2. Chapter 5 Gases

4. Properties of a gas - Uniformly fills any container. - Mixes completely with any other gas - Exerts pressure on its surroundings. - Compressible - Gas pressure varies with altitudes and storms 5.1 Pressure

5. Measuring atmospheric pressure Torricellian barometer  Torricelli (1608-1647) studied the problem using mercury rather than H 2 O.  Mercury is denser than water, so the column wasn’t quite so high.  Gas Pressure  Liquid Pressure =  Liquid Pressure = g ·h ·d P (Pa) = Area (m 2 ) Force (N) Pascal (SI units)

6. Pascal and Torricelli Blaise Pascal (1623-1662) Evangelista Torricelli (1608-1647)

7. Barometer 760 mmHg 760 mmHg atmospheric atmospheric pressure pressure P = d·g·h d - density g - acc. of gravity h atmosphericpressure

8. Units of Pressure One atmosphere (1 atm)  Is the average pressure of the atmosphere at sea level  Is the standard atmospheric pressure Standard Atmospheric Pressure: 1 atm = 76 cm Hg = 760 mm Hg = 760 Torr = 101,325 Pa Very small unit, thus it is not commonly used

9. Example A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm 760 mm Hg 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 10 3 mmHg 14.7 psi 1.00 atm 14.7 psi 1.00 atm

10. Manometer Device for Measuring the Pressure of a Gas in a Container

11. 5.2 The Gas Laws of Boyle, Charles and Avogadro  Boyle’s Law:PV = const at constant n, T at constant n, T  Charles’ Law:V/T = const at constant n, P at constant n, P  Avogadro’s Law:V/n = const at constant P, T at constant P, T

12. Boyle’s Law P 1 V 1 = P 2 V 2 PV =k (at constant T and n) Slope= 1/k VV

13. A Plot of PV Versus P for Several Gases at Pressure Below 1 atm Boyle’s holds Only at very Low pressures A gas strictly obeys Boyle’s law is called Ideal gas

14. Charles’s Law V/T = b V = bT (constant P & n) V 1 /T 1 = V 2 /T 2

15. Plots of V Versus T(Celsius) for Several Gases Volume of a gas Changes by When the temp. Changes by 1 o C. I.e., at -273 o C, V=0 ???

16. All gases will solidify or liquefy before reaching zero volume.

17. Avogadro’s Law V  n V = an (constant P& T)

18. 5.3 The Ideal Gas Law Boyle’s law Charles's law Avogadro’s law Universal gas constant

19. The Ideal Gas Law PV = nRT R = 0.0821 atm L mol -1 K -1 The Ideal Gas Law can be used to derive the gas laws as needed!

20. Molar Volume At STP 4.0 g He 16.0 g CH 4 44.0 g CO 2 1 mole 1 mole1mole (STP) (STP)(STP) V = 22.4 L V = 22.4 L V = 22.4 L

21. The value of R  What is R for 1.00 mol of an ideal gas at STP (25 o C and 1.00 atm)?Given that V of 1 mol of gas at STOP= 22.4L V of 1 mol of gas at STOP= 22.4L

22. Example  A reaction produces enough CO 2 (g) to fill a 500 mL flask to a pressure of 1.45 atm at a temperature of 22 o C. How many moles of CO 2 (g) are produced? PV = nRT

23. The Ideal Gas Law: Final and initial state problems

24. Ideal Gas Law  The ideal gas law is an equation of state.  Independent of how you end up where you are at. Does not depend on the path.  The state of the gas is described by:  The state of the gas is described by: P, V, T and n  Given 3 you can determine the fourth.  Ideal gas equation is an empirical equation - based on experimental evidence.

25. Example A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

26. Data Table Set up Data Table P 1 = 0.800 atm V 1 = 0.180 L T 1 = 302 K P 2 = 3.20 atm V 2 = 90.0 mL T 2 = ?? ??

27. Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K - 273 = °C

28. Calculation Solve for T 2 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T 2 = 604 K - 273 = 331 °C

29. Example A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg? A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

30. Solution T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = 0.315 L = 315 mL P 1 = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL 646 mm Hg 675 mL = 178 K - 273 = - 95°C = 178 K - 273 = - 95°C

31. 5.4 Gas Stoichiometry  Reactions happen in moles  At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.4 L.  If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

32. Example A.What is the volume at STP of 4.00 g of CH 4 ? 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 16.0 g CH 4 1 mole CH 4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He 22.4 He 1 mole He

33. A 12.25 L cylinder contains 75.5 g of neon at 24.5 o C. Determine the pressure in the cylinder. PV = nRT P = V = n = R = T = ? 12.25 L 75.5 g = mol 20.18 g mol 3.74 0.082 Latm molK 24.5 + 273 = 297.5 K P = nRT V = (3.74 mol)(0.082Latm)(297.5K) (12.25 L) molK = 1.009 atm = 5670 torr Example

34. Example  30.2 mL of 1.00 M HCl are reacted with excess FeS. What volume of gas is generated at STP? STP means standard temperature and pressure... 0 o C and 1 atm. HCl + FeS  FeCl 2 + H 2 S2 HCl + FeS  FeCl 2 + H 2 S Now... Go for moles,

35. 2 HCl + FeS  FeCl 2 + H 2 S PV = nRT V = nRT/P

36. Example The decomposition of sodium azide, NaN 3, at high temperatures produces N 2 (g). What volume of N 2 (g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN 3 is decomposed. 2 NaN 3 (s) → 2 Na(l) + 3 N 2 (g)

37. Determine moles of N 2 : Determine volume of N 2 : n N 2 = 70 g NaN 3 1 mol NaN 3 65.01 g N 3 /mol N 3 3 mol N 2 2 mol NaN 3 = 1.62 mol N 2 = 41.1 L P nRT V =V = = (735 mm Hg) (1.62 mol)(0.08206 L atm mol -1 K -1 )(299 K) 760 mm Hg 1.00 atm X X 2 NaN 3 (s) → 2 Na(l) + 3 N 2 (g)

38. Molar mass of a gas  P x V = m x R x T M  m = mass, in grams  M = molar mass, in g/mol  Molar mass = m R T P V P x V = n x R x T

39. Density  Density (d) is mass divided by volume  P x V = m x R x T M  P = m x R x T V x M  P = d x R x T M  d = m V 

40. Example A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (d=0.9970 g cm -3 ) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene?  Volume of the vessel

41. V flask = = 0.1654 g m gas = m filled - m empty = (40.2959 g – 40.1305 g) = 98.41 cm 3 = 0.09841 L

42. PV = nRT PV = m M RT M = m PV RT M = (0.9741 atm)(0.09841 L) (0.6145 g)(0.08206 L atm mol -1 K -1 )(297.2 K) M = 42.08 g/mol

43. Example Calculate the density in g/L of O 2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L d = PXM d = PXM R x T R x T

44. The density of O 2 gas at STP is 1.43 grams per liter

45. Example  2.00 g sample of SX 6 (g) has a volume of 329.5 Cm 3 at 1.00 atm and 20 o C. Identify the element X. Name the compound  P= 1.00 atm  T = 273+20 = 293K M = m PV RT

46. f = 146 g SX 6 /mol Molar mass of (X 6 )= 146- 32 = 114 g/mol Molar mass of X = (114 g/mol X 6 ) /6 = 19 X = with a molar mass of 19 = F The compound is SF 6

47. 5.5 Dalton’s Law of Partial Pressures  For a mixture of gases in a container, the total pressure is the sum of the pressure each gas would exert if it were alone in the container.  The total pressure is the sum of the partial pressures.  P Total = P 1 + P 2 + P 3 + P 4 + P 5...  For each gas:

48.  P Total = n 1 RT + n 2 RT + n 3 RT +... V V V  In the same container R, T and V are the same.  P Total = (n 1 + n 2 + n 3 +...)RT V Thus, Partial pressure –Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone

50. A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H 2 at 25.0 o C. Calculate the total gas pressure in the flask in atmospheres. The total pressure is due to the partial pressures of each of these gases. so: For He: _____________________ = mol He 1.00 x 10 -3 g He 4.00 g mol 2.50 x 10 -4 For H 2 : ______________________ = mol H 2 2.00 x 10 -3 g H2H2 2.016 g mol 9.92 x 10 -4

51. A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H 2 at 25.0 o C. Calculate the total gas pressure in the flask in atmospheres. so: For He: _____________________ = mol He 1.00 x 10 -3 g He 4.00 g mol 2.50 x 10 -4 For H 2 : ______________________ = mol H 2 2.00 x 10 -3 g H 2 2.016 g mol 9.92 x 10 -4 And: P total = (2.50 x 10 -4 + 9.92 x 10 -4 )(RT/V) = (0.001242 mol)(0.0821 Latm)(25 + 273)K molK (0.2500 L) P total = 0.1216 atm

52. A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H 2 at 25.0 o C. Calculate the total gas pressure in the flask in atmospheres. so: For He: _____________________ = mol He 1.00 x 10 -3 g He 4.00 g mol 2.50 x 10 -4 For H 2 : ______________________ = mol H 2 2.00 x 10 -3 g H 2 2.016 g mol 9.92 x 10 -4 Calculate the pressure due just to He (???): = 0.0245 atm and P hydrogen = ? 0.1216 - 0.0245 = 0.0971 atm

53. Magnesium is an active metal that replaces hydrogen from an acid by the following reaction: Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) How many g of Mg are needed to produce 5.0 L of H 2 at a temperature of 25 o C and a pressure of 745 mmHg? Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) 5.0 L? g Hint: find moles of H 2 using PV = nRT then work as a stoichiometry problem. n = PV RT n = 0.20 mol =____________________________________ 745 mmHg5.0 L 62.4 LmmHg molK 298 K

54. The mole fraction  Mole fraction: number of moles of one component in a mixture relative to the total number of moles in the mixture  symbol is Greek letter chi  =

55. Mole fraction expressed in pressures

56. Example  A 1.00 L sample of dry air at 786 Torr and 25 o C contains 0.925 g N 2 plus other gasses (such as O 2, Ar and CO 2.) a) What is the partial pressure of N 2 ? b) What is the mole fraction of N 2 ?

57. Collecting gas over water An insoluble gas is passed into a container of water, the gas rises because its density is much less than that of water and the water must be displaced KClO 3 O 2 gas

58. Collection of Gases over Water Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water. P gas = P total – P H 2 O(g) P total = P gas + P H 2 O(g)

59. Oxygen gas generated was collected over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25 o C, what is the volume of the “dry” oxygen gas at STP? (P water = 23.8 torr at 25 o C) P O2 = P Total - P water = (758 - 23.8) torr = 734 torr Example P 1 = P O2 = 734 torr; P 2 = SP = 760. torr V 1 = 245mL T 1 = 298K; T 2 = 273K; V 2 = ?  (245mL)(734torr)(273K) (298K)(760.torr) V 2 = =217 mL

60. 5.6 The Kinetic Molecular Theory of Gases  It explains why ideal gases behave the way they do.  Assumptions are made to simplify the theory, but don’t work in real gases.  Postulates of the kinetic Theory:  Gas particles (atoms or molecules) are so small compared with distances between them, thus we can ignore their volume.  The particles are in constant motion and their collisions with walls cause pressure exerted by the gas

61. Kinetic Molecular Theory  The particles do not affect each other, neither attracting or repelling  The average kinetic energy (KE ) is proportional to the Kelvin temperature.  KE = 1/2 mu 2 m=mass of gas particle m=mass of gas particle v=average velocity of particles v=average velocity of particles

62. KMT explains ideal gas laws P&V: P = (nRT). (1/V)  P  1/V –# collisions increases when V decreases P & T: P = (nR/V).T  P  T –When T increases hits with walls become stronger and more frequent V & T: V=(nR/P).T  V  T –When T increases hits with walls become stronger and more frequent. To keep P constant, V must increase to compensate for particles speeds V & n: V= (RT/P). N  V  n –When n increase P would increase if the volume to be kept constant. V must increase to return P to its original value Dalton ’ s law: Individual particles are independent of each other and their volumes are negligible. Thus identities of gas particles do not matter

63. Driving the ideal gas law from KMT The following expression was derived for pressure of an ideal gas:

65. The meaning of temperature = =

66. Root mean square velocity  (KE) avg = N A (1/2 mu 2 )  (KE) avg = 3/2 RT  Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol.  The velocity will be in m/s  Combine these two equations

67. same gas Molecular speed for same gas at two different temperatures

68. Molecular speed for two different gases at two different temperatures

69. Effects of Molar Mass on u rms At constant T, 273 K, the most probable speed for O 2 > CH 4 > H 2 EOS u rms  M –1/2 so smaller molar masses result in higher molecular speeds

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

71. Effects of Temperature on u rms EOS u rms  T 1/2 so higher temperatures result in higher molecular speeds At constant mass the most probable speeds for O 2 increase with temperature Summary of Behaviors

72. number of particles Molecular Velocity 273 K

73. number of particles Molecular Velocity 273 K 1273 K

74. number of particles Molecular Velocity 273 K 1273 K

75. Molecular Velocity Molecular Velocity  Average increases as temperature increases.  Spread increases as temperature increases.   Smaller molar masses result in higher molecular speeds

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

77. 5.7 Effusion and Diffusion  Passage of gas through a small hole, into a vacuum.  The effusion rate measures how fast this happens.  Graham’s Law : the rate of effusion is inversely proportional to the square root of the mass of its particles.

78. Deriving  The rate of effusion should be proportional to u rms  Effusion Rate 1 u rms for gas 1 Effusion Rate 2 u rms for gas 2 Effusion Rate 2 u rms for gas 2 =

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

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

81. 5.8 Real Gases  Real molecules do take up space and they do interact with each other (especially polar molecules).  Need to add correction factors to the ideal gas law to account for these.

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

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

84. 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

85. Corrected Corrected Pressure Volume Corrected Corrected Pressure Volume Van der Wall’s equation

86.  a and b are determined by experiment.  Different for each gas.  Bigger molecules have larger b.  a depends on both size and polarity.

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