無機化學課件：08 Ch11 Gases

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 11GasesGases1111.1 Properties of Gases11.2 The Kinetic Molecular Theory of GasesMolecular SpeedDiffusion and Effusion11.3 Gas PressureDefinition and Units of PressureCalculation of PressureMeasurement of Pressure11.4 The Gas LawsBoyles Law: Th

2、e Pressure-Volume RelationshipCharless and Gay-Lussacs Law: The Temperature-Volume RelationshipAvogadros Law: The Amount-Volume RelationshipThe Gas Laws and Kinetic Molecular TheoryThe Combined Gas Law: The Pressure-Temperature-Amount-Volume Relationship11.5 The Ideal Gas EquationApplications of the

3、 Ideal Gas EquationGases1111.6 Real GasesFactors That Cause Deviation from Ideal BehaviorThe van der Waals Equationvan der Waals Constants11.7 Gas MixturesDaltons Law of Partial PressuresMole Fractions11.6 Reactions with Gaseous Reactants and ProductsCalculating the Required Volume of a Gaseous Reac

4、tantDetermining the Amount of Reactant Consumed Using Change in PressureUsing Partial Pressures to Solve ProblemsProperties of GasesGases differ from solids and liquids in the following ways: A sample of gas assumes both the shape and volume of the container. Gases are compressible. The densities of

5、 gases are much smaller than those of liquids and solids and are highly variable depending on temperature and pressure. Gases form homogeneous mixtures (solutions) with one another in any proportion.11.1The Kinetic Molecular TheoryThe kinetic molecular theory explains how the molecular nature of gas

6、es gives rise to their macroscopic properties.The basic assumptions of the kinetic molecular theory are as follows: A gas is composed of particles that are separated by large distances. The volume occupied by individual molecules is negligible. Gas molecules are constantly in random motion, moving i

7、n straight paths, colliding with perfectly elastic collisions. Gas molecules do not exert attractive or repulsive forces on one another. The average kinetic energy of a gas molecules in a sample is proportional to the absolute temperature: 11.2The Kinetic Molecular TheoryGases are compressible becau

8、se molecules in the gas phase are separated by large distances (assumption 1).Pressure is the result of the collisions of gas molecules with the walls of their container (assumption 2).Decreasing volume increases the frequency of collisions.Pressure increases as collision frequency increases.The Kin

9、etic Molecular TheoryHeating a sample of gas increases its average kinetic energy (assumption 4).Gas molecules must move faster.Faster molecules collide more frequently and at a greater speed.Pressure increases as collision frequency increases.The Kinetic Molecular TheoryThe total kinetic energy of

10、a mole of gas is equal to:The average kinetic energy of one molecule is:For one mole of gas:m is the mass is the mean square speedrearrange and Take the square root (m x NA = M)The Kinetic Molecular TheoryThe root-mean-square (rms) speed (urms) is the speed of a molecule with the average kinetic ene

11、rgy in a gas sample.urms is directly proportional to temperatureThe Kinetic Molecular TheoryThe root-mean-square (rms) speed (urms) is the speed of a molecule with the average kinetic energy in a gas sample.urms is inversely proportional to the square root of M.The Kinetic Molecular TheoryWhen two g

12、ases are at the same temperature, it is possible to compare the the urms values of the different gases. Worked Example 11.1Strategy Use equation and the molar masses of He and CO2 to determine the ratio of their root-mean-square speeds. When solving a problem such as this, it is generally best to la

13、bel the lighter of the two molecules as molecule 1 and the heavier as molecule 2. This ensures that the result will be greater than 1, which is relatively easy to interpret.Determine how much faster a helium atom moves, on average, than a carbon dioxide molecule at the same temperature.Solution The

14、molar masses of He and CO2 are 4.003 and 44.02 g/mol, respectively.On average, He atoms move 3.316 times as fast as CO2 molecules at the same temperature.44.02 g1 mol= 3.3164.003 g1 molurms(He)urms(CO2)=Think About It Remember that the relationship between molar mass and molecular speed is reciproca

15、l. A CO2 molecule has approximately 10 times the mass of an He atom. Therefore, we should expect an He atom, on average, to be moving approximately 10 times (3.2 times) as fast as a CO2 molecule.The Kinetic Molecular TheoryDiffusion is the mixing of gases as the result of random motion and frequent

16、collisions.擴散Effusion is the escape of gas molecules from a container to a region of vacuum.溢散The Kinetic Molecular TheoryGrahams law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass.The Kinetic Molecular TheoryDetermine the molar

17、mass and identity of a diatomic gas that moves 4.67 times as fast as CO2.Solution:Step 1: Use the equation below to determine the molar mass of the unknown gas:urms(unknown gas) = 4.67 x urms(CO2)The gas must be H2.Gas PressurePressure is defined as the force applied per unit area:The SI unit of for

18、ce is the newton (N), whereThe SI unit of pressure is the pascal (Pa), defined as 1 newton per square meter.1 N = 1 kgm/s21 Pa = 1 N/m211.3Gas PressureGas PressureA barometer is an instrument that is used to measure atmospheric pressure.Standard atmospheric pressure (1 atm) was originally defined as

19、 the pressure that would support a column of mercury exactly 760 mm high.h = height in md = is the density in kg/m3g = is the gravitational constant (9.80665 m/s2)Gas PressureA barometer is an instrument that is used to measure atmospheric pressure.1 atm*= 101,325 Pa= 760 mmHg*= 760 torr*= 1.01325 b

20、ar= 14.7 psi* Represents an exact numberGas PressureA manometer is a device used to measure pressures other than atmospheric pressure. Worked Example 11.2Strategy Use P = hdg to calculate pressure. Remember that the height must be expressed in meters and density must be expressed in kg/m3.Calculate

21、the pressure exerted by a column of mercury 70.0 cm high. Express the pressure in pascals, in atmospheres, and in bars. The density of mercury is 13.5951 g/cm3.Solution h = 70.0 cm = 0.700 md = g = 9.80665 m/s21 m100 cm13.5951 gcm31 kg1000 g100 cm1 m3= 1.35951104 kg/m3 Worked Example 11.2 (cont.)Sol

22、ution pressure = 0.700 m = 9.33104 kg/ms2 9.33104 Pa 9.33104 Pa = 0.921 atm 0.921 atm = 0.933 bar1.35951104 kgm39.80665 ms21 atm101,325 Pa1.01325 bar1 atmThink About It Make sure your units cancel properly in this type of problem. Common errors include forgetting to express height in meters and dens

23、ity in kg/m3. You can avoid these errors by becoming familiar with the value of atmospheric pressure in the various units. A column of mercury slightly less than 760 mm is equivalent to slightly less than 101,325 Pa, slightly less than 1 atm, and slightly less than 1 bar.The Gas LawsBoyles law state

24、s that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas.11.4(a)(b)(c)P (mmHg)76015202280V (mL)1005033P1V1 = P2V2at constant temperatureThe Gas LawsCalculate the volume of a sample of gas at 5.75 atm if it occupies 5.14 L at 2.49 atm.

25、(Assume constant temperature.)Solution:Step 1: Use the relationship below to solve for V2:P1V1 = P2V2 Worked Example 11.3Strategy Use P1V1 = P2V2 to solve for V2.If a skin diver takes a breath at the surface, filling his lungs with 5.82 L of air, what volume will the air in his lungs occupy when he

26、dives to a depth where the pressure of 1.92 atm? (Assume constant temperature and that the pressure at the surface is exactly 1 atm.)Solution P1 = 1.00 atm, V1 = 5.82 L, and P2 = 1.92 atm.V2 = = = 3.03 L P1 V1P21.00 atm 5.82 L1.92 atmThink About It At higher pressure, the volume should be smaller. T

27、herefore, the answer makes sense.The Gas LawsCharless and Gay-Lussacs law, (or simply Charless Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.Higher temperatureLower temperatureThe Gas LawsCharless and Gay-Lussacs

28、law, (or simply Charless Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.at constant pressure Worked Example 11.4Strategy Use V1/T1 = V2/T2 to solve for V2. Remember that temperatures must be expressed in kelvin.A s

29、ample of argon gas that originally occupied 14.6 L at 25C was heated to 50.0C at constant pressure. What is its new volume?Solution T1 = 298.15 K, V1 = 14.6 L, and T2 = 323.15 K.V2 = = = 15.8 L V1 T2T114.6 L 323.15 K298.15 KThink About It When temperature increases at constant pressure, the volume o

30、f a gas sample increases.The Gas LawsAvogadros law states that the volume of a sample of gas is directly proportional to the number of moles in the sample at constant temperature and pressure.at constant temperature and pressure Worked Example 11.5Strategy Apply Avogadros law to determine the volume

31、 of a gaseous product.If we combine 3.0 L of NO and 1.5 L of O2, and they react according to the balanced equation 2NO(g) + O2(g) 2NO2(g), what volume of NO2 will be produced? (Assume that the reactants and products are all at the same temperature and pressure.)Solution Because volume is proportiona

32、l to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometric amounts.According to the balanced equation, the volume of NO2 formed will be equal to the volume

33、 of NO that reacts. Therefore, 3.0 L of NO2 will form.Think About It When temperature increases at constant pressure, the volume of a gas sample increases.The Gas LawsA sample of gas originally occupies 29.1 L at 0.0C. What is its new volume when it is heated to 15.0C? (Assume constant pressure.)Sol

34、ution:Step 1:Use the relationship below to solve for V2: (Remember that temperatures must be expressed in kelvin.The Gas LawsWhat volume in liters of water vapor will be produced when 34 L of H2 and 17 L of O2 react according to the equation below:2H2(g) + O2(g) 2H2O(g)Assume constant pressure and t

35、emperature.Solution:Step 1:Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometeric amounts.34 L of H2O will form.The Gas L

36、awsHeating at constant volume: pressure increasesCooling at constant volume:pressure decreasesThe Gas LawsHeating at constant pressure: volume increasesCooling at constant pressure:volume decreasesThe Gas LawsThe presence of additional molecules causes an increase in pressure.The Gas LawsThe combine

37、d gas law can be used to solve problems where any or all of the variables changes. The Gas LawsThe volume of a bubble starting at the bottom of a lake at 4.55C increases by a factor of 10 as it rises to the surface where the temperature is 18.45C and the air pressure is 0.965 atm. Assume the density

38、 of the lake water is 1.00 g/mL. Determine the depth of the lake.Solution:Step 1:Use the combined gas law to find the pressure at the bottom of the lake; assume constant moles of gas.The Gas LawsThe volume of a bubble starting at the bottom of a lake at 4.55C increases by a factor of 10 as it rises

39、to the surface where the temperature is 18.45C and the air pressure is 0.965 atm. Assume the density of the lake water is 1.00 g/mL. Determine the depth of the lake.Solution:Step 2:Use the equation P = hdg to determine the depth of the lake. The pressure represents the difference in pressure from th

40、e surface to the bottom of the lake. Pressure exerted by the lake = Pbottom of lake PairP = 9.19 atm 0.965 atm = 8.225 atmThe Gas LawsThe volume of a bubble starting at the bottom of a lake at 4.55C increases by a factor of 10 as it rises to the surface where the temperature is 18.45C and the air pr

41、essure is 0.965 atm. Assume the density of the lake water is 1.00 g/mL. Determine the depth of the lake.Solution:Step 2 (Cont): Convert pressure to pascals and density to kg/m3. P = hdg 833,398 Pa = h(1000 kg/m3)(9.81 m/s2)h = 85.0 m Worked Example 11.6Strategy In this case, because there is a fixed

42、 amount of gas, we use P1V1/T1 = P2V2/T2. The only value we dont know is V2. Temperatures must be expressed in kelvins. We can use any units of pressure, as long as we are consistent.If a child releases a 6.25-L helium balloon in the parking lot of an amusement park where the temperature is 28.50C a

43、nd the air pressure is 757.2 mmHg, what will the volume of the balloon be when it has risen to an altitude where the temperature is -34.35C and the air pressure is 366.4 mmHg?Think About It Note that the solution is essentially multiplying the original volume by the ratio of P1 and P2, and by the ra

44、tio of T2 to T1. The effect of decreasing external pressure is to increase the balloon volume. The effect of decreasing temperature is to decrease the volume. In this case, the effect of decreasing pressure predominates and the balloon volume increases significantly.Solution T1 = 301.65 K, T2 = 238.

45、80 K.V2 = = = 10.2 L P1T2V1P2T1757.2 mmHg 238.80 K 6.25 L366.4 mmHg 301.65 KThe Ideal Gas EquationThe gas laws can be combined into a general equation that describes the physical behavior of all gases.11.5Boyles lawAvogadros lawCharless lawPV = nRTrearrangementR is the proportionality constant, call

46、ed the gas constant.The Ideal Gas EquationThe ideal gas equation (below) describes the relationship among the four variables P, V, n, and T.PV = nRTAn ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.The Ideal Gas

47、EquationThe gas constant (R) is the proportionality constant and its value and units depend on the units in which P and V are expressed.PV = nRTThe Ideal Gas EquationStandard Temperature and Pressure (STP) are a special set of conditions where:Pressure is 1 atmTemperature is 0C (273.15 K)The volume

48、occupied by one mole of an ideal gas is then 22.41 L:PV = nRT Worked Example 11.7Strategy Convert the temperature in C to kelvins, and use the ideal gas equation to solve for the unknown volume.Calculate the volume of a mole of ideal gas at room temperature (25C) and 1 atm.Think About It With the pr

49、essure held constant, we should expect the volume to increase with increased temperature. Room temperature is higher than the standard temperature for gases (0C), so the molar volume at room temperature (25C) should be higher than the molar volume at 0Cand it is.Solution The data given are n = 1 mol

50、, T = 298.15 K, and P = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = 0.08206 Latm/Kmol in order to solve for volume in liters.V = = = 24.5 L nRTP(1 mol)(0.08206 Latm/Kmol)(298.15 K)1 atmMolar Volume of Some Common Gases at STP (00C and 1 atm)GasMolar Volume(L/mol)Condensati

51、on Point(0C) HeH2NeIdeal gasArN2O2COCl2NH322.43522.43222.42222.41422.39722.39622.39022.38822.18422.079-268.9-252.8-246.1-185.9-195.8-183.0-191.5-34.0-33.4The behavior of several real gases with increasing external pressure.The Ideal Gas EquationUsing algebraic manipulation, it is possible to solve f

52、or variables other than those that appear explicitly in the ideal gas equation.PV = nRTd is the density (in g/L)M is the molar mass (in g/mol)The Ideal Gas EquationWhat pressure would be required for helium at 25C to have the same density as carbon dioxide at 25C and 1.00 atm?Solution:Step 1:Use the

53、 equation below to calculate the density of CO2 at25C and 1 atm.The Ideal Gas EquationWhat pressure would be required for helium at 25C to have the same density as carbon dioxide at 25C and 1 atm?Solution:Step 2:Use the density of CO2 found in step 1 to calculate thepressure for He at 25C.P = 11.0 a

54、tm Worked Example 11.8Strategy Use d = PM/RT to solve for density. Because the pressure is expressed in atm, we should use R = 0.08206 Latm/Kmol. Remember the express temperature in kelvins.Carbon dioxide is effective in fire extinguishers partly because its density is greater than that of air, so C

55、O2 can smother the flames by depriving them of oxygen. (Air has a density of approximately 1.2 g/L at room temperature and 1 atm.) Calculate the density of CO2 at room temperature (25C) and 1.0 atm.Think About It The calculated density of CO2 is greater than that of air under the same conditions (as

56、 expected). Although it may seem tedious, it is a good idea to write units for each and every entry in a problem such as this. Unit cancellation is very useful for detecting errors in your reasoning or your solution setup.Solution The molar mass of CO2 is 44.01 g/mol.d = =PMRT(1 atm)(44.01 g/mol)0.0

57、8206 Latm/Kmol)(298.15 K)= 24.5 L Real GasesThe van der Waals equation is useful for gases that do not behave ideally.11.6Experimentally measured pressure corrected pressure termcorrected volume termContainer volumeReal GasesThe van der Waals equation is useful for gases that do not behave ideally.R

58、eal GasesCalculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 32C using (a) the ideal gas equation and(b) the van der Waals equation.Solution:Step 1:Use the ideal gas equation to calculate the pressure of O2.PV = nRTReal GasesCalculate the pressure exerted by 0.35 mole

59、of oxygen gas in a volume of 6.50 L at 32C using (a) the ideal gas equation and(b) the van der Waals equation.Solution:Step 2:Use table 11.6 to find the values of a and b for O2.Real GasesCalculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 32C using (a) the ideal gas e

60、quation and (b) the van der Waals equation.SolutionStep 3:Use the van der Waals equation to calculate P.P = 1.3 atm Worked Example 11.10Strategy (a) Use the ideal gas equation, PV = nRT. (b) Use (P + an2/V2)(V nb) = nRT and a and b values for NH3 from Table 11.5A sample of 3.50 moles of NH3 gas occu