1. 1 Chapter 5 The Gas Laws

2. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container. These collisions cause the pressure. n Measured with a barometer or a manometer.

3. 3 Barometer n The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. n 1 atm = 760 mm Hg 1 atm Pressure 760 mm Hg Vacuum

4. 4 Manometer Gas h n Column of mercury to measure pressure. n ‘h’ is how much lower the pressure is than outside.

5. 5 Manometer n h is how much higher the gas pressure is than the atmosphere. h Gas

6. 6 Pressure Readings The difference in the height of mercury onboth sides is the difference in pressurebetween the two gases.The gas with the higher pressure will be ableto move the mercury more than the gas withthe lower pressure.Remember our “Law of Phyics” The strongerforce always wins !!

7. 7 Units of pressure n 1 atmosphere = 760 mm Hg = 760 torr n 1 atm = 101,325 Pascals = 101.325 kPa n The above values are Standard Pressure and need to be known.

8. 8 The Gas Laws n Boyle’s Law n Pressure and volume are inversely related at constant temperature. OR As one goes up, the other goes down. n PV= k was Boyle’s data and it turned into P 1 V 1 = P 2 V 2

9. 9 Boyle’s Law n The following two graphs show the data when Pressure v. Volume are plotted. n In the first graph, it shows Pressure graphed v. Volume. n The second graph shows the straight line plot of Volume v. 1 / Pressure

10. 10 V P (at constant T)

11. 11 V 1/P (at constant T) Slope = k

12. 12 Molar Volume n Standard Pressure = 1 atmosphere n Standard Temperature = 273 K n Together they from STP conditions n 1 mole of any gas at STP occupies 22.4 Liters of Volume

13. 13 Examples n 20.5 L of nitrogen at 742 torr are compressed to 9.8 atm. What is the new volume? n Pressure is changing and a new volume is needed = Boyle’s Law n Make sure the units match up as torr does not equal atm. n (20.5 L) (742 torr) = (7448 torr) V 2 n V 2 = 2.042 L

14. 14 More examples n 30.6 mL of carbon dioxide at 740 torr is to 750 mL What is the final pressure in kPa ? n (30.6 mL) (740 torr) = (750 mL) P 2 n P 2 = 30.19 torr (needs to be in kPa) 30.19 torr = 4.025 kPa

15. 15 Charles’ Law n Volume of a gas varies directly with the absolute temperature at constant pressure. n V = kT (if T is in Kelvin) n V 1 = V 2 T 1 = T 2 n Graphically it looks like this.

16. 16 V (L) T (ºC) He H2OH2O CH 4 H2H2 -273.15ºC Charles’ Law Graph

17. 17 Examples n What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC ? n 247 / 295 = V 2 / 371 V 2 = 310.6 mL n Remember to convert °C to K

18. 18 Avogadro's Law n At constant temperature and pressure, the volume of gas is directly related to the number of moles. When air is added to a ball to pump it up, it is putting more moles of air in it !! n V = k n (n is the number of moles) n V 1 = V 2 n 1 = n 2

19. 19 Gay- Lussac Law n At constant volume, pressure and absolute temperature are directly related. n P = k T n P 1 = P 2 T 1 = T 2

20. 20 Combined Gas Law n If the moles of gas remains constant, use this formula and cancel out the other things that don’t change. n P 1 V 1 = P 2 V 2. T 1 T 2

21. 21 Examples n A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC? n What volume of gas could the can release at 22ºC and 743 torr?

22. 22 Ideal Gas Law n PV = nRT n R is the ideal gas law constant. n R = ? It depends on the pressure unit. 0.08205 L atm / mol K OR 0.08205 L atm / mol K OR 62.36 L torr / mol K OR 62.36 L torr / mol K OR 8.314 J / mol K 8.314 J / mol K n The other laws tell you about a gas when it changes.

23. 23 Examples n A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature? n Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC. What is the mass of Kr? n A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?

24. 24 Solved Answers 1. PV = nRT (1.85)(47.3) = 1.62 R T (1.85)(47.3) = 1.62 R T T = 658.32 K or 385.32 °C T = 658.32 K or 385.32 °C 2. PV = nRT (8.61) (18.5) = n (.08205)(297.8) (8.61) (18.5) = n (.08205)(297.8) n = 6.52 moles = 546.28 grams n = 6.52 moles = 546.28 grams

25. 25 Gas Density and Molar Mass n Sometimes we have to combine our gas laws with formulas that we learned earlier, such as : n Density = m / V OR moles = g / M n PV = nRT substitute for n. M = g RT / PV Another algebra manipulation is M = g RT / PV Another algebra manipulation is P x M = gRT/ V = P x M = DRT P x M = gRT/ V = P x M = DRT

26. 26 Examples n What is the density of ammonia at 23ºC and 735 torr?

27. 27 n A compound has the empirical formula CHCl. A 256 mL flask at 100.ºC and 750 torr contains.80 g of the gaseous compound. What is the empirical formula?

28. 28 Gases and Stoichiometry n Reactions happen in moles n At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.42 L. n If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

29. 29 Examples n Mercury can be achieved by the following reaction: n What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at STP? n At 400.ºC and 740 torr?

30. 30 Examples n Using the following reaction n Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm. n If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added what is the concentration of HCl?

31. 31 Examples n Consider the following reaction What volume of NO at 1.0 atm and 1000º C can be produced from 10.0 L of NH 3 and excess O 2 at the same temperature and pressure? n What volume of O 2 measured at STP will be consumed when 10.0 kg NH 3 is reacted?

32. 32 The Same reaction n What mass of H 2 O will be produced from 65.0 L of O 2 and 75.0 L of NH 3 both measured at STP? n What volume of NO would be produced? n What mass of NO is produced from 500. L of NH3 at 250.0ºC and 3.00 atm?

33. 33 Dalton’s Law n The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. n The total pressure is the sum of the partial pressures. n P Total = P 1 + P 2 + P 3 + P 4 + P 5... n Make sure that the units match up.

34. 34 The mole fraction n Ratio of moles of the substance to the total moles. Symbol is Greek letter chi  Symbol is Greek letter chi  n It is the same as % except you don’t multiply by 100 %    = n 1 = P 1 n Total P Total    = n 1 = P 1 n Total P Total

35. 35 Examples n The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen? n What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

36. 36 Examples 3.50 L O 2 1.50 L N 2 2.70 atm n When these valves are opened, what is each partial pressure and the total pressure? 4.00 L CH 4 4.58 atm 0.752 atm

37. 37 Vapor Pressure n Water evaporates! n When that water evaporates, the vapor has a pressure. n Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure.

38. 38 Example n N 2 O can be produced by the following reaction n What volume of N 2 O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH 4 NO 3 ? ( the vapor pressure of water at 22ºC is 21 torr)

39. 39 Kinetic Molecular Theory n Explains why ideal gases behave the way they do. n Assumptions that simplify the theory, but don’t work in real gases.  The particles are so small we can ignore their volume.  The particles are in constant motion and their collisions cause pressure.

40. 40 Kinetic Molecular Theory  The particles do not affect each other, neither attracting or repelling.  The average kinetic energy is proportional to the Kelvin temperature. n Appendix 2 shows the derivation of the ideal gas law and the definition of temperature. n We need the formula KE = 1/2 mv 2

41. 41 What it tells us n (KE) avg = 3/2 RT R is in Joules ! n This the meaning of temperature. n u is the particle velocity. n u is the average particle velocity. n u 2 is the average particle velocity squared. n the root mean square velocity is  u 2 = u rms  u 2 = u rms

42. 42 Combine these two equations n (KE) avg = N A (1/2 mu 2 ) n (KE) avg = 3/2 RT

43. 43 Combine these two equations n (KE) avg = N A (1/2 mu 2 ) n (KE) avg = 3/2 RT Where M is the molar mass in kg/mole, and R has the units 8.3145 J/K mol. n The velocity (u) will be in m/s

44. 44 Example n Calculate the root mean square velocity of carbon dioxide at 25ºC. n Calculate the root mean square velocity of hydrogen at 25ºC. n Calculate the root mean square velocity of chlorine; Cl 2 at 25ºC.

45. 45 Range of velocities n The average distance a molecule travels before colliding with another is called the mean free path and is small (near 10 -7 ) n Temperature is an average. There are molecules of many speeds in the average. n Shown on a graph called a velocity distribution

46. 46 number of particles Molecular Velocity 273 K

47. 47 number of particles Molecular Velocity 273 K 1273 K

48. 48 number of particles Molecular Velocity 273 K 873 K 1273 K

49. 49 Velocity n Average increases as temperature increases. n Spread increases as temperature increases.

50. 50 Effusion n Passage of gas through a small hole, into a vacuum. n The effusion rate measures how fast this happens. n Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

51. 51 Effusion n Passage of gas through a small hole, into a vacuum. n The effusion rate measures how fast this happens. n Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

52. 52 Deriving n The rate of effusion should be proportional to u rms n Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2

53. 53 Deriving n The rate of effusion should be proportional to u rms n Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2

54. 54 Diffusion n The spreading of a gas through a room. n Slow considering molecules move at 100’s of meters per second. n Collisions with other molecules slow down diffusions. n Best estimate is Graham’s Law.

55. 55 Examples n A compound effuses through a porous cylinder 3.20 time faster than helium. What is it’s molar mass? n If 0.00251 mol of NH 3 effuse through a hole in 2.47 min, how much HCl would effuse in the same time? n A sample of N 2 effuses through a hole in 38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions?

56. 56 Diffusion n The spreading of a gas through a room. n Slow considering molecules move at 100’s of meters per second. n Collisions with other molecules slow down diffusions. n Best estimate is Graham’s Law.

57. 57 Real Gases n Real molecules do take up space and they do interact with each other (especially polar molecules). n Need to add correction factors to the ideal gas law to account for these.

58. 58 Volume Correction n The actual volume free to move in is less because of particle size. n More molecules will have more effect. n Corrected volume V’ = V - nb n b is a constant that differs for each gas. n P’ = nRT (V-nb)

59. 59 Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared.

60. 60 Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared. P observed = P’ - a 2 () V n

61. 61 Altogether n P obs = nRT - a n 2 V-nb V n Called the Van der Waal’s equation if rearranged n Corrected Corrected Pressure Volume ()

62. 62 Where does it come from n a and b are determined by experiment. n Different for each gas. n Bigger molecules have larger b. n a depends on both size and polarity. n once given, plug and chug.

63. 63 Example n Calculate the pressure exerted by 0.5000 mol Cl 2 in a 1.000 L container at 25.0ºC n Using the ideal gas law. n Van der Waal’s equation –a = 6.49 atm L 2 /mol 2 –b = 0.0562 L/mol