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Gases & Liquids

Gases & Liquids. Ch.12. (12-1) Properties of Gases. Fluids Low density Compressible Fill a container & exert P equally in all directions Influenced by T. Kinetic-Molecular Theory. Explains behavior of gases 2 major assumptions: Collisions are elastic

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Gases & Liquids

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  1. Gases & Liquids Ch.12

  2. (12-1) Properties of Gases • Fluids • Low density • Compressible • Fill a container & exert P equally in all directions • Influenced by T

  3. Kinetic-Molecular Theory • Explains behavior of gases • 2 major assumptions: • Collisions are elastic • V of individual gas molecules is negligible KE is lost KE is maintained

  4. Ideal Gas • Describes the behavior of gases under most conditions • High T & low P gases act more ideal

  5. Kinetic Energy • T of a gas determines the avg. KE of its particles • KE = ½ mv2 • Where, m = mass, v = speed

  6. Pressure • Pressure (P) = force (F) area (A) • SI units: • F: newtons (N) • P: pascal (Pa) = 1 N/m2

  7. Standard Temp. & Pressure • STP: std. conditions for a gas • Temp. (T): 0 °C (273 K) • Pressure (P): 1 atm (101.325 kPa) • See Table 12-1, p.428 for more P units

  8. Greenhouse Effect • Inc. in the T of Earth caused by reflected solar radiation that’s trapped in the atmosphere • Inc. in greenhouse gases such as CO2 & CFCs

  9. Free Radical • Atom or molecule that has 1 or more unpaired e- & is very reactive • UV radiation breaks apart CFCs, making Cl• • Chain rxn: self-sustaining rxn in which the product from 1 step acts as a reactant for the next step • Cl• + O3 ClO + O2 + O • ClO + O  Cl• + O2

  10. (12-2) The Gas Laws • Symbols: • P = pressure • T = temp in K • V = volume • n = # of moles

  11. Boyle’s Law • At constant T: • Inc. P, dec. V • Dec. P, inc. V • P1V1 = P2V2

  12. Boyle’s Law Practice If the P exerted on a 300 mL sample of H2 gas at constant T is inc. from 0.500 atm to 0.750 atm, what will be the final V of the sample? • List known V1 = 300 mL P1 = 0.500 atm V2 = ? mL P2 = 0.750 atm

  13. Boyle’s Law Practice • Write eq. P1V1 = P2V2 V2 = P1V1 P2 3. Substitute & solve V2= (0.500 atm)(300 mL) = 200 mL 0.750 atm

  14. Dalton’s Law of Partial P’s • Total P in a gas mixture is the sum of the partial P’s of the individual components • Ptotal = PA + PB + PC… • Where, Ptotal = total P, PA = partial P of A

  15. Dalton’s Law Practice A mixture of O2, N2, & H2 gases exerts a total P of 278 kPa. If the partial P’s of O2 & H2 are 112 kPa & 101 kPa respectively, what would be the partial P of the N2? • List the known Ptotal = 278 kPa PN2 = ? kPa PO2 = 112 kPa PH2 = 101 kPa

  16. Dalton’s Law Practice • List the eq. & rearrange Ptotal = PO2 + PH2 + PN2 PN2 = Ptotal - PO2 - PH2 • Substitute & solve PN2 = 278 kPa – 112 kPa – 101 kPa = 65 kPa

  17. Mole Fraction • # of moles of 1 component compared w/ the total # of moles in the mixture • Mole fraction (X) = ____mol A___ total mols • To calc. partial P: • PA = PTXA

  18. Mole Fraction Example The total P of a mixture of gases is 0.97 atm. The mole fraction for N2 is 0.78. What’s the partial P of N2? • List the known Ptotal = 0.97 atm XN2 = 0.78

  19. Mole Fraction Example • Write the eq. PN2 = Ptotal XN2 • Substitute & solve PN2 = (0.97 atm)(0.78) = 0.76 atm

  20. Charles’ Law • At constant P: • V inc., T inc. • V dec., T dec. • V1 = V2 T1 T2

  21. Charles’ Law Practice Gas in a balloon occupies 2.5 L at 300 K. The balloon is dipped into liquid N2 at 80 K. What V will the gas in the balloon occupy at this T? • List known V1 = 2.5 L T1 = 300 K V2 = ? L T2 = 80 K

  22. Charles’ Law Practice 2. Write eq. V1 = V2 V2 = V1T2 T1 T2 T1 3. Substitute & solve V2 = (2.5 L)(80 K) = 0.67 L (300 K)

  23. Pressure & Temp. • P inc. w/ inc. in T at constant V • P1 = P2 T1 T2

  24. P & T Practice Gas in a sealed can has a P of 3.00 atm at 25°C. A warning says not to store the can in a place where the T will exceed 52°C. What would the gas P in the can be at 52°C? • List known P1 = 3.00 atm T1 = 25°C = 298 K P2 = ? atm T2 = 52°C = 325 K

  25. P & T Practice 2. Write eq. P1 = P2 P2 = P1T2 T1 T2 T1 3. Substitute & solve P2 = (3.00 atm)(325 K) = 3.27 atm (298 K)

  26. Avogadro’s Law • V’s of different gases under the same T & P’s have the same # of molecules • V1 = V2 n1 n2

  27. Gay-Lussac’s Law • Law of Combining Volumes: at constant T & P, gases react in whole # V proportions • H2 + Cl2 2 HCl • 1 V + 1 V  2 V

  28. Effusion • Motion of a gas through a small opening • Diffusion: gas particles disperse from areas of high to low conc.

  29. Graham’s Law of Effusion • At the same T & P, 2 gases rates of effusion can be measured by: • ½ MAvA2 = ½ MBvB2 or vA = MB vB MA • Where: • v = speed of effusion (2 gases, A & B) • M = molar mass

  30. Graham’s Law Practice O2 has an avg. speed of 480 m/s at room T. On avg., how fast is SO3 traveling at the same T? • List known vO2 = 480 m/s MO2 = 32 g/mol vSO3 = ? m/s MSO3 = 80.07 g/mol

  31. Graham’s Law Practice • Write eq. vSO3 = MO2 vSO3 =vO2MO2 vO2 MSO3 MSO3 3. Substitute & solve vSO3 = (480 m/s) (32 g/mol) (80.07 g/mol) = 300 m/s

  32. More Graham’s Practice Compare the rate of effusion of H2O vapor w/ O2 gas at the same T & P. 1. List known MH2O = 18.02 g/mol MO2 = 32 g/mol

  33. More Graham’s Practice 2. Write eq. vH2O = MO2 vO2 MH2O 3. Substitute & solve vH2O = 32 g/mol = 1.33 vO2 18.02g/mol H2O effuses 1.33X faster than O2

  34. (12-3) Ideal Gas Law • PV = nRT • Where: • R = 8.314 L•kPa / mol•K or • R = 0.0821 L•atm / mol•K

  35. Ideal Gas Law Practice Calculate the V of 1.00 mol of CO2 gas at STP. • List known P = 1.00 atm V = ? L n = 1.00 mol R = 0.0821 L•atm/mol•K T = 273 K

  36. Ideal Gas Law Practice • Write eq. PV = nRT  V = nRT P • Substitute & solve V = (1.00 mol)(0.0821 L•atm/mol•K)(273 K) (1.00 atm) = 22.4 L

  37. Combined Gas Law • Moles remain constant, but other conditions change • P1V1 = P2V2 T1 T2

  38. Combined Gas Law Practice A sample of CO2 gas occupies 45 L at 750 K & 500 kPa. What’s the V of this gas at STP? • List known P1 = 500 kPa P2 = 101.325 kPa V1 = 45 L V2 = ? L T1 = 750 K T2 = 273 K

  39. Combined Gas Law Practice • Write eq. P1V1 = P2V2 V2 = P1V1T2 T1 T2 T1P2 3. Substitute & solve V2 = (500 kPa)(45 L)(273 K) (750 K)(101.325 kPa) = 81 L

  40. Gas Stoichiometry • Gas V’s can be determined from mole ratios in bal. eqs. • 3H2 + N2 2NH3 3 L 1 L 2 L 22 L N2 x 3 L H2 = 66 L H2 1 L N2

  41. (12-4) Changes of State • Evaporation: l g • Condensation: g  l • Sublimation: s  g

  42. Vapor Pressure • P exerted by a vapor in equilibrium w/ its liquid state at a given T • H2O(l) H2O(g)

  43. Phase Diagrams • Shows T’s & P’s at which a substance exists in different phases • Phases are at equilibrium along the lines • Phase: substance has uniform composition & properties

  44. Phase Diagrams (cont.) • Normal bp: boiling T at 1 atm • Critical point: T & P above which the properties of vapor can’t be distinguished from a liquid • Supercritical fluids • Triple point: T & P where 3 phases exist in equil.

  45. C.P. N.M.P. N.B.P.

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