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BEHAVIOR OF GASES Chapter 12

BEHAVIOR OF GASES Chapter 12. General Properties of Gases. There is a lot of “free” space in a gas. The particles of gas are considered to have insignificant volume. Gases can be expanded infinitely. Gases occupy containers uniformly and completely.

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BEHAVIOR OF GASES Chapter 12

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  1. BEHAVIOR OF GASESChapter 12

  2. General Properties of Gases • There is a lot of “free” space in a gas. The particles of gas are considered to have insignificant volume. • Gases can be expanded infinitely. Gases occupy containers uniformly and completely. • Gases diffuse and mix rapidly. Perfectly elastic collisions.

  3. KINETIC MOLECULAR THEORY(KMT) Theory used to explain gas laws. KMT assumptions are • Gases consist of 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.

  4. THREE STATES OF MATTER

  5. 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)

  6. IDEAL GAS LAW P V = n R T Brings together gas properties. Can be derived from experiment and theory.

  7. Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, that P(pressure) goes up as V(volume) goes down. Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

  8. Boyle’s Law If (nRT) are constant, = to 1, then Boyles Law becomes: P1V1 = P2V2 Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

  9. 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.

  10. Charles’s original balloon Modern long-distance balloon

  11. V1 x P1 V2 = P2 Sample problem 14.1 A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? P1V1 = P2V2 P1 = 103 kPa P2 = 25.0 kPa V1 = 30.0 L V2 = ?????

  12. 30.0 L x 103 kPa V2 = 25.0 kPa V1 x P1 V2 = P2 Sample problem 14.1 A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? = 124 L

  13. Question 1and 2 CPS

  14. 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.

  15. V1V2 = T1 T2 Charles’s Law If (nR/P) are constant, = to 1, then Charles’s Law becomes: Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  16. Charles’s Law

  17. V1 x T2 V2 = T1 Sample problem 14.2 A balloon inflated in a room at 24oC has a volume of 4.00 L. The ballon is then heated to a temperature of 58C. What is the new volume if the pressure remains constant? V1 = V2 T1 T2 T1 = 24oC T2 = 58oC V1 = 30.0 L V2 = ?????

  18. 4.00 L x 331 K V2 = 297 K Sample problem 14.2 A balloon inflated in a room at 24oC has a volume of 4.00 L. The ballon is then heated to a temperature of 58C. What is the new volume if the pressure remains constant? V1 = V2 T1 T2 Change the temperature to Kelvins = 4.46 L

  19. Question 3 CPS

  20. The Combined Gas Law Bigger combines use more gas The combined gas law is when the only variable held constant is the amount (mol) of gas.

  21. Sample problem 14.4 The volume of a gas-filled balloon is 30.0 L at 313 K and 152 kPa pressure. What would the volume be at standard temperature and pressure? P1V1 = P2V 2 T1 T2 T1 = 313 K T2 = 273 K V1 = 30.0 L P2 = 101.3 kPa P1 = 153 kPa V2 = ????

  22. P1V1 = P2V 2 P1V1T2 = V2 152 kPa • 30.0L • 273 KT2 = V2 T1 T2 T1P2 313 K • 101.3 kPa Sample problem 14.4 The volume of a gas-filled balloon is 30.0 L at 313 K and 152 kPa pressure. What would the volume be at standard temperature and pressure? 39.5 L = V2

  23. Question 4 CPS

  24. IDEAL GAS LAW P V = n R T R is the ideal gas constant. R = 8.314 L•kPa/K•mol or 0.0821 L•atm/K•mol

  25. 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.

  26. Avogadro’s Hypothesis The gases in this experiment are all measured at the same T and P.

  27. 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 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

  28. 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 30 kg of gas)

  29. 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? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

  30. 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.

  31. 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

  32. 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 P of O2 = 0.16 atm

  33. 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

  34. Question 1 Exam View CPS

  35. 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.

  36. Sample problem 14.6 Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.30 kPa of total pressure if the partial pressure of nitrogen, carbon dioxide, and other gases arte 79.10 kPa, 0.040 kPa, and 0.94 kPa, respectively? Ptotal = PO2 + PN2 + PCO2 + Pother Rearrange Dalton’s Law to solve for Po2 PO2 = Ptotal - (PN2 + PCO2 + Pother) = 101.30 Kpa – (79.10 kPa + 0.040 kPa + 0.94 kPa) = 21.22 kPa

  37. Question 1 Exam View CPS

  38. Dalton’s Law John Dalton 1766-1844

  39. 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, KE also increases — and so does speed.

  40. Kinetic Molecular Theory At the same T, all gases have the same average KE. As T goes up, KE also increases — and so does speed.

  41. Velocity of Gas Molecules Average velocity decreases with increasing mass.

  42. diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container. GAS DIFFUSION AND EFFUSION

  43. 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

  44. 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.

  45. 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.

  46. Using KMT to Understand Gas Laws Recall that KMT assumptions are • Gases consist of 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.

  47. Avogadro’s Hypothesis and Kinetic Molecular Theory P proportional to n

  48. Gas Pressure, Temperature, and Kinetic Molecular Theory P proportional to T

  49. Boyle’s Law and Kinetic Molecular Theory P proportional to 1/V

  50. Deviations from Ideal Gas Law • Real molecules have volume. • There are intermolecular forces. • Otherwise a gas could not become a liquid. Fig. 12.20

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