  34. PT Calculation ? P1 = 2.0 atm T1 = 18°C + 273 = 291 K P2 = ? T2 = 62°C + 273 = 335 K What happens to P when T increases? P increases (directly related to T) P2 = P1 x T2 T1 P2 = 2.0 atm x K = atm K

  35. Check Yourself Complete with 1) Increases 2) Decreases 3) Does not change A. Pressure _____, when V decreases B. When T decreases, V _____. C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)

  36. Solution 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 (constant n and T) D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)

  37. 13.4 Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume?

  38. Avogadro’s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V1 = V2 n1 n2 initial final

  39. Combined Gas Law P1V1 = P2V2 T1 T2 Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1

  40. Combined Gas Law P1V1 = P2V2 T1 T2 Isolate V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1

  41. Check Yourself Solve the combined gas laws for T2.

  42. Solution Solve the combined gas law for T2. (Hint: cross-multiply first.) P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1

  43. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

  44. Data Table List given information: P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ??

  45. Solution Solve for T2 Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C

  46. Calculation Solve for T2 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

  47. Check Yourself A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

  48. Solution T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K - 273 = - 95°C

  49. Check Yourself True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

  50. Solution True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

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