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Unit 6 Gases and Gas Laws

Unit 6 Gases and Gas Laws. Gases in the Atmosphere. The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity. By volume, dry air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO 2 , and small amounts of other gases. Air Pollution.

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Unit 6 Gases and Gas Laws

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  1. Unit 6Gases and Gas Laws

  2. Gases in the Atmosphere • The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity. • By volume, dry airis 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases.

  3. Air Pollution • Human activity has polluted the air with other gases: • Sulfur Oxides (SO2 & SO3) – produced from coal burning. Contribute to acid rain. • Nitrogen Oxides (NO & NO2) – produced by burning fossil fuels. Contribute to acid rain. • Carbon Monoxide (CO) – emitted by motor vehicles. • Ground-level Ozone (O3) – produced when products offossil fuel combustion reactin the presence of sunlight.

  4. The Ozone Layer • O3 in the troposphere (ground-level ozone) is a pollutant, but O3 in the stratosphere is a necessary part of our atmosphere. • Stratospheric O3 protectsus by absorbing UV light. • CFCs destroy stratospheric O3, and have been banned in the US. • Ozone: Good up high,bad nearby.

  5. Atmospheric Pressure • Atmospheric pressureis the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it. Force Pressure = Area

  6. Measuring Pressure • A common unit of pressure is millimeters of mercury(mm Hg). • 1 mm Hg is also called 1torr in honor of Evangelista Torricelli whoinvented the barometer (used tomeasure atmospheric pressure). • The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm)of pressure is 760 mm Hg.

  7. Measuring Pressure (continued) • The pressure of a gas sample in the laboratory is often measured with a manometer. • the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere. For this sample, the gas has a larger pressure than the atmosphere.

  8. Measuring Pressure (continued) • Pressure can also be measured in pascals (Pa):1 Pa = 1 N/m2. • One pascal is verysmall, so usually kilopascals (kPa) are used instead. • Oneatm is equal to 101.3 kPa. 1 atm = 760 mm Hg (Torr) = 101.3 kPa

  9. Units of Pressure

  10. Converting PressureSample Problem The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in: a. millimeters of mercury (mm Hg) b. kilopascals (kPa) 760 mm Hg x 0.830 atm 631 mm Hg = 1 atm 101.3 kPa x 0.830 atm 84.1 kPa = 1 atm

  11. Dalton’s Law of Partial Pressures • Dalton’s law of partial pressures- the total pressure of a gas mixture is the sum of the partial pressures of the component gases. PT = P1 + P2 + P3 …

  12. Dalton’s Law of Partial PressuresSample Problem A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm. • What is the total pressure of this system? b. What is the total pressure in mm Hg? PT = P1 + P2 + P3 … PT = + 0.7 atm = 2.4 atm 0.5 atm + 1.2 atm 760 mm Hg x 2.4 atm 1800mm Hg = 1 atm

  13. The Kinetic-Molecular Theory: A Model for Gases • Matter is composed of particles which are constantly moving. • The average kinetic energyof a particle is proportionalto its Kelvin temperature. • The size of a particle is negligibly small. • Collisions are completelyelastic – energy may beexchanged, but not lost(like billiard balls.)

  14. Ideal Gases • The kinetic-molecular theory assumes: • no attractions between gas molecules • gas molecules do not take up space • An Ideal Gas is a hypothetical gas that perfectly fits the assumptions of the kinetic-molecular theory. • Many gases behave nearlyideally if pressure is not veryhigh and temperature is not very low.

  15. Properties of Gases: Fluidity • Gas particles glide easily past one another.Because liquids and gases flow, they are both referred to as fluids.

  16. Properties of Gases: Expansion • Since there’s no significantattraction between gasmolecules, they keep moving around and spreading out until they fill their container. • As a result, gases take the shape and the volume of the container they are in.

  17. Properties of Gases: Low Density • Gas particles are very far apart.There is a lot of unoccupied space in the structure of a gas. • Since gases do not have a lot of mass in a given volume, they have a very low density • The density of a gas is about 1/1000 the density of the same substance as a liquid or solid.

  18. Properties of Gases: Compressibility • Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can easily be squeezed closer together.

  19. Diffusion and Effusion • Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion. • Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

  20. Rate of Diffusion • Light molecules move faster than heavy ones. • The greater the molar mass of a gas,the slower it will diffuse and/or effuse.

  21. Graham’s Law of Effusion • Graham’s Law of Effusion states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas. • The ratio of effusion rates of two different gases is given by the following equation:

  22. Graham’s Law of EffusionSample Problem Calculate the molar mass of a gas that effuses at a rate 0.462 times N2 . Solution: 28.0 g/mol (0.462)2 = MMunknown 28.0 g/mol 28.0 g/mol = = = 131 g/mol MMunknown (0.462)2 (0.213)

  23. Gases and Pressure • Gas pressure is caused by collisions of the gas molecules with each other and with the walls of their container. • The greater the number of collisions, the higher the pressure will be.

  24. Pressure – Volume Relationship • When the volume of a gas is decreased, more collisions will occur. • Pressure is caused by collisions. • Therefore, pressure will increase.

  25. Boyle’s Law • Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature. • P1 and V1 representinitial conditions, andP2 and V2 representanother set of conditions. P1V1 = P2V2

  26. Boyle’s LawSample Problem A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Solution: P1V1 = P2V2 (0.947 atm) = (150.0 mL) (0.987 atm) V2 (0.947 atm) (150.0 mL) = = V2 144 mL (0.987 atm)

  27. Volume – Temperature Relationship • the pressure of gas inside and outside the balloon are the same. • at low temperatures, the gas molecules don’t move as much – therefore the volume is small. • at high temperatures, the gas molecules move more – causing the volume to become larger.

  28. Charles’s Law • Charles’s Law– The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. V1 V2 = T1 T2 • V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions.

  29. The Kelvin Temperature Scale • Absolute zero – The theoretical lowest possible temperature where all molecular motion stops. • The Kelvin temperature scale starts at absolute zero (-273oC.) • This gives the followingrelationship between the two temperature scales: K = oC + 273

  30. Charles’s LawSample Problem A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Solution: K = oC + 273 V1 V2 T1 = 25 + 273 = 298 = T1 T2 T2 = 50 + 273 = 323 752 mL V2 = 298 K 323 K 752 mL V2 = x = 323 K 815 mL 298 K

  31. Pressure – Temperature Relationship • Increasing temperature means increasing kinetic energy of the particles. • The energy and frequency of collisions depend on the average kinetic energy of the molecules. • Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.

  32. Gay-Lussac’s Law • Gay-Lussac’s Law– The pressure of a fixed mass of gas varies directly with the Kelvin temperature. • P1 and T1 representinitial conditions.P2 and T2 representanother set of conditions. P1 P2 = T1 T2

  33. Gay-Lussac’s LawSample Problem The gas in a container is at a pressure of 3.00 atm at 25°C. What would the gas pressure in the container be at 52°C? Solution: K = oC + 273 P1 P2 T1 = 25 + 273 = 298 = T1 T2 T2 = 52 + 273 = 325 3.00 atm P2 = 298 K 325 K 3.00 atm P2 = x = 325 K 3.27 atm 298 K

  34. The Combined Gas Law • The combined gas law is written as follows: • Each of the other simple gas laws can be obtained from the combined gas law when the proper variable is kept constant. P1 V1 P2 V2 = T1 T2

  35. The Combined Gas LawSample Problem A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Solution: K = oC + 273 T1 = 25 + 273 = 298 P1 V1 P2 V2 T2 = 10 + 273 = 283 = T1 T2 (1.08 atm) (50.0 L) (0.855 atm) V2 = 298 K 283 K (1.08 atm) (50.0 L) (283 K) V2 = = 60.0 L (298 K) (0.855 atm)

  36. Avogadro’s Law • In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.) • Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or: V1 V2 = n1 n2

  37. The Ideal Gas Law • All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law: • Rrepresents the ideal gas constant which has a value of 0.0821 (L•atm)/(mol•K). PV = nRT

  38. The Ideal Gas LawSample Problem What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? Solution: PV = nRT P (10.0 L) = (298 K) (0.500 mol) (0.0821 L•atm/mol•K) (0.500 mol) (0.0821 L•atm/mol•K) (298 K) 1.22 atm = = P (10.0 L)

  39. Molar Mass of a Gas • The ideal gas law can beused in combination withmass measurements to calculate the molar mass of an unknown gas. • Molar mass is calculated by dividing the mass (in grams) by the amount of gas (in moles.) g Molar Mass = mol

  40. Molar Mass of a GasSample Problem 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. Solution: 1 atm = 886 mmHg P 1.17 atm = PV = nRT 760. mmHg T = 55 + 273 = 328 K (1.17 atm) (0.225 L) PV = 0.00978 mol n = = RT (0.0821 L•atm/mol•K) (328 K) 0.311 g grams MM = = = 31.8 g/mol mole 0.00978 mol

  41. Standard Molar Volume • Standard Temperature and Pressure (STP) is 0oC (273 K) and 1 atm. • The Standard Molar Volumeof a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

  42. Molar Volume Conversion Factor • Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

  43. Molar Volume ConversionSample Problem a. What quantity of gas, in moles, is contained in 5.00 L at STP? b. What volume does 0.768 moles of a gas occupy at STP? 1 mol x 5.00 L 0.223 mol = 22.4 L 22.4 L x 0.768 mol 17.2 L = 1 mol

  44. The Mole Map Revisited • As you recall, you can convert between number of particles, mass (g), and volume (L) by going through moles.

  45. Stoichiometry Revisited • Remember that you can use mole ratios and volume ratios (gases only) as conversion factors: 2CO(g) + O2(g) → 2CO2(g) 2 molecules 1 molecule 2 molecules 2 mole 1 mole 2 mol 2 volumes 1 volume 2 volumes • Example: What volume of O2 is needed to react completely with 0.626 L of CO at the same temperature & pressure conditions to form CO2? 1 L O2 x 0.626 L CO 0.313 L O2 = 2 L CO

  46. Gas StoichiometrySample Problem What volume (in L) of H2 at 355 K and 738 mmHg is required to synthesize 35.7 g of methanol, given: CO(g) + 2H2(g) → CH3OH(g) Solution: First, use stoichiometry to solve for moles of H2: Then, use the ideal gas law to find the volume of H2: 1 mol CH3OH 2 mol H2 35.7 g CH3OH 2.23 mol H2 = 1 mol CH3OH 32.0 g CH3OH (0.0821 L•atm/mol•K) (2.23mol) (355 K) nRT = 66.9 L H2 V = = P (0.971 atm)

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