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Gases - Chemistry

Gases - Chemistry. Macroscopic vs. Microscopic Representation. Kinetic Molecular Theory of Gases. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. Kinetic Molecular Theory of Gases.

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Gases - Chemistry

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  1. Gases - Chemistry

  2. Macroscopic vs. Microscopic Representation

  3. Kinetic Molecular Theory of Gases • Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

  4. Kinetic Molecular Theory of Gases • The average kinetic energy of the molecules is proportional to the temperature of the gas in Kelvin. Any two gases at the same temperature will have the same average kinetic energy • A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. • Gas molecules exert neither attractive nor repulsive forces on one another. • Each gas molecule “behaves” as if it were alone in container (due to #3 and #4) 5.7

  5. Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids. 5.1

  6. Temperature K = ºC + 273 ºF -459 32 212 ºC -273 0 100 K 0 273 373 Always use absolute temperature (Kelvin) when working with gases.

  7. Pressure

  8. Pressure Aneroid Barometer Mercury Barometer Barometer measures atmospheric pressure

  9. Pressure U-tube Manometer Bourdon-tube gauge Manometer • measures contained gas pressure

  10. Pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

  11. STP Standard Temperature & Pressure 0°C273 K 1 atm101.325 kPa -OR- STP

  12. Boyle’s Law P V PV = k

  13. Boyle’s Law P V The pressure and volume of a gas are inversely related at constant mass & temp PV = k

  14. Boyle’s Law The pressure and volume of a gas are inversely related Animations for Boyle’s Law http://www.physics-chemistry-interactive-flash-animation.com/matter_change_state_measurement_mass_volume/pressure_volume_boyle_mariotte_law_ideal_gas_closed_system_MCQ.htm http://www.grc.nasa.gov/WWW/k-12/airplane/aboyle.html

  15. Charles’ Law V T

  16. Charles’ Law V T The volume and absolute temperature (K) of a gas are directly related • at constant mass & pressure

  17. Charles’ Law The volume and absolute temperature (K) of a gas are directly related Animations for Charle’s Law http://www.grc.nasa.gov/WWW/k-12/airplane/aglussac.html

  18. Gay-Lussac’s Law P T

  19. Gay-Lussac’s Law P T The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume

  20. Combined Gas Law P1V1 T1 P2V2 T2 = P T V T PV T PV = k P1V1T2 =P2V2T1

  21. Gas Law Problems A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

  22. E. Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1T2 = P2V2T1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  23. Gas Law Problems A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P T V GIVEN: V1=7.84 cm3 P1=71.8 kPa T1=25°C = 298 K V2=? P2=101.325 kPa T2=273 K WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa)V2 (298 K) V2 = 5.09 cm3

  24. Gas Law Problems A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1V1T2 = P2V2T1 (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C

  25. Avogadro’s Principle V n

  26. A. Avogadro’s Principle V n Equal volumes of gases contain equal numbers of moles • at constant temp & pressure • true for any gas

  27. Ideal Gas Law = k V n PV T PV nT = R UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK

  28. B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK

  29. Ideal Gas Law • Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

  30. Ideal Gas Law WORK: 85 g 1 mol = 2.7 mol 32.00 g • Find the volume of 85 g of O2 at 25°C and 104.5 kPa. IDEAL GAS LAW GIVEN: V=? n=85 g T=25°C = 298 K P=104.5 kPa R=8.315dm3kPa/molK = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molKK V = 64 dm3

  31. Gas Stoichiometry at Non-STP Conditions

  32. Gas Stoichiometry Moles  Liters of a Gas • STP - use 22.4 L/mol • Non-STP - use ideal gas law Non-STP Problems • Given liters of gas? • start with ideal gas law • Looking for liters of gas? • start with stoichiometry conv.

  33. Gas Stoichiometry Problem • What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3 CaO + CO2 5.25 g ? Lnon-STP Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g CaCO3 1 mol CO2 1 mol CaCO3 = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters.

  34. Gas Stoichiometry Problem • What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P=103 kPa V = ? n=1.26 mol T=25°C = 298 K R=8.315dm3kPa/molK WORK: PV = nRT (103 kPa)V=(1mol)(8.315dm3kPa/molK)(298K) V = 1.26 dm3 CO2

  35. Gas Stoichiometry Problem • How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 2 Al2O3 15.0 L non-STP ? g GIVEN: P=97.3 kPa V = 15.0 L n=? T=21°C = 294 K R=8.315dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

  36. Gas Stoichiometry Problem • How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 101.96 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

  37. Two More Laws

  38. Dalton’s Law Ptotal = P1 + P2 + ... Patm = PH2 + PH2O The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

  39. Dalton’s Law Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH2 + 2.72 kPa PH2 = 91.7 kPa Look up water-vapor pressure on p.899 for 22.5°C. Sig Figs: Round to least number of decimal places.

  40. Dalton’s Law A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor. DALTON’S LAW GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr WORK: Ptotal = Pgas + PH2O 742.0 torr = PH2 + 42.2 torr Pgas = 699.8 torr Look up water-vapor pressure on p.899 for 35.0°C. Sig Figs: Round to least number of decimal places.

  41. Graham’s Law Diffusion • Spreading of gas molecules throughout a container until evenly distributed. Effusion • Passing of gas molecules through a tiny opening in a container

  42. Graham’s Law Speed of diffusion/effusion • Kinetic energy is determined by the temperature of the gas. • At the same temp & KE, heavier molecules move more slowly. • Larger m smaller v because… KE = ½mv2

  43. Graham’s Law Ratio of gas A’s speed to gas B’s speed Graham’s Law • Rate of diffusion of a gas is inversely related to the square root of its molar mass.

  44. Graham’s Law Determine the relative rate of diffusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. Krdiffuses 1.381 times faster than Br2.

  45. Graham’s Law A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Put the gas with the unknown speed as “Gas A”.

  46. Graham’s Law An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. Square both sides to get rid of the square root sign.

  47. TEAM PRACTICE! Work the following problems in your book. Check your work using the answers provided in the margin. • p. 324 • SAMPLE PROBLEM 10-6 • PRACTICE 1 & 2 • p. 355 • SAMPLE PROBLEM 11-10 • PRACTICE 1, 2, & 3

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