Download
chapter 5 gases n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 5 Gases PowerPoint Presentation
Download Presentation
Chapter 5 Gases

Chapter 5 Gases

174 Views Download Presentation
Download Presentation

Chapter 5 Gases

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro Chapter 5Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

  2. Air Pressure & Shallow Wells • water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface • the pump removes air from the pipe, decreasing the air pressure in the pipe • the outside air pressure then pushes the water up the pipe • the maximum height the water will rise is related to the amount of pressure the air exerts Tro, Chemistry: A Molecular Approach

  3. Atmospheric Pressure • pressure is the force exerted over an area • on average, the air exerts the same pressure that a column of water 10.3 m high would exert • 14.7 lbs./in2 • so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m Tro, Chemistry: A Molecular Approach

  4. Gases Pushing • gas molecules are constantly in motion • as they move and strike a surface, they push on that surface • push = force • if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting • pressure = force per unit area Tro, Chemistry: A Molecular Approach

  5. The Effect of Gas Pressure • the pressure exerted by a gas can cause some amazing and startling effects • whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure • the bigger the difference in pressure, the stronger the flow of the gas • if there is something in the gas’s path, the gas will try to push it along as the gas flows Tro, Chemistry: A Molecular Approach

  6. Atmospheric Pressure Effects • differences in air pressure result in weather and wind patterns • the higher up in the atmosphere you climb, the lower the atmospheric pressure is around you • at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi • rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum Tro, Chemistry: A Molecular Approach

  7. Pressure Imbalance in Ear If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a “popped eardrum.” Tro, Chemistry: A Molecular Approach

  8. The Pressure of a Gas • result of the constant movement of the gas molecules and their collisions with the surfaces around them • the pressure of a gas depends on several factors • number of gas particles in a given volume • volume of the container • average speed of the gas particles Tro, Chemistry: A Molecular Approach

  9. gravity Measuring Air Pressure • use abarometer • column of mercury supported by air pressure • force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury Tro, Chemistry: A Molecular Approach

  10. Common Units of Pressure Tro, Chemistry: A Molecular Approach

  11. psi atm mmHg Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? Given: Find: 132 psi mmHg Concept Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg Solution: Check: since mmHg are smaller than psi, the answer makes sense

  12. Manometers • the pressure of a gas trapped in a container can be measured with an instrument called a manometer • manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other • a competition is established between the pressure of the atmosphere and the gas • the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere Tro, Chemistry: A Molecular Approach

  13. Manometer for this sample, the gas has a larger pressure than the atmosphere, so Tro, Chemistry: A Molecular Approach

  14. Boyle’s Law • pressure of a gas is inversely proportional to its volume • constant T and amount of gas • graph P vs V is curve • graph P vs 1/V is straight line • as P increases, V decreases by the same factor • P x V = constant • P1 x V1 = P2 x V2 Tro, Chemistry: A Molecular Approach

  15. Boyle’s Experiment • added Hg to a J-tube with air trapped inside • used length of air column as a measure of volume Tro, Chemistry: A Molecular Approach

  16. Tro, Chemistry: A Molecular Approach

  17. Tro, Chemistry: A Molecular Approach

  18. Boyle’s Experiment, P x V Tro, Chemistry: A Molecular Approach

  19. When you double the pressure on a gas, the volume is cut in half (as long as the temperature and amount of gas do not change) Tro, Chemistry: A Molecular Approach

  20. Boyle’s Law and Diving if your tank contained air at 1 atm pressure you would not be able to inhale it into your lungs • since water is denser than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm • at 20 m the total pressure is 3 atm Tro, Chemistry: A Molecular Approach

  21. V1, P1, P2 V2 Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? Given: Find: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm V2, L Concept Plan: Relationships: P1∙ V1 = P2∙ V2 Solution: Check: since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

  22. Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally? Tro, Chemistry: A Molecular Approach

  23. V1, P1, P2 V2 A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally? Given: Find: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm V1, mL Concept Plan: Relationships: P1∙ V1 = P2∙ V2 , 1 atm = 760 torr (exactly) Solution: Check: since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

  24. Charles’ Law • volume is directly proportional to temperature • constant P and amount of gas • graph of V vs T is straight line • as T increases, V also increases • Kelvin T = Celsius T + 273 • V = constant x T • if T measured in Kelvin Tro, Chemistry: A Molecular Approach

  25. Charles’ Law – A Molecular View • the pressure of gas inside and outside the balloon are the same • at high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger • the pressure of gas inside and outside the balloon are the same • at low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small Tro, Chemistry: A Molecular Approach

  26. The data fall on a straight line. If the lines are extrapolated back to a volume of “0,” they all show the same temperature, -273.15°C, called absolute zero

  27. T(K) = t(°C) + 273.15, V1, V2, T2 T1 Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L? Given: Find: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C t1, K and °C Concept Plan: Relationships: Solution: Check: since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

  28. Practice – The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? Tro, Chemistry: A Molecular Approach

  29. T(K) = t(°C) + 273.15, V1, T1, T2 V2 The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? Given: Find: V1 =10.0 L, t1 = 25.0°C L, t2 = 250.0°C V2, L Concept Plan: Relationships: Solution: Check: since T and V are directly proportional, when the temperature increases, the volume should increase, and it does

  30. Avogadro’s Law • volume directly proportional to the number of gas molecules • V = constant x n • constant P and T • more gas molecules = larger volume • count number of gas molecules by moles • equal volumes of gases contain equal numbers of molecules • the gas doesn’t matter Tro, Chemistry: A Molecular Approach

  31. mol added = n2 – n1, V1, V2, n1 n2 Example 5.4 – A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? Given: Find: V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol n2, and added moles Concept Plan: Relationships: Solution: Check: since n and V are directly proportional, when the volume increases, the moles should increase, and it does

  32. By combing the gas laws we can write a general equation • R is called the gas constant • the value of R depends on the units of P and V • we will use 0.08206 and convert P to atm and V to L • the other gas laws are found in the ideal gas law if • two variables are kept constant • allows us to find one of the variables if we know the other 3 Ideal Gas Law Tro, Chemistry: A Molecular Approach

  33. 1 atm = 14.7 psi T(K) = t(°C) + 273.15 P, V, T, R n Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? Given: Find: V = 3.24 L, P = 24.3 psi, t = 25 °C, n, mol Concept Plan: Relationships: Solution: Check: 1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

  34. Standard Conditions • since the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions • STP • standard pressure = 1 atm • standard temperature = 273 K • 0°C Tro, Chemistry: A Molecular Approach

  35. Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions? Tro, Chemistry: A Molecular Approach

  36. 1 atm = 14.7 psi T(K) = t(°C) + 273.15 P1, V1, T1, R n P2, n, T2, R V2 A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions? Given: Find: V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C V2, L Concept Plan: Relationships: Solution: Check: 1 mole at STP occupies 22.4 L, since there is more than 1 mole, we expect more than 22.4 L of gas

  37. Molar Volume • solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L • 6.022 x 1023 molecules of gas • notice: the gas is immaterial • we call the volume of 1 mole of gas at STP the molar volume • it is important to recognize that one mole of different gases have different masses, even though they have the same volume Tro, Chemistry: A Molecular Approach

  38. Molar Volume Tro, Chemistry: A Molecular Approach

  39. Density at Standard Conditions • density is the ratio of mass-to-volume • density of a gas is generally given in g/L • the mass of 1 mole = molar mass • the volume of 1 mole at STP = 22.4 L Tro, Chemistry: A Molecular Approach

  40. Gas Density • density is directly proportional to molar mass Tro, Chemistry: A Molecular Approach

  41. 1 atm = 760 mmHg, MM = 28.01 g T(K) = t(°C) + 273.15 P, MM, T, R d Example 5.7 – Calculate the density of N2at 125°C and 755 mmHg Given: Find: P = 755 mmHg, t = 125 °C, dN2,g/L Concept Plan: Relationships: Solution: since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is Check:

  42. Molar Mass of a Gas • one of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law Tro, Chemistry: A Molecular Approach

  43. n, m P, V, T, R MM n 1 atm = 760 mmHg, T(K) = t(°C) + 273.15 Example 5.8 – Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg Given: Find: m=0.311g, V=0.225 L, P=886 mmHg, t=55°C, molar mass,g/mol m=0.311g, V=0.225 L, P=1.1658 atm, T=328 K, molar mass,g/mol Concept Plan: Relationships: Solution: Check: the value 31.9 g/mol is reasonable

  44. Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Tro, Chemistry: A Molecular Approach

  45. V, m P, n, T, R d V 1 atm = 760 mmHg, T(K) = t(°C) + 273.15 Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Given: Find: m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C, density,g/L m=9.988g, n=0.250 mol, P=1.0197 atm, T=300. K density,g/L Concept Plan: Relationships: Solution: Check: the value 1.65 g/L is reasonable

  46. Mixtures of Gases • when gases are mixed together, their molecules behave independent of each other • all the gases in the mixture have the same volume • all completely fill the container  each gas’s volume = the volume of the container • all gases in the mixture are at the same temperature • therefore they have the same average kinetic energy • therefore, in certain applications, the mixture can be thought of as one gas • even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance • we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules Tro, Chemistry: A Molecular Approach

  47. Partial Pressure • the pressure of a single gas in a mixture of gases is called its partial pressure • we can calculate the partial pressure of a gas if • we know what fraction of the mixture it composes and the total pressure • or, we know the number of moles of the gas in a container of known volume and temperature • the sum of the partial pressures of all the gases in the mixture equals the total pressure • Dalton’s Law of Partial Pressures • because the gases behave independently Tro, Chemistry: A Molecular Approach

  48. Composition of Dry Air Tro, Chemistry: A Molecular Approach

  49. The partial pressure of each gas in a mixture can be calculated using the ideal gas law Tro, Chemistry: A Molecular Approach

  50. Ptot, PHe, PNe PAr PAr, V, T nAr mAr Ptot = Pa + Pb + etc., 1 atm = 760 mmHg, MMAr = 39.95 g/mol Example 5.9 – Determine the mass of Ar in the mixture PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg, V = 1.00 L, T=298 K massAr,g PAr = 0.275 atm, V = 1.00 L, T=298 K massAr,g Given: Find: Concept Plan: Relationships: PAr = Ptot – (PHe + PNe) Solution: Check: the units are correct, the value is reasonable