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Chapter 12

Chapter 12. BEHAVIOR OF GASES. BEHAVIOR OF GASES. Importance of Gases. Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3 . 2 NaN 3  2 Na + 3 N 2 if bag ruptures 2 Na + 2 H 2 O  2 NaOH + H 2. THREE STATES OF MATTER.

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Chapter 12

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  1. Chapter 12 BEHAVIOR OF GASES Dr. S. M. Condren

  2. BEHAVIOR OF GASES Dr. S. M. Condren

  3. Importance of Gases • Airbags fill with N2 gas in an accident. • Gas is generated by the decomposition of sodium azide, NaN3. • 2 NaN3 2 Na + 3 N2 • if bag ruptures 2 Na + 2 H2O  2 NaOH + H2 Dr. S. M. Condren

  4. THREE STATES OF MATTER Dr. S. M. Condren

  5. THREE STATES OF MATTER Dr. S. M. Condren

  6. General Properties of Gases • There is a lot of “free” space in a gas. • Gases can be expanded infinitely. • Gases occupy containers uniformly and completely. • Gases diffuse and mix rapidly. Dr. S. M. Condren

  7. Properties of Gases Gas properties can be modeled using math. Model depends on— • V = volume of the gas (L) • T = temperature (K) • n = amount (moles) • P = pressure (atmospheres) Dr. S. M. Condren

  8. Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) Dr. S. M. Condren

  9. Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to • Hg density • column height Dr. S. M. Condren

  10. Pressure Column height measures P of atmosphere • 1 standard atm= 760 mm Hg = 29.9 inches Hg = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = 101.325 kPa Dr. S. M. Condren

  11. Effect of Pressure Differential Dr. S. M. Condren

  12. IDEAL GAS LAW P V = n R T Brings together gas properties. Can be derived from experiment and theory. Dr. S. M. Condren

  13. Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, for example, that P goes up as V goes down. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland. Dr. S. M. Condren

  14. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. Dr. S. M. Condren

  15. Boyle’s Law Dr. S. M. Condren

  16. Charles’s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist. Dr. S. M. Condren

  17. Charles’s original balloon Modern long-distance balloon Dr. S. M. Condren

  18. Charles’s Law Balloons immersed in liquid N2 (at -196 ˚C) will shrink as the air cools (and is liquefied). Dr. S. M. Condren

  19. Charles’s Law Dr. S. M. Condren

  20. twice as many molecules Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. Dr. S. M. Condren

  21. Avogadro’s Hypothesis The gases in this experiment are all measured at the same T and P. 2 H2(g) + O2(g) 2 H2O(g) Dr. S. M. Condren

  22. Combining the Gas Laws • V proportional to 1/P • V prop. to T • V prop. to n • Therefore, V prop. to nT/P • V = 22.4 L for 1.00 mol when Standard pressure and temperature (STP) ST = 273 K SP = 1.00 atm Dr. S. M. Condren

  23. Using PV = nRT How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? R = 0.082057 L•atm/K•mol memorize Solution 1. Get all data into proper units V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm Dr. S. M. Condren

  24. Using PV = nRT How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? R = 0.082057 L•atm/K•mol Solution 2. Now calc. n = PV / RT n = 1.1 x 103 mol (or about 22 kg of gas) Dr. S. M. Condren

  25. Ideal Gas Constant R = 0.082057 L*atm/mol*K R has other values for other sets of units. R = 82.057 mL*atm/mol*K = 8.314 J/mol*K = 1.987 cal/mol*K Dr. S. M. Condren

  26. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V. Dr. S. M. Condren

  27. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution #mol H2O2 = 1.1g H2O2 (1mol/ 34.0g H2O2) = 0.032 mol H2O2 #mol O2 = (0.032mol H2O2)(1mol O2/2mol H2O2) = 0.016mol O2 P of O2 = nRT/V = (0.016mol)(0.0821L*atm/K*mol)(298K) 2.50L = 0.16 atm Dr. S. M. Condren

  28. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) What is P of H2O? Could calculate as above. But recall Avogadro’s hypothesis. V n at same T and P P n at same T and V There are 2 times as many moles of H2O as moles of O2. P is proportional to n. Therefore, P of H2O is twice that of O2. P of H2O = 0.32 atm Dr. S. M. Condren

  29. Dalton’s Law John Dalton 1766-1844 Dr. S. M. Condren

  30. Dalton’s Law of Partial Pressures • 2 H2O2(liq) ---> 2 H2O(g) + O2(g) • 0.32 atm 0.16 atm What is the total pressure in the flask? Ptotal in gas mixture = PA + PB + ... Therefore, Ptotal = P(H2O) + P(O2) = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. Dr. S. M. Condren

  31. Collecting Gases over Water Dr. S. M. Condren

  32. ExampleA student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25oC, what is the volume of the “dry” oxygen gas at STP? Pwater = 23.8 torr at 25oC PO2 = Pbar - Pwater = (758 - 23.8) torr = 734 torr P1= PO2 = 734 torr; P2= SP = 760. torr V1= 245mL; T1= 298K; T2= 273K; V2= ? (V1P1/T1) = (V2P2/T2) V2= (V1P1T2)/(T1P2) = (245mL)(734torr)(273K) (298K)(760.torr) = 217mL Dr. S. M. Condren

  33. Low density helium Higher Density air GAS DENSITY PV = nRT n = P V RT m = P MV RT Where m => mass M => molar mass and density (d) = m/V d = m/V = PM/RT d and M are proportional Dr. S. M. Condren

  34. USING GAS DENSITY The density of air at 15 oC and 1.00 atm is 1.23 g/L. What is the molar mass of air? What is air? 79% N2; M a 28g/mol 20% O2; M a 32g/mol 1. Calc. moles of air. V = 1.00 L P = 1.00 atm T = 288 K n = PV/RT = 0.0423 mol 2. Calc. molar mass mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol Reasonable? Dr. S. M. Condren

  35. KINETIC MOLECULAR THEORY(KMT) Theory used to explain gas laws. KMT assumptions are • Gases consist of atoms or molecules in constant, random motion. • P arises from collisions with container walls. • No attractive or repulsive forces between molecules. Collisions elastic. • Volume of molecules is negligible. Dr. S. M. Condren

  36. Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed)2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases – and so does the speed. Dr. S. M. Condren

  37. Kinetic Molecular Theory Maxwell’s equation where u is the speed and M is the molar mass. • speed INCREASES with T • speed DECREASES with M Dr. S. M. Condren

  38. Velocity of Gas Molecules Molecules of a given gas have a rangeof speeds. Dr. S. M. Condren

  39. Velocity of Gas Molecules Average velocity decreases with increasing mass. All gases at the same temperature Dr. S. M. Condren

  40. DIFFUSION is the gradual mixing of molecules of different gases. GAS DIFFUSION AND EFFUSION Dr. S. M. Condren

  41. EFFUSION is the movement of molecules through a small hole into an empty container. GAS EFFUSION Dr. S. M. Condren

  42. GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is • proportional to T • inversely proportional to M. Therefore, He effuses more rapidly than O2 at same T. He Dr. S. M. Condren

  43. GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Rate of effusion is inversely proportional to its molar mass. Thomas Graham, 1805-1869. Professor in Glasgow and London. Dr. S. M. Condren

  44. Gas Diffusionrelation of mass to rate of diffusion • HCl and NH3 diffuse from opposite ends of tube. • Gases meet to form NH4Cl • HCl heavier than NH3 • Therefore, NH4Cl forms closer to HCl end of tube. Dr. S. M. Condren

  45. Deviations from Ideal Gas Law • Real molecules have volume. • There are intermolecular forces. • Otherwise a gas could not become a liquid. Dr. S. M. Condren

  46. Measured V = V(ideal) Measured P ( ) 2 n a nRT V - nb P + ----- J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. 2 V vol. correction intermol. forces Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with VAN DER WAALS’s EQUATION. Dr. S. M. Condren

  47. Deviations from Ideal Gas Law Cl2 gas has a = 6.49, b = 0.0562 For 8.0 mol Cl2 in a 4.0 L tank at 27 oC. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm Dr. S. M. Condren

  48. Dr. S. M. Condren

  49. Dr. S. M. Condren

  50. Carbon Dioxide and Greenhouse Effect Dr. S. M. Condren

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