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  1. Chapter 5Gases Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

  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. Practice – What happens to the height of the column of mercury in a mercury barometer as you climb to the top of a mountain? • The height of the column increases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure increases with increasing altitude Tro: Chemistry: A Molecular Approach, 2/e 10

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

  12. 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

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

  14. 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

  15. 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 pressures 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, 2/e

  16. Manometer for this sample, the gas has a larger pressure than the atmosphere, so Tro: Chemistry: A Molecular Approach, 2/e

  17. 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

  18. 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, 2/e

  19. Tro: Chemistry: A Molecular Approach, 2/e

  20. Tro: Chemistry: A Molecular Approach, 2/e

  21. Boyle’s Experiment, P x V Tro: Chemistry: A Molecular Approach, 2/e

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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 2.78 x 103 mL, what was it originally? Tro: Chemistry: A Molecular Approach, 2/e

  27. 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

  28. 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

  29. If you plot volume vs. temperature for any gas at constant pressure, the points will all 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 Tro: Chemistry: A Molecular Approach, 2/e

  30. 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

  31. 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

  32. 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, 2/e

  33. 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

  34. 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

  35. 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

  36. Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy? Tro: Chemistry: A Molecular Approach, 2/e

  37. Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy? V1, n1, n2 V2 Given: Find: V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol V2 Conceptual Plan: Relationships: Solution: Check: because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does Tro: Chemistry: A Molecular Approach, 2/e

  38. 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

  39. 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

  40. 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

  41. 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, 2/e

  42. 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

  43. Practice — Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C Tro: Chemistry: A Molecular Approach, 2/e

  44. 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

  45. 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

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

  47. Practice — How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2?2 PbO2(s)→2 PbO(s) + O2(g)(PbO2 =239.2, O2 = 32.00) Tro: Chemistry: A Molecular Approach, 2/e

  48. Practice — How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2?2 PbO2(s)→2 PbO(s) + O2(g) g PbO2 mol PbO2 mol O2 L O2 Given: Find: 100.0 g PbO2, 2 PbO2→ 2 PbO + O2 L O2 Conceptual Plan: Relationships: 1 mol O2 = 22.4 L, 1 mol PbO2 = 239.2g, 1 mol O2≡ 2 mol PbO2 Solution: because less than 1 mole PbO2, and ½ moles of O2 as PbO2, the number makes sense Check: Tro: Chemistry: A Molecular Approach, 2/e

  49. 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, 2/e

  50. Practice – Calculate the density of N2(g) at STP Tro: Chemistry: A Molecular Approach, 2/e