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Chapter 5

Chapter 5. GASES. What we’ve had so far!. Different ways of calculating moles of substances Solids: Moles = grams molar mass Liquids: Molarity = moles Liter. Ideal Gas Law. In Chapter 5, everything comes down to another way of solving for moles.

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Chapter 5

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  1. Chapter 5 GASES

  2. What we’ve had so far! • Different ways of calculating moles of substances • Solids: Moles = grams molar mass • Liquids: Molarity = moles Liter

  3. Ideal Gas Law • In Chapter 5, everything comes down to another way of solving for moles. PV = nRT where n = number of moles

  4. Atmospheric Pressure • Pressure exerted by a gas on its surroundings

  5. Pressure • Units of Pressure • mm Hg • Atm • Torr (in honor of Evangelista Torricelli) • Pascal (Pa)

  6. Pressure • Conversions: • 1 atm = 760 Torr = 760 mm Hg • 1 atm = 101,325 Pa

  7. The Gas Laws • Boyle’s Law • PV = k • the product of pressure (P) and volume (V) of a trapped gas is constant

  8. Pressure vs. Volume • P is inversely proportional to V • Since PV = k • If k = 1, • then P = 1 or V = 1 V P

  9. Boyle’s Law • only holds true at very low pressures • at high pressures, PV is not constant

  10. Application of Boyle’s Law • commonly used to predict the new volume when pressure is changed • If: PV = k • Then: (PV)1 = (PV)2 for same gas at same temperature

  11. Sample Problem on Boyle’s Law • An aerosol can contains 400 mL of compressed gas at 5.20 atm pressure. When all the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. If the T is constant, what is the pressure of gas inside the plastic bag?

  12. Boyle’s Law • A gas that strictly follows Boyle’s Law is an IDEAL GAS.

  13. Charles’s Law • filled balloon with H2 and made the first solo balloon flight • discovered that the volume of a gas at constant P increases linearly with the T of the gas

  14. Charles’s Law • The V of a gas at constant P is directly proportional to T. • Charles’s Law Equation: • V = kT where k is the proportionality constant

  15. Units of Temperature • Celsius • Kelvin • Conversion: • 0 oC = 273 K

  16. Application of Charles’s Law • To predict new volume of a gas when T is changed • Thus if : V = kT • Then: (V1/T1) = (V2/T2)

  17. Sample Problem • If a gas at 15 oC and 1 atm has a volume of 2.58 L, what volume will this gas occupy at 38 oC and 1 atm? [Note: Convert all T to K]

  18. Avogadro’s Law • Avogadro’s Postulate: [Chapter 2] • Equal volumes of gases at the same T and P contain the same number of particles. Closely obeyed by gases at low P.

  19. Avogadro’s Law • V = a x n • where n = number of moles • a = proportionality constant • So for a gas at constant T and P, the volume is directly proportional to the number of moles of gas.

  20. Application of Avogadro’s Law • To predict changes in V when the number of moles changes. • V1/n1 = V2/n2

  21. Charles’ Law • the volume of a gas at constant P increases linearly with the T of the gas

  22. Summary of 3 Laws • Boyle’s Law: V = K/P [at constant T and n] • Charles’s Law: V = bT • [at constant P and n] • Avogadro’s Law: V = an [at constant T and P]

  23. Sample Problem • A 1-L container contains 2.75 moles of H2 at 400 oC. What would be the pressure in the container at this temperature?

  24. Sample Problem • 3.5 moles of N2 has a pressure of 3.3 atm at 375 oC. What would be the pressure of the 5.3 moles of gas at 900 oC?

  25. Boyle’s Law • A gas that strictly follows Boyle’s Law is an IDEAL GAS.

  26. Ideal Gas Law • expresses behavior that real gases approach at low P’s and high T’s • An Ideal Gas is a hypothetical substance!

  27. Ideal Gas Law • Since most gases approach close to ideal behavior anyway, we will assume that the gases we encounter in this course are all ideal gases.

  28. Ideal Gas Law • PV = nRT • where R = universal gas constant • = .08206 L-atm/K

  29. Sample Problem • A sample of methane gas that has a volume of 3.8 L at 5 oC is heated to 86 oC at constant P. Calculate its new volume.

  30. Another Sample Problem • A sample of hydrogen gas has a volume of 10.6 L at a T of 0 oC and a P of 2.5 atm. Calculate the moles of hydrogen gas present in this gas sample.

  31. Molar Mass and Density • Molar Mass = gmRT PV • Density = Molar Mass x Pressure RT

  32. Dalton’s Law of Partial Pressures • For a mixture of gases in a container, the total P exerted is the sum of the pressures that each gas would exert if it were alone. • Ptotal = P1 + P2 + P3 …….

  33. Dalton’s Law • Ptotal = P1 + P2 + P3 ……. • If P1 = n1RT/V P2 = n2RT/V P3= n3RT/V • Then: Ptotal = (n1RT/V) + (n2RT/V) + (n3RT/V)

  34. Gas Stoichiometry • PV = nRT • For 1 mole of an ideal gas at 0 oC, the molar volume at STP is 22.42 L. • STP = standard T and P • where T = 0 oC and P = 1 atm

  35. Sample Problem • A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

  36. Kinetic Molecular Theory • Is a model that attempts to explain the behavior of ideal gases • Based on speculations about the behavior of the gas particles

  37. Postulates of KMT • The gas particles are assumed to: • 1] be so small and have V = 0 • 2] be in constant motion. • 3] exert no force on each other • 4] have average KE that is a T (in Kelvin)

  38. KMT on P and V • Since P a1 (Boyle’s Law) V • As V increases, P decreases. • As V decreases, P increases. KMT: As V is decreased, P increases because particles hit walls of container more often.

  39. KMT on P and T • As T is increased, the speed of the particles increases resulting in more collisions and stronger collisions. Thus P increases.

  40. KMT on V and T • Charles’s Law: V a T • KMT: As T is increased, P normally increases because particles collide more. To keep P constant, V has to increase to compensate for the increased collisions.

  41. KMT on V and n • Avogadro’s Law: V a n • KMT: Normally, an increase in n (moles of particles) increases P if V was held constant. To return P back to normal, V has to be increased.

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