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Chapter 5

Chapter 5. Gases and the Kinetic-Molecular Theory. If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

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Chapter 5

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  1. Chapter 5 Gases and the Kinetic-Molecular Theory If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

  2. Why do we need to learn about gases? To protect our earth!

  3. Name Origin and Use Methane (CH4) Ammonia (NH3) Chlorine (Cl2) Oxygen (O2) Ethylene (C2H4) natural deposits; domestic fuel from N2+H2; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Table 5.1 from 4th ed. Some Important Industrial Gases Atmosphere-Biosphere Redox Interconnections

  4. An Overview of the Physical States of Matter The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure. 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity. 4. Most gases have relatively low densities under normal conditions. 5. Gases are miscible.

  5. The three states of matter. Figure 5.1 Bromine as a gas, a liquid and a solid.

  6. Four measurementss define the state of a gas 1. Quantity in moles, n 2. Volume in Liters 3. Temperature in Kelvin 4. Pressure in Atm

  7. Gas Pressure Pressure = Force/unit area Force causes something to move a distance D in work Gravity is a weak force, F = ma On earth F = mg, where g is the acceleration due to earth’s gravity

  8. A BAROMETER Measures pressure of atmosphere above Essential principle of the barometer – force of air on surface of water keeps water in the tube If we had a long enough tube, about 34 ft., we'd reach the limit of air pressure to hold the water in

  9. Figure 5.3 A mercury barometer. Patm is due to weight of atmosphere above the earth

  10. Barometers Why do our barometers contain mercury? Because of its density. Hg is 13.6 g/mL vs. water at ~1.0 g/mL. P = Force/Area = mass of Hg column * grav. constant/area of column = vol of Hg column * density of Hg/area of column = height of Hg*area of column* mass/area of column = height of Hg column * density of Hg Normal atmospheric pressures is measured at sea level on a calm day: it is 760 mm Hg = one standard atmosphere pressure (both are exact quantities)

  11. Figure 5.2 Effect of atmospheric pressure on objects at the Earth’s surface. When the air inside the can is removed, atmospheric pressure crushes the can. A metal can filled with air has equal pressure on the inside & outside.

  12. Gas Pressure MEMORIZE: 1 atm = 760 mm Hg exactly = 760 torr exactly = 101.325 kPa = 14.7 psi Practice: convert 29.5 inches of Hg to mm Hg, torr, atm and psi. (Hint – do you remember how to convert inches to cm?)

  13. MANOMETERS • Definition: A manometer is a scientific instrument used to measure gas pressures. Open manometers measure gas pressure relative to atmospheric pressure. A mercury or oil manometer measures gas pressure as the height of a fluid column of mercury or oil that the gas sample supports.

  14. closed-end open-end Figure 5.4 from 4th ed. Pgas = ht of Hg, Dh Two types of manometers

  15. The Simple Gas Laws Memorize Boyle’s, Charles’, Avogadro’s, and Amonton’s (Gay-Lussac’s) laws

  16. Figure 5.4 The relationship between the volume and pressure of a gas. Boyle’s Law See notes below or look in textbook.

  17. Boyle’s Law As shown in Fig 5.4, as Pressure  the Volume , which means V ~ 1/P (inversely proportional) P*V is a constant P1V1 = Cb and P2V2 = Cb Therefore, P1V1 = P2V2 Example: gaseous CO2 at 55 torr occupies 125mL. It is transferred to new flask and P increases to 78 mm of Hg. What is the new volume? First think about whether V inc or dec? Dec. Why? P Inc. V2 = P1V1/P2 = 55 mm*125 mL/78 mm = 88 mL

  18. Figure 5.5 The relationship between the volume and temperature of a gas. Notice all lines extrapolate to -273.15 oC! What is that? Charles’ Law See notes below or in textbook.

  19. Charles’ Law As shown in Fig 5.5, as Temperature  the Volume , i.e., directly proportional. Therefore: V/T = Cc BUT Temperature has to be in positive increments, never negative. This is why the absolute scale, Kelvin, was developed. V1/T1 = Cc. V2/T2 = Cc. Therefore, V1/T1 = V2/T2 Example: A balloon is filled with He at 4.5 L and 25.0°C. Take it outside in winter in Chicago, -10.0°C. What is new volume? First: is new V larger or smaller? Smaller, because T dec. V2 = V1*T2/T1 = 4.5 L*263 K/298 K = 4.0 L

  20. Figure 5.6 An experiment to study the relationship between the volume and amount of a gas comparing 0.1 mol and 0.2 mol. Avogadro’s Law At a given external P & T, a given amt of CO2 is placed in tube. When it changes from solid to gas, it pushes the piston up until Pgas = Patm. When twice as much CO2 is used, twice the volume is occupied, showing V is proportional to amt of gas.

  21. Avogadro’s Law As shown in Fig 5.6, as the quantity of a gas increases, the volume increases. V1/n1 = Ca = V2/n2 therefore, V1/n1 = V2/n2 n = moles, which we know represents #'s of molecules

  22. Amonton’s Law (also Gay-Lussac) P & T are directly related P1/T1 = P2/T2 if n and V are constant Does T have to be in Kelvin? Why?

  23. Practice with the Simple Gas Laws Work on problems 8a, 16a and 18. 8a: Convert 76.8 cm Hg to atm 16a: What is the effect on the volume of 1 mole of an ideal gas if the pressure is reduced by a factor of four at constant temperature? 18: A 93 L sample of dry air is cooled from 145oC to -22oC at constant pressure. What is its final volume?

  24. Video Lecture on Ideal Gas Law • Caveat: you will use R = 0.082057 L-atm/mol-K, because we want 5 sig figs for many of our calcs. You will also use more sig figs for temperatures, so add 273.15 to convert to Kelvin. (Also her final answer’s sig figs should be at least 3.)

  25. IDEAL GAS LAW: Combining the work of Boyle, Charles, Amonton, and Avogadro: V=Cb/P V=Cc*T V=Ca*n V=(Cb*Cc*Ca)*(n*T/P) V = R*nT/P which rearranges to PV = nRT R = 0.082057 L-atm/mol-K (memorize)

  26. EXAMPLE of Ideal Gas Law A big balloon is filled with 1300. moles H2, T is 20.0°C, P = 750. torr. What is Volume of balloon? P = 750. torr*(1 atm/760 torr) =0.98684 atm V = nRT/P = 1300. mol* 0.082057 * 293.15K 0.98684 atm = 3.17 x 104 L

  27. Standard molar volume. Figure 5.7 One mole of any ideal gas occupies 22.414 L at STP. Notice these three gases behave ideally. Masses are different, therefore density will be different.

  28. STP: standard temperature & pressure Defined as 273.15 K and 1.0000 atm At STP, one mole of any gas occupies 22.414 L, called the standard molar volume of a gas Try V = nRT/P using this STP for one mole of gas

  29. Simple combined gas law: P1V1/T1 = P2V2/T2 Works if moles of a gas are constant.

  30. Example If you have 20.0 L of He at 150. atm and 30.0°C, then how many balloons can be filled with 5.00 L each at 755 mm and 22.0°C? (Really asking for V2 at new T & P, note that n is constant.) Convert P2 to atm: 755 mm (1 atm/760 mm) = 0.99342 atm V2 = P1V1T2/P2T1 = 150atm*20.0L*295.15K 0.99342atm*303.15K = 2940 L 2940L/5.00L = 588 balloons

  31. MY SUPER-COMBINED GAS LAW: Rearrange R = P1V1 = P2V2 n1T1 n2T2 Then "uncombine" these to get the simple gas laws: @ const T & n : cross out all n & T to get Boyle's law @ const P & n : cross out all n & P to get Charles' law @ const P & T : cross out all P & T to get Avogadro's law @ const V & n : cross out all V & n to get Amonton’s Law

  32. IDEAL AND COMBINED GAS LAWS: Practice: A sample of O2 gas occupies 255.5 mL at 25.5oC and 757.7 torr. (First determine the number of moles present.) If the temperature drops to 15.5oC and the pressure drops to 745.0 torr, and you add 0.00100 moles of gas, what will the new volume read? (You should get 275.3 mL)

  33. Practice with Combined Gas Laws and Ideal Gas Law Work on problems 20, 24 and 83. 20: Calculate V of a sample of CO that is at -14oC and 367 torr if it first occupies 3.65 L at 298K and 745 torr. 24: A 75.0 g sample of N2O is held in a 3.1 L vessel at 115oC. What is its pressure? 83: (look in book)

  34. The Density of a Gas Recall that Density = m/V Recall we can find moles: n = m/M PV = nRT; substitute PV = (m/M)RT Rearrange: m/V = M P/ RT or D = M P/RT • The density of a gas is directly proportional to its molar mass. • The density of a gas is inversely proportional to the temperature. Rearrange again to find molar mass: M = DiRT/P

  35. Example: An unknown gas has a density of 1.429 g/L at STP. Find its molar mass. (Hard way & easy way if at STP.) MA = DRT/P = 1.429 g/L*0.082057*273.15K/1 atm = 32.03 g/mol or: MA = 1.429 g/L * 22.414 L/mol = 32.03 g/mol

  36. Mixtures of Gases and Dalton’s Law of Partial Pressures • Gases mix homogeneously in any proportions. • Each gas in a mixture behaves as if it were the only gas present. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + …

  37. Mixtures of Gases and Mole Fraction: cA = nA/nT = the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture PO2= nO2RT/V = nO2 = cO2 PT nTRT/V nT Therefore, PO2 = cO2 * PT

  38. Example: We react N2 and H2 to make NH3. We start with 10.3 moles of H2 and 3.71 moles of N2. We use up the H2 to make 6.87 mol of NH3. Find the total pressure in the vessel and the partial pressure of each gas, assuming that the product is collected in a 125 L tank at 25.0°C. N2(g) + 3 H2(g) 2 NH3(g) Step 1: Determine moles NH3 produced and also find how much excess N2 remains: 10.3 mol H2 * 2NH3/3H2 = 6.87 mol NH3 produced 10.3 mol H2 * 1N2/3H2 = 3.43 moles N2 used 3.71moli - 3.43 molused = 0.28 moles N2 remaining Add that to moles of ammonia produced for total moles: nT = mol NH3 + mol N2 =6.87 + 0.28 = 7.15 moles total in tank

  39. Example continued Step 2: Determine total pressure, then pressure of each gas. PT=nTRT/V = 7.15 mol * 0.082057 * 298K/125L = 1. 399 atm Partial pressure for each gas could be determined using moles of each in the ideal gas law, or by using mole fraction, as shown below. cN2 = nN2/nT = 0.28/7.15 = 0.0392, cNH3 = 0.961 (from 1.000-.0392) PN2 = cN2*PT = 0.0392*1.399 = 0.0548 atm PNH3 = cNH3*PT 0.961*1.399 = 1.344 atm

  40. Figure 5.10 Collecting a water-insoluble gaseous reaction product and determining its pressure.

  41. Example: We collected 0.500 L of hydrogen gas over water at a room temperature of 26.0°C and atmospheric pressure of 745 torr. How many moles of hydrogen gas were collected? Look up PH2O at 26.0oC and find it is 25.2 mm Hg. Find PH2: Pbar = PH2 + PH2O 745 torr = PH2 + 25.2 torr PH2 = 720. torr Convert: 720. torr (1 atm/760 torr) = 0.94737 atm Rearrange Ideal Gas Law: n = PV/RT n = (0.94737 atm) * 0.500 L 0.082057 L.atm/mol.K*299.15K n = 0.0193 mol of H2

  42. Practice Problems Work on problems 32, 34, 38, and 72ab. 32: What is the density of Freon gas (CFCl3) at 120.oC and 1.50 atm? 34: The density of a noble gas is 2.71 g/L at 3.00 atm and 0.00oC. Name the gas. 38: A 355 mL container holds 0.146 g of Ne and some amt of Ar, at 35.0oC and total pressure of 626 torr. Calculate moles of Ar present.

  43. STOICHIOMETRY OF GASES: Stoichiometry problems that you will love to do! At STP, one mole of any gas occupies 22.414 L. WE CAN USE THIS A LOT! This means that equal volumes of gases at STP have the same # of moles!!!! 11.2 L of N2 = 0.5 moles = 11.2 L of O2 = 0.5 moles = 11.2 L of He, etc. We can relate volumes through stoich coeff just like moles

  44. Stoichiometry of Gases Example: N2(g) + 3 H2(g) 2 NH3(g) (over Fe catalyst at 500°C) Given 355 L of hydrogen, how many L of nitrogen required? 355L H2 * 1 N2/3H2 = 118 L How many L of ammonia produced? 355L H2 * 2NH3/3H2 = 237 L

  45. Stoichiometry at STP: regular? Example 2: 12.0 g Zn react with excess sulfuric acid, how many liters of gas will you have at STP? Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) 12.0 g/(65.4g/mol) = 0.183 mol Zn * 1 H2/1 Zn = 0.183 mol H2(g) 0.183 mol gas * 22.414 L/mol = 4.11 L Note: if at other conditions, must use Ideal Gas Law to find volume, etc.

  46. You Try “Regular” Stoichiometry: During ammonia production, with 355 L of hydrogen @ 25.0°C & 542 mm of Hg plus 105 L of nitrogen @ 20.0°C & 645 torr. Find volume and mass of ammonia produced at STP. (154 L, 117 g)

  47. Postulates of the Kinetic-Molecular Theory Postulate 1: Particle Volume The volume of an individual gas particle is so small compared to the volume of its container, that the gas particles are considered to have mass, but essentially no volume. Postulate 2: Particle Motion Gas particles are in constant, rapid, random, straight-line motion except when they collide with each other or with the container walls. Postulate 3: Particle Collisions Collisions are elastic therefore the total kinetic energy(Ek) of the particles is constant (although energy can be transferred).

  48. Distribution of molecular speeds at three temperatures. Figure 5.12 See note below or in textbook.

  49. Figure at end of section 5.6 Diffusion of a gas particle through a space filled with other particles. distribution of molecular speeds mean free path collision frequency A bizarre and dangerous molecular highway!

  50. GRAHAM'S LAWS OF DIFFUSION AND EFFUSION: Diffusion is the gradual mixing of molecules of 2 or more gases in a container, owing to the continual molecular motion. Effusion is the escape of molecules through a tiny hole in a pressurized container into a vacuum or lower pressure system. Rate of effusion is related inversely to the molar mass of the gas, because of its relation to average speed of the gas molecules. Rategas1/Rategas2 = (MAgas2/MAgas1)½

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