1 / 56

Chapter 5 Gases

Chapter 5 Gases. Chemistry: A Molecular Approach , 2nd Ed. Nivaldo Tro. April Senger MALMSTROM Park University Great Falls Montana. The Structure of a Gas. Gases are composed of particles that are flying around very fast in their container(s)

sybill-brooks
Download Presentation

Chapter 5 Gases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 5Gases Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro April Senger MALMSTROM Park University Great Falls Montana

  2. The Structure of a Gas • Gases are composed of particles that are flying around very fast in their container(s) • The particles in straight lines until they encounter either the container wall or another particle, then they bounce off • If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there Tro: Chemistry: A Molecular Approach, 2/e

  3. 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, 2/e

  4. 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 an area of high pressure to an area of 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, 2/e

  5. Air Pressure • The atmosphere exerts a pressure on everything it contacts • the atmosphere goes up about 370 miles, but 80% is in the first 10 miles from the earth’s surface • This is the same pressure that a column of water would exert if it were about 10.3 m high Tro: Chemistry: A Molecular Approach, 2/e

  6. Atmospheric Pressure Effects • Differences in air pressure result in weather and wind patterns • The higher 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 Tro: Chemistry: A Molecular Approach, 2/e

  7. Pressure Imbalance in the 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, 2/e

  8. The Pressure of a Gas • Gas pressure is a 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, 2/e

  9. gravity Measuring Air Pressure • We measure air pressure with abarometer • Column of mercury supported by air pressure • Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury Tro: Chemistry: A Molecular Approach, 2/e

  10. Common Units of Pressure Tro: Chemistry: A Molecular Approach, 2/e

  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 Conceptual Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg Solution: Check: because mmHg are smaller than psi, the answer makes sense Tro: Chemistry: A Molecular Approach, 2/e

  12. Practice—Convert 45.5 psi into kPa Tro: Chemistry: A Molecular Approach, 2/e 13

  13. psi atm kPa Practice—Convert 45.5 psi into kPa Given: Find: 645.5 psi kPa Conceptual Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 101.325 kPa Solution: Check: because kPa are smaller than psi, the answer makes sense Tro: Chemistry: A Molecular Approach, 2/e

  14. Boyle’s Law Robert Boyle (1627–1691) • 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, 2/e

  15. Boyle’s Law: A Molecular View • Pressure is caused by the molecules striking the sides of the container • When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant • This results in increasing the pressure Tro: Chemistry: A Molecular Approach, 2/e

  16. Boyle’s Law and Diving • Because water is more dense 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 • If your tank contained air at 1 atm of pressure, you would not be able to inhale it into your lungs • you can only generate enough force to overcome about 1.06 atm Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you. This allows you to take in air even when the outside pressure is large. Tro: Chemistry: A Molecular Approach, 2/e

  17. Boyle’s Law and Diving • If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm • According to Boyle’s law, what should happen to the volume of air in the lungs? • Because the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!! Tro: Chemistry: A Molecular Approach, 2/e

  18. 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 Conceptual Plan: Relationships: P1∙ V1 = P2∙ V2 Solution: Check: because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does Tro: Chemistry: A Molecular Approach, 2/e

  19. V2, P1, P2 V1 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 2.78x 103 mL, what was it originally? Given: Find: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm V1, mL Conceptual Plan: Relationships: P1∙ V1 = P2∙ V2 , 1 atm = 760 torr (exactly) Solution: Check: because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does Tro: Chemistry: A Molecular Approach, 2/e

  20. Charles’s Law Jacques Charles (1746–1823) • 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, 2/e

  21. Charles’s 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, 2/e

  22. 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 Conceptual Plan: Relationships: Solution: Check: because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does Tro: Chemistry: A Molecular Approach, 2/e

  23. 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? V1, T1, T2 V2 Given: Find: V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C V2, L Conceptual Plan: Relationships: Solution: Check: when the temperature increases, the volume should increase, and it does Tro: Chemistry: A Molecular Approach, 2/e

  24. Avogadro’s Law Amedeo Avogadro (1776–1856) • 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, 2/e

  25. 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 Conceptual Plan: Relationships: Solution: Check: because n and V are directly proportional, when the volume increases, the moles should increase, and they do Tro: Chemistry: A Molecular Approach, 2/e

  26. Ideal Gas Law • By combining 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 three Tro: Chemistry: A Molecular Approach, 2/e

  27. 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 Conceptual Plan: Relationships: Solution: 1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas Check: Tro: Chemistry: A Molecular Approach, 2/e

  28. Standard Conditions • Because 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, 2/e

  29. 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 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C V2, L Conceptual Plan: Relationships: Solution: Check: 1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas Tro: Chemistry: A Molecular Approach, 2/e

  30. Practice—Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C. P, n, T, R g V n Given: Find: mSO2 = 637 g, P = 6.08 x 104 mmHg, t= −23 °C, V, L Conceptual Plan: Relationships: Solution: Tro: Chemistry: A Molecular Approach, 2/e

  31. 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 measures of different gases have different masses, even though they have the same volume Tro: Chemistry: A Molecular Approach, 2/e

  32. Molar Volume Tro: Chemistry: A Molecular Approach, 2/e

  33. 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, 2/e

  34. 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 n, m P, V, T, R MM n 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 Conceptual Plan: Relationships: Solution: Check: the unit is correct, the value is reasonable Tro: Chemistry: A Molecular Approach, 2/e 34

  35. Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 380 torr and 127 °C? n, m P, V, T, R MM n Given: Find: m=12.0 g, V= 197 L, P=380 torr, t=127°C, molar mass,g/mol m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K, molar mass,g/mol Conceptual Plan: Relationships: Solution: Check: the unit is correct and the value is reasonable Tro: Chemistry: A Molecular Approach, 2/e

  36. 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, 2/e

  37. Composition of Dry Air Tro: Chemistry: A Molecular Approach, 2/e

  38. 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, 2/e

  39. The partial pressure of each gas in a mixture can be calculated using the ideal gas law Tro: Chemistry: A Molecular Approach, 2/e

  40. Ptot, PHe, PNe PAr PAr, V, T nAr mAr 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: Conceptual Plan: Relationships: PAr = Ptot – (PHe + PNe) Solution: Check: the units are correct, the value is reasonable Tro: Chemistry: A Molecular Approach, 2/e

  41. nXe, V, T, R PXe Ptot, PXe PNe Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe Given: Find: Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm Conceptual Plan: Relationships: Solution: the unit is correct, the value is reasonable Check: Tro: Chemistry: A Molecular Approach, 2/e

  42. Properties of Gases • Expand to completely fill their container • Take the shape of their container • Low density • much less than solid or liquid state • Compressible • Mixtures of gases are always homogeneous • Fluid Tro: Chemistry: A Molecular Approach, 2/e

  43. Kinetic Molecular Theory • The particles of the gas (either atoms or molecules) are constantly moving • The attraction between particles is negligible • When the moving gas particles hit another gas particle or the container, they do not stick; but they bounce off and continue moving in another direction • like billiard balls Tro: Chemistry: A Molecular Approach, 2/e

  44. Kinetic Molecular Theory • There is a lot of empty space between the gas particles • compared to the size of the particles • The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature • as you raise the temperature of the gas, the average speed of the particles increases • but don’t be fooled into thinking all the gas particles are moving at the same speed!! Tro: Chemistry: A Molecular Approach, 2/e

  45. Gas Properties Explained – Pressure • Because the gas particles are constantly moving, they strike the sides of the container with a force • The result of many particles in a gas sample exerting forces on the surfaces around them is a constant pressure Tro: Chemistry: A Molecular Approach, 2/e

  46. Gas Properties Explained – Indefinite Shape and Indefinite Volume Because the gas molecules have enough kinetic energy to overcome attractions, they keep moving around and spreading out until they fill the container. As a result, gases take the shape and the volume of the container they are in. Tro: Chemistry: A Molecular Approach, 2/e

  47. Gas Properties Explained - Compressibility Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can be squeezed closer together Tro: Chemistry: A Molecular Approach, 2/e

  48. Gas Properties Explained – Low Density Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume; the result is they have low density Tro: Chemistry: A Molecular Approach, 2/e

  49. Density & Pressure • Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them • When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure • also higher density Tro: Chemistry: A Molecular Approach, 2/e

  50. Gas Laws Explained – Boyle’s Law • Boyle’s Law says that the volume of a gas is inversely proportional to the pressure • Decreasing the volume forces the molecules into a smaller space • More molecules will collide with the container at any one instant, increasing the pressure Tro: Chemistry: A Molecular Approach, 2/e

More Related