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Chapter 11

Chapter 11. The Behavior of Gases. Kinetic Theory. Kinetic Theory – all molecules are in constant motion. Collisions between gas molecules are perfectly elastic. Diffusion – movement of molecules from areas of high concentration to low concentration.

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Chapter 11

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  1. Chapter 11 The Behavior of Gases

  2. Kinetic Theory • Kinetic Theory – all molecules are in constant motion. • Collisions between gas molecules are perfectly elastic. • Diffusion – movement of molecules from areas of high concentration to low concentration. • Rate of diffusion – the size and mass of the molecule. • Smaller, lighter molecules move faster.

  3. Pressure • Gas pressure – due to collisions of gas molecules on an object. • Atmospheric pressure – due to collisions of air molecules on an object. • 1 atm = 760 mm Hg = 30 in Hg = 14.7 psi • Partial pressure – the portion of pressure that one gas contributes to the total pressure in a mixture of gases.

  4. Dalton’s Law of Partial Pressure • The total pressure of a mixture of gases is equal to the sum of the partial pressures. • PT = P1 + P2 + P3 • Pair = PN2 + PO2 + PCO2

  5. Directly proportional # moles, P 2 mol. 1mol: Pressure vs. Moles (at constant volume) • Same volume containers at constant temperature: • If 1 mole of gas exerts 1 atm of pressure and we add another mol of gas  twice as many particle will have twice as many collisions  exert twice the pressure (2atm)

  6. P V 1 1 .5 2 2 .5 V P ½ volume  2x P V P 2x volume  ½ P Pressure vs. Volume (at constant Temperature) Start with 1 L of gas at 1 atm.

  7. Pressure vs. Volume (at constant Temperature) • As volume decreases, the pressure increases proportionally. • As volume increases, the pressure decreases proportionally. • As one goes up, the other goes down: P and V are Inversely Proportional. • P1V1 = P2V2

  8. Boyle’s Law • For a given mass of gas, at constant temperature, the pressure of the gas varies inversely with the volume. P1V1 = P2V2

  9. 200 K = 2 L 2x T  2x V T = V Volume vs. Temperature (at constant Pressure) Start with 1 L of gas at 100 K and 1 atm. K = oC + 273 Heat the gas the molecules speed up and hit the top, pushing it to maintain constant pressure.

  10. 50 K = ½ L ½x T ½x V T = V Volume vs. Temperature (at constant Pressure) Start with 1 L of gas at 100 K and 1 atm. K = oC + 273 Cool the gas the molecules slow down, fewer collisions w/the top  so it falls.

  11. Charles’ Law • For a given mass of gas, at constant pressure, the volume of the gas varies directly with its Kelvin temperature. V1T2 = V2T1

  12. T = P Pressure vs. Temperature(at constant volume) Start w/ 1 L at 100 K and 1 atm. Heat the gas  the moles speed up and increase the # of collisions, which increases the pressure. 2x T = 2x P

  13. T = P Pressure vs. Temperature(at constant volume) Start w/ 1 L at 100 K and 1 atm. Cool the gas  the moles slow down and decrease the # of collisions,which decreases the pressure. ½x T = ½x P

  14. Gay-Lusaac’s Law • For a given mass of gas, at constant volume, the pressure of the gas varies directly with its Kelvin temperature. P1T2 = P2T1

  15. Combined Gas Law • Combines Boyle’s, Charles’, and Gay-Lusaac’s Laws into one equation. • P1V1T2 = P2V2T1 • When using the combined gas law, UNIT MUST AGREE and all temperatures must be in Kelvin.

  16. Moles Meets Gas Laws • We know that the volume of a gas is proportional to its number of particles and the pressure of a gas is proportional to its number of particles, which means: • V~ # mol and P ~ # mol or V ~ n and P ~ n

  17. Moles Meets Gas Laws • We also know that if the temperature of a gas increases, its pressure increases and if the temperature of a gas increases, its volume increases. This means: • T ~ P and T ~ V • so we can write PV ~ nT

  18. ( atm) atm L ( L) mol K ( mol) ( K) Moles Meet Gas Laws • In order to make this proportion useful as a mathematical expression we can derive a constant by solving PV/nT using the values for 1 mole of a gas at STP. This constant will be called “R”. • Substituting into the equation we get: 1 22.4 = .0821 = R 1 273

  19. Ideal Gas Law • PV = nRT • When using this equation, units MUST be the same as those of the R value therefore: • Pressure must be in ________ • Volume must be in _____ • n must be in ________ • Temperature must be in _____ atm L mol K The Ideal Gas Law applies to real and Ideal gases under ALL conditions.

  20. Pressure Conversions 1 atm = 760 mm Hg = 30 in Hg = 14.7 psi = 101.3 kPa

  21. Problem 1 • .05 moles of a gas at a temperature of 20oC is contained in a 150 mL vessel. What is the pressure of this gas inside the vessel? P = V = n = R = T = 150 mL = .150 L PV = nRT .05 mol .0821 atmL/mol K 20oC + 273 = 293 K

  22. Problem 1: Answer P = V = n = R = T = PV = nRT .150 L .05 mol .0821 P(.150) = (.05)(.0821)(293) 293 K P = atm 8.02

  23. Problem 2 • How many grams of bromine gas at – 10oC and 1277 mm Hg would be contained in a 3000 mL vessel? P = V = n = R = T = 1277 mm x (1 atm/760 mm) = 1.68 atm 3000 mL = 3 L .0821 atmL/mol K -10oC + 273 = 263 K

  24. 160 g 1 mol Problem 2: Answer PV = nRT P = V = n = R = T = 1.68 atm 3 L (1.68)(3) = n(.0821)(264) .0821 5.04 = 21.59n 263 K n = .23 mol Br2 36.8 g .23 mol x = Br2 = 2(80) = 160 g

  25. 1 mol 110 g CO x ---------- 28 g Problem 3 • 110 g of carbon monoxide at a pressure of 35.4 in Hg and a volume of 782 mL would be at what temperature? Express your answer in degrees Celsius. P = V = n = R = T = 35.4 in x (1 atm/30 in) = 1.18 atm 782 mL = .782 L C 1 x 12 = 12 O 1 x 16 = 16 = 28 3.93 mol .0821 atmL/mol K

  26. Problem 3: Answer P = V = n = R = T = 1.18 atm PV = nRT .782 L 3.93 mol .0821 (1.18)(.782) = (3.93)(.0821)T .92 = .32T T = K 2.86 -270.14 oC oC = 2.86 K – 273 =

  27. Ideal Gas Follows the gas laws at all conditions of temp. and pressure. Particles are infinitely small (have no vol.) Particles are not attracted to one another. DO NOT EXIST! Real Gas Do not follow gas laws at all conditions of temp. and pressure. Particles have volume. Particles may attract one another when very close. Real vs. Ideal Gases

  28. Real Gases • Conditions at which real gases do NOT behave as ideal gases and therefore do not obey the gas laws: • At extremely high pressures  do not obey Boyle’s Law. • At extremely low temperatures  do not obey Charles’ Law.

  29. Reasons: • This occurs because under these two conditions the gas molecules are close enough together that they begin to exert forces on one another and behave similarly to a liquid. • Gas Law equations are still extremely useful because under common conditions the behavior of a real gas is the same as the behavior of an ideal gas.

  30. Density and Molecular Weight of Gases • Density (D) = mass/volume = m/V = g/L • Molecular Weigh (MW) = gram/mol • For gases we know that at STP : • 1 mol = gfm = 22.4 L = 6.02x1023 molecules • STP is defined as _________ and _________. 1 atm 0 oC

  31. 28 g 28 g 22.4 L D = M = 31 L 31 L 1 mol V MW = g = x mol Problem 1 • What is the density of a gas with a mass of 28 g and a volume 31 L? What is its MW? = .9 g/L = 20.23 g/mol

  32. 45 g 22.4 L 6.8 L 1 mol MW = g = x mol Problem 2 • Calculate the molecular weight of a gas with a mass of 45 g and a volume of 6.8 L. = 148.24 g/mol

  33. 32 g D = M 22.4 L V D = Problem 3 • What is the density of oxygen at STP? O2: mass = gfm mass = 2(16) mass = 32 g volume = 22.4 L = 1.43 g/L

  34. 80 g D = M S 1 x 32 = 32 O 3 x 16 = 48 22.4 L V D = Problem 4 • What is the density of sulfur trioxide at STP? SO3: mass = gfm 80 volume = 22.4 L = 3.57 g/L

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