2. 9/22/2015 1 Chapter 12 The Behavior of Gases Chemistry

3. 9/22/2015 2 The Kinetic Theory Revisited Have you ever noticed that a balloon shrinks when it gets cold and expands when it is warmed? You saw a balloon shrink to nothing when placed in liquid nitrogen and then expand when removed from the liquid nitrogen!

4. 9/22/2015 3 The Kinetic Theory Revisited Gases consist of hard spherical particles These particles are so small that in relation to the distances between them that their individual volumes can be assumed to be insignificant. There is considerable empty space between the particles Between the particles is empty space. No attractive or repulsive forces exist between the particles.

5. 9/22/2015 4 The Kinetic Theory Revisited The particles move in constant random motion. They travel in straight paths, independently of each other. The gas particles change direction only from collision with one another or with other objects. Gases fill their containers regardless of the shape and volume of the containers. Uncontained gases diffuse into space without limit.

6. 9/22/2015 5 The Kinetic Theory Revisited All collisions are perfectly elastic. This means that during collisions kinetic energy is transferred without loss from one particle to another and the total kinetic energy remains the same.

7. 9/22/2015 6 The Kinetic Theory Revisited Compressibility is a measure of how much the volume of matter decreased under pressure. Gases are easily compressed (unlike solids or liquids) Compressibility is why air bags work.

8. 9/22/2015 7 The Kinetic Theory Revisited Remember that the average kinetic energy of a collection of gas particles is directly proportional to the Kelvin temperature of a gas

9. 9/22/2015 8 The Kinetic Theory Revisited The four variables that describe a gas are: –Pressure in kilopascals, kPa –Volume in liters, L –Temperature in Kelvin degrees – K has no° sign –Number of moles, n

10. 9/22/2015 9 Factors Affecting Gas Pressure The limitation of how many gas particles can be added to a container is the strength of the container. Halving the number of gas particles, halves the pressure.

11. 9/22/2015 10 Factors Affecting Gas Pressure Raising the temperature of a gas in a closed container increases the pressure. The speed and kinetic energy increases as the temperature increases. Faster moving particles impact the walls of the container with more energy, exerting greater pressure. Doubling the Kelvin temperature doubles the gas pressure.

12. 9/22/2015 11 Factors Affecting Gas Pressure Heating an aerosol can can cause the can to burst. Why? As the temperature increases the gas pressure increases. At some point, the aerosol can cannot withstand the increased pressure and it bursts open.

13. 9/22/2015 12 Factors Affecting Gas Pressure You can raise the pressure of a contained gas by decreasing the volume. The more gas is compressed, the greater the gas pressure. Reducing the volume by half doubles the gas pressure. Doubling the volume halves the gas pressure.

14. 9/22/2015 13 Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P 1 V 1 = P 2 V 2 Robert Boyle (1627-1691).

15. 9/22/2015 14 The Gas Laws –Boyle’s Law Boyle’s Law is the pressure-volume relationship Boyle’s Law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. In an inverse relationship the product of the two variable quantitative is constant. P 1  V 1 = P 2  V 2

16. 9/22/2015 15 The Gas Laws –Boyle’s Law Now, let's look at how these ideas relate to diving. It is well known among divers that diving at the surface can be more dangerous than deeper diving. We can understand this by first noting that for every 10 meters you descend in the water, the pressure increases by about 1 atm.

17. 9/22/2015 16 The Gas Laws –Boyle’s Law

18. 9/22/2015 17 The Gas Laws –Boyle’s Law Therefore, when you hold your breath, you create a closed system with your lungs and thus Boyle's law will hold. If you are down at 90 meters (at 10 atm) and you rise 10 meters to 80 meters (at 9 atm), the pressure has decreased by about 10%, and since PV is a constant your lungs expand by about 10% (probably not too bad).

19. 9/22/2015 18 The Gas Laws –Boyle’s Law Now if you are at 10 meters (at 2 atm), and you rise 10 meters to the surface (at 1 atm) the pressure had decreased by 50% and expanding your lungs by this factor could cause significant damage, maybe death!

20. 9/22/2015 19 The Gas Laws –Boyle’s Law

21. 9/22/2015 20 The Gas Laws –Boyle’s Law A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure in only 25.0 kPa?

22. 9/22/2015 21 The Gas Laws –Boyle’s Law Knowns: P 1 = 103 kPaV 1 = 30.0 L P 2 = 25.0 kPaV 2 = UNK P 1  V 1 = P 2  V 2 (Rearrange for V 2 ) V 2 = P 1  V 1 V 2 = 103 kPa 30.0 L = P 2 25.0 kPa 124 L 1.24  10 2 L

23. 9/22/2015 22 Charles’s Law Timberlake, Chemistry 7 th Edition, page 259

24. Raising the temperature of a gas increases the pressure if the volume is held constant. The molecules hit the walls harder. Temperature

25. If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to 600 K one of 2 things happens 300 K

26. Either the volume will increase to 2 liters at 1 atm 300 K 600 K

27. 300 K 600 K ….or the pressure will increase to 2 atm.

28. 9/22/2015 27 Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V 1 V 2 = T 1 T 2 If one temperature goes up, the volume goes up! If one temperature goes up, the volume goes up! Jacques Charles (1746-1823).

29. 9/22/2015 28 Charles’s Law Timberlake, Chemistry 7 th Edition, page 259

30. 9/22/2015 29 Charles’s Law

31. 9/22/2015 30 Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P 1 P 2 = T 1 T 2 If one temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac (1778-1850)

32. 9/22/2015 31 Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 P 2 V 2 = T 1 T 2

33. 9/22/2015 32 Kelvin vs. Celsius Remember that all gas laws must be applied in Kelvin If your given a problem in Celsius Temp you must convert it to Kelvin K = C + 273 degrees

34. 9/22/2015 33 And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 101.3 kPa, 1 atm, or 760 mm Hg Standard Temperature = 0 deg C (273 K)

35. 9/22/2015 34 Ideal Gas Equation 5.4 Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

36. 9/22/2015 35 The conditions 0 0 C and 101.3 kPa are called standard temperature and pressure (STP). PV = nRT R = PV nT = (101.3kPa)(22.414L) (1 mol)(273.15 K) R = 8.31 L kPa / (mol K) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

37. 9/22/2015 36 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

38. 9/22/2015 37 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? 5.3 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 101.3 kPa x 8.31 x 300.15 K LkPa molK M = 54.6 g/mol

39. 9/22/2015 38 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 8.31 x 310.15 K LkPa molK 101.3 kPa = = 4.76 L 5.5

40. 9/22/2015 39 Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 5.6

41. 9/22/2015 40 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT

42. 9/22/2015 41 A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm 5.6

43. 9/22/2015 42 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl

44. 9/22/2015 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.

45. 9/22/2015 44 GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv 2 Thomas Graham, 1805-1869. Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass.

46. 9/22/2015 45 GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to T proportional to T inversely proportional to M. inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He

47. 9/22/2015 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.

48. 9/22/2015 47 Graham’s Law Problem 1 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 degrees celsius according to the equation: 4 NH 3 (g) + 5 O 2 (g) --> 4 NO(g) + 6 H 2 O(g) Using Graham's Law, what is the ratio of the effusion rates of NH 3 (g) to O 2 (g)?

49. 9/22/2015 48 Graham’s Law Problem 2 What is the rate of effusion for H 2 if 15.00 ml of CO 2 takes 4.55 sec to effuse out of a container?

50. 9/22/2015 49 Graham’s Law Problem 3 What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2 (g)?