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Chapters 10 and 11

Chapters 10 and 11 . Kinetic molecular theory and states of matter. Gases . A pure substance in its gaseous state is often referred to as a vapor. The molecules of a vapor are not tightly bound together and are move freely through space.

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Chapters 10 and 11

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  1. Chapters 10 and 11 Kinetic molecular theory and states of matter

  2. Gases • A pure substance in its gaseous state is often referred to as a vapor. • The molecules of a vapor are not tightly bound together and are move freely through space. • Gases form homogeneous mixtures regardless of the identities or relative proportions of the component gases. • The three most readily measured properties of a gas are, temperature, volume, and pressure.

  3. Pressure • Pressure is measured as the force applied over a given area. • P = F/A • Standard atmospheric pressure at sea level is measured as 1 atmosphere (atm). • Other units for pressure are: • Torr • mm Hg • kPa • 1 atm = 760 torr = 760 mm Hg = 101.3 kPa

  4. Boyles Law • Boyles law relates the pressure of a gas and the volume of a gas at constant temperature. • Boyles law states that, the volume of a fixed quantity of gas maintained at constant temperature is inversely proportional to the pressure. • Boyle’s Law is represented by the equation: • P1V1= P2V2

  5. Charles’s Law • Charles’s law relates the temperature of a gas and the volume of a gas. • This law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to its absolute temperature (Kelvin) • Charles’s law is represented by: • V1/T1 = V2/T2 or V2/V1 = T2/T1 or V1T2 = V2T1

  6. Avogadro’s Law • We know that if we add more air into a balloon it will get bigger. • Avogadro’s Law relates the amount of gas and the volume of a gas. • Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. • What Avogadro found was that one mole of any gas at standard pressure (1 atm) and standard pressure (273.15 K) will have a volume of 22.4 L.

  7. The Ideal Gas Equation • The ideal gas equation is written as: • PV = nRT • The term R is called the ideal gas constant. • The values of R are… • 0.0821 L-atm/mol-K • 8.13 J/mol-K • 62.4 L-torr/mol-K • 8.31 volt-coulomb/mol-K

  8. Using the Ideal Gas Equation • Calcium carbonate decomposes upon heating to give calcium oxide and carbon dioxide gas. A sample of calcium carbonate is decomposed, and the carbon dioxide is collected in a 250-mL flask. The pressure of the gas collected is measured at 1.3 atm at a temperature of 31o C. How many moles of gas were collected? • 0.013 mol CO2

  9. Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure, in atmospheres, inside the ball at 24oC ? • 2.0 atm

  10. Gas Densities and Molar Mass • The ideal gas equation allows us to calculate gas density from the molar mass, pressure, and temperature. • Remember density is defined as mass/volume. • What is the density of carbon tetrachloride vapor at 714 torr and 125oC? • 4.43 g/L

  11. A large flask is evacuated and found to weigh 134.567 g. It is then filled with a gas to a pressure of 735 torr at 31oC and reweighed. Its mass is now 137.456 g. Finally the flask is filled with water at 31oC and found to weigh 1067.9 g. (The density of water at this temperature is 0.997 g/mL) calculate the molar mass of the unknown gas. • 79.9 g/mol

  12. Gas Mixtures and Partial Pressure • All gases form homogeneous mixtures. • The pressure exerted by a particular component of a mixture of gases is called the partial pressure of that gas. • Pt = P1 + P2 + P3 + …… • If P = n(RT/V) than P1 = n1(RT/V) • So Pt = (n1 + n2 + n3 + ……) (RT/V)

  13. Partial Pressure and Mole Fractions • If we have a mixture of gasses the mole ratio of gas number 1 is simply n1/nt. • A mixture of gas is composed of 1.5 mol percent CO2, 18.0 mol percent O2, and 80.5 mole percent Ar. • (a) Calculate the partial pressure of O2 if the total pressure of the mixture is 745 torr. • 134 torr.

  14. A gaseous mixture made from 6.00 g of O2 and 9.00 g of CH4 is placed ina 15. 0 L vessel at 0o C. What is the partial pressure of each gas and what is the total pressure in the vessel? • P(O2) = 0.281 atm • P (CH4) = 0.841 atm • Pt = 1.122 atm

  15. Assuming ideal conditions, calculate the total pressure in 3.0 Liter container that contains 0.015 moles of carbon dioxide combined with 2.0 moles of oxygen at a temperature of 30 degrees Celsius? • (A) 1.65 atm • (B) 22.4 atm • (C) 16.7 atm • (D) 17.8 atm • (E) None of the Above

  16. Kinetic Molecular Theory • The ideal gas equation explains how gases but it does not explain why they behave as they do. • The Kinetic Molecular Theory states that • Gases consist of large numbers of molecules that are in continuous, random motion. • The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. • Attractive and repulsive forces between gas molecules are negligible. • Energy can be transferred between molecules during collisions but the average kinetic energy does not change. As long as the temperature remains constant. • The average kinetic energy of the molecules is proportional to the absolute temperature.

  17. Distributions of Molecular Speed • Although the molecules in a sample of gas have and average kinetic energy (and therefore and average speed) some molecules may be moving faster than others. • At higher temperatures a larger fraction of molecules are moving at greater speeds.

  18. A constant temperature means that the average kinetic energy of the gas molecules remains unchanged. This means that u is unchanged as well. • An increase of temperature will result in an increase of u and in result an increase in E.

  19. Mass and average kinetic energy • As the molar mass of a gas increases the average kinetic energy of the sample will decrease. (If the temperature remains constant) • We use the equation below to find the average speed of gas molecules in a sample. • Calculate the average speed of a molecule of N2 at 25oC.

  20. Effusion vs. Diffusion • Effusion is the escape of gas molecules through a tiny hole into an evacuated space. • Diffusion is the spreading of one substance through a space or throughout a second substance.

  21. Effusion

  22. Diffusion

  23. Comparing solids, liquids, and gases • Solids: • Retain their own shape • Virtually incompressible • Do not flow • Diffusion within a solid occurs extremely slowly • Liquids: • Assume the shape of the container they occupy • Do not expand to fill container • Virtually incompressible • Flow readily • Diffusion with in a liquid occurs slowly • Gases: • Assume both the volume and shape of their container • Are compressible • Flow readily • Diffusion within a gas happens rapidly

  24. These properties can be explained in terms of the kinetic energy of the particles. • Gases: • In gases the average kinetic energy of the particles is much greater than the forces of attraction between the particles. • This allows gases to expand to fill their container. • Liquids: • Liquids have much stronger intermolecular attractions than gases. This results in liquids being far less compressible than gases. • These intermolecular attractions are not strong enough to keep the molecules from moving around one another. This results in liquids being able to flow and be poured. • Solids: • Solids have strong intermolecular attractive forces. • These forces are greater than the average kinetic energy of the particles. • This causes solids to have a fixed shape and be incompressible.

  25. Solid • Gas: Total disorder, a large amount of empty space, particles can move freely, particles are far apart. • Liquid: Disorder, particles or clusters of particles are free to move relative to each other, particles close together • Solid: Ordered arrangement, particles are in fixed positions, particles are close together.

  26. Intermolecular Forces • The strengths of intermolecular forces for different substances vary greatly. • The strength of intermolecular forces are much weaker than the bond forces that hold molecules together. • Example: • The energy required to vaporize liquid HCL is only 16 kJ/mol where as the energy required to break the bond between the hydrogen and the chlorine atoms is 431 kJ/mol

  27. Effects of Intermolecular Interactions • The properties of substances reflect the strengths of their intermolecular attractions. • A liquids boiling point is directly related to the intermolecular attractions. • Example: • Water boils at 100o C. Liquid HCl boils at -85o C. Which substance has greater intermolecular interactions?

  28. Types of Intermolecular Interactions • Ion-Dipole Forces • Dipole-Dipole Forces • London Dispersion Forces • Hydrogen Bonding

  29. Ion-Dipole Forces • Ion-Dipole Forces happen between an ion and a polar molecule. • These are the forces that are present in aqueous solutions of ionic compounds. • In which of the following mixtures do you encounter ion-dipole forces: CH3OH in water, or Ca(NO3)2 in water?

  30. Dipole-Dipole Forces • Dipole-Dipole forces happen between two polar molecules. • These forces are generally weaker than Ion-Dipole forces. • For molecules with approximately the same mass these forces get stronger with increasing polarity.

  31. For which of the substances in the table below are the dipole-dipole attractive forces greatest?

  32. London Dispersion Forces • All substances experience London Dispersion Forces. • No dipole-dipole interactions happen between non-polar molecules. • The movement of molecules can cause momentary dipoles. • These temporarily created dipoles cause attractive forces between molecules.

  33. London Dispersion Forces Cont. • The strength of London Dispersion Forces depend on an atom or a molecules polarizability. • Polarizability is the measure of how easily the electron cloud of an atom or molecule can be distorted. • The more polarizable a molecule or atom is the stronger London Forces it will experience. • Dispersion forces tend to increase with an increasing molecular weight. Why?

  34. Pentane (C5H12) n-Pentante: bp = 309.4 K Neopentante: bp = 282.7

  35. Hydrogen Bonds • Some molecules were found to have abnormally high boiling points considering their low molecular weights. • This lead scientists to believe that there were other, stronger, forces acting on these molecules.

  36. Hydrogen Bonds • Hydrogen bonds occur between a hydrogen atom in a polar bond and a non-bonding pair of electrons on a small electronegative ion or atom. (F, O, N) • Examples: • Water • DNA • Hydrogen bonds can be thought of as extremely strong dipole-dipole attractions.

  37. Properties of Liquids • Viscosity • Viscosity is the resistance of a liquid to flow. • Viscosity is related to how easily molecules of a liquid are allowed to flow around one another.

  38. Surface Tension • Surface tension • Molecules in the center of a liquid experience attractive forces from all directions. • Molecules on the surface experience a net inward force, pulling them into the bulk of the liquid and packing the surface molecules tighter together. • Stronger intermolecular forces result in higher surface tension.

  39. Phase Changes

  40. Energy of Phase Changes • Heat of fusion (ΔHfus) • The energy required to melt a solid • the heat of fusion for ice is 6.01 kJ/mol. • Heat of vaporization (ΔHVap) • The energy required to turn a liquid into a gas. • The heat of vaporization for liquid water is 40.7 kJ/mol. • Heat of sublimation (ΔHsub) • A solid can be transformed directly into a gas. • The heat of sublimation for ice is 47 kJ/mol.

  41. Generally the hear of fusion is les than the heat of vaporization. Why? • It takes more energy to completely separate the particles than to just partially separate them. • Endothermic Process • Heat solid  Melt  Heat liquid  Boil  Heat gas • Exothermic Process • Cool gas  Condense  Cool Liquid  Freeze  cool solid

  42. Heating Curves • A heating curve is a plot of temperature change versus heat added. • During the plateaus the temperature does not change but we continue adding energy. • The extra energy is used to break intermolecular attractions rather than cause a temperature change.

  43. Example Problem • Calculate the enthalpy change (total ΔH) to convert 1.00 mol of ice at -250 C to water vapor at 1250 C under a constant pressure of 1atm. The specific heats of ice, water, and steam are 2.03 J/g-K, 4.18 J/g-K, 1.84 J/g-K. • For H2O: • ΔHfus = 6.01 kJ/mol • ΔHvap = 40.67 kJ/mol

  44. AB: heating solid ice. • AB represents heating ice from -250 C to 0.000 C. • We can use the specific heat of water to calculate ΔH • We have 18g of ice and the specific heat of ice is 2.03 J/g-K. • (18g)(2.03 J/g-k)(25K) • 914J or 0.914 kJ

  45. BC: For this section of the heating curve we are converting solid ice to liquid water at 0.00o C. We can use the heat of fusion. • (1.00mol H2O)(6.01 kJ/mol) • 6.01 kJ

  46. CD: For CD we are heating liquid water. • We need to use the specific heat of water. (4.18 J/g-k) • We sill have 18g of liquid water. • (18g)(4.18 J/g-k)(100K) • 752 J or 0.752 kJ

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