Ch5 z5e gases

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  • Z5e 188 Section 5.1 Pressure
  • Z5e 189 Fig. 5.2
  • Z5e 189 Fig. 5.3
  • Z5e 189 Units of Pressure 96.5 kPa 724 torr 0.952 atm
  • Z5e 190 Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro
  • Hrw 314 ref. Fig 10-10
  • Z5e 192 Fig. 5.6 Plot of PV vs P at P below 1 atm Ideal gas expected to have constant value of PV Boyle’s law is good approximation at relatively low pressures
  • (20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x); so x = 2.04 L (30.6)(740) = (x)(750); x = 30.2 torr = 402 kPa
  • (20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x); so x = 2.04 L (30.6)(740) = (x)(750); x = 30.2 torr = 402 kPa
  • Z5e 193 Charles’ Law
  • Hrw 316 and rf hrw 318 fig. 10-12.
  • Z5e 194 Fig. 5.8 Plot V vs. T Gas samples contain different numbers of moles
  • (247)/(22 + 273) = (x)(273 + 98); x = 311 ml
  • (40.5)/(26.4 + 273) = (81.0)/x; x = 593 K = 320. Celsius
  • Z5e 195 Avogadro’s Law
  • Z5e 195 Fig. 5.10
  • V is constant, so (3.8)/(295) = (x)/373; x = 4.8 atm (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22); x = 680 ml
  • V is constant, so (3.8)/(295) = (x)/373; x = 4.8 atm (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22); x = 680 ml
  • Z5e 196 Section 5.3 The Ideal Gas Law Also 62.396 l torr/k mol = 8.3145 J/k mol
  • PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
  • PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
  • PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
  • Z5e Section 5.4 Gas Stoichiometry
  • Z5e 202
  • Balance 1st! Stoich MW HgO 216.59; Hg 200.59; O2 32; ans. 0.212 L Use ideal gas law; ans. 0.537 L
  • Balance 1st! Stoich MW HgO 216.59; Hg 200.59; O2 32; ans. 0.212 L Use ideal gas law; ans. 0.537 L
  • MW NaHCO 3 84 Not at STP so use PV=nRT; ans. 21.5 g Use stoich to get moles HCl, then mole ratio --> moles CO 2 , 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 CO 2 ; so concentration is 0.415 M
  • MW NaHCO 3 84 Not at STP so use PV=nRT; ans. 21.5 g Use stoich to get moles HCl, then mole ratio --> moles CO 2 , 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 CO 2 ; so concentration is 0.415 M
  • T & P constant so cancel out. L proportional to moles, so 10 L watch Kg! Ans. Rounded to 16 500 (from calc ans of 16 470)
  • T & P constant so cancel out. L proportional to moles, so 10 L watch Kg! Ans. Rounded to 16 500 (from calc ans of 16 470)
  • LR = O 2 , ans. 313 g ans. 52 L correct to STP = 783 L; ans. 895 g.
  • Z5e 205
  • Hrw 345
  • Hrw additional sample problem 11-5
  • Hrw additional sample problem 11-5
  • Hrw additional sample problem 11-6, modified so that density is used, where density is rounded down to 3.36 g/L from 3.365 g/L.
  • D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
  • D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
  • D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
  • Z5e 205 Section 5.5 Dalton’s Law of Partial Pressures
  • Z5e 206 Fig 5.12
  • Z5e 205 Section 5.5 Dalton’s Law of Partial Pressures
  • Z5e 207
  • 592/752 = 0.787 0.787 x 5.25 = 4.13 atm Note: can use mole fraction regardless of units (don’t have to convert). Also, Air pressure is 752 (so this is the total pressure in the denominator)
  • 592/752 = 0.787 0.787 x 5.25 = 4.13 atm Note: can use mole fraction regardless of units (don’t have to convert). Also, Air pressure is 752 (so this is the total pressure in the denominator)
  • P1V1 = P2V2; total V = 9 L CH4 (4)(2.70) = (9)(x); x = 1.2 atm N2 ans. 0.763 O2 = 0.294 Sum = 2.127 = ans.
  • Z5e 209
  • Z5e 209 Fig. 5.13
  • Make sure reaction is balanced MW of ammonium nitrate = 80. G Use spider to get moles of N20 = .0325, then use PV = nRT, but first have to: Convert 21 torr to 2.8 kPa 94.0 - 2.8 = 91.2 kPa N2O convert P to atm d/t R ( or use 62.4)!!! Convert C to K Ans. 6.6 L
  • Make sure reaction is balanced MW of ammonium nitrate = 80. G Use spider to get moles of N20 = .0325, then use PV = nRT, but first have to: Convert 21 torr to 2.8 kPa 94.0 - 2.8 = 91.2 kPa N2O convert P to atm d/t R ( or use 62.4)!!! Convert C to K Ans. 6.6 L
  • Z5e 211 Section 5.6 The Kinetic Molecular Theory of Gases
  • Z5e 216, 217
  • M rms = sq root 3RT/u Must express R using joules and M = Kg/mol (not g, d/t R) R = 8.3145 J/K*Mol 411 m/s 1928 m/s 324 m/s
  • Z5e 218 Fig. 5.18: Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of its container.
  • Z5e Section 5.7 Effusion and Diffusion
  • Z5e 219 Fig 5.21: Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.
  • Z5e 220
  • Z5e 221 Diffusion
  • a. ans. 40.96 b. rate NH3/rate HCl = 2.47/2.47 = 1/1 = sq rt (.0025*17/x*36.5) = .00177 mol HCl. Alternate method: (0.251/2.47 divided by (x)/2.47) = (sq rt 36.5)/(sq rt 17) = 0.00177 mol HCl. c. Rate N2is faster than rate x by 55/38 times = 1.45 times faster. Then, 1.45/1 = Sq Rt M x divided by sq rt N 2 and answer is 59 g/mol (sig figs)
  • Z5e Section 5.8 Real Gases
  • Z5e 224
  • Z5e 222: Behavior is close to ideal ONLY at low pressures (less than 1 atm)
  • Z5e 222
  • a. P = nRT/V = 12.2 atm b. P = nRT/V - nb - a (n/v) 2 = 11.0 atm
  • Z5e 225
  • Ch5 z5e gases

    1. 1. Chapter 5 The Gas Laws pp
    2. 2. 5.1 Pressure <ul><li>Force per unit area. </li></ul><ul><li>Gas molecules fill container. </li></ul><ul><li>Molecules move around and hit sides. </li></ul><ul><li>Collisions are the force. </li></ul><ul><li>Container has the area. </li></ul><ul><li>Measured with a barometer. </li></ul>
    3. 3. Barometer <ul><li>The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. </li></ul><ul><li>1 atm = 760 mm Hg </li></ul>1 atm Pressure 760 mm Hg Vacuum
    4. 4. Manometer <ul><li>Column of mercury to measure pressure. </li></ul><ul><li>h is how much lower the pressure is than outside. </li></ul>Gas h
    5. 5. Manometer <ul><li>h is how much higher the gas pressure is than the atmosphere. </li></ul>h Gas
    6. 6. Units of pressure <ul><li>1 atmosphere = 760 mm Hg </li></ul><ul><li>1 mm Hg = 1 torr </li></ul><ul><li>1 atm = 101,325 Pascals = 101.325 kPa </li></ul><ul><li>Can make conversion factors from these. </li></ul><ul><li>What is 724 mm Hg in kPa? . . . </li></ul><ul><li>96.5 kPa </li></ul><ul><li>In torr? </li></ul><ul><li>724 torr </li></ul><ul><li>In atm? </li></ul><ul><li>0.952 atm. </li></ul>
    7. 7. 5.2 The Gas Laws of Boyle, Charles, and Avogadro <ul><li>Boyle’s Law </li></ul><ul><li>Pressure and volume are inversely related at constant temperature. </li></ul><ul><li>PV= k </li></ul><ul><li>As one goes up, the other goes down. </li></ul><ul><li>P 1 V 1 = P 2 V 2 </li></ul><ul><li>Graphically </li></ul>
    8. 8. <ul><li>As the pressure on a gas increases </li></ul>1 atm 4 Liters
    9. 9. <ul><li>As the pressure on a gas increases the volume decreases </li></ul><ul><li>Pressure and volume are inversely related </li></ul>2 atm 2 Liters
    10. 10. A Boyle’s Law Relationship
    11. 11. V P (at constant T)
    12. 12. V 1/P (at constant T) Slope = k
    13. 13. PV P (at constant T) CO 2 O 2 22.41 L atm Ne
    14. 14. Example pp <ul><li>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 . . . </li></ul><ul><li>P 1 V 1 = P 2 V 2 </li></ul><ul><li>(20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x) </li></ul><ul><li>x = 2.04 L </li></ul>
    15. 15. You try it! <ul><li>30.6 mL of CO 2 at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa? Ans. . . </li></ul><ul><li>4.02 kPa . Steps . . . </li></ul><ul><li>P 1 V 1 = P 2 V 2 </li></ul><ul><li>( 30.6 mL)(740 torr) = (750 mL) (x torr) </li></ul><ul><li>x = 30.2 torr </li></ul><ul><li>(30 torr)(101.325 kPa/ 760 torr) = 4.02 kPa </li></ul>
    16. 16. Charles’ Law <ul><li>Volume of a gas varies directly with the absolute temperature at constant pressure. </li></ul><ul><li>V = kT (if T is in Kelvin) </li></ul><ul><li>V 1 = V 2 T 1 = T 2 </li></ul><ul><li>Graphically </li></ul>
    17. 17. Tr 51A Charles’ Law Relationship
    18. 18. V (L) T (ºC) He H 2 O CH 4 H 2 -273.15ºC
    19. 19. Example pp <ul><li>What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant ? Steps. </li></ul><ul><li>V 1 /T 1 = V 2 /T 2 </li></ul><ul><li>(247 mL)/(22 + 273) K = (x mL)/(273 + 98) K </li></ul><ul><li>x = 311 ml </li></ul>
    20. 20. Examples (do this) <ul><li>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 </li></ul><ul><li>320. ºC (change K to ºC!) </li></ul>
    21. 21. Avogadro's Law <ul><li>At constant temperature and pressure, the volume of gas is directly related to the number of moles. </li></ul><ul><li>V = k n (n is the number of moles) </li></ul><ul><li>V 1 = V 2 n 1 = n 2 </li></ul>
    22. 22. Figure 5.10 p. 195 Balloons Holding 1.0 L of Gas at 25º C and 1 atm . Each contains 0.041 mol of gas, or 2.5 x 10 22 molecules, even though their masses are different ( equal numbers, different masses ).
    23. 23. Gay- Lussac Law <ul><li>At constant volume, pressure and absolute temperature are directly related. </li></ul><ul><li>P = k T </li></ul><ul><li>P 1 = P 2 T 1 = T 2 </li></ul>
    24. 24. <ul><li>Tr 52A Gay-Lussac’s Law Relationship </li></ul>
    25. 25. Combined Gas Law pp <ul><li>If the moles of gas remain constant , use this formula and cancel out the other things that don’t change. </li></ul><ul><li>P 1 V 1 = P 2 V 2 . T 1 T 2 </li></ul>
    26. 26. Examples pp <ul><li>A deodorant can has volume of 175 mL & pressure of 3.8 atm at 22ºC. What pressure if heated to 100.ºC? </li></ul><ul><li>V is constant, so (P 1 V 1 )/T 1 = (P 2 V 2 /T 2 ) </li></ul><ul><li>(3.8)/(295) = (x)/373 </li></ul><ul><li>x = 4.8 atm </li></ul>
    27. 27. Examples <ul><li>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 . . . </li></ul><ul><li>680 mL Steps. . . </li></ul><ul><li>( 3.8 )( 175 )/(273 + 22) = ( 743/760 )(x)/( 273 + 22 ) </li></ul>
    28. 28. 5.3 Ideal Gas Law pp <ul><li>PV = nRT </li></ul><ul><li>V = 22.41 L at 1 atm, 0ºC, n = 1 mole, what is R? </li></ul><ul><li>R is the ideal gas constant. </li></ul><ul><li>R = 0.08206 L atm/ mol K </li></ul><ul><li>R also = 62.396 L torr/k mol or 8.3145 J/k mol </li></ul><ul><li>Tells you about a gas is NOW . </li></ul><ul><li>The other laws tell you about a gas when it changes ( exam hint! ) . </li></ul>
    29. 29. Ideal Gas Law <ul><li>An equation of state . </li></ul><ul><li>Independent of how you end up where you are at. Does not depend on the path. </li></ul><ul><li>Given 3 measurements you can determine the fourth. </li></ul><ul><li>An Empirical Equation - based on experimental evidence. </li></ul>
    30. 30. Ideal Gas Law <ul><li>A hypothetical substance - the ideal gas </li></ul><ul><li>Think of it as a limit. </li></ul><ul><li>Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature. </li></ul><ul><li>Use the laws anyway, unless told to do otherwise. </li></ul><ul><li>They give good estimates. </li></ul>
    31. 31. Examples pp <ul><li>A 47.3 L container with 1.62 mol of He heated until pressure reaches 1.85 atm . What is the temperature? Steps . . . </li></ul><ul><li>PV = nR T </li></ul><ul><li>T = ( 1.85 )( 47.3 )/(. 0821 )( 1.62 ) = 658 K or 385 ºC </li></ul>
    32. 32. Examples pp <ul><li>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. . . </li></ul><ul><li>546 g Steps . . . </li></ul><ul><li>n = PV/RT = 6.51 mol </li></ul><ul><li>Since M = m/mol = m/n, m = M • n </li></ul><ul><li>Get M from periodic table </li></ul><ul><li>m = 83.8 • 6.51 = 546 g </li></ul>
    33. 33. Examples <ul><li>A gas is 4.18 L at 29 ºC & 732 torr. What volume at 24.8ºC & 756 torr? Ans. . . </li></ul><ul><li>Use combined law. Why? . . . </li></ul><ul><li>The conditions are changing </li></ul><ul><li>x = 3.99 L (can convert torr to atm when calculating, but pressure units cancel out so don’t need to for this problem). </li></ul>
    34. 34. 5.4 Gases and Stoichiometry <ul><li>Reactions happen in moles </li></ul><ul><li>At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.41 L. </li></ul><ul><li>If not at STP, use the ideal gas law to calculate moles of reactant or volume of product. </li></ul>
    35. 35. Figure 5.11 A Mole of Any Gas Occupies a Volume of Approximately 22.4 L at STP How many particles are in each box? 6.022 x 10 23 particles
    36. 36. Examples pp <ul><li>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. </li></ul><ul><li>Balance first! 2HgO  2Hg + O 2 </li></ul><ul><li>Since at STP can then use stoich (spider) </li></ul><ul><li>4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O 2 /2 mol HgO • 22.4 L O 2 /1 mol O 2 </li></ul><ul><li>= 0.212 L or 212 mL </li></ul>
    37. 37. Examples pp <ul><li>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?) . . . </li></ul><ul><li>1st, find molar mass of mercury(II) oxide. </li></ul><ul><li>2nd, balance reactions & spider under STP </li></ul><ul><li>3rd, then use the combined law to convert. </li></ul><ul><li>(1)(212)/273 = (740/760)(x)/(273 + 400) </li></ul><ul><li>0.537 L </li></ul>
    38. 38. Examples <ul><li>Using the following reaction </li></ul><ul><li>calculate the mass of sodium hydrogen carbonate (M = 84.01) that produces 2.87 L of carbon dioxide at 25ºC & 2.00 atm. </li></ul><ul><li>Not at STP so use PV=nRT; ans. . . </li></ul><ul><li>19.7 g </li></ul>
    39. 39. Examples <ul><li>Using the following reaction </li></ul><ul><li>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? </li></ul><ul><li>Use stoich to get moles HCl, then mole ratio  moles CO 2 , 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 CO 2 ; so concentration is 0.415 M </li></ul>
    40. 40. Examples <ul><li>Consider the following reaction What volume NO at 1.0 atm & 1000ºC produced from 10.0 L of NH 3 & xs O 2 at same temperature & pressure? </li></ul><ul><li>T & P constant & cancel out so volume is proportional to moles = 10 L </li></ul>
    41. 41. You try it! <ul><li>Consider the following reaction What volume of O 2 at STP will be consumed when 10.0 kg NH 3 is reacted? (Use “rounded” molar masses). Ans. . . </li></ul><ul><li>watch Kg! Ans. Rounded to 16 500 L (from calc ans of 16 470) </li></ul>
    42. 42. The Same reaction with some twists pp <ul><li>What mass of H 2 O produced from 65.0 L O 2 & 75.0 L of NH 3 measured at STP? . . . Ans. </li></ul><ul><li>As before, but must find LR = O 2 ans. 313 g </li></ul><ul><li>What volume of NO would be produced? </li></ul><ul><li>Above + mole ratio. Ans. = 52 L </li></ul><ul><li>What mass of NO is produced from 500. L of NH 3 at 250.0ºC and 3.00 atm? Ans. </li></ul><ul><li>non-STP so use combined gas law to get to STP (783 L), then “spider” Ans. = 895 g </li></ul>
    43. 43. Gas Density and Molar Mass <ul><li>D = m/V </li></ul><ul><li>Let M stand for molar mass </li></ul><ul><li>M = m/n </li></ul><ul><li>n= PV/RT </li></ul><ul><li>M = m PV/RT </li></ul><ul><li>M = mRT = m RT = DRT PV V P P </li></ul>
    44. 44. Finding M or D from ideal gas law <ul><li>M = m (PV/RT) </li></ul><ul><li>M = m RT V P </li></ul><ul><li>Since m / V = density (D) </li></ul><ul><li>M = DRT = molar mass P </li></ul><ul><li>Density varies directly with M and P </li></ul><ul><li>Density varies indirectly with T </li></ul>
    45. 45. Quick Example pp <ul><li>What mass of ethene gas, C 2 H 4 , is in a 15.0 L tank at 4.40 atm & 305 K ? (steps) </li></ul><ul><li>PV = mRT/M and M = 28 g/mol </li></ul><ul><li>m = PVM/RT </li></ul><ul><li>m = ( 4.40 atm )( 15.0 L )(28 g/mol) (0.0821 L atm/mol K)( 305 K ) </li></ul><ul><li>m = 73.8 g C 2 H 4 </li></ul>
    46. 46. Another way to solve it (quick) pp <ul><li>What mass of ethene gas, C 2 H 4 , is in a 15.0 L tank at 4.40 atm & 305 K? (steps) </li></ul><ul><li>PV = nRT, solve for n </li></ul><ul><li>n = PV/RT </li></ul><ul><li>n = (4.40 atm)(15.0 L) (0.0821 L atm/mol K)(305 K) </li></ul><ul><li>n = 2.64 mol C 2 H 4 </li></ul><ul><li>Use spider to get g C 2 H 4 (M = 28 g/mol) </li></ul><ul><li>Answer is 73.8 g C 2 H 4 </li></ul>
    47. 47. Quick Example <ul><li>A gas has a density of 3.36 g/L at 14 ºC and 1.09 atm. What is its molar mass? </li></ul><ul><li>Use M = DRT/P to get an answer of . . . </li></ul><ul><li>72.6 g/mol. Calculation . . . </li></ul><ul><li>(3.36)(0.0821)(14 + 273)/(1.09) = 72.6 g/mol </li></ul>
    48. 48. AP Examples <ul><li>Density of ammonia at 23ºC & 735 torr? Ans. </li></ul><ul><li>D = MP/RT = 0.68 g/L </li></ul>
    49. 49. AP Examples <ul><li>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. . . </li></ul><ul><li>What do you need to get the molecular formula? . . . </li></ul><ul><li>The molecular mass, M. </li></ul><ul><li>How do you get this? </li></ul><ul><li>M = g/mol. You are given the grams, so find the number of moles using . . . </li></ul><ul><li>PV = nRT and solve for n . . . Try it! . . . </li></ul>
    50. 50. AP Examples cont. <ul><li>EF of CHCl. 0.80 g are in 256 mL flask at 100.ºC & 750 torr. What is the MF? </li></ul><ul><li>n = 0.0083 from n = PV/RT </li></ul><ul><li>What’s M? . . . </li></ul><ul><li>M  97 from 0.80 g/0.0083 moles = g/mol </li></ul><ul><li>How do you find molecular formula from EF and MF? </li></ul><ul><li>Find how many Efs are in the MF (CHCl) then multiply the subscripts of the EF by that number. Try it! . . . </li></ul><ul><li>97/49  2, so formula is C 2 H 2 Cl 2 </li></ul>
    51. 51. 5.5 Dalton’s Law of Partial Pressures <ul><li>The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. </li></ul><ul><li>The total pressure is the sum of the partial pressures. </li></ul>
    52. 52. Figure 5.12 p. 206 Partial Pressure of Each Gas in a Mixture
    53. 53. <ul><li>We can find out the pressure in the fourth container by adding up the pressure in the first 3. </li></ul>2 atm 1 atm 3 atm 6 atm
    54. 54. Dalton’s Law <ul><li>P Total = P 1 + P 2 + P 3 + P 4 + P 5 ... </li></ul><ul><li>For each, P still = nRT/V </li></ul>
    55. 55. Dalton's Law <ul><li>P Total = n 1 RT + n 2 RT + n 3 RT +... V V V </li></ul><ul><li>In the same container R, T and V are the same. </li></ul><ul><li>P Total = (n 1 + n 2 + n 3 +...)RT V </li></ul><ul><li>P Total = (n Total )RT V </li></ul>
    56. 56. Examples pp <ul><li>In a balloon the total pressure is 1.3 atm. What is the pressure of oxygen if the pressure of nitrogen is 720 mm Hg? . . . </li></ul><ul><li>P Total = P 1 + P 2 + P 3 . . . </li></ul><ul><li>1.3 atm = Po 2 + 720 mm Hg of N 2 </li></ul><ul><li>Convert to same units (we’ll use mm Hg) </li></ul><ul><li>1.3 atm(760 mm Hg/1 atm) = Po 2 + 720 mm Hg </li></ul><ul><li>988 mm Hg = Po 2 + 720 mm Hg so, . . . </li></ul><ul><li>Po 2 = 988 mm Hg - 720 mm Hg = 268 mm Hg </li></ul><ul><li>Can also express as 0.35 atm. </li></ul>
    57. 57. The mole fraction <ul><li>Ratio of moles of the substance to the total moles. </li></ul><ul><li>symbol is Greek letter chi  </li></ul><ul><li>   = n 1 = P 1 n Total P Total </li></ul>
    58. 58. Examples pp <ul><li>The partial pressure of nitrogen in air is 592 torr. The partial pressure of oxygen in air is 160 torr. What is the mole fraction of oxygen? Ans. . . </li></ul><ul><li>0.213 Why isn’t it 0.270? . . . </li></ul><ul><li>Because you have to divide by the total pressure, which is 752 torr. </li></ul>
    59. 59. Examples <ul><li>The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen? Steps. . . </li></ul><ul><li> = 592 torr/752 torr = 0.787 </li></ul><ul><li>What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm? . . . </li></ul><ul><li>0.787 x 5.25 atm = 4.13 atm </li></ul><ul><li>Note: can use mole fraction regardless of units (don’t have to convert). </li></ul>
    60. 60. Examples pp <ul><li>When these valves are opened, what is each partial pressure and the total pressure? Hint. </li></ul><ul><li>Use mole fraction to get the new partial pressures (or Boyle’s law), then add to get total pressure. Ans. </li></ul><ul><li>CH 4 = 1.2 atm N 2 = 0.763 O 2 = 0.294 Sum = 2.127 </li></ul>3.50 L O 2 1.50 L N 2 2.70 atm 4.00 L CH 4 4.58 atm 0.752 atm
    61. 61. Vapor Pressure <ul><li>Water evaporates! </li></ul><ul><li>When that water evaporates, the vapor has a pressure. </li></ul><ul><li>Gases are often collected over water so the vapor. pressure of water must be subtracted from the total pressure. </li></ul><ul><li>It must be given. </li></ul>
    62. 62. Figure 5.13 The Production of Oxygen by Thermal Decomposition of KCIO 3
    63. 63. Example pp <ul><li>N 2 O can be produced by the following reaction what volume of N 2 O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH 4 NO 3 ? (the vapor pressure of water at 22ºC is 21 torr). Steps . . . </li></ul>
    64. 64. Example <ul><li>What vol. N 2 O over water at total pressure of 94 kPa & 22ºC can be produced from 2.6 g NH 4 NO 3 ? (Vp water at 22ºC is 21 torr ) </li></ul><ul><li>Make sure reaction is balanced (it is) </li></ul><ul><li>M of ammonium nitrate = 80. g </li></ul><ul><li>Use spider to get moles of N 2 0 = .0325, then use PV = nRT, but first have to: </li></ul><ul><li>Convert 21 torr to 2.8 kPa (or vice versa ) </li></ul><ul><li>94.0 - 2.8 = 91.2 kPa N 2 O </li></ul><ul><li>convert P to atm d/t R ( or use 8.31 for kPa)!!! </li></ul><ul><li>Convert C to K </li></ul><ul><li>Ans. 6.6 L </li></ul>
    65. 65. 5.6 Kinetic Molecular Theory <ul><li>Theory tells why the things happen. </li></ul><ul><li>explains why ideal gases behave the way they do. </li></ul><ul><li>Assumptions that simplify the theory, but don’t work in real gases. </li></ul><ul><li>The particles are so small we can ignore their volume . </li></ul><ul><li>The particles are in constant motion and their collisions cause pressure . </li></ul>
    66. 66. <ul><li>The Average speed of an oxygen molecule is 1656 km/hr at 20ºC </li></ul><ul><li>The molecules don’t travel very far without hitting each other so they move in random directions. </li></ul>
    67. 67. Kinetic Molecular Theory <ul><li>The particles do not affect each other, neither attracting or repelling . </li></ul><ul><li>The average kinetic energy is proportional to the Kelvin temperature. </li></ul><ul><li>Appendix 2 shows the derivation of the ideal gas law and the definition of temperature. </li></ul><ul><li>We need the formula KE = 1/2 mv 2 </li></ul>
    68. 68. What it tells us <ul><li>(KE) avg = 3/2 RT </li></ul><ul><li>This the meaning of temperature. </li></ul><ul><li>u is the particle velocity. </li></ul><ul><li>u is the average particle velocity. </li></ul><ul><li>u 2 is the average particle velocity squared. </li></ul><ul><li>the root mean square velocity is  u 2 = u rms </li></ul>
    69. 69. Combine these two equations <ul><li>(KE) avg = n A (1/2 mu 2 ) (KE) avg = 3/2 RT </li></ul>
    70. 70. Combine these two equations <ul><li>(KE) avg = n A (1/2 mu 2 ) </li></ul><ul><li>(KE) avg = 3/2 RT Where M is the molar mass in kg /mole ( watch for test question ! ), and R has the units 8.3145 J /K mol. </li></ul><ul><li>The velocity will be in m/s </li></ul>
    71. 71. Example pp <ul><li>Calculate the root mean square velocity of carbon dioxide at 25ºC. </li></ul><ul><li>Must express R using joules & M = Kg/mol (not g, d/t R) </li></ul><ul><li>R = 8.3145 J/K•Mol, M = 0.0 44 Kg /mol ans. . . </li></ul><ul><li>411 m/sec </li></ul><ul><li>If time, do for hydrogen, for chlorine. Ans. . . </li></ul><ul><li>Hydrogen is 1928 m/s; Chorine is 324 m/s </li></ul><ul><li>Notice that the less massive hydrogen has a higher velocity. </li></ul>
    72. 72. Range of velocities <ul><li>The average distance a molecule travels before colliding with another is called the mean free path and is small (around 10 -7 m) </li></ul>
    73. 73. Figure 5.18 Path of One Particle in a Gas. Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of its container.
    74. 74. Range of velocities <ul><li>Temperature is an average. There are molecules of many speeds in the average. </li></ul><ul><li>This is shown on a graph called a velocity distribution . . . </li></ul>
    75. 75. number of particles Molecular Velocity 273 K
    76. 76. number of particles Molecular Velocity 273 K 1273 K
    77. 77. number of particles Molecular Velocity 273 K 1273 K 2273 K
    78. 78. Velocity <ul><li>Average increases as temperature increases. </li></ul><ul><li>Spread increases as temperature increases. </li></ul>
    79. 79. 5.7 Effusion & Diffusion <ul><li>Effusion : Passage of gas through a small hole, into a vacuum. </li></ul><ul><li>The effusion rate measures how fast this happens. </li></ul><ul><li>Graham’s Law : The rate of effusion is inversely proportional to the square root of the mass of its particles. </li></ul>
    80. 80. Figure 5.21 p. 219The Effusion of a Gas into an Evacuated Chamber Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.
    81. 81. Deriving <ul><li>The rate of effusion should be proportional to u rms </li></ul><ul><li>Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2 </li></ul>
    82. 82. Deriving Graham’s Law <ul><li>The rate of effusion should be proportional to u rms </li></ul><ul><li>Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2 </li></ul>
    83. 83. Diffusion <ul><li>The spreading of a gas through a room. </li></ul><ul><li>Slow, considering molecules move at 100’s of meters per second. </li></ul><ul><li>Collisions with other molecules slow down diffusions. </li></ul><ul><li>Best estimate of diffusion is Graham’s Law. </li></ul><ul><li>Do not confuse rate with time. </li></ul><ul><li>Time is in seconds (or min). Rate is in 1/s (or 1/min), etc. </li></ul><ul><li>The following questions test your understanding of this. </li></ul>
    84. 84. Figure 5.23 HCI( g ) and NH 3 ( g ) Meet in a Tube
    85. 85. Whiteboard Examples pp <ul><li>An element effuses through a porous cylinder 3.20 times slower than helium. What is it’s molar mass? Steps on whiteboard to get . . . </li></ul><ul><li>40.96 (Probably Argon = 39.948 g/mol) </li></ul><ul><li>0.00251 mol NH 3 effuses via a hole in 2.47 min, how much HCl would effuse in the same time? Your ans.? </li></ul><ul><li>0.0017 mol HCl </li></ul><ul><li>Some N 2 effuses via a hole in 38 seconds. What is molecular weight of gas that effuses in 55 seconds under identical conditions? Your answer? . . . (hint: think if N 2 is faster or slower. If faster then the rate must be > 1, if slower, then the rate is a fraction. </li></ul><ul><li>59 g/mol </li></ul>
    86. 86. 5.8 Real Gases <ul><li>Real molecules do take up space and they do interact with each other (especially polar molecules). </li></ul><ul><li>Need to add correction factors to the ideal gas law to account for these. </li></ul>
    87. 87. Volume Correction <ul><li>The actual volume free to move in is less because of particle size. </li></ul><ul><li>More molecules ( n ) will have more effect. </li></ul><ul><li>Corrected volume V’ = V - nb. Substitute this for V in PV = nRT </li></ul><ul><li>b is a constant that differs for each gas. </li></ul><ul><li>P ’ = nRT (V- n b ) </li></ul>
    88. 88. Pressure correction <ul><li>Because the molecules are attracted to each other, the pressure on the container will be less than ideal </li></ul><ul><li>Depends on the number of molecules per liter ( n ÷ V ). “a” is a unique constant. </li></ul><ul><li>Since two molecules interact, the effect must be squared . </li></ul>
    89. 89. Pressure correction <ul><li>Because the molecules are attracted to each other, the pressure on the container will be less than ideal </li></ul><ul><li>Depends on the number of molecules per liter ( n ÷ V ). “a” is a unique constant </li></ul><ul><li>Since two molecules interact, the effect must be squared. </li></ul>P observed = P’ - a 2 ( ) V n
    90. 90. Altogether <ul><li>P obs = nRT - a n 2 V-nb V </li></ul><ul><li>Called the Van der Waal’s equation if rearranged </li></ul><ul><li>Corrected Corrected Pressure Volume “P” x “V” = nRT </li></ul>( )
    91. 91. Where does it come from <ul><li>a and b are determined by experiment. </li></ul><ul><li>Different for each gas. </li></ul><ul><li>Bigger molecules have larger b. </li></ul><ul><li>a depends on both size and polarity. </li></ul><ul><li>once given, plug and chug. </li></ul>
    92. 92. Figure 5.24 p. 222 Plots of PV/nRT Versus P for Several Gases (200 K) Behavior is close to ideal ONLY at low pressures (less than 1 atm)
    93. 93. Figure 5.25 Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures Deviations are smaller at higher temperatures.
    94. 94. Real vs Ideal Behavior (test question) <ul><li>Gases act most ideally if at: </li></ul><ul><li>Low pressure </li></ul><ul><li>High temperature, and have . . . </li></ul><ul><li>Low mass and . . . </li></ul><ul><li>Low polarity </li></ul><ul><li>This provides least deviation from the expected collisions derived from the kinetic molecular theory. </li></ul><ul><li>Be able to explain why this is so. </li></ul>
    95. 95. Example <ul><li>Calculate the pressure exerted by 0.5000 mol Cl 2 in a 1.000 L container at 25.0ºC </li></ul><ul><li>Using the ideal gas law. . . </li></ul><ul><li>P = nRT/V = 12.2 atm </li></ul><ul><li>Using Van der Waal’s equation, where </li></ul><ul><ul><li>a = 6.49 atm L 2 /mol 2 </li></ul></ul><ul><ul><li>b = 0.0562 L/mol . . . </li></ul></ul><ul><li>P = nRT/V - nb - a (n/v) 2 = 11.0 atm </li></ul><ul><li>Note the difference. </li></ul>
    96. 96. 5.9 Chemistry in the Atmosphere <ul><li>Acid Rain </li></ul><ul><li>There will be AP questions on this. . . </li></ul>
    97. 97. S ources of Acid Rain <ul><li>Power stations </li></ul><ul><li>Oil refineries </li></ul><ul><li>Coal with high S content </li></ul><ul><li>Car and truck emissions </li></ul><ul><li>Bacterial decomposition, and lightning hitting N 2 in the atmosphere. </li></ul>
    98. 98. Acid Rain <ul><li>Unpolluted rain has a pH of 5.6 </li></ul><ul><li>Rain with a pH below 5.6 is “acid rain“ </li></ul><ul><li>CO 2 in the air forms carbonic acid CO 2 + H 2 O H 2 CO 3 </li></ul><ul><li>Adds to H + of rain H 2 CO 3 H + (aq) + HCO 3 -(aq) </li></ul>
    99. 99. <ul><li>SO 2 26 million tons in 1980 </li></ul><ul><li>NO and NO 2 22 million tons in 1980 </li></ul><ul><li>Mt. St Helens (1980) 400,000 tons SO 2 </li></ul><ul><li>Reactions with oxygen in air form SO 3 </li></ul><ul><li>2SO 2 + O 2 2 SO 3 </li></ul><ul><li>Reactions with water in air form acids </li></ul><ul><li>SO 3 + H 2 O H 2 SO 4 sulfuric acid </li></ul><ul><li>NO + H 2 O HNO 2 nitrous acid </li></ul><ul><li> HNO 2 + H 2 O HNO 3 nitric acid </li></ul>Sources of Acid Rain
    100. 100. Effects of Acid Rain <ul><li>Leaches Al from soil, which kills fish </li></ul><ul><li>Fish kills in spring from runoff due to accumulation of large amounts of acid in snow </li></ul><ul><li>Dissolves waxy coatings that protect leaves from bacteria </li></ul><ul><li>Corrodes metals, textiles, paper and leather </li></ul><ul><li>Dissolves minerals Mg, Ca, and K from the soil and waxy coatings that protect leaves from bacteria. </li></ul><ul><li>Read 5.9 in book for the rest of the story </li></ul>

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