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Chapter 14 Properties of Gases

Chapter 14 Properties of Gases

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Chapter 14 Properties of Gases

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  1. Chapter 14Properties of Gases

  2. The Properties of Gases • Gas can expand to fill its container • Gases are easily compressed, or squeezed into a smaller volume. • Gases occupy far more space than a liquid or a solid • Compressibility – measure of how much the volume of matter decreases under pressure.

  3. Compressibility of Gases When an airbag triggers due to a sudden stop, a chemical reaction inside the airbag occurs. One product of the reaction is nitrogen gas, which causes the bag to inflate. When a person collides with an inflated air bag, the impact forces the molecules of gas in the bag closer together The compression of the gas absorbs the energy of the impact.

  4. Compressibility of Gases What factors do you think affect the pressure of the air inside the soccer ball? Temperature of the air inside the ball Volume of the ball

  5. Kinetic Theory & Gases What is kinetic energy The energy of motion How are temperature and kinetic energy related? Temperature is a measure of average kinetic energy.

  6. Kinetic Theory & Gases Gases are easily compressed because of the space between the particles in a gas. Under pressure, the particles in a gas are forced closer together, or compressed.

  7. Factors Affecting Gas Pressure Pressure (P) - kPa Volume (V) - liters Temperature (T) - Kelvin Number of moles (n) The amount of gas, volume, and temperature are factors that affect gas pressure

  8. Amount of Gas and Gas Pressure When you inflate an air raft, the pressure inside the raft will increase. (this is a container with a volume that can vary. A balloon is another example) Collisions of particles with the inside walls of the raft result in the pressure that is exerted by the gas. By adding gas, you increase the number of particles. Increasing the number of particles increases the number of collisions, which is why the gas pressure increases.

  9. Amount of Gas & Gas Pressure When a gas is put into a closed rigid container, the pressure increases as more particles are added Because the container is rigid, the volume of the gas is constant. Assume the temperature doesn't change Doubling the number of particles of gas, doubles the pressure. As gas is removed, the pressure inside the container is reduced.

  10. Cause and Effect If the pressure of the gas in a sealed container is lower than the outside air pressure, air will rush into the container when the container is opened. When the pressure of the gas in a sealed container is higher than the outside air pressure, the gas will flow out of the container when the container is unsealed.

  11. Cause and Effect Gas pressure inside a new spray paint can is greater than the air pressure outside the can. As the can is used, the pressure inside the can decreases until there is not enough pressure inside the can to force the paint out.

  12. Volume & Gas Pressure You can raise the pressure exerted by a contained gas by reducing its volume. The more gas is compressed, the greater is the pressure the gas exerts inside the container.

  13. Volume & Gas Pressure When cylinder has a volume of 1 L, the pressure is 100 kPa If volume is halved to 0.5 L, the pressure doubles to 200kPa If volume is doubled to 2.0 L, the pressure of the volume is cut in half to 50 kPa.

  14. Temperature & Gas Pressure A sealed bag of potato chips may bulge at the seams if placed in the sun. Bag bulges because an increase in the temperature of the gas inside the bag causes an increase in its pressure. As gas inside bag is heated, the temperature increases, increasing kinetic energy of the particles, and causing more collisions, thus more pressure.

  15. Temperature & Gas Pressure If volume and amount of gas are constant, when the Kelvin temperature of gas doubles, the gas pressure doubles. Gas in sealed container may generate enormous pressure when heated. For that reason, an aerosol can, even an “empty” one, may explode if thrown onto a fire. As the temperature of an enclosed gas decreases by half, the pressure decreases by half.

  16. Questions What effect would tripling the number of particles of a gas in a closed container have on the pressure exerted? Gas pressure would triple What effect would doubling the volume of an enclosed gas have on the pressure? Gas pressure would decrease by half How does the pressure of an enclosed gas change with increasing temperature? The number and force of collisions increase with temperature, and the pressure increases

  17. Questions Why is a gas easy to compress? Because of the space between particles in a gas List three factors that can affect gas pressure? Temperature, pressure, & amount of gas Why does a collision with an air bag cause less damage than a collision with a steering wheel? Gas in the inflated airbag can be compressed, and absorbs some of the energy from the impact. The solid steering wheel cannot.

  18. Questions If temperature is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one quarter of its original value? The volume would need to increase by a factor of four How does a decrease in temperature affect the pressure of a contained gas If temperature decreases, the pressure will decrease If gas temperature in a container is constant, how could you increase the pressure one hundredfold? Increase the amount of gas in the container one hundredfold.

  19. End of Section 14.1

  20. Boyle’s Law (Pressure & Volume) If the temperature is constant, as the pressure of a gas increase, the volume decreases. Conversely, if the temperature is constant, as the pressure of a gas decreases, the volume increases. Robert Boyl was the first person to study this pressure-volume relationship. In 1662, Boyle proposed a law to describe the relationship.

  21. Boyle’s Law (Pressure & Volume) Boyle’s Law – states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. P1V1 = P2V2

  22. Sample Problem UsingBoyle’s Law A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume the temperature remains constant) What do you think will happen to the volume at a higher temperature knowing what you know already about gases?

  23. Sample Problem UsingBoyle’s Law P1 = 103 kPa P2 = 25.0 kPa V1 = 30.0 L V2 = ? L P1V1 = P2V2 orP1V1 / P2 = V2 V2 = (30.0 L) (103 kPa) 25.0 KPa V2 = 1.24 x 102 L (3 sig figs)

  24. Sample Problem UsingBoyle’s Law Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.50 L of N2O changes from 105 KPa to 40.5 KPa. It the temperature does not change, what will the new volume be? P1 = 105 kPa P2 = 40.5 kPa V1 = 2.50 L V2 = ? L P1V1 = P2V2 orP1V1 / P2 = V2 V2 = (2.50 L) (105 kPa) 40.5 KPa V2 = 6.48 L (3 sig figs)

  25. Sample Problem UsingBoyle’s Law A gas with a volume of 4.00 L at a pressure of 205 KPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant? P1 = 205 kPa P2 = ? kPa V1 = 4.00 L V2 = 12.0 L P1V1 = P2V2 orP1V1 / V2 = P2 P2 = (4.00 L) (205 kPa) 12.0 L P2 = 68.3 KPa (3 sig figs)

  26. Sample Problem UsingBoyle’s Law The volume of a gas at 99.6 KPa and 24ºC is 4.23L. What volume will it occupy at 93.3 KPa and 24ºC? P1 = 99.6 kPa P2 = 93.3 kPa T1 = 24ºC V1 = 4.23 L V2 = ? L T2 = 24ºC P1V1 = P2V2 orP1V1 / P2 = V2 V2 = (4.23 L) (99.6 kPa) 93.3 kPa V2 = 4.52 L (3 sig figs)

  27. Charles’s Law Temperature and Volume As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant. In 1787, French physicist Jacques Charles studies the effect of temperature on the volume of a gas at constant pressure. Charles’s Law – states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. V1 = V2 T1 T2

  28. Sample Problem UsingCharles’s Law A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant? T1 = 24ºC or 297 K V1 = 4.00 L T2 = 58ºC or 331 K V2 = ? L When using gas laws always express the temperatures in kelvins! T1 = 24ºC + 273 = 297 K T2 = 58ºC + 273 = 331 K

  29. Sample Problem UsingCharles’s Law A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant? T1 = 24ºC or 297 K V1 = 4.00 L T2 = 58ºC or 331 K V2 = ? L V1 = V2 or V1T2 = V2 T1 T2 T1 V2 = (4.00 L) (331 K) = 4.46 L 297 K

  30. Sample Problem UsingCharles’s Law If a sample of gas occupies 6.80 L at 325ºC, what will its volume be at 25ºC if the pressure does not change? T1 = 325ºC or 598 K V1 = 6.80 L T2 = 25ºC or 298 K V2 = ? L V1 = V2 or V1T2 = V2 T1 T2 T1 V2 = (6.80 L) (298 K) = 3.39 L 598 K

  31. Gay-Lussac’s LawPressure and Temperature As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant. Joseph Gay-Lussac discovered the relationship between the pressure and the temperature of gas in 1802. Gay-Lussac’s Law – states that the pressure of a gas is directly proportional to the Kelvin temperature if the pressure if the volume remains constant. P1 = P2 T1 T2

  32. Sample Problem UsingGay-Lusaac’s Law A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K? P1 = 6.58 kPa T1 = 539 K P2 = ? kPa T2 = 211 K P1 = P2 or P1T2 = P2 T1 T2 T1 P2 = (6.58 K) (211 K) = 2.58kPa 539 K

  33. Sample Problem UsingGay-Lusaac’s Law The pressure in a car tire is 198 kPa at 27ºC. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. P1 = 198 kPa T1 = 300 K P2 = 225 kPa T2 = ? K P1 = P2 or P2T1 = T2 T1 T2 P1 T2 = (225 kPa) (300 K) = 341 K 198 kPa

  34. Gases If the gas is heated (T2 > T1), the new pressure is greater. (volume constant) If the gas is heated (T2 > T1), the new volume is greater because the gas expands. (pressure constant) If the gas is cooled (T2 < T1), the new pressure is less. (volume constant) If the gas is cooled (T2 < T1), the new volume is smaller because the gas contracts. (pressure constant)

  35. Combined Gas Law There is a single expression that combines Boyle’s, Charles’s and Gay-Lusaac’s Law. The combined gas law describes the relationship among the pressure, temperature, and volume of an enclosed gas. The combined gas law allows you to do calculation for situations in which only the amount of gas is constant P1V1 = P2 V2 T1 T2

  36. Sample Problem UsingCombined Gas Law A gas at 155 kPa and 25º C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125º C. What is the new volume? P1 = 155 kPa T1 = 298 K V1 = 1.00 L P2 = 605 kPa T2 = 398 K V2 = ? P1V1 = P2 V2 or P1V1 T2 = V2 T1 T2 T1 P2 V2 = (155kPa)(1.00 L)(398 K) = 0.342 L (298 K)(605 kPa)

  37. Sample Problem UsingCombined Gas Law A 5.00 L air sample has a pressure of 107 kPa at a temp of -50 º C. If the temperature is raised to 102 º C and the volume expands to 7.00 L, what will the new pressure be? P1 = 107 kPa T1 = 223 K V1 = 1.00 L P2 = ? kPa T2 = 375 K V2 = 7.00 L P1V1 = P2 V2 or P1V1 T2 = P2 T1 T2T1 V2 P2 = (107 kPa)(5.00 L)(375 K) = 1.29 x 102kPa (223K)(7.00 L)

  38. Questions How are the pressure and volume of a gas related at constant temperature? The volume of a gas decreases as the pressure increases. (Boyle’s Law) If pressure is constant, how does a change in temperature affect the volume of a gas? As the temperature increases, the volume increases. (Charles’s Law)

  39. Questions What is the relationship between the temperature and pressure of a contained gas at constant volume? As the temperature increases, the pressure increases. (Gay-Lusaac’s Law) In what situations is the combined gas law useful? Allows you to do calculations when the only constant is the amount of gas.

  40. Question Explain how Charles’s law can be derived from the combined gas law. When the pressure is constant, P1 = P2, the pressure terms cancel, leaving an equation for Charles’s Law. P1V1 = P2 V2 T1 T2

  41. Question The volume of a weather balloon increases as the balloon rises in the atmosphere. Why doesn’t the drop in temperature at higher altitudes cause the volume to decrease? The outside pressure decreases, causing a greater increase in the balloon’s volume The higher it rises, the colder the temperature and the lower the volume. At the same time, atmospheric pressure decreases, allowing the gas to expand.

  42. End of Section 14.2

  43. Ideal Gas Law With the combined gas law, you can solve problems with three variables: temperature, volume & pressure The combined gas law assumes that the amount of gas does not vary. To calculate the number of moles of a contained gas requires an expression that contains the variable n. The number of moles of gas is directly proportional to the number of particles.

  44. Ideal Gas Law The volume occupied by a gas at a specified temperature and pressure also must depend on the number of particles. So… moles must be directly proportional to volume. P1V1 = P2 V2 T1 nT2 n This equation shows that P1V1 is a constant. T1 n This constant holds for ideal gases – gases that conform to the gas laws.

  45. Ideal Gas Law If you know the values for P,V, T and n for one set of condition, you can calculate a value for the constant. 1 mole of every gas occupies 22.4 L at STP. (101.3kPa and 273K) Ideal gas constant uses the symbol R R = P1V1 R= (101.3kPa)(22.4L) T1 n (273 K)(1mol) R = 8.31 L · kPa / mole · K or R = 0.0831 L · atm / mole · K

  46. Ideal Gas Law PV = nRT pressure volume moles constant temperature(K) 8.31L · kPa / mole · K

  47. Sample Problem Using Ideal Gas Law When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 103 kPa. How many moles of helium does the sphere contain? P = 1.89 x 103 V = 685 L T = 621 K PV = nRT or PV / RT = n n = (1.89 x 103 kPa) (685 L)mol · K (8.31L · kPa) (621K) n = 251 mol He

  48. Sample Problem Using Ideal Gas Law A child’s lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a body temperature of 37ºC? Use a molar mass of 29 g for air. P = 102 kPaV = 2.20 L T = 310 K PV = nRT or PV / RT = n n = (102kPa) (2.20 L)mol · K (8.31L · kPa) (310K) n = 0.087 mol air 0.087 mol air x 29g air / mol air = 2.5 g air

  49. Ideal Gases & Real Gases Ideal gas – one that follows the gas laws at all conditions of pressure and temperature. Such a gas would have to conform precisely to the assumptions of kinetic theory. Its particles could have no volume, and there could be no attraction between particles in the gas. There is no gas for which these assumptions are true.

  50. Ideal Gases & Real Gases At many conditions of temperature and pressure, real gases behave very much like an ideal gas. Particles of a real gas do have volume and there are attractions between the particles. Because of these attractions, a gas can condense or solidify when it is compressed or cooled. Example – if water vapor is cooled below 100ºC at standard atmospheric pressure, it condenses to a liquid.