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Gases

Gases. What do we know?. Gases have mass. Gases are easily compressed. Gases uniformly and completely fill their containers. Different gases move through each other quite rapidly. Gases exert pressure. The pressure of a gas depends on its temperature. Measuring Gases.

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Gases

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  1. Gases

  2. What do we know? • Gases have mass. • Gases are easily compressed. • Gases uniformly and completely fill their containers. • Different gases move through each other quite rapidly. • Gases exert pressure. • The pressure of a gas depends on its temperature.

  3. Measuring Gases • Amount of gas (moles = n) n = mass (g) / molar mass (g/mol) • Volume (V) Vgas = Vcontainer Measured in liters

  4. Measuring Gases • Temperature (T) Always measured in Kelvin (K) T(K) = T(oC) + 273 • Pressure (P) Gases exert pressure when particles collide with walls and the force is spread over the area of the container STP = 1 atm pressure, 0oC (273 K)

  5. Measuring Pressure • Pressure of air is measured with a barometer (developed by Torricelli in 1643) • Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up).

  6. Measuring Pressure 1.000 atm (standard atmosphere) = 760.0 mm Hg (millimeters of mercury) = 760 torr (named after Torricelli) = 14.69 psi (pounds per square inch) = 101,325 Pa (pascals) The Pascal is the SI unit for pressure, but it is used sparingly because it is so small. Just use the equalities as conversion factors to get from unit to another.

  7. The Gas Laws • Boyle’s Law (pressure-volume relationship) • Charles’s Law (volume-temperature relationship) • Gay-Lussac’s Law (pressure-temperature relationship) • Combined Gas Law (relates pressure, volume, and temperature) • Dalton’s Law of Partial Pressures • The Ideal Gas Law

  8. Boyle’s Law The pressure of a given sample of a gas is inversely proportional to the volume of the gas at constant temperature. If we know the volume of a gas at a given pressure, we can predict the new volume if the pressure is changed, as long as the temperature and the amount of the gas remain the same. P1V1 = P2V2

  9. Boyle’s Law Freon-12 (the common name for the compound CCl2F2) was once widely used in refrigeration systems, but has now been replaced by other compounds that do not lead to the breakdown of the protective ozone in the upper atmosphere. Consider a 1.5-L sample of gaseous CCl2F2 at a pressure of 56 torr. If the pressure is changed to 150 torr at a constant temperature, will the volume of the gas increase or decrease? What will be the new volume of the gas?

  10. Boyle’s Law • Unknown: • V2 = ? L Given: • P1 = 56 torr • V1 = 1.5 L • P2 = 150 torr • Temperature is constant Pressure increases so volume decreases. Boyle’s Law: P1V1 = P2V2 (56 torr)(1.5 L) = (150 torr) V2 V2 = (56 torr)(1.5 L) / (150 torr) V2 = 0.56 L

  11. Boyle’s Law In an automobile engine the gaseous fuel-air mixture enters the cylinder and is compressed by a moving piston before it is ignited. In a certain engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The fuel-air mixture initially has a pressure of 1.00 atm. Calculate the pressure of the compressed fuel-air mixture, assuming that both the temperature and the amount of gas remain constant.

  12. Boyle’s Law • Unknown: • P2 = ? atm Given: • P1 = 1.00 atm • V1 = 0.725 L • V2 = 0.075 L • Temperature and amount of gas are constant Volume decreases so pressure increases. Boyle’s Law: P1V1 = P2V2 (1.00 atm)(0.725 L) = P2(0.075 L) P2 = (1.00 atm)(0.725 L) / (0.075 L) P2 = 9.7 atm

  13. Charles’s Law The volume of a given sample of a gas is directly proportional to the temperature of the gas at constant pressure. Initial values can also be used to find final values V1/T1 = V2/T2

  14. Charles’s Law A 2.0-L sample of air is collected at 298 K and then cooled to 278 K. The pressure is held constant at 1.0 atm. Does the volume increase or decrease? Calculate the volume of the air at 278 K.

  15. Charles’s Law • Unknown: • V2 = ? L Given: • T1 = 298 K • V1 = 2.0 L • T2 = 278 K • Pressure is constant Temperature decreases so volume decreases.

  16. Charles’s Law V1/T1 = V2/T2 2.0 L V2 = 298 K 278 K (2.0 L)(278 K) = V2 (298K) 278 K 2.0 L x V2 = 298 K V2 = 1.9 L

  17. Charles’s Law A sample of gas at 15oC (at 1 atm) has a volume of 2.58 L. The temperature is then raised to 38oC (at 1 atm). Does the volume of the gas increase or decrease? Calculate the new volume.

  18. Charles’s Law Given: • T1 = 15oC • V1 = 2.58 L • T2 = 38oC • Pressure is constant Temperature increases so volume increases • Unknown: • V2 = ? L + 273 = 288 K + 273 = 311 K

  19. Charles’s Law V1/T1 = V2/T2 2.58 L V2 = 288 K 311 K (2.58 L)(311 K) = V2 (288K) x 2.58 L 311 K V2 = 288 K V2 = 2.79 L

  20. Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain an equal number of particles. Initial values can also be used to find final values V1/n1 = V2/n2

  21. Gay-Lussac’s Law The pressure of a given sample of a gas is directly proportional to the temperature of the gas at constant volume. Initial values can also be used to find final values P1/T1 = P2/T2

  22. Gay-Lussac’s Law Calculate the final pressure inside a scuba tank after it cools from 1000 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm.

  23. Gay-Lussac’s Law • Unknown: • P2 = ? atm Given: • P1 = 130 atm • T1 = 1000 oC • T2 = 25 oC • Volume is constant Temperature decreases so pressure decreases. + 273 = 1273 K + 273 = 298 K

  24. Gay-Lussac’s Law P1/T1 = P2/T2 130.0 atm P2 = 1273 K 298 K (130 atm)(298 K) = P2 (1273 K) 130 atm x 298 K P2 = 1273 K P2 = 30.4 atm

  25. Gay-Lussac’s Law If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be?

  26. Gay-Lussac’s Law • Unknown: • T2 = ? K Given: • P1 = 15 atm • P2 = 16 atm • T1 = 25.0 oC • Volume is constant Pressure increases so temperature increases. + 273 = 298 K

  27. Gay-Lussac’s Law P1/T1 = P2/T2 15.0 atm 16.0 atm = 298 K T2 (15 atm)T2 = (16 atm)(298 K) 16 atm x 298 K T2 = 15 atm T2 = 318 K

  28. Summary of Laws

  29. Combined Gas Law

  30. Combined Gas Law • Combines Boyle’s, Charles’s, and Gay-Lussac’s laws to relate the pressure, volume, and temperature of constant amounts of gases. Initial values can also be used to find final values P1V1/T1 = P2V2/T2

  31. Combined Gas Law A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature above where the balloon is released is 20oC, what will happen to the volume when the ballooon rises to an altitude where the pressure is 0.65 atm and the temperature is -15oC?

  32. Combined Gas Law • Unknown: • V2 = ? L Given: • P1 = 1.05 atm • V1 = 5.0 L • T1 = 20 oC • P2 = 0.65 atm • T2 = -15 oC + 273 = 293 K + 273 = 258 K

  33. Combined Gas Law P1V1/T1 = P2V2/T2 (1.05 atm)(5.0 L) (0.65 atm)V2 = 293 K 258 K (1.05 atm)(5.0 L)(258 K) = (0.65 atm)V2(293 K) (1.05 atm)(5.0 L)(258 K) V2 = (0.65 atm)(293 K) V2 = 7.1 L

  34. Combined Gas Law If I initially have 4.0 L of gas at a pressure of 1.1 atm and a temperature of 25oC, what will happen to the pressure if I decrease the volume to 3.0 L and decrease the temperature to 20oC?

  35. Combined Gas Law • Unknown: • P2 = ? L Given: • V1 = 4.0 L • P1 = 1.1 atm • T1 = 25 oC • V2 = 3.0 L • T2 = 20 oC + 273 = 298 K + 273 = 293 K

  36. Combined Gas Law P1V1/T1 = P2V2/T2 (1.1 atm)(4.0 L) P2(3.0 L) = 298 K 293 K (1.1 atm)(4.0 L)(293 K) = P2(3.0 L)(298 K) (1.1 atm)(4.0 L)(293 K) P2 = (3.0 L)(298 K) P2 = 1.4 atm

  37. The Ideal Gas Law

  38. The Ideal Gas Law Combines Boyle’s, Charles’s, and Avogadro’s Laws to describe the behavior of gases as dependent upon volume, pressure, temperature, and the number of moles present. PV = nRT where R is the combined constant called the universal gas constant R = 0.08206 L atm / mol K

  39. The Ideal Gas Law A gas that obeys this equation is said to behave ideally, thus the name “Ideal Gas Law” There is no such thing as an ideal gas, but many real gases behave ideally at pressures of approximately 1 atm and lower and temperatures of approximately 273 K or higher.

  40. The Ideal Gas Law A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0oC and a pressure of 1.5 atm. Calculate the number of moles of H2 present in this gas sample. Assume that the gas behaves ideally.

  41. The Ideal Gas Law Given: • P = 1.5 atm • T = 0oC = 273 K • V = 8.56 L PV = nRT (1.5 atm)(8.56 L) = n(0.08206 L-atm/mol-K)(273 K) • Unknown: • n = ? (1.5 atm) (8.56 L) n = (0.08206 L-atm/mol-K) (273 K) n = 0.57 mol

  42. The Ideal Gas Law What volume is occupied by 0.250 mol of carbon dioxide gas at 25oC and 371 torr? Because the gas constant is measured in L-atm/mol-K, volume must be measured in liters, pressure in atmospheres, amount of gas in moles, and temperature in Kelvin. 25oC + 273 = 298 K 1.00 atm 371 torr x = 0.488 atm 760 torr

  43. The Ideal Gas Law Given: • P = 0.488 atm • T = 298 K • n = 0.250 mol CO2 PV = nRT (0.488 atm)V = (0.250 mol)(0.08206 L-atm/mol-K)(298K) • Unknown: • V = ? L (0.08206 L-atm/mol-K) (0.250 mol) (298 K) V = (0.488 atm) V = 12.5 L

  44. Dalton’s Law of Partial Pressures

  45. Dalton’s Law of Partial Pressures For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. Ptotal = P1 + P2 + P3 Each gas is responsible for only part of the total pressure.

  46. Dalton’s Law of Partial Pressures What is the atmospheric pressure if the partial pressure of nitrogen, oxygen, and argon are 604.5 mm Hg, 162.8 mm Hg, and 0.5 mm Hg, respectively? PT = PN2 + PO2 + PAr PT = 604.5 mm Hg + 162.8 mm Hg + 0.5 mm Hg PT = 767.8 mm Hg

  47. Dalton’s Law of Partial Pressures A gas mixture contains hydrogen, helium, neon, and argon. The total pressure of the mixture is 93.6 kPa. The partial pressures of helium, neon, and argon are 15.4 kPa, 25.7 kPa, and 35.6 kPa, respectively. What is the pressure exerted by hydrogen? PT = PH2 + PHe + PNe + PAr 93.6 kPa = PH2 + 15.4 kPa + 25.7 kPa + 35.6 kPa PH2 = 93.6 – (15.4 + 25.7 +35.6) = 16.9 kPa

  48. Dalton’s Law of Partial Pressures Mixtures of helium and oxygen are used in the “air” tanks of underwater divers for deep dives. For a particular dive, 12 L of O2 at 25oC and 1.0 atm and 46 L of He at 25oC and 1.0 atm were pumped into a 5.0-L tank. Calculate the partial pressure of each gas and the total pressure in the tank at 25oC.

  49. Dalton’s Law of Partial Pressures Assuming that each gas behaves ideally we can calculate the partial pressure of each gas from the ideal gas law: P = (nRT)/V By combining the formulas, we get P = ntotal(RT/V) We need to calculate the number of moles of each gas present in the mixture.

  50. Dalton’s Law of Partial Pressures OXYGEN: V = 12 L P = 1 atm *note: this pressure is not what the oxygen is exerting T = 25oC = 298 K PV=nRT (1 atm)(12 L) = n(0.08206 L-atm/mol-K)(298 K) n = [(1atm)(12 L)] / [0.08206 L-atm/mol-K)(298 K)] n = 0.49 mol

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