2. BEHAVIOR OF GASES Chapter 12

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

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

5. THREE STATES OF MATTER

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

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

8. 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.

9. Boyle’s Law If (nRT) are constant, = to 1, then Boyles Law becomes: P 1 V 1 = P 2 V 2 Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

10. 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.

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

12. 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? P 1 V 1 = P 2 V 2 P 1 = 103 kPaP 2 = 25.0 kPa V 1 = 30.0 LV 2 = ????? V 1 x P 1 P2P2 V 2 =

13. 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? 30.0 L x 103 kPa 25.0 kPa V 2 = V 1 x P 1 P2P2 V 2 = = 124 L

14. Question 1and 2 CPS

15. 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.

16. Charles’s Law If (nR/P) are constant, = to 1, then Charles’s Law becomes: V 1 V 2 = T1T1 T2T2 Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

17. Charles’s Law

18. Sample problem 14.2 A balloon inflated in a room at 24 o C has a volume of 4.00 L. The ballon is then heated to a temperature of 58 C. What is the new volume if the pressure remains constant? V 1 = V 2 T 1 = 24 o CT 2 = 58 o C V 1 = 30.0 LV 2 = ????? V 1 x T 2 T1T1 V 2 = T 1 T 2

19. Sample problem 14.2 A balloon inflated in a room at 24 o C has a volume of 4.00 L. The ballon is then heated to a temperature of 58 C. What is the new volume if the pressure remains constant? V 1 = V 2 4.00 L x 331 K 297 K V 2 = T 1 T 2 Change the temperature to Kelvins = 4.46 L

20. Question 3 CPS

21. 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.

22. 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? P 1 V 1 = P 2 V 2 T 1 = 313 KT 2 = 273 K V 1 = 30.0 LP 2 = 101.3 kPa P 1 = 153 kPaV 2 = ???? T 1 T 2

23. 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? P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = V 2 T 1 P 2 152 kPa 30.0L 273 KT 2 = V 2 313 K 101.3 kPa 39.5 L = V 2

24. Question 4 CPS

25. IDEAL GAS LAW P V = n R T R is the ideal gas constant. R = 8.314 LkPa/Kmol or 0.0821 Latm/Kmol

26. 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. twice as many molecules

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

28. Using PV = nRT How much N 2 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 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/KmolSolution 1. Get all data into proper units V = 27,000 L V = 27,000 L T = 25 o C + 273 = 298 K T = 25 o C + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm

29. Using PV = nRT How much N 2 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 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/KmolSolution 2. Now calc. n = PV / RT n = 1.1 x 10 3 mol (or about 30 kg of gas)

30. Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

31. Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc. P from n, R, T, and V.

32. Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution

33. Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution P of O 2 = 0.16 atm

34. Gases and Stoichiometry What is P of H 2 O? 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 H 2 O as moles of O 2. P is proportional to n. Therefore, P of H 2 O is twice that of O 2. P of H 2 O = 0.32 atm 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g)

35. Question 1 Exam View CPS

36. Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P(H 2 O) + P(O 2 ) = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

37. Sample problem 14.6 Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (P O2 ) 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? P total = P O2 + P N2 + P CO2 + P other Rearrange Dalton’s Law to solve for P o2 P O2 = P total - (P N2 + P CO2 + P other ) = 101.30 Kpa – (79.10 kPa + 0.040 kPa + 0.94 kPa) = 21.22 kPa

38. Question 1 Exam View CPS

39. Dalton’s Law John Dalton 1766-1844

40. 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.

41. 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.

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

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

44. GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He

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

46. Gas Diffusion relation of mass to rate of diffusion HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube.

47. Using KMT to Understand Gas Laws Recall that KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

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

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

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

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