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Chapter 5

Chapter 5. The Gas Laws pp. 5.1 Pressure. Force per unit area. Gas molecules fill container. Molecules move around and hit sides. Collisions are the force. Container has the area. Measured with a barometer. Barometer. Vacuum.

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Chapter 5

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  1. Chapter 5 The Gas Laws pp

  2. 5.1 Pressure • Force per unit area. • Gas molecules fill container. • Molecules move around and hit sides. • Collisions are the force. • Container has the area. • Measured with a barometer.

  3. Barometer Vacuum • The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. • 1 atm = 760 mm Hg 760 mm Hg 1 atm Pressure

  4. Manometer • Column of mercury to measure pressure. • h is how much lower the pressure is than outside. h Gas

  5. Manometer • h is how much higher the gas pressure is than the atmosphere. h Gas

  6. Units of pressure • 1 atmosphere = 760 mm Hg • 1 mm Hg = 1 torr • 1 atm = 101,325 Pascals = 101.325 kPa • Can make conversion factors from these. • What is 724 mm Hg in kPa? . . . • 96.5 kPa • In torr? • 724 torr • In atm? • 0.952 atm.

  7. 5.2 The Gas Laws of Boyle, Charles, and Avogadro • Boyle’s Law • Pressure and volume are inversely related at constant temperature. • PV= k • As one goes up, the other goes down. • P1V1 = P2 V2 • Graphically

  8. 1 atm • As the pressure on a gas increases 4 Liters

  9. As the pressure on a gas increases the volume decreases • Pressure and volume are inversely related 2 atm 2 Liters

  10. A Boyle’s Law Relationship

  11. V P (at constant T)

  12. Slope = k V 1/P (at constant T)

  13. Ne 22.41 L atm O2 PV CO2 P (at constant T)

  14. Example pp • 20.5 L of nitrogen at 25.0ºC and 742 torr are compressed to 9.80 atm at constant T. What is the new volume? Steps . . . • P1V1 = P2 V2 • (20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x) • x = 2.04 L

  15. You try it! • 30.6 mL of CO2 at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa? Ans. . . • 4.02 kPa. Steps . . . • P1V1 = P2 V2 • (30.6 mL)(740 torr) =(750 mL) (x torr) • x = 30.2 torr • (30 torr)(101.325 kPa/ 760 torr) = 4.02 kPa

  16. Charles’ Law • Volume of a gas varies directly with the absolute temperature at constant pressure. • V = kT (if T is in Kelvin) • V1 = V2 T1 = T2 • Graphically

  17. Tr 51A Charles’ Law Relationship

  18. He CH4 H2O V (L) H2 -273.15ºC T (ºC)

  19. Example pp • What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant? Steps. • V1/T1 = V2/T2 • (247 mL)/(22 + 273) K = (x mL)/(273 + 98) K • x = 311 ml

  20. Examples (do this) • At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 L at constant pressure? a) 173 ºC b) 280 ºC c) 320 ºC d) 593 ºC • 320. ºC (change K to ºC!)

  21. Avogadro's Law • At constant temperature and pressure, the volume of gas is directly related to the number of moles. • V = k n (n is the number of moles) • V1 = V2 n1 = n2

  22. Figure 5.10 p. 195Balloons Holding 1.0 L of Gas at 25º C and 1 atm. Each contains 0.041 mol of gas, or 2.5 x 1022 molecules, even though their masses are different (equal numbers, different masses).

  23. Gay- Lussac Law • At constant volume, pressure and absolute temperature are directly related. • P = k T • P1 = P2 T1 = T2

  24. Tr 52A Gay-Lussac’s Law Relationship

  25. Combined Gas Law pp • If the moles of gas remain constant, use this formula and cancel out the other things that don’t change. • P1 V1 = P2 V2. T1 T2

  26. Examples pp • A deodorant can has volume of 175 mL & pressure of 3.8 atm at 22ºC. What pressure if heated to 100.ºC? • V is constant, so (P1V1)/T1 = (P2V2/T2) • (3.8)/(295) = (x)/373 • x = 4.8 atm

  27. Examples • A deodorant can has volume of 175 mL & pressure of 3.8 atm at 22ºC. What volume of gas could the can release at 22ºC and 743 torr? Answer . . . • 680 mLSteps. . . • (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22)

  28. 5.3 Ideal Gas Law pp • PV = nRT • V = 22.41 L at 1 atm, 0ºC, n = 1 mole, what is R? • R is the ideal gas constant. • R = 0.08206 L atm/ mol K • R also = 62.396 L torr/k mol or 8.3145 J/k mol • Tells you about a gas NOW. • The other laws tell you about a gas when it changes.

  29. Ideal Gas Law • An equation of state. • Independent of how you end up where you are at. Does not depend on the path. • Given 3 measurements you can determine the fourth. • An Empirical Equation - based on experimental evidence.

  30. Ideal Gas Law • A hypothetical substance - the ideal gas • Think of it as a limit. • Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature. • Use the laws anyway, unless told to do otherwise. • They give good estimates.

  31. Examples pp • A 47.3 L container with 1.62 mol of He heated until pressure reaches 1.85 atm. What is the temperature? Steps . . . • PV = nRT • T = (1.85)(47.3)/(.0821)(1.62)=658 K or 385 ºC

  32. Examples pp • Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC. What is the mass of Kr? Ans. . . • 546 g Steps . . . • n = PV/RT = 6.51 mol • Since M = m/mol = m/n, m = M • n • Get M from periodic table • m = 83.8 • 6.51 = 546 g

  33. Examples • A gas is 4.18 L at 29 ºC & 732 torr. What volume at 24.8ºC & 756 torr? Ans. . . • Use combined law. Why? . . . • The conditions are changing • x = 3.99 L (can convert torr to atm when calculating, but pressure units cancel out so don’t need to for this problem).

  34. 5.4 Gases and Stoichiometry • Reactions happen in moles • At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.41 L. • If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

  35. Figure 5.11A Mole of Any Gas Occupies a Volume of Approximately 22.4 L at STP How many particles are in each box? 6.022 x 1023 particles

  36. Examples pp • Can get mercury by the following: What volume oxygen gas produced from 4.10 g of mercury (II) oxide (M = 217 g/mol) at STP? Steps. • Balance first! 2HgO  2Hg + O2 • Since at STP can then use stoich • 4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O2/2 mol HgO • 22.4 L O2/1 mol O2 • = 0.212 L or 212 mL

  37. Examples pp • Can get mercury by the following: What volume oxygen gas can be produced from 4.10 g of mercury (II) oxide at 400.ºC & 740. torr? (Hint: conditions are changed, so what equation do you then use?) . . . • 1st, find molar mass of mercury(II) oxide. • 2nd, balance reactions & spider under STP • 3rd, then use the combined law to convert. • (1)(212)/273 = (740/760)(x)/(273 + 400) • 0.537 L

  38. Examples • Using the following reaction calculate the mass of sodium hydrogen carbonate (M = 84.01) that produces 2.87 L of carbon dioxide at 25ºC & 2.00 atm. • Not at STP so use PV=nRT; ans. . . • 19.7 g

  39. Examples • Using the following reaction • If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added what is the concentration of HCl? • Use stoich to get moles HCl, then mole ratio  moles CO2, then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = 1.08 mole CO2; so concentration is 0.415 M

  40. Examples • Consider the following reaction What volume NO at 1.0 atm & 1000ºC produced from 10.0 L of NH3 & xs O2 at same temperature & pressure? • T & P constant & cancel out so volume is proportional to moles = 10 L

  41. You try it! • Consider the following reaction What volume of O2 at STP will be consumed when 10.0 kg NH3 is reacted? (Use “rounded” molar masses). Ans. . . • watch Kg! Ans. Rounded to 16 500 L (from calc ans of 16 470)

  42. The Same reaction with some twists pp • What mass of H2O produced from 65.0 L O2 & 75.0 L of NH3 measured at STP? . . . Ans. • As before, but must find LR = O2 ans. 313 g • What volume of NO would be produced? • Above + mole ratio. Ans. = 52 L • What mass of NO is produced from 500. L of NH3 at 250.0ºC and 3.00 atm? Ans. • non-STP so use combined gas law to get to STP (783 L), then “spider” Ans. = 895 g

  43. Gas Density and Molar Mass • D = m/V • n= grams of gas/molar mass or m/molar mass • Let M stand for molar mass • M = m/n • n= PV/RT • M = m PV/RT • M = mRT = mR T = DRT PV V P P

  44. Finding M or D from ideal gas law • M = m (PV/RT) • M =m RT V P • Since m/V = density (D) • M = DRT = molar mass P • Density varies directly with M and P • Density varies indirectly with T

  45. Quick Example pp • What mass of ethene gas, C2H4, is in a 15.0 L tank at 4.40 atm & 305 K? (steps) • PV = mRT/M and M = 28 g/mol • m = PVM/RT • m = (4.40 atm)(15.0 L)(28 g/mol) (0.0821 L atm/mol K)(305 K) • m = 73.8 g C2H4

  46. Another way to solve it (quick) pp • What mass of ethene gas, C2H4, is in a 15.0 L tank at 4.40 atm & 305 K? (steps) • PV = nRT, solve for n • n = PV/RT • n = (4.40 atm)(15.0 L) (0.0821 L atm/mol K)(305 K) • n = 2.64 mol C2H4 • C2H4 (M = 28 g/mol) • Answer is 73.8 g C2H4

  47. Quick Example • A gas has a density of 3.36 g/L at 14 ºC and 1.09 atm. What is its molar mass? • Use M = DRT/P to get an answer of . . . • 72.6 g/mol. Calculation . . . • (3.36)(0.0821)(14 + 273)/(1.09) = 72.6 g/mol

  48. AP Examples • Density of ammonia at 23ºC & 735 torr? Ans. • D = MP/RT = 0.68 g/L

  49. AP Examples • A compound has empirical formula CHCl. A 256 mL flask at 100.ºC & 750 torr contains 0.80 g of the gaseous compound. What is the molecular formula? Steps. . . • What do you need to get the molecular formula? . . . • The molecular mass, M. • How do you get this? • M = g/mol. You are given the grams, so find the number of moles using . . . • PV = nRT and solve for n . . . Try it! . . .

  50. AP Examples cont. • EF of CHCl. 0.80 g are in 256 mL flask at 100.ºC & 750 torr. What is the MF? • n = 0.0083 from n = PV/RT • What’s M? . . . • M  97 from 0.80 g/0.0083 moles = g/mol • How do you find molecular formula from EF and MF? • Find how many Efs are in the MF (CHCl) then multiply the subscripts of the EF by that number. Try it! . . . • 97/49  2, so formula is C2H2Cl2

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