Ch 14 Ideal Gas Law & Kinetic Theory


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Ch 14 Ideal Gas Law & Kinetic Theory

  1. 1. Chapter 14 The Ideal Gas Law & Kinetic Theory
  2. 2. AP Learning Objectives Kinetic theory  Ideal gases  Students should understand the kinetic theory model of an ideal gas, so they can:  State the assumptions of the model.  State the connection between temperature and mean translational kinetic energy, and apply it to determine the mean speed of gas molecules as a function of their mass and the temperature of the gas.  State the relationship among Avogadro’s number, Boltzmann’s constant, and the gas constant R, and express the energy of a mole of a monatomic ideal gas as a function of its temperature.  Explain qualitatively how the model explains the pressure of a gas in terms of collisions with the container walls, and explain how the model predicts that, for fixed volume, pressure must be proportional to temperature.
  3. 3. Table of Contents 1. Molecular Mass, the Mole, Avogadro’s Number 2. Ideal Gas law 3. Kinetic Theory of Gases 4. Diffusion (AP?)
  4. 4. Chapter 14: Ideal Gas Law & Kinetic Theory Section 1: Molecular Mass, the Mole, and Avogadro’s Number
  5. 5. Atomic Mass  To facilitate comparison of the mass of one atom with another, a mass scale know as the atomic mass scale has been established.  The unit is called the atomic mass unit (symbol u). The reference element is chosen to be the most abundant isotope of carbon, which is called carbon-12  Carbon-12 is defined at 12 u  The atomic mass is given in atomic mass units. For example, a Li atom has a mass of 6.941u. kg106605.1u1 27− ×=
  6. 6. 123 mol10022.6 − ×=AN AN N n = number of moles number of atoms The Mole & Avogadro’s Number  One mole of a substance contains as many particles as there are atoms in 12 grams of the isotope cabron-12.  The number of atoms per mole is known as  Avogadro’s number, NA.
  7. 7. M µ µ µ == AN N n Molar Mass  μ is mass of the molecule  M is molar mass  The molar mass (in g/mol) of a substance has the same numerical value as the atomic or molecular mass of the substance (in atomic mass units).  For example Hydrogen has an atomic mass of 1.00794 g/mol, while the mass of a single hydrogen atom is 1.00794 u.
  8. 8. Example 1 The Hope Diamond and the Rosser Reeves Ruby The Hope diamond (44.5 carats) is almost pure carbon. The Rosser Reeves ruby (138 carats) is primarily aluminum oxide (Al2O3). One carat is equivalent to a mass of 0.200 g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al2O3 molecules in the ruby. M µ =n(a) (b) ( ) ( ) ( )[ ] molg96.101 carat1g200.0carats138 = AnNN = AnNN = ( ) ( ) ( )[ ] molg011.12 carat1g200.0carats5.44 = mol741.0= ( )( )123 mol10022.6mol741.0 − ×= atoms1046.4 23 ×= M µ =n ( ) ( )  99.15398.262 + mol271.0= ( )( )123 mol10022.6mol271.0 − ×= atoms1063.1 23 ×=
  9. 9. 14.1.1. In 1865, Loschmidt calculated the number of molecules in a cubic centimeter of a gas under standard temperature and pressure conditions. He later used this number to estimate the size of an individual gas molecule. Calculate Loschmidt’s number for helium using the density, the atomic mass, and Avogadro’s number. a) 2.7 × 1019 atoms/cm3 b) 3.5 × 1020 atoms/cm3 c) 4.1 × 1021 atoms/cm3 d) 5.4 × 1022 atoms/cm3 e) 6.2 × 1023 atoms/cm3
  10. 10. 14.1.2. Suppose that molecules of water (molecular mass = 0.01802 kg/m3 ) completely fill a container so that there is no empty space within the container. Using the density of water and Avogadro’s number, estimate the size of the water molecule. Hint: assume the water molecule fits within a cube and that these cubes are stacked to fill the volume of the container. a) 4 × 10−11 m b) 2 × 10−9 m c) 3 × 10−10 m d) 5 × 10−9 m e) 6 × 10−10 m
  11. 11. 14.1.3. The standard for determining atomic masses is the carbon-12 atom, so that the mass of one mole of carbon-12 is exactly twelve grams. What would Avogadro’s number and the atomic mass of oxygen-16 be if the standard were that one mole of hydrogen is exactly one gram? a) 6.020 × 1023 mol−1 , 15.9898 grams b) 6.069 × 1023 mol−1 , 16.1200 grams c) 5.844 × 1023 mol−1 , 15.7845 grams d) 5.975 × 1023 mol−1 , 15.8707 grams e) 6.122 × 1023 mol−1 , 16.3749 grams
  12. 12. 14.1.4. Under which of the following circumstances does a real gas behave like an ideal gas? a) The gas particles move very slowly. b) The gas particles do not collide with each other very often. c) The interaction between the gas particles is negligible. d) The interaction between the gas particles and the walls of the container is negligible. e) There are only one kind of particles in the container.
  13. 13. Chapter 14: Ideal Gas Law & Kinetic Theory Section 2: The Ideal Gas Law
  14. 14. TP ∝ The Ideal Gas Law  An ideal gas is an idealized model for real gases that have sufficiently low densities.  The condition of low density means that the molecules are so far apart that they do not interact except during collisions, which are effectively elastic.  At constant volume the pressure is proportional to the temperature.
  15. 15. At constant temperature, the pressure is inversely proportional to the volume. VP 1∝ The pressure is also proportional to the amount of gas. nP ∝ The Ideal Gas Law
  16. 16. The absolute pressure of an ideal gas is directly proportional to the Kelvin temperature and the number of moles of the gas and is inversely proportional to the volume of the gas. V nRT P = nRTPV = ( )KmolJ31.8 ⋅=R The Ideal Gas Law V nT P ∝
  17. 17. TNkB= AN N n = ( ) KJ1038.1 mol106.022 KmolJ31.8 23 123 − − ×= × ⋅ == A B N R k The Ideal Gas Law nRTPV = T N R N A       = Boltzmann’s Constant TNknRTPV B==
  18. 18. Example 2 Oxygen in the Lungs In the lungs, the respiratory membrane separates tiny sacs of air (pressure 1.00x105 Pa) from the blood in the capillaries. These sacs are called alveoli. The average radius of the alveoli is 0.125 mm, and the air inside contains 14% oxygen. Assuming that the air behaves as an ideal gas at 310K, find the number of oxygen molecules in one of these sacs. TNkPV B= Tk PV N B = ( )( ) 2 1314 Oofmolecules107.214.0109.1 ×=×=N ( ) ( )[ ] ( )( )K310KJ1038.1 m10125.0Pa1000.1 23 33 3 45 − − × ×× = π airofmolecules109.1 14 ×=N
  19. 19. Conceptual Example 3 Soda Bubbles on the Rise Watch the bubbles rise in a glass of soda. If you look carefully, you’ll see them grow in size as they move upward, often doubling in volume by the time they reach the surface. Why does the bubble grow as it ascends?
  20. 20. Consider a sample of an ideal gas that is taken from an initial to a final state, with the amount of the gas also changing. nRTPV = ii ii ff ff Tn VP Tn VP =constant== R nT PV Other Gas Laws Constant T, constant n: iiff VPVP = Boyle’s law Constant P, constant n: i i f f T V T V = Charles’ law Constant P, constant T: i i f f n V n V = Avogadro’s law Constant V, constant n: i i f f T P T P = Gay-Lussac’s law
  21. 21. 14.2.1. Using the ideal gas law, estimate the approximate number of air particles within an otherwise empty room that has a height of 2.5 m, a width of 4.0 m, and a length of 5.0 m. a) 1.2 × 1027 b) 6.8 × 1025 c) 3.0 × 1026 d) 2.5 × 1028 e) 9.1 × 1024
  22. 22. 14.2.2. An ideal gas is enclosed within a container by a moveable piston. If the final temperature is two times the initial temperature and the volume is reduced to one-fourth of its initial value, what will the final pressure of the gas be relative to its initial pressure, P1? a) 8P1 b) 4P1 c) 2P1 d) P1/2 e) P1/4
  23. 23. 14.2.3. Consider a commercial sightseeing hot air balloon that carries a basket with more than 20 passengers. Assume that balloon contains 1.5 × 104 m3 of air. Estimate the order of magnitude of the number of air molecules inside the balloon. a) 1023 b) 1029 c) 1035 d) 1018 e) 104
  24. 24. Chapter 14: Ideal Gas Law & Kinetic Theory Section 3: The Kinetic Theory of Gases
  25. 25. Postulates of Kinetic(-Molecular) Theory  All gases are made up of particles  Usually molecules  The particles are in constant, random motion, colliding with each other and with the walls of the container.  All collisions are perfectly elastic  Volume of the particles is insignificant  There are no interactions between particles (attraction/repulsion)  The average kinetic energy of the particles is a function of only absolute temperature
  27. 27. collisionssuccessivebetweenTime momentumInitial-momentumFinal F = ( ) t mv t v mmaF ∆ ∆ = ∆ ∆ ==∑ Kinetic Theory L mv2 − = ( ) ( ) vL mvmv 2 F +−− =
  28. 28. L mv F 2 = For a single molecule, the average force is: For N molecules in 3 dimensions, the average force is:               = L vmN F 2 3 root-mean-square speed A F P = volume Kinetic Theory 2 L F =               = 3 2 3 L vmN
  29. 29.               = V vmN P 2 3 ( )2 3 1 rmsmvNPV = TNkB K TkB2 3 K = Kinetic Theory ( )2 2 1 3 2 rmsmvN= µ Tk M RT v B rms 33 == Molar mass
  30. 30. Conceptual Example 5 Does a Single Particle Have a Temperature? Each particle in a gas has kinetic energy. On the previous page, we have established the relationship between the average kinetic energy per particle and the temperature of an ideal gas. Is it valid, then, to conclude that a single particle has a temperature? No, the temperature relates to the average of the whole sample, as there is one temperature for the sample.
  31. 31. Example 6 The Speed of Molecules in Air Air is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0 u) and oxygen O2 molecules (molecular mass 32.0 u). Assume that each behaves as an ideal gas and determine the rms speeds of the nitrogen and oxygen molecules when the temperature of the air is 293K. M RT vrms 3 = ( )( ) molM RT vrms kg8002.0 K293KmolJ31.833 ⋅ == For nitrogen… ( )( ) molM RT vrms kg2003.0 K293KmolJ31.833 ⋅ == For oxygen… sm511= sm478=
  32. 32. Tkmv Brms 2 32 2 1 K == nRTkTNU 2 3 2 3 == Internal Energy of a Monatomic Ideal Gas
  33. 33. 14.3.1. Two sealed containers, labeled A and B as shown, are at the same temperature and each contain the same number of moles of an ideal monatomic gas. Which one of the following statements concerning these containers is true? a) The rms speed of the atoms in the gas is greater in B than in A. b) The frequency of collisions of the atoms with the walls of container B are greater than that for container A. c) The kinetic energy of the atoms in the gas is greater in B than in A. d) The pressure within container B is less than the pressure inside container A. e) The force that the atoms exert on the walls of container B are greater than in for those in container A.
  34. 34. 14.3.2. Two identical, sealed containers have the same volume. Both containers are filled with the same number of moles of gas at the same temperature and pressure. One of the containers is filled with helium gas and the other is filled with neon gas. Which one of the following statements concerning this situation is true? a) The speed of each of the helium atoms is the same value, but this speed is different than that of the neon atoms. b) The average kinetic energy of the neon atoms is greater than that of the helium atoms. c) The pressure within the container of helium is less than the pressure in the container of neon. d) The internal energy of the neon gas is greater than the internal energy of the helium gas. e) The rms speed of the neon atoms is less than that of the helium atoms.
  35. 35. 14.3.3. A monatomic gas is stored in a container with a constant volume. When the temperature of the gas is T, the rms speed of the atoms is vrms. What is the rms speed when the gas temperature is increased to 3T? a) vrms/9 b) c) 3vrms d) vrms e) 9vrms rms / 3v 3
  36. 36. 14.3.4. Closed containers A and B both contain helium gas at the same temperature. There are n atoms in container A and 2n atoms in container B. At time t = 0 s, all of the helium atoms have the same kinetic energy. The atoms have collisions with each other and with the walls of the container. After a long time has passed, which of the following statements will be true? a) The atoms in both containers have the same kinetic energies they had at time t = 0 s. b) The atoms in both containers have a wide range of speeds, but the distributions of speeds are the same for both A and B. c) The average kinetic energy for atoms in container B is higher than that for container A. d) The average kinetic energy for atoms in container A is higher than that for container B. e) The atoms in both containers have a wide range of speeds, but the distributions of speeds has a greater range for container B than that for container A.
  37. 37. 14.3.5. Assume that you have a container with 0.25 kg of helium gas at 20 °C. How much energy must be added to the gas to increase its temperature to 70 °C? a) 4 × 104 J b) 2 × 105 J c) 5 × 106 J d) 1 × 107 J e) 3 × 108 J
  38. 38. Chapter 14: Ideal Gas Law & Kinetic Theory Section 4: Diffusion
  39. 39. The process in which molecules move from a region of higher concentration to one of lower concentration is called diffusion. Diffusion
  40. 40. Conceptual Example 7 Why Diffusion is Relatively Slow A gas molecule has a translational rms speed of hundreds of meters per second at room temperature. At such speed, a molecule could travel across an ordinary room in just a fraction of a second. Yet, it often takes several seconds, and sometimes minutes, for the fragrance of a perfume to reach the other side of the room. Why does it take so long?
  41. 41. A Transdermal Patch
  42. 42. ( ) L tCDA m ∆ = FICK’S LAW OF DIFFUSION The mass m of solute that diffuses in a time t through a solvent contained in a channel of length L and cross sectional area A is concentration gradient between ends diffusion constant SI Units for the Diffusion Constant: m2 /s
  43. 43. Example 8 Water Given Off by Plant Leaves Large amounts of water can be given off by plants. Inside the leaf, water passes from the liquid phase to the vapor phase at the walls of the mesophyll cells. The diffusion constant for water is 2.4x10-5 m2 /s. A stomatal pore has a cross sectional area of about 8.0x10-11 m2 and a length of about 2.5x10-5 m. The concentration on the interior side of the pore is roughly 0.022 kg/m3 , while that on the outside is approximately 0.011 kg/m3 . Determine the mass of water that passes through the stomatal pore in one hour.
  44. 44. ( ) ( )( )( )( ) kg100.3 m102.5 s3600mkg011.0mkg022.0m100.8sm104.2 9 5- 3321125 − −− ×= × −×× = ∆ = L tCDA m
  45. 45. 14.4.1. Sealed containers with a valve in the middle contain equal amounts of two different monatomic gases at room temperature as shown. A few of the gas atoms are illustrated with arrows representing their velocities. Which of the following statements concerning the gases after the valve has been opened is correct? a) Because there is no pressure difference, the two gases will remain separated for the most part. Only a few atoms of each gas will be exchanged. b) As soon as the valve is opened, the two gases will mix completely. c) The two gases will collide in the narrow tube and be scattered back into their original chambers, so no mixing will occur. d) After a relatively long period of time, the two gases will be well mixed in both containers. e) The lighter of the two gases will occupy both spherical containers, but the heavier gas atom will mostly remain in their original container.