1. Intermolecular forces and phases of matter Why does matter exist in different phases? What if there were no intermolecular forces? The ideal gas Chapter 10

2. Physical phases of matter Gas Liquid Solid Plasma

3. Physical properties of the states of matter Gases: 1.Highly compressible 2.Low density 3.Fill container completely 4.Assume shape of container 5.Rapid diffusion 6.High expansion on heating

4. Liquid (condensed phase) 1.Slightly compressible 2.High density 3.Definite volume, does not expand to fill container 4.Assumes shape of container 5.Slow diffusion 6.Low expansion on heating

5. Solid (condensed phase) 1.Slightly compressible 2.High density 3.Rigidly retains its volume 4.Retains its own shape 5.Extremely slow diffusion; occurs only at surfaces 6.Low expansion on heating

6. Why water exists in three phases? Kinetic energy(the state of substance at room temperature depends on the strength of attraction between its particles) Intermolecular forces stick molecules together (heating and cooling)

7. Intermolecular forces London Force or dispersion forces Dipole-dipole Hydrogen bond

8. London Force Weak intermolecular force exerted by molecules on each other, caused by constantly shifting electron imbalances. This forces exist between all molecules. Polar molecules experience both dipolar and London forces. Nonpolar molecules experience only London intermolecular forces

9. Dipole-dipole Intermolecular force exerted by polar molecules on each other. The name comes from the fact that a polar molecule is like an electrical dipole, with a + charge at one end and a - charge at the other end. The attraction between two polar molecules is thus a "dipole-dipole" attraction.

10. Hydrogen bond Intermolecular dipole-dipole attraction between partially positive H atom covalently bonded to either an O, N, or F atom in one molecule and an O, N, or F atom in another molecule.

11. To form hydrogen bonds, molecules must have at least one of these covalent bonds: H-N or H-N= H-O- H-F

14. Nonmolecular substances Solids that don’t consist of individual molecules. Ionic compounds(lattices of ions) They are held together by strong ionic bonds Melting points are high

15. Other compounds Silicon dioxide(quartz sand) and diamond (allotrope of carbon) These are not ionic and do not contain molecules They are network solids or network covalent substances

16. Real Gas Molecules travel fast Molecules are far apart Overcome weak attractive forces

17. Ideal Gas Gas that consists of particles that do not attract or repel each other. In ideal gases the molecules experience no intermolecular forces. Particles move in straight paths. Does not condense to a liquid or solid.

18. Kinetic Molecular Theory Particles in an ideal gas… –have no volume. –have elastic collisions. –are in constant, random, straight-line motion. –don’t attract or repel each other. –have an avg. KE directly related to Kelvin temperature.

19. Ideal Gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. Gases consist of tiny particles that are far apart relative to their size. Collisions between gas particles and between particles and the walls of the container are elastic collisions No kinetic energy is lost in elastic collisions

20. Ideal Gases Ideal Gases (continued) Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion There are no forces of attraction between gas particles The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

21. Measurable properties used to describe a gas: Pressure (P) P=F/A Volume (V) Temperature (T) in Kelvins Amount (n) specified in moles

22. Pressure Is caused by the collisions of molecules with the walls of a container is equal to force/unit area SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere = 101.3 kPa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

23. Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure

24. An Early Barometer The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

25. Manometer –measures contained gas pressure U-tube Manometer

26. measures the pressure of a confined gas manometer: CONFINED GAS AIR PRESSURE Hg HEIGHT DIFFERENCE SMALL + HEIGHT = BIG differential manometer manometers can be filled with any of various liquids

27. Atmosphere Pressure and the Barometer The pressures of gases not open to the atmosphere are measured in manometers. A manometer consists of a bulb of gas attached to a U-tube containing Hg: –If P gas < P atm then P gas + P h2 = P atm. –If P gas > P atm then P gas = P atm + P h2. Pressure

28. Chapter 10 The Pressure-Volume Relationship: Boyle’s Law Weather balloons are used as a practical consequence to the relationship between pressure and volume of a gas. As the weather balloon ascends, the volume decreases. As the weather balloon gets further from the earth’s surface, the atmospheric pressure decreases. Boyle’s Law: the volume of a fixed quantity of gas is inversely proportional to its pressure. Boyle used a manometer to carry out the experiment. The Gas Laws

30. 760 mm Hg X mm Hg 112.8 kPa 0.78 atm BIG small height BIG = small + height 101.3 kPa = 846 mm Hg 0.78 atm 760 mm Hg 1 atm = 593 mm Hg height = BIG - small X mm Hg = 846 mm Hg - 593 mm Hg X mm Hg = 253 mm Hg STEP 1) Decide which pressure is BIGGER STEP 2) Convert ALL numbers to the unit of unknown STEP 3) Use formula Big = small + height 253 mm Hg

31. Atmospheric pressure is 96.5 kPa; mercury height difference is 233 mm. Find confined gas pressure, in atm. X atm 96.5 kPa 233 mm Hg B 96.5 kPa S SMALL + HEIGHT = BIG 0.953 atm0.307 atm+= 1.26 atm 233 mm Hg + X atm =

32. X mm Hg 112.8 kPa 0.78 atm 760 mm Hg 101.3 kPa = 846 mm Hg 0.78 atm 760 mm Hg 1 atm = 593 mm Hg KEY 0 mm Hg X atm 125.6 kPa 1 atm 101.3 kPa = 1.24 atm 125.6 kPa 1. 2. Because no difference in height is shown in barometer, You only need to convert “kPa” into “atm”. Convert all units into “mm Hg” Use the formula Big = small + height Height = Big - small X mm Hg = 846 mm Hg - 593 mm Hg X = 253 mm Hg

33. 98.4 kPa X mm Hg 0.58 atm 760 mm Hg 1 atm = 441 mm Hg 98.4 kPa 760 mm Hg 101.3 kPa = 738 mm Hg KEY 135.5 kPa 208 mm Hg X atm 760 mm Hg 1 atm = 0.28 atm 135.5 kPa 1 atm 101.3 kPa = 1.34 atm 3. 4. Height = Big - small X mm Hg = 738 mm Hg - 441 mm Hg X = 297 mm Hg small = Big - height X atm = 1.34 atm - 0.28 atm X = 1.06 atm

34. Units of Pressure UnitSymbolDefinition/Relationship PascalPaSI pressure unit 1 Pa = 1 newton/meter 2 Millimeter of mercury mm HgPressure that supports a 1 mm column of mercury in a barometer AtmosphereatmAverage atmospheric pressure at sea level and 0  C Torrtorr1 torr = 1 mm Hg

35. Standard Temperature and Pressure “STP” P = 1 atmosphere, 760 torr, 101.3 kPa T =  C, 273 Kelvins The molar volume of an ideal gas is 22.4 liters at STP

36. Behavior of gases Rule 1: P is proportional to 1/V Rule 2: P is proportional to T Rule 3: P is proportional to n Combining all three: P is proportional to nT/V P=constant x nT/v R=constant= 0.0821 L atm/K mole

37. Pressure - Temperature - Volume Relationship P T V Gay-Lussac’s P T  CharlesV T  P T V Boyle’s P 1V1V  ___

38. Boyle’s Law P inversely proportional to V PV= k Temperature and number of moles constant

39. Chapter 10 The Pressure-Volume Relationship: Boyle’s Law Mathematically: A plot of V versus P is a hyperbola. Similarly, a plot of V versus 1/P must be a straight line passing through the origin. The Gas Laws

40. Boyle’s Law Pressure is inversely proportional to volume when temperature is held constant.

41. A Graph of Boyle’s Law

42. Charles’s Law V directly proportional to T T= absolute temperature in kelvins V/T =k 2 Pressure and number of moles constant

43. Chapter 10 The Temperature-Volume Relationship: Charles’s Law We know that hot air balloons expand when they are heated. Charles’s Law: the volume of a fixed quantity of gas at constant pressure increases as the temperature increases. Mathematically: The Gas Laws

44. Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. (P = constant) Temperature MUST be in KELVINS!

45. A Graph of Charles’ Law

46. Chapter 10 The Temperature-Volume Relationship: Charles’s Law A plot of V versus T is a straight line. When T is measured in  C, the intercept on the temperature axis is -273.15  C. We define absolute zero, 0 K = -273.15  C. Note the value of the constant reflects the assumptions: amount of gas and pressure. The Gas Laws

47. Gay Lussac’s Law The pressure and temperature of a gas are directly related, provided that the volume remains constant. Temperature MUST be in KELVINS!

48. Prentice Hall © 2003Chapter 10 The Quantity-Volume Relationship: Avogadro’s Law Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers. The Gas Laws

49. A Graph of Gay-Lussac’s Law

50. Chapter 10 The Quantity-Volume Relationship: Avogadro’s Law Avogadro’s Hypothesis: equal volumes of gas at the same temperature and pressure will contain the same number of molecules. Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas. The Gas Laws

51. Chapter 10 The Quantity-Volume Relationship: Avogadro’s Law Mathematically: We can show that 22.4 L of any gas at 0  C contain 6.02  10 23 gas molecules. The Gas Laws

52. Chapter 10 The Quantity-Volume Relationship: Avogadro’s Law The Gas Laws

53. The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas. Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

54. Standard Molar Volume Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro

55. Avogadro’s Law V directly proportional to n V/n = k 3 Pressure and temperature are constant

56. Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +... This is particularly useful in calculating the pressure of gases collected over water.

57. 3.24 atm 2.82 atm 1.21 atm 0.93 dm 3 1.23 dm 3 1.42 dm 3 1.51 dm 3 2.64 atm 1.74 atm 1.14 atm 5.52 atm TOTAL A B C PxPx VxVx PDPD VDVD 1.Container A (with volume 1.23 dm 3 ) contains a gas under 3.24 atm of pressure. Container B (with volume 0.93 dm 3 ) contains a gas under 2.82 atm of pressure. Container C (with volume 1.42 dm 3 ) contains a gas under 1.21 atm of pressure. If all of these gases are put into Container D (with volume 1.51 dm 3 ), what is the pressure in Container D? Dalton’s Law of Partial Pressures P T = P A + P B + P C (3.24 atm)(1.23 dm 3 ) = (x atm)(1.51 dm 3 ) (P A )(V A ) = (P D )(V D ) (P A ) = 2.64 atm (2.82 atm)(0.93 dm 3 ) = (x atm)(1.51 dm 3 ) (P B )(V B ) = (P D )(V D ) (P B ) = 1.74 atm (1.21 atm)(1.42 dm 3 ) = (x atm)(1.51 dm 3 ) (P C )(V A ) = (P D )(V D ) (P C ) = 1.14 atm 1.51 dm 3

58. PAPA 628 mm Hg 437 mm Hg 250 mL 150 mL 350 mL 300 mL 406 mm Hg 523 mm Hg 510 mm Hg 1439 mm Hg TOTAL A B C PxPx VxVx PDPD VDVD Dalton’s Law of Partial Pressures 3.Container A (with volume 150 mL) contains a gas under an unknown pressure. Container B (with volume 250 mL) contains a gas under 628 mm Hg of pressure. Container C (with volume 350 mL) contains a gas under 437 mm Hg of pressure. If all of these gases are put into Container D (with volume 300 mL), giving it 1439 mm Hg of pressure, find the original pressure of the gas in Container A. (P A )(150 mL) = (406 mm Hg)(300 mL) (P A )(V A ) = (P D )(V D ) (P A ) = 812 mm Hg STEP 1) STEP 2) STEP 3) STEP 4) (437)(350) = (x)(300) (P C )(V C ) = (P D )(V D ) (P C ) = 510 mm Hg (628)(250) = (x)(300) (P B )(V B ) = (P D )(V D ) (P B ) = 523 mm Hg P T = P A + P B + P C 1439 -510 -523 406 mm Hg STEP 1) STEP 2) STEP 3) STEP 4) 812 mm Hg 300 mL

59. Chapter 10 Consider the three gas laws. We can combine these into a general gas law: The Ideal Gas Equation Boyle’s Law: Charles’s Law: Avogadro’s Law:

60. Chapter 10 If R is the constant of proportionality (called the gas constant), then The ideal gas equation is: R = 0.08206 L·atm/mol·K = 8.314 J/mol·K The Ideal Gas Equation

62. Ideal Gas Law PV = nRT P = pressure in atm V = volume in liters n = moles R = proportionality constant = 0.0821 L atm/ mol·  T = temperature in Kelvins Holds closely at P < 1 atm

63. = 22.4 L The Ideal Gas Law P V = n R T P = pres. (in kPa) V = vol. (in L or dm 3 ) T = temp. (in K) n = # of moles of gas (mol) R = universal gas constant = 8.314 L-kPa/mol-K 32 g oxygen at 0 o C is under 101.3 kPa of pressure. Find sample’s volume. P V = n R T T = 0 o C + 273 = 273 K PP

64. 54.0 kPa 0.25 g carbon dioxide fills a 350 mL container at 127 o C. Find pressure in mm Hg. = 54.0 P V = n R T T = 127 o C + 273 = 400 K VV V = 0.350 L kPa = 405 mm Hg

65. Gas Density … so at STP…

66. Density and the Ideal Gas Law Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically: M = Molar Mass P = Pressure R = Gas Constant T = Temperature in Kelvins

67. Density of Gases Density formula for any substance: For a sample of gas, mass is constant, but pres. and/or temp. changes cause gas’s vol. to change. Thus, its density will change, too. ORIG. VOL. NEW VOL. ORIG. VOL. NEW VOL. If V (due to P or T ), then… D If V (due to P or T ), then… D

68. Density of Gases Equation: ** As always, T’s must be in K. A sample of gas has density 0.0021 g/cm 3 at –18 o C and 812 mm Hg. Find density at 113 o C and 548 mm Hg. 812(386)(D 2 ) = 255(0.0021)(548) 255 K 386 K 812 (0.0021) = 255 548 (D 2 ) 386 D 2 = 9.4 x 10 –4 g/cm 3 (386) 812 (386) 812

69. Find density of nitrogen dioxide at 75 o C and 0.805 atm. 348 K NO 2 D of NO 2 @ STP… 1(348)(D 2 ) = 273(2.05)(0.805) 1 (2.05) = 273 0.805 (D 2 ) 348 D 2 = 1.29 g/L (348) 1 1 NO 2 participates in reactions that result in smog (mostly O 3 )

70. A gas has mass 154 g and density 1.25 g/L at 53 o C and 0.85 atm. What vol. does sample occupy at STP? 326 K Find D @ STP… 0.85(273)(D 2 ) = 326(1.25)(1) 0.85 (1.25) = 326 1 (D 2 ) 273 D 2 = 1.756 g/L (273) 0.85 (273) 0.85 Find vol. when gas has that density. = 87.7 L

71. Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Diffusion

72. Effusion Effusion: describes the passage of gas into an evacuated chamber.

73. Graham’s Law KE = ½mv 2 Speed of diffusion/effusion –Kinetic energy is determined by the temperature of the gas. –At the same temp & KE, heavier molecules move more slowly. Larger m  smaller v

74. Derivation of Graham’s Law The average kinetic energy of gas molecules depends on the temperature: where m is the mass and v is the speed Consider two gases:

75. Graham’s Law Consider two gases at same temp. Gas 1: KE 1 = ½ m 1 v 1 2 Gas 2: KE 2 = ½ m 2 v 2 2 Since temp. is same, then… KE 1 = KE 2 ½ m 1 v 1 2 = ½ m 2 v 2 2 m 1 v 1 2 = m 2 v 2 2 Divide both sides by m 1 v 2 2 … Take square root of both sides to get Graham’s Law:

76. On average, carbon dioxide travels at 410 m/s at 25 o C. Find the average speed of chlorine at 25 o C. **Hint: Put whatever you’re looking for in the numerator.

77. Effusion: Diffusion: Graham’s Law Rates of Effusion and Diffusion