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Chapter 10 and 11

Chapter 10 and 11. Intermolecular forces and phases of matter Why does matter exist in different phases? What if there were no intermolecular forces? The ideal gas. Physical phases of matter . Gas Liquid Solid Plasma. Physical properties of the states of matter. Gases:

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Chapter 10 and 11

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  1. Chapter 10 and 11 Intermolecular forces and phases of matter Why does matter exist in different phases? What if there were no intermolecular forces? The ideal gas

  2. Physical phases of matter • Gas • Liquid • Solid • Plasma

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

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

  5. Solid (condensed phase) Slightly compressible High density Rigidly retains its volume Retains its own shape Extremely slow diffusion; occurs only at surfaces 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

  12. 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

  13. 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

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

  15. 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.

  16. 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.

  17. 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

  18. 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.

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

  20. Pressure • Is caused by the collisions of molecules with the walls of a container • is equal to force/unit area • SI units = Newton/meter2 = 1 Pascal (Pa) • 1 standard atmosphere = 101.3 kPa • 1 standard atmosphere = 1 atm = • 760 mm Hg = 760 torr

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

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

  23. U-tube Manometer • Manometer • measures contained gas pressure

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

  25. BIG BIG = small + height 760 mm Hg 112.8 kPa 846 mm Hg = height = BIG - small 101.3 kPa X mm Hg 846 mm Hg 593 mm Hg - = X mm Hg = 253 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 small 0.78 atm height X mm Hg 760 mm Hg 0.78 atm 593 mm Hg = 1 atm

  26. 96.5 kPa X atm 233 mm Hg Atmospheric pressure is 96.5 kPa; mercury height difference is 233 mm. Find confined gas pressure, in atm. S B SMALL + HEIGHT = BIG + = 96.5 kPa 233 mm Hg X atm 0.953 atm + 0.307 atm = 1.26 atm

  27. Units of Pressure

  28. Standard Temperature and Pressure“STP” • P = 1 atmosphere, 760 torr, 101.3 kPa • T = 0°C, 273 Kelvins • The molar volume of an ideal gas is 22.4 liters at STP

  29. 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

  30. PTV PT V P T V 1 V Boyle’s P a ___ a Charles V T Gay-Lussac’s P T a Pressure - Temperature - Volume Relationship

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

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

  33. A Graph of Boyle’s Law

  34. Charles’s Law • V directly proportional to T • T= absolute temperature in kelvins • V/T =k2 • Pressure and number of moles constant

  35. 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!

  36. A Graph of Charles’ Law

  37. 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!

  38. A Graph of Gay-Lussac’s Law

  39. 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.

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

  41. Avogadro’s Law • V directly proportional to n • V/n = k3 • Pressure and temperature are constant

  42. Dalton’s Law of Partial Pressures • For a mixture of gases in a container, • PTotal = P1 + P2 + P3 + . . . This is particularly useful in calculating the pressure of gases collected over water.

  43. Dalton’s Law of Partial Pressures • Container A (with volume 1.23 dm3) contains a gas under 3.24 atm of pressure. Container B (with volume 0.93 dm3) contains a gas under 2.82 atm of pressure. Container C (with volume 1.42 dm3) contains a gas under 1.21 atm of pressure. If all of these gases are put into Container D (with volume 1.51 dm3), what is the pressure in Container D? Px Vx PD VD 1.51 dm3 3.24 atm 2.64 atm 1.51 dm3 A 1.23 dm3 B 1.51 dm3 2.82 atm 0.93 dm3 1.74 atm C 1.51 dm3 1.42 dm3 1.21 atm 1.14 atm PT = PA + PB + PC TOTAL 5.52 atm (PA)(VA) = (PD)(VD) (PB)(VB) = (PD)(VD) (PC)(VA) = (PD)(VD) (3.24 atm)(1.23 dm3) = (x atm)(1.51 dm3) (2.82 atm)(0.93 dm3) = (x atm)(1.51 dm3) (1.21 atm)(1.42 dm3) = (x atm)(1.51 dm3) (PA) = 2.64 atm (PB) = 1.74 atm (PC) = 1.14 atm

  44. Dalton’s Law of Partial Pressures • 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. Px Vx PD VD STEP 3) STEP 4) 300 mL PA 406 mm Hg 300 mL A 150 mL STEP 2) B 300 mL 628 mm Hg 250 mL 523 mm Hg STEP 1) C 300 mL 350 mL 437 mm Hg 510 mm Hg PT = PA + PB + PC TOTAL 1439 mm Hg STEP 4) STEP 2) STEP 3) STEP 1) 1439 -510 -523 406 mm Hg (PC)(VC) = (PD)(VD) (PB)(VB) = (PD)(VD) (PA)(VA) = (PD)(VD) (437)(350) = (x)(300) (628)(250) = (x)(300) (PA)(150 mL) = (406 mm Hg)(300 mL) 812 mm Hg (PC) = 510 mm Hg (PB) = 523 mm Hg (PA) = 812 mm Hg

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

  46. P P P V = n R T The Ideal Gas Law P = pres. (in kPa) T = temp. (in K) V = vol. (in L or dm3) n = # of moles of gas (mol) R = universal gas constant = 8.314 L-kPa/mol-K 32 g oxygen at 0oC is under 101.3 kPa of pressure. Find sample’s volume. T = 0oC + 273 = 273 K P V = n R T = 22.4 L

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

  48. Gas Density … so at STP…

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