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Chapter 14

Chapter 14. The Ideal Gas Law and Kinetic Theory of Gases. 14.1 Molecular Mass, the Mole, and Avogadro’s Number. The atomic mass scale facilitates comparison of the atomic mass (protons, neutrons and electrons) of one atom with another.

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Chapter 14

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  1. Chapter 14 The Ideal Gas Law and Kinetic Theory of Gases

  2. 14.1 Molecular Mass, the Mole, and Avogadro’s Number • The atomic mass scale facilitates comparison of the atomic mass (protons, neutrons and electrons) of one atom with another. • The unit is called the atomic mass unit (symbol u). 1 u = 1.6605 x 10-27 kg. • The atomic mass number given in the period table is a weighted average of the mass of all isotopes of an element.

  3. 14.1 Molecular Mass, the Mole, and Avogadro’s Number • One mole of a substance contains as many particles as there are atoms in 12 grams of the isotope carbon-12 (atomic mass =12 u). • The number of atoms or molecules per mole is a constant number known as Avogadro’s number, (NA ). • NA = 6.022 x 1023 particles/mole of substance • The number of moles, n, of a substance is: • Where N is the total number of particles, and NA is Avogadro’s Number.

  4. 14.1 Molecular Mass, the Mole, and Avogadro’s Number The molar mass (M) of a substance is the mass per mole in grams/mole. Note grams are not SI units! The number of moles, n, is then: Where n = number of moles, m = mass (in grams) NA = Avagadro’s number M = mass in grams per mole

  5. 14.1 Molecular Mass, the Mole, and Avogadro’s Number • The mass per mole (M, in grams/mole) of a substance has the same numerical value as the atomic or molecular mass of the substance (in atomic mass units). • For example, the atomic mass of hydrogen is 1.00794 u (1 proton plus the occasional neutron or two). Therefore, hydrogen has an molar mass (Mhydrogen ) of 1.00794 g/mol. Likewise, lithium has a molar mass of 6.941 g/mol, and magnesium has a molar mass of 24.305 g/mol. • If SI units are necessary, change grams to kg.

  6. EXAMPLE Moles of oxygen How many moles of oxygen gas, O2, are there in 100 g of the gas? Assume that each molecule of this diatomic gas consists of two O-16 atoms.

  7. EXAMPLE Moles of oxygen QUESTION: How many moles of oxygen gas, O2, are there in 100g? • Known: • Molecular mass of oxygen gas (O2) = 32u. Therefore molar mass (M) = 32g • m = 100g • Find: n • What do we know? The number of moles in a sample equals the mass of the sample divided by its molar mass. • n = m/M

  8. EXAMPLE Moles of oxygen QUESTION: How many moles of oxygen gas, O2, are there in 100g? • Known: • Atomic mass of oxygen gas (O2) = 32u. Therefore molar mass (M) = 32g • m = 100g • Find: n • Pertinent relationship: The number of moles in a sample equals the mass of the sample divided by its molar mass. • n = m/M = 100 g/32g/mol = 3.13 moles

  9. Moles, molar mass and particles Argon gas has an atomic mass = 39.948 u, neon has an atomic mass = 20.180 u, and helium has an atomic mass = 4.0026 u. You have a mole of each substance in 3 separate containers. Which container has the greatest mass? • Argon • Helium • All have the same • No way to tell

  10. Moles, molar mass and particles Argon gas has an atomic mass = 39.948 u, neon has an atomic mass = 20.180 u, helium has an atomic mass = 4.0026 u. You have a mole of each substance in 3 separate containers. Which container has the most gas atoms? • Argon • Helium • All have the same • No way to tell

  11. How many moles? Consider a mixture of three different monatomic gases: 1.10 g of argon (molar mass = 39.948 g/mol), 2.48 g of neon (molar mass = 20.180 g/mol), and 3.12 g of helium (molar mass = 4.0026 g/mol). How many moles of gas atoms are in the container?

  12. How many moles? Consider a mixture of three different monatomic gases: 1.10 g of argon (molar mass = 39.948 g/mol), 2.48 g of neon (molar mass = 20.180 g/mol), and 3.12 g of helium (molar mass = 4.0026 g/mol). How many moles of gas atoms are in the container? n = m/M (do separately for each) na = .0275 mol, nn = .123 mol, nh = .779 mol ntotal = .930 moles

  13. EOC #8 percentages of atoms Consider a mixture of three different monatomic gases: 1.10 g of argon (molar mass = 39.948 g/mol), 2.48 g of neon (molar mass = 20.180 g/mol), and 3.12 g of helium (molar mass = 4.0026 g/mol). For this mixture, determine the percentage of the total number of atoms that corresponds to each of the components.

  14. EOC #8 percentages of atoms Consider a mixture of three different monatomic gases: 1.10 g of argon (molar mass = 39.948 g/mol), 2.48 g of neon (molar mass = 20.180 g/mol), and 3.12 g of helium (molar mass = 4.0026 g/mol). For this mixture, determine the percentage of the total number (N) of atoms that corresponds to each of the components. Starting with argon: fa = Na /Ntot but n = N/NA so use n (number of moles) instead. fa = na /ntot = (ma /Ma )/ntot and so forth. Multiply answers by 100 to convert into percentages.

  15. An ideal gas is an idealized model for real gases with low densities (i.e. the space that the gas molecules occupy is much smaller than the container). The pressure from the gas comes from the force of gas molecules when they collide against the wall of the container. Properties of ideal gases were determined experimentally by investigators in the 17th-19th centuries. We can’t use: Pb = Pt + ρgh because a gas is not an incompressible fluid, and density is not constant.

  16. 14.2 The Ideal Gas Law Boyle’s Law (circa 1660s): It can be shown experimentally that, at constant temperature, the pressure of an ideal gas is inversely proportional to the volume. PV = constant, for a fixed number of moles at a given temperature. This is the shape of the graph for an inverse relationship: P = k(1/V) where k is a constant (analogous to a slope) PV is equal to the same value (k) anywhere on the curve. This value changes if the temperature changes. Isotherm – curve or line of equal temperature.

  17. Gay-Lussac Law (circa 1700): Experimentally, it can be shown that at constant volume the pressure of an ideal gas is proportional to the temperature, i.e. the relationship plots as a line: or P = cT where c is a constant. A temperature of absolute zero theoretically corresponds to where pressure is zero, i.e. the molecules are not moving, so they cannot hit the container.

  18. Charles Law(circa 1780s) Experimentally, it can be shown that at a constant pressure , the volume of an ideal gas is proportional to the temperature: Or V = cT where c is a constant http://www.grc.nasa.gov/WWW/K-12/airplane/aglussac.html

  19. Avogadro’s Law (circa 1811) • If two containers of different ideal gases are at the same pressure, volume and temperature, then they contain the same number of particles, regardless of the mass of those particles. Atomic mass = 2u Atomic mass = 32u

  20. The Ideal-Gas Law – A combination of Gay-Lussac, Charles, Avogadro’s and Boyle’s Law The pressure p, the volume V, the number of moles n and the temperature T of an ideal gas are related by the ideal-gas law as follows: where R is the universal gas constant, R = 8.31 J/mol K, and T must be in Kelvin (Celsius + 273) The ideal gas law may also be written as where N is the number of molecules in the gas rather than the number of moles n. The Boltzmann’s constant is kB = 1.38 × 10−23 J/K, and is equal to R/NA.

  21. The mass of a gas (!) These 2 containers containing H2 and O2 are at the same pressure, volume and temperature. Use the gas law to determine which container has greater mass: • O2 • H2 • Both have the same mass • Unable to determine without numerical values for the other variables Atomic mass = 2u Atomic mass = 32u

  22. Calculating a gas temperature using ratios A cylinder of gas is at 0◦ C. A piston compresses the gas to half its original volume and 3 times it’s original pressure. What is the final temperature of the gas?

  23. Calculating a gas temperature using ratios A cylinder of gas is at 0◦ C. A piston compresses the gas to half its original volume and 3 times it’s original pressure. What is the final temperature of the gas? T1 = (P1 V1 )/nR and T2 = (P2 V2 )/nR T2 = (P2 V2 )/nR so T2 = T1 (P2 /P1 ) (V2 /V1 ) T1 = (P1 V1 )/nR (P2 /P1 ) = 3, and (V2 /V1 ) = .5 T2 = 273 K (3) (.5) = 409 K

  24. EOC #17 – party time A clown at a birthday party has brought along a helium cylinder, with which he intends to fill balloons. When full, each balloon contains 0.036 m3 of helium at an absolute pressure of 1.20 x 105 Pa. The cylinder contains helium at an absolute pressure of 1.80 x 107 Pa and has a volume of 0.0031 m3. The temperature of the helium in the tank and in the balloons is the same and remains constant. What is the maximum number of balloons that can be filled (rounded to a whole number)? Hint: There is the same amount of moles of gas (n) in x balloons as there is in the whole cylinder.

  25. Moveable Piston A frictionless gas-filled cylinder (V = πr2 h) is fitted with a movable piston, as the drawing shows. The height h is 0.110 m when the temperature is 271 K and increases as the temperature increases. What is the value of h when the temperature reaches 315 K? Another ratio problem? PV = nRT P(πr2 h) = nRT (πr2 h) = nRT/P πr2 h2 =(nRT2 )/P2 πr2 h1 =(nRT1 )/P1 And if we divide out quantities that don’t change h2 = (T2/P2 ) h1 =(T1/P1) Too many unknowns?

  26. Moveable Piston FBD of piston and mass Use Newton’s Law to find an expression for pressure and see if pressure depends on volume, temperature or number of moles if the container is capped by a moveable piston. PgasA PaA Fg ΣF = 0 Pgas A = PatmA + mg Pgas = Patm + mg/A Where cross-sectional area of cap

  27. Moveable Piston FBD of piston and mass Pgas = Patm + mg/A If T changes does P change? If V changes does P change? If n changes does P change? NO! P is constant for a moveable piston PgasA PaA Fg ΣF = 0 Pgas A = PatmA + mg Pgas = Patm + mg/A Where cross-sectional area of cap

  28. Moveable Piston A frictionless gas-filled cylinder (V = πr2 h) is fitted with a movable piston, as the drawing shows. The height h is 0.110 m when the temperature is 271 K and increases as the temperature increases. What is the value of h when the temperature reaches 315 K? Another ratio problem? P(πr2 h) = nRT (πr2 h) = nRT/P πr2 h2 =(nRT2 )/P2 πr2 h1 =(nRT1 )/P1 And if we divide out quantities that don’t change h2 = T2 h1 = T1

  29. Two different gases at the same temperature are in cylinders with identical moveable pistons. Which of the following is different for the gases (besides volume)? • Pressure • Number of moles • Both pressure and number of moles • Either pressure and/or number of moles, but it is impossible to determine which without more information.

  30. The diving bell (#24) A primitive diving bell consists of a cylindrical tank with one end open and one end closed. The tank is lowered into a freshwater lake, open end downward. Water rises into the tank, compressing the trapped air, whose temperature remains constant during the descent. The tank is brought to a halt when the distance between the surface of the water in the tank and the surface of the lake is 40.0 m. Atmospheric pressure at the surface of the lake is 1.01 x 105 Pa. Find the fraction of the tank's volume that is filled with air.

  31. Another Planet – Number density On a hypothetical planet, the atmospheric pressure is 3.0 x 106 Pa, and the temperature is 900 K. On the earth's surface the atmospheric pressure is 1.0 x 105 Pa, while the surface temperature can reach 320 K. Researchers can determine how “thick” the atmosphere of this planet is compared to that of Earth, by comparing their number densities (N/V) . Number density is the number of particles per unit volume (units of m-3 ) Find the ratio: (N/V)Planet / (N/V)Earth. It is necessary to use the “particle version” of the Ideal Gas Law: PV = NkT.

  32. PV diagrams • Each point on the graph represents unique values of V,P. • If the number of moles are known, T can be determined with the ideal gas law.

  33. PV Diagram Rank the temperature of one mole of an ideal gas on different points on the PV diagram

  34. What is the ratio Tf /Ti for this process? • 4 • 2 • 1 (no change)

  35. Kinetic Gas Theory • The pressure of the gas comes from the force due to the collisions with the container wall, divided by the surface area. • Collisions with the container wall do not change the speed of the ideal gas (only direction) molecules, but collisions with other gas molecules change both speed and direction.

  36. Speed distribution in an ideal gas • A fixed amount of an ideal gas in a container has gas molecules moving at a number of speeds. • If the temperature increases, the speeds increase. • Note that the range of speeds (width) increases as well. Maxwell distribution curves for O2

  37. Speed distribution in an ideal gas http://www.chm.davidson.edu/ChemistryApplets/KineticMolecularTheory/Maxwell.html Maxwell distribution curves for O2

  38. Kinetic Gas Theory The molecules of an ideal gas travel at many different speeds. Use vrms as an “average speed”. Starting with Newton’s 2nd Law (what else?) it can be shown that: PV = 2/3 N (1/2 mvrms 2 ) Where: • N = number of molecules in the container • m = mass of one molecule of the ideal gas (really small number) • vrms = rms speed, • P = pressure, • V = volume • 1/2 mvrms 2 is the kinetic energy = K of the “ average molecule” in the container

  39. Kinetic Gas Theory Newton’s Law predicts: PV = 2/3 N (1/2 mv 2rms) Experimental evidence (ideal gas law) states: PV = NkT (molecular version) The N term cancels out and leaves: KEav/molecule = (3/2)kT This shows that: • The temperature of an ideal gas is determined only by its temperature • The product PV is an energy term: PV = (2/3N)KE

  40. Example root mean squared speed of a cesium atom What is the rms speed for cesium atoms at a temperature of 1 x 10-6 K (a very low temperature!)

  41. 14.3 Kinetic Theory of Gases Example Total microscopic kinetic energy What is the total microscopic kinetic energy of a mole of an ideal gas at 0˚ C and standard atmospheric pressure ?

  42. Which system has the largest average translational kinetic energy per molecule? • 2 mol of He at p = 2 atm, T = 300 K • 2 mol of N2 at p = 0.5 atm, T = 450 K • 1 mol of He at p = 1 atm, T = 300 K • 1 mol of N2 at p = 0.5 atm, T = 600 K • 1 mol of Ar at p = 0.5 atm, T = 450 K

  43. Microscopic Kinetic energy of a volume of gas Ktot = 3/2 N k T where N is the number of molecules of gas in the volume: Nk = nR where n is the number of moles of gas. The total internal energy, U, of a gas, consists of the kinetic energy AND the potential energy of its molecular bonds In a monatomic gas (molecules composed of a single atom), there are no molecular bonds: Total internal energy U = 3/2 N k T

  44. Uint of a monatomic ideal gas In a monatomic ideal gas, all energy is kinetic. The average kinetic energy per molecule is: Monatomic gases include helium, neon, argon, krypton, xenon and radon. Watch out Superman!

  45. Example Problem using number density • A cylinder of monatomic gas has a pressure that is twice as great as standard amospheric pressure and a number density (N/V) of 4.2 x 1025 m-3. The rms speed of the atoms is 660 m/s. Identify the gas.

  46. EOC # 37 Initially, the translational rms speed of a molecule of an ideal gas is 578 m/s. The pressure and volume of this gas are kept constant, while the number of molecules is doubled. What is the final translational rms speed of the molecules?

  47. EOC #38 Two gas cylinders are identical. One contains the monatomic gas neon (Ne), and the other contains an equal mass of the monatomic gas xenon (Xe). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio KEAvg,xenon / KEAvg,neon of the average kinetic energy of a xenon atom to the average kinetic energy of a neon atom.

  48. EOC #42 Helium (He), a monatomic gas, fills a 0.009 m3 container. The pressure of the gas is 5.60 x 105 Pa. How long would a 0.25 horsepower engine have to run (1 hp = 746 W) to produce an amount of energy equal to the internal energy of this gas?

  49. EOC #42 Helium (He), a monatomic gas, fills a 0.009 m3 container. The pressure of the gas is 5.60 x 105 Pa. How long would a 0.25 horsepower engine have to run (1 hp = 746 W) to produce an amount of energy equal to the internal energy of this gas? Ans: 40.5 seconds

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