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Chapter 11 Gases and Their Properties

Chapter 11 Gases and Their Properties. Properties of Gases. Gas properties are characterized by the following quantities: V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres).

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Chapter 11 Gases and Their Properties

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  1. Chapter 11Gases and Their Properties

  2. Properties of Gases Gas properties are characterized by the following quantities: V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres)

  3. Torricelli’sWeather InstrumentsWhich barometer reads a higher atmospheric pressure?

  4. Why is mercury used for barometers? Normal atmospheric pressure = 30 mm of Hg = 76 cm of Hg = 760 mm of Hg =760 Torr = 10,000 Pa = 1 atmosphere

  5. Gas: Temperature Scales For gas calculations, we need an absolute scale, one that does not take on negative values. The conversion between °C and Kelvins is: K = °C + 273.15 When performing calculations with absolute temperatures, on must use the Kelvin scale.

  6. Boyle’s Law The pressure of a system of gas particles is inversely proportional to the volume of fixed number of moles at constant temperature. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

  7. Problem: A sample of nitrogen gas has a pressure of 67.5 mm Hg in a 500.-mL flask. What is the pressure of this gas sample when it is transferred to a 125-mL flask at the same temperature?

  8. Charles’s Law The volume of a system of gas particles is directly proportional to the absolute temperature of fixed number of moles at constant pressure. Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  9. Charles’s Law Balloons immersed in liquid N2 (at -196 °C) will shrink as the air cools (and is liquefied).

  10. Charles’s Law

  11. Problem: A 5.0-mL sample of CO2 gas is enclosed in a gas-tight syringe at 22 °C. If the syringe is immersed in an ice bath (0 °C), what is the new gas volume, assuming that the pressure is held constant?

  12. General Gas Law Combining Charles’s and Boyle’s Laws… So at two sets of conditions:

  13. Problem: You have a sample of CO2 in flask A with a volume of 25.0 mL. At 20.5 °C, the pressure of the gas is 436.5 mm Hg. To find the volume of another flask, B, you move the CO2 to that flask and find that its pressure is now 94.3 mm Hg at 24.5 °C. What is the volume of flask B?

  14. Avogadro’s Hypothesis twice as many molecules Equal amounts of gases (moles) at the same temperature (T) and pressure (P) occupy equal volumes (V).

  15. Problem: A sample of ethane (C2H6) undergoes combustion. What volume of O2 (L) is required for complete reaction with 5.2 L of ethane? What volume of H2O (g) in liters is produced? Assume all gases are measured at the same temperature and pressure. Write the balanced chemical equation for the reaction. 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

  16. Problem: A sample of ethane (C2H6) undergoes combustion. What volume of O2 (L) is required for complete reaction with 5.2 L of ethane? What volume of H2O (g) in liters is produced? Assume all gases are measured at the same temperature and pressure. 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

  17. Problem: A sample of ethane (C2H6) undergoes combustion. What volume of O2 (L) is required for complete reaction with 5.2 L of ethane? What volume of H2O (g) in liters is produced? Assume all gases are measured at the same temperature and pressure. 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

  18. Combining Avogadro’s Hypothesis with the General Gas Law This in known as the “Ideal Gas Law” PV = nRT

  19. Rules for Ideal Gas Law Calculations • Always convert the temperature to Kelvin (K = 273.15 + °C) • Convert from grams to moles if necessary. • Be sure to convert to the units of “R” (L, atm, mol & K).

  20. Standard Temperature & Pressure STP 1 atm= 760 torr or mm Hg & 0 C = 273.15K Problem:Calculate the standard molar volume of a gas. PV = nRT 1.00 mol  273.15K = 22.4 L V = 1.00 atm

  21. Standard Temperature & Pressure In order to provide a reference point for the comparison of gasses, standard temperature and pressureare set at: 1 atm (760 torr or mm Hg & 0 °C (273.15K) Problem:Calculate the volume occupied by 43.7 g of hydrogen at STP. PV = nRT  273.15K 43.7 g H2 = 485 L V = 1.00 atm

  22. Gas Density The density of a gas (grams / L) can be obtained from the ideal gas law. Rearranging:

  23. Example Problem for finding Gas Density What is the density of oxygen at STP? as we found before: The molar mass of O2 is 32.00 g/mol. Find its density in grams/liter . STP = 1 atm & 273 K°C

  24. Problem: A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 °C. What is the molecular formula of the compound?

  25. Problem: A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 °C. What is the molecular formula of the compound?

  26. Problem: A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 °C. What is the molecular formula of the compound? Find its density Convert p to atmospheres Use formula for M

  27. Problem: A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 °C. What is the molecular formula of the compound?

  28. Problem: A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 °C. What is the molecular formula of the compound? The molecular formula is (CHF2)2 or C2H2F4.

  29. Gas Density Balloon filled with helium (lower density) Balloons filled with air

  30. Gas Laws & Chemical Reactions If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

  31. Problem: If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O? Step 1: Write the balanced chemical reaction. Step 2: Calculate the moles of each product. Step 3: Find the pressure of each via PV = nRT

  32. Problem: If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O? Step 1: Write the balanced chemical reaction. Step 2: Calculate the moles of each product. Step 3: Find the pressure of each via PV = nRT 2 H2O2(liq)  2 H2O(g) + O2(g) Step 1:

  33. Problem: If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O? 2 H2O2(liq)  2 H2O(g) + O2(g) Step 2:

  34. Problem: If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O? 2 H2O2(liq)  2 H2O(g) + O2(g) Step 2:

  35. Problem: If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O? 2 H2O2(liq)  2 H2O(g) + O2(g) Step 3:

  36. Problem: If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25°C, what is the pressure of O2 & H2O? 2 H2O2(liq)  2 H2O(g) + O2(g) Step 3:

  37. Gas Mixtures & Partial Pressures: Dalton’s Law John Dalton 1766-1844

  38. Dalton’s Law of Partial Pressure Each gas can be represented as a fraction of the total moles in the mixture. This is called theMole Fraction. Since each gas adds together to create a total pressure (Ptot). (ex: 0.4 atm + 0.6 atm = 1.0 atm) The molar fraction that represents each of the individual gases is akin to a concentration.

  39. The Mole Fraction For a mixture of gases 1,2,3,4…i with moles n1, n2, n3, n4…ni moles of an individual gas “Mole Fraction” = = Xi total moles of gas in the mixture The sum of all of the mole fractions in the mixture is 1 (exactly) X1 + X2 + X3… = 1

  40. The Kinetic-Molecular Theory of Gases Postulates: Clausius (1857)  A gas is a collection of a very large number of particles that remains in constant random motion.  The pressure exerted by a gas is due to collisions with the container walls  The particles are much smaller than the distance between them.

  41. The Kinetic-Molecular Theory of Gases  The particles move in straight lines between collisions with other particles and between collisions with the container walls. (i.e. the particles do not exert forces on one another between collisions.)  The average kinetic energy (½ mv2) of a collection of gas particles is proportional to its Kelvin temperature.  Gas particles collide with the walls of their container and one another without a loss of energy.

  42. The Kinetic-Molecular Theory of Gases Gas pressure at the particle level:

  43. The Kinetic-Molecular Theory of Gases The “root mean square velocity” for a gas is: Find the speed of an oxygen molecule at room temperature (20 C) Find the speed of a hydrogen molecule at room temperature (20 C)

  44. Kinetic-Molecular Theory At the same T, all gases have the same average KE. As T goes up, KE also increases — and so does speed.

  45. Kinetic-Molecular Theory ½ m/s = 1.103103 mph What is the RMS velocity of a nitrogen molecule in miles per hr at STP? pretty zippy eh?

  46. Kinetic Molecular Theory • For a given temperature, heavier gases move slower than lighter gases. • The velocities are described by a distribution.

  47. Velocity of Gas Particles Average velocity decreases with increasing mass.

  48. Distribution of Gas Molecule Speeds • Boltzmann Distribution • Named for Ludwig Boltzmann doubted the existence of atoms.

  49. Gas Diffusion & Effusion • Diffusion is the process of gas migration due to the random motions and collisions of gas particles. • It is diffusion that results in a gas completely filling its container. • After sufficient time gas mixtures become homogeneous.

  50. Gas Diffusion: Relation of mass to Rate of Diffusion • HCl and NH3 diffuse from opposite ends of tube. • Gases meet to form NH4Cl • HCl heavier than NH3 • Therefore, NH4Cl forms closer to HCl end of tube.

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