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Chapter 5 Gases.

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1 Chapter 5 Gases

2 Chapter 5: Gases 5.1 Early Experiments
The gas laws of Boyle, Charles, and Avogadro The Ideal Gas Law Gas Stiochiometry Dalton’s Law of Partial Pressures The Kinetic molecular Theory of Gases Effusion and Diffusion Collisions of Gas Particles with the Container Walls Intermolecular Collisions Real Gases Chemistry in the Atmosphere

3 Hurricanes, such as this one off the coast of Florida, are evidence of the powerful forces present in the earth's atmosphere.

4 Important Characteristics of Gases
1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase. 2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases. 3) Gases have low viscosity Gases flow much easier than liquids or solids. 4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater. 5) Gases are infinitely miscible Gases mix in any proportion such as in air, a mixture of many gases.

5 Substances that are Gases under Normal Conditions
Substance Formula MM(g/mol) Helium He Neon Ne Argon Ar Hydrogen H Nitrogen N Nitrogen Monoxide NO Oxygen O Hydrogen Chloride HCL Ozone O Ammonia NH Methane CH

6 Some Important Industrial Gases
Name - Formula Origin and use Methane (CH4) Natural deposits; domestic fuel Ammonia (NH3) From N2 + H2 ; fertilizers, explosives Chlorine (Cl2) Electrolysis of seawater; bleaching and disinfecting Oxygen (O2) Liquified air; steelmaking Ethylene (C2H4) High-temperature decomposition of natural gas; plastics

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8 Pressure of the Atmosphere
Called “Atmospheric pressure,” or the force exerted upon us by the atmosphere above us. A measure of the weight of the atmosphere pressing down upon us. Measured using a Barometer! - A device that can weigh the atmosphere above us! Force Area Pressure =

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10 Figure 5.1: A torricellian barometer.

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13 Construct a Barometer using Water!
Density of water = 1.00 g/cm3 Density of Mercury = 13.6 g/cm3 Height of water column = Hw Hw = Height of Hg x Density of Mercury Hw = 760 mm Hg x 13.6/1.00 = 1.03 x 104 mm Hw = 10.3 m = __________ ft HeightWater HeightMercury DensityMercury DensityWater = Density of Water

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15 Common Units of Pressure
Unit Atmospheric Pressure Scientific Field Used Pascal (Pa); x 105 Pa SI unit; physics, kilopascal (kPa) kPa chemistry Atmosphere (atm) atm Chemistry Millimeters of mercury mmHg Chemistry, medicine (mmHg) biology Torr torr Chemistry Pounds per square inch lb/in Engineering (psl or lb/in2) Bar bar Meteorology, chemistry

16 Converting Units of Pressure
Problem: A chemist collects a sample of Carbon dioxide from the decomposition of Lime stone (CaCO3) in a closed end manometer, the height of the mercury is mm Hg. Calculate the CO2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmHg, so we use the conversion factors from Table 5.2(p.178) to find the pressure in the other units. Solution: converting from mmHg to torr: 1 torr 1 mm Hg PCO2 (torr) = mm Hg x = torr converting from torr to atm: 1 atm 760 torr PCO2( atm) = torr x = atm converting from atm to kPa: kPa 1 atm PCO2(kPa) = atm x = _____________ kPa

17 Figure 5.2: A simple manometer, a device for measuring the pressure of a gas in a container.

18 Figure 5.3: A J-tube similar to the one used by Boyle.

19 Boyle’s Law : P - V relationship
Pressure is inversely proportional to Volume P = or V = or PV=k Change of Conditions Problems if n and T are constant ! P1V1 = k P2V2 = k’ k = k’ Then : P1V1 = P2V2 k V k P

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21 Figure 5.4: Plotting Boyle’s data from Table 5.1.

22 Figure 5.5: Plot of PV versus P for several gases.

23 Applying Boyles Law to Gas Problems
Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm3, what will be the volume if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm3 to ml and then to liters, then we do the pressure change to obtain the final volume! Solution: V1 (cm3) P1 = 1.23 atm P2 = 3.16 atm V1 = 15.8 cm V2 = unknown T and n remain constant 1cm3 = 1 mL V1 (ml) 1 mL 1 cm3 1 L 1000mL 1000mL = 1L V1 = 15.8 cm3 x x = L V1 (L) x P1/P2 P1 P2 1.23 atm 3.16 atm V2 = V1 x = L x = ________ L V2 (L)

24 Boyle’s Law - A gas bubble in the ocean!
A bubble of gas is released by the submarine “Alvin” at a depth of 6000 ft in the ocean, as part of a research expedition to study under water volcanism. Assume that the ocean is isothermal( the same temperature through out ),a gas bubble is released that had an initial volume of 1.00 cm3, what size will it be at the surface at a pressure of 1.00 atm?(We will assume that the density of sea water is g/cm3, and use the mass of Hg in a barometer for comparison!) Initial Conditions Final Conditions V 1 = 1.00 cm3 V 2 = ? P 1 = ? P 2 = 1.00 atm

25 Calculation Continued
m 1 ft 100 cm 1 m 1.026 g SH2O 1 cm3 Pressure at depth = 6 x 103 ft x x x Pressure at depth = 187, g pressure from SH2O For a Mercury Barometer: 760 mm Hg = 1.00 atm, assume that the cross-section of the barometer column is 1 cm2. The mass of Mercury in a barometer is: 1.00 atm 760 mm Hg 10 mm 1 cm Area 1 cm2 1.00 cm3 Hg 13.6 g Hg Pressure = x x x x 187,635 g = Pressure = _____ atm Due to the added atmospheric pressure = ___ atm! V1 x P1 P2 1.00 cm3 x ____ atm 1.00 atm V2 = = = ____ cm3 = liters

26 Boyle’s Law : Balloon A balloon has a volume of 0.55 L at sea level
(1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant(which obviously is not true), what is the final volume of the balloon? P1 = 1.0 atm P2 = 0.40 atm V1 = 0.55 L V2 = ? V2 = V1 x P1/P2 = (0.55 L) x (1.0 atm / 0.40 atm) V2 = __________ L

27 Figure 5.6: PV plot verses P for 1 mole of ammonia

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31 Charles Law - V - T- relationship
Temperature is directly related to volume T proportional to Volume : T = kV Change of conditions problem: Since T/V = k or T1 / V1 = T2 / V or: T1 = V1 x T2 V2 T1 V1 T2 = V2 Temperatures must be expressed in Kelvin to avoid negative values!

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34 Figure 5.7: Plots of V versus T (c) for several gases.

35 Figure 5. 8: Plots of V versus T as in Fig. 5
Figure 5.8: Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature.

36 Charles Law Problem A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. V1 = 3.20 L T1 = 125oC = 398 K V2 = 1.54 L T2 = ? T2 = T1 x ( V2 / V1) T2 = 398 K x =______K T2 = ___ K oC = K = ______ - 273 oC = ______oC 1.54 L 3.20 L

37 Charles Law Problem - I A balloon in Antarctica in a building is at room temperature ( 75o F ) and has a volume of 20.0 L . What will be its volume outside where the temperature is -70oF ? V1 = 20.0 L V2 = ? T1 = 75o F T2 = -70o F o C = ( o F - 32 ) 5/9 T1 = ( )5/9 = 23.9o C K = 23.9o C = __________ K T2 = ( ) 5/9 = o C K = o C = ___________ K

38 Antarctic Balloon Problem - II
V1 / T1 = V2 / T V2 = V1 x ( T2 / T1 ) V2 = 20.0 L x V2 = __________ L The Balloon shrinks from 20 L to 15 L !!!!!!! Just by going outside !!!!! 216.4 K 297.0 K

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40 Applying the Temperature - Pressure Relationship (Amonton’s Law)
Problem: A copper tank is compressed to a pressure of 4.28 atm at a temperature of oF. What will be the pressure if the temperature is raised to 95.6 oC? Plan: The volume of the tank is not changed, and we only have to deal with the temperature change, and the pressure, so convert to SI units, and calculate the change in pressure from the Temp.and Pressure change. Solution: P P2 T T2 = T1 = (0.185 oF oF)x 5/9 = oC T1 = oC K = K T2 = oC K = K P2 = P1 x = ? T2 T1 368.8 K K P2 = 4.28 atm x = _______ atm

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42 Avogadro’s Law - Amount and Volume
The Amount of Gas (Moles) is directly proportional to the volume of the Gas. n  V or n = kV For a change of conditions problem we have the initial conditions, and the final conditions, and we must have the units the same. n1 = initial moles of gas V1 = initial volume of gas n2 = final moles of gas V2 = final volume of gas n1 V1 n2 V2 V1 V2 = or: n1 = n2 x

43 Avogadro’s Law: Volume and Amount of Gas
Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes in the atmosphere, if the volume of 2.67 g of SF6 at atm and 28.5 oC is 2.93 m3, what will be the mass of SF6 in a container whose volume is 543.9 m3 at atm and 28.5 oC? Plan: Since the temperature and pressure are the same it is a V - n problem, so we can use Avogadro’s Law to calculate the moles of the gas, then use the molecular mass to calculate the mass of the gas. Solution: Molar mass SF6 = g/mol 2.67g SF6 146.07g SF6/mol = mol SF6 V2 V1 n2 = n1 x = mol SF6 x = 3.40 mol SF6 543.9 m3 2.93 m3 mass SF6 = 3.40 mol SF6 x g SF6 / mol = __________ g SF6

44 Volume - Amount of Gas Relationship
Problem: A balloon contains 1.14 moles(2.298g H2) of Hydrogen and has a volume of L. What mass of Hydrogen must be added to the balloon to increase it’s volume to Liters. Assume T and P are constant. Plan: Volume and amount of gas are changing with T and P constant, so we will use Avogadro’s law, and the change of conditions form. Solution: V1 = L V2 = L T = constant P = constant n1 = 1.14 moles H2 n2 = 1.14 moles + ? moles n n2 V V2 V2 V1 L 28.75 L = n2 = n1 x = 1.14 moles H2 x mass = moles x molecular mass mass = 4.46 moles x g / mol mass = ___________ g H2 gas n2 = moles = 4.46 moles added mass = 8.99g g = 6.69g

45 Change of Conditions, with no change in the amount of gas !
= constant Therefore for a change of conditions : T T2 P x V T P1 x V1 P2 x V2 =

46 Change of Conditions :Problem -I
A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10 ml what is the pressure? P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm P2 = ? V1 = 45.9 L V2 = 3.10 ml = L T1 = 25 oC = 298 K T2 = 155 oC = 428 K

47 Change of Condition Problem I : continued
P1 x V1 P2 x V2 = T1 T2 P2 ( L) ( atm) ( 45.9 L) = ( 298 K) ( 428 K) ( 428 K) ( atm) ( 45.9 L) P2 = = ________ atm ( 298 K) ( L)

48 Change of Conditions : Problem II
A weather balloon is released at the surface of the earth. If the volume was 100 m3 at the surface ( T = 25 oC, P = 1 atm ) what will its volume be at its peak altitude of 90,000 ft where the temperature is - 90 oC and the pressure is 15 mm Hg ? Initial Conditions Final Conditions V1 = 100 m V2 = ? T1 = 25 oC T2 = -90 oC = 298 K = 183 K P1 = 1.0 atm P2 = 15 mm Hg 760 mm Hg/ atm P2= ___________ atm

49 Change of Conditions Problem II: continued
P1 x V P2 x V2 V2 = V2 = = V2 = m3 = __________ m3 or 3.1 times the volume !!! P1V1T2 = T1 T2 T1P2 ( 1.0 atm) ( 100 m3) (183 K) (298 K) (0.198 atm)

50 Change of Conditions :Problem III
How many liters of CO2 are formed at 1.00 atm and 900 oC if 5.00 L of Propane at 10.0 atm, and 25 oC is burned in excess air? C3H8 (g) + 5 O2 (g) = 3 CO2 (g) + 4 H2O(g) 25 oC = 298 K 900 oC = 1173 K

51 Change of Conditions Problem III:continued
V1 = 5.00 L V2 = ? P1 = 10.0 atm P2 = 1.00 atm T1 = 298K T2 = 1173 K P1V1/T1 = P2V2/T V2 = V1P1T2/ P2T1 V2 = = 197 L VCO2 = (197 L C3H8) x (3 L CO2 / 1 L C3H8) = VCO2 = _________ L CO2 ( 5.00 L) (10.00 atm) (1173 K) ( 1.00 atm) ( 298 K)

52 Standard Temperature and Pressure (STP)
A set of Standard conditions have been chosen to make it easier to understand the gas laws, and gas behavior. Standard Temperature = 00 C = K Standard Pressure = 1 atmosphere = 760 mm Mercury At these “standard” conditions, if you have 1.0 mole of a gas it will occupy a “standard molar volume”. Standard Molar Volume = Liters = 22.4 L

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55 Table 5.2 (P 149) Molar Volumes for Various Gases at 00 and 1 atm
Gas Molar Volume (L) Oxygen (O2) Nitrogen (N2) Hydrogen (H2) Helium (He) Argon (Ar) Carbon dioxide (CO2) Ammonia (NH3)

56 IDEAL GASES An ideal gas is defined as one for which both the volume of molecules and forces between the molecules are so small that they have no effect on the behavior of the gas. The ideal gas equation is: PV=nRT R = Ideal gas constant R = J / mol K = J mol-1 K-1 R = l atm mol-1 K-1

57 Variations on the Gas Equation
During chemical and physical processes, any of the four variables in the ideal gas equation may be fixed. Thus, PV=nRT can be rearranged for the fixed variables: for a fixed amount at constant temperature P V = nRT = constant Boyle’s Law for a fixed amount at constant volume P / T = nR / V = constant Amonton’s Law for a fixed amount at constant pressure V / T = nR / P = constant Charles’ Law for a fixed volume and temperature P / n = R T / V = constant Avogadro’s Law

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59 Evaluation of the Ideal Gas Constant, R
PV nT Ideal gas Equation PV = nRT R = at Standard Temperature and Pressure, the molar volume = 22.4 L P = 1.00 atm (by Definition!) T = 0 oC = K (by Definition!) n = 1.00 moles (by Definition!) (1.00 atm) ( L) ( 1.00 mole) ( K) L atm mol K R = = L atm mol K or to three significant figures R =

60 Values of R in Different Units
Atm x L Mol x K R* = R = 62.36 R = 8.314 Torr x L Mol x K kPa x dm3 Mol x K J# Mol x K * Most calculations in this text use values of R to 3 significant figures. # J is the abbreviation for joule, the SI unit of energy. The joule is a derived unit composed of the base units kg x m2/s2

61 Gas Law: Solving for Pressure
Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC. Plan: Convert all information into the units required, and substitute into the Ideal Gas equation ( PV=nRT ). Solution: 5038 g Xe 131.3 g Xe / mol nXe = = mol Xe T = 18.8 oC K = K PV = nRT P = nRT V (38.37 mol )( L atm)( K) P = = _______ atm = atm 87.5 L (mol K)

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63 Ideal Gas Calculation - Nitrogen
Calculate the pressure in a container holding 375 g of Nitrogen gas. The volume of the container is m3 and the temperature is 36.0 oC. n = 375 g N2/ 28.0 g N2 / mol = 13.4 mol N2 V = m3 x 1000 L / m3 = 150 L T = 36.0 oC = K PV=nRT P= nRT/V P = P = atm ( 13.4 mol) ( L atm/mol K) ( K) 150 L

64 Mass of Air in a Hot Air Balloon - Part I
Calculate the mass of air in a spherical hot air balloon that has a volume of 14,100 cubic feet when the temperature of the gas is 86 oF and the pressure is 748 mm Hg? P = 748 mm Hg x 1atm / 760 mm Hg= atm V = 1.41 x 104 ft3x (12 in/1 ft)3x(2.54 cm/1 in)3 x x (1ml/1 cm3) x ( 1L / 1000 cm3) =3.99 x 105 L T = oC =(86-32)5/9 = 30 oC 30 oC = 303 K

65 Mass of Air in a Hot Air Balloon - Part II
PV = nRT n = PV / RT n = = 1.58 x 104 mol mass = 1.58 x 104 mol air x 29 g air/mol air = 4.58 x 105 g Air = 458 Kg Air ( atm) ( 3.99 x 105 L) ( L atm/mol K) ( 303 K )

66 Inflated dual air bag

67 Sodium Azide Decomposition-I
Sodium Azide (NaN3) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 . 2 NaN3 (s) Na (s) + 3 N2 (g) mol NaN3 = 60.0 g NaN3 / g NaN3 / mol = = mol NaN3 mol N2= mol NaN3 x 3 mol N2/2 mol NaN3 = ___________mol N2

68 Sodium Azide Calc: - II PV = nRT V = nRT/P V =
V = __________ liters ( 1.38 mol) ( L atm / mol K) (294 K) ( 823 mm Hg / 760 mmHg / atm )

69 Ammonia Density Problem
Calculate the Density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC. Density = mass per unit volume = g / L P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm T = 55 oC = 328 K n = mass / Molar mass = g / M d = = d = __________ g / L P x M R x T ( atm) ( g/mol) ( L atm/mol K) ( 328 K)

70 Like Example 5.6 (P 149) Problem: Calculate the volume of carbon dioxide produced by the thermal decomposition of 36.8 g of calcium carbonate at 715mm Hg, and 370 C according to the reaction below: CaCO3(s) CaO(s) + CO2 (g) Solution: 1 mol CaCO3 100.1g CaCO3 36.8g x = mol CaCO3 Since each mole of CaCO3 produces one 1 mole of CO2 , moles of CO2 at STP, we now have to convert to the conditions stated: nRT P (0.368 mol)( L atm/mol K)(310.15K) (715mmHg/760mm Hg/atm) V = = V = __________ L

71 Calculation of Molar Mass
P x V Mass R x T Molar Mass Mass x R x T Molar Mass = = MM P x V

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73 Dumas Method of Molar Mass
Problem: A volatile liquid is placed in a flask whose volume is ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of oC and 736 mm Hg pressure. If the mass of the flask before and after the experiment was g and g, what is the molar mass of the liquid? Plan: Use the gas law to calculate the molar mass of the liquid. Solution: 1 atm 760 mm Hg Pressure = 736 mm Hg x = atm mass = g g = g (1.082 g)( Latm/mol K)(373.2 K) Molar Mass = = g/mol ( atm)(0.590 L) note: the compound is acetone C3H6O = MM = 58g mol !

74 Calculation of Molecular Weight of a Gas Natural Gas - Methane
Problem: A sample of natural gas is collected at 25.0 oC in a ml flask. If the sample had a mass of g at a pressure of Torr, what is the molecular weight of the gas? Plan: Use the Ideal gas law to calculate n, then calculate the molar mass. Solution: 1mm Hg 1 Torr 1.00 atm 760 mm Hg P = Torr x x = atm 1.00 L 1000 ml V = ml x = L P V R T n = T = 25.0 oC K = K (0.724 atm)(0.250 L) n = = mol ( L atm/mol K) (298.2 K) MM = g / mol = ____________ g/mol

75 Gas Mixtures Gas behavior depends on the number, not the identity, of gas molecules. Ideal gas equation applies to each gas individually and the mixture as a whole. All molecules in a sample of an ideal gas behave exactly the same way.

76 Figure 5.9: Partial pressure of each gas in a mixture of gases depends on the number of moles of that gas.

77 Dalton’s Law of Partial Pressures - I
Definiton: In a mixture of gases, each gas contributes to the total pressure the amount it would exert if the gas were present in the container by itself. To obtain a total pressure, add all of the partial pressures: P total = p1+p2+p3+...pi

78 Dalton’s Law of Partial Pressure - II
Pressure exerted by an ideal gas mixture is determined by the total number of moles: P=(ntotal RT)/V n total = sum of the amounts of each gas pressure the partial pressure is the pressure of gas if it was present by itself. P = (n1 RT)/V + (n2 RT)/V + (n3RT)/V + ... the total pressure is the sum of the partial pressures.

79 Dalton’s Law of Partial Pressures - Prob #1
A 2.00 L flask contains 3.00 g of CO2 and g of Helium at a temperature of 17.0 oC. What are the Partial Pressures of each gas, and the total Pressure? T = 17 oC = 290 K nCO2 = 3.00 g CO2/ g CO2 / mol CO2 = mol CO2 PCO2 = nCO2RT/V PCO2 = PCO2 = _____________ atm ( mol CO2) ( L atm/mol K) ( 290 K) (2.00 L)

80 Dalton’s Law Problem - #1 cont.
nHe = 0.10 g He / g He / mol He = mol He PHe = nHeRT/V PHe = PHe = 0.30 atm PTotal = PCO2 + PHe = atm atm PTotal = 1.11 atm (0.025 mol) ( L atm / mol K) ( 290 K ) ( 2.00 L )

81 Dalton’s Law Problem #2using mole fractions
A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ? Total # moles = = 7.35 mol XNe = 4.46 mol Ne / 7.35 mol = 0.607 PNe = XNe PTotal = ( 2.00 atm) = 1.21 atm for Ne XAr = 0.74 mol Ar / 7.35 mol = 0.10 PAr = XAr PTotal = 0.10 (2.00 atm) = 0.20 atm for Ar XXe = 2.15 mol Xe / 7.35 mol = 0.293 PXe = XXe PTotal = (2.00 atm) = atm for Xe

82 Relative Humidity Pressure of Water in Air Rel Hum = x 100%
Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the Relative Humidity? Rel Hum =(6.54 mm Hg/ mm Hg )x100% = ___________ % Pressure of Water in Air Maximum Vapor Pressure of Water

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84 Figure 5.10: Production of oxygen by thermal decomposition of KCIO3.

85 Molecular sieve framework of titanium (blue), silicon (green), and oxygen (red) atoms contract on heating–at room temperature (left) d = 4.27 Å; at 250°C (right) d = 3.94 Å.

86 Collection of Hydrogen gas over Water - Vapor pressure - I
2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g) Calculate the mass of Hydrogen gas collected over water if 156 ml of gas is collected at 20oC and 769 mm Hg. PTotal = P H2 + PH2O PH2 = PTotal - PH2O PH2 = 769 mm Hg mm Hg = 752 mm Hg T = 20oC = 273 = 293 K P = 752 mm Hg /760 mm Hg /1 atm = atm V =___________ L

87 COLLECTION OVER WATER CONT.-II
PV = nRT n = PV / RT n = n = mol mass = mol x 2.01 g H2 / mol H2 mass = ______________ g Hydrogen (0.989 atm) (0.156 L) ( L atm/mol K) (293 K)

88 Chemical Equation Calc - III
Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles Molarity PV = nRT moles / liter Solutions Gases

89 Gas Law Stoichiometry - I - NH3 + HCl
Problem: A slide separating two containers is removed, and the gases are allowed to mix and react. The first container with a volume of 2.79 L contains Ammonia gas at a pressure of atm and a temperature of 18.7 oC. The second with a volume of 1.16 L contains HCl gas at a pressure of atm and a temperature of 18.7 oC. What mass of solid ammonium chloride will be formed, and what will be remaining in the container, and what is the pressure? Plan: This is a limiting reactant problem, so we must calculate the moles of each reactant using the gas law to determine the limiting reagent. Then we can calculate the mass of product, and determine what is left in the combined volume of the container, and the conditions. Solution: Equation: NH3 (g) + HCl(g) NH4Cl(s) TNH3 = 18.7 oC = K

90 Gas Law Stoichiometry - II - NH3 + HCl
PV n = RT (0.776 atm) (2.79 L) nNH3 = = mol NH3 ( L atm/mol K) (291.9 K) limiting (0.932 atm) (1.16 L) nHCl = = mol HCl ( L atm/mol K) (291.9 K) reactant Therefore the product will be mol NH4Cl or g NH4Cl Ammonia remaining = mol mol = mol NH3 V = 1.16 L L = 3.95 L nRT ( mol) ( L atm/mol K) (291.9 K) PNH3 = = = ____ atm V (3.95 L)

91 Postulates of Kinetic Molecular Theory
I The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). II The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. III The particles are assumed to exert no forces on each other; they are assumed to neither attract nor repel each other. IV The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

92 Figure 5.11: An ideal gas particle in a cube wholse sides are of length L (in meters).

93 Figure 5.12: (a) The Cartesian coordinate axes.

94 Figure 5.12: (b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, ux, uy, and uz.

95 Figure 5.12: (c) In the xy plane, ux2 + uy2 = uyx2 by the Pythagorean theorem.

96 Figure 5.13: (a) Only the x component of the gas particle’s velocity affects the frequency of impacts on the shaded walls. (b) For an elastic collision, there is an exact reversal of the x component of the velocity and of the total velocity.

97 Figure 5.14: Path of one particle in a gas.

98 A balloon filled with air at room temperature.

99 The balloon is dipped into liquid nitrogen at 77 K.

100 The balloon collapses as the molecules inside slow down because of the decreased temperature.

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102 Velocity and Energy Kinetic Energy = 1/2mu2
Average Kinetic Energy (KEavg) add up all the individual molecular energies and divide by the total number of molecules! The result depends on the temp. of the gas KE avg=3RT/2Na T=temp. in Kelvin , Na =Avogadro’s number, R=new quantity (gas constant) kEtotal = (No. of molecules)(KEavg) = (NA)(3RT/2NA) = 3/2RT So, 1 mol of any gas has a total molecular Kinetic Energy = KE of 3/2RT!!!

103 Figure 5.15: A plot of the relative number of O2 molecules that have a given velocity at STP.

104 Figure 5.16: A plot of the relative number of N2 molecules that have a given velocity at three temperatures.

105 Figure 5.17: A velocity distribution for nitrogen gas at 273 K.

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107 Molecular Mass and Molecular Speeds - I
Problem: Calculate the molecular speeds of the molecules of Hydrogen, Methane, and carbon dioxide at 300K! Plan: Use the equations for the average kinetic energy of a molecule, and the relationship between kinetic energy and velocity to calculate the average speeds of the three molecules. Solution: for Hydrogen, H2 = g/mol 3 R 2 NA 8.314 J/mol K EK = x T = 1.5 x x 300K = 6.022 x 1023 molecules/mol EK = x J/molecule = 1/2 mu2 2.016 x kg/mole 6.022 x 1023 molecules/mole m = = x kg/molecule 6.213 x J/molecule = x kg/molecule(u2) u = _______________ m/s = ___________ m/s

108 Molecular Mass and Molecular Speeds - II
for methane CH4 = g/mole 3 R 2 NA 8.314 J/mol K 6.022 x 1023 molecules/mole EK = x T = 1.5 x x 300K = EK = x J/molecule = 1/2 mu2 16.04 x kg/mole 6.022 x 1023 moleules/mole m = = x kg/molecule 6.213 x J/molecule = x kg/molecule (u2) u = ___________________ m/s = __________ m/s

109 Molecular Mass and Molecular Speeds - III
for Carbon dioxide CO2 = g/mole EK = x T = 1.5 x x 300K = 3 R 2 NA 8.314 J/mol K 6.022 x 1023 molecules/mole EK = x J/molecule = 1/2mu2 44.01 x kg/mole 6.022 x 1023 molecules/mole m = = x kg/molecule 6.213 x J/molecule = x kg/molecule (u2) u = ___________________ m/s = _________ m/s

110 Molecular Mass and Molecular Speeds - IV
Molecule H CH CO2 Molecular Mass (g/mol) Kinetic Energy (J/molecule) 6.213 x x x Velocity (m/s) 1,

111 Important Point ! At a given temperature, all gases have the same molecular Kinetic energy distributions. or The same average molecular Kinetic Energy!

112 Diffusion vs Effusion Diffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules. Effusion - A gas escaping from a container into a vacuum. There are no other (or few ) for collisions.

113 Figure 5.18: The effusion of a gas into an evacuated chamber.

114

115 Relative Diffusion of H2 versus O2 and N2 gases
Average Molecular weight of air: 20% O g/mol x = 80% N g/mol x = 28.80 28.80 g/mol or approximately 29 g/mol

116 Graham's Law calc. RateHydrogen = RateAir x (MMAir / MMHydrogen)1/2
Or RateHydrogen = RateAir x 4 !!!!!!

117

118

119 NH3 (g) + HCl(g) = NH4Cl (s)
HCl = g/mol NH3 = g/mol RateNH3 = RateHCl x ( / )1/ 2 RateNH3 = RateHCl x 1.463

120 Figure 5.19: (a) demonstration of the relative diffusion rates of NH3 and HCL molecules through air.

121 Figure 5.19: (b) When HCL(g) and NH3 (g) meet in the tube, a white ring of NH4CL(s) forms.

122

123 Figure 5.20: Uranium-enrichment converters from the Paducah gaseous diffusion plant in Kentucky.

124 Gaseous Diffusion Separation of Uranium 235 / 238
235UF6 vs 238UF6 Separation Factor = S = after Two runs S = after approximately runs 235UF6 is > 99% Purity !!!!! Y Plant at Oak Ridge National Lab ( (6 x 19))0.5 ( (6 x 19))0.5

125 Example 5.9 (P167) Calculate the impact rate on a 1.00-cm2 section of a vessel containing oxygen gas at a pressure of 1.00 atm and 270C. Solution: N RT V 2pM Calculate ZA : ZA = A A = 1.00 x 10-4 m2 1.00 atm N P V RT ( ) L atm K mol = = (300. K) N V Mol L molecules mol 1000 L m3 = x x x x = = 2.44 x 1025 molecules/m3 1 kg 1000g M = 32.0 g/mol x = 3.2 x 10-2 kg/mol J K mol 8.3145 (300. K) ZA = (1.00 x 10-4 m2) (2.44 x 1025 m-3) x 2(3.14)( 3.20 x 10-2 kg/mol) ZA = 2.72 x 1023 collisions/s

126 Figure 5.21: The cylinder swept out by a gas particle of diameter d.

127 Like Example 5.10(P 169) Problem: Calculate the collision frequency for a Nitrogen molecule in a sample of pure nitrogen gas at 280C and 2.0 atm. Assume that the diameter of an N2 molecule is 290 pm. Solution: 2.0 atm n V P RT = = = x 10-2 mol/L ( ) L atm K mol (301. K) N V = (8.097 x 10-2 mol/L)( x 1023 molecules/mol)(1000L/m3) N V = x 1025 molecules/m3 d= 290 pm = 2.90 x m Also, for N2, M = 2.80 x 10-2 kg/mol p( J K-1mol-1)(301 K) Z = 4(4.876 x 1025 m-3)(2.90 x m)2 x Z = collisions/sec 2.80 x 10-2 kg/mol

128 Like Example 5.11 (P171) Problem: Calculate the mean free path in a sample of nitrogen gas at 280 C and 2.00 atm. Solution: Using data from the previous example, we have: 1 l = 2 (N/V)(pd2) 1 l = = 5.49 x 10-8 2 ( x 1025 )(p)(2.90 x )2 Note that a nitrogen molecule only travels 0.5 nano meters before it collides with another nitrogen atom.

129 Figure 5.22: Plots of PV/nRT versus P for several gases (200 K).

130 Figure 5.23: Plots of PV/nRT versus P for nitrogen gas at three temperatures.

131 Figure 5.24: (a) gas at low concentration - relatively few interactions between particles (b) gas at high concentration - many more interactions between particles.

132 Figure 5.25: Illustration of pairwise interactions among gas particles.

133 ( ) ( ) Table 5.3 (P172) Values of Van der Waals
Constants for some Common Gases ( ) ( ) atm L2 mol2 L mol Gas a b He Ne Ar Kr Xe H N O Cl CO CH NH H2O

134 Figure 5.26: The volume occupied by the gas particles themselves is less important at (a) large container volumes (low pressure) than at (b) small container volumes (high pressure).

135 Van der Waals Calculation of a Real gas
Problem: A tank of 20.0 liters contains Chlorine gas at a temperature of C at a pressure of atm. if the tank is pressureized to a new volume of L and a temperature of C. What is the new pressure using the ideal gas equation, and the Van der Waals equation? Plan: Do the calculations! Solution: PV (2.000 atm)(20.0L) RT ( Latm/molK)( K) n = = = mol nRT (1.663 mol) ( Latm/molK) ( K) V (1.000 L) P = = = atm nRT n2a (1.663 mol) ( Latm/molK)( K) (V-nb) V (1.00 L) - (1.663 mol)(0.0562) P = = (1.663 mol)2(6.49) (1.00 L)2 = = atm

136 Table 5.4(P173) Atmospheric Composition near Sea Level (dry air)#
Component Mole Fraction N O Ar CO Ne He CH Kr H NO Xe # The atmosphere contains various amounts of water vapor, depending on conditions.

137 Figure 5.27: The variation of temperature and pressure with altitude.

138 Severe Atmospheric Environmental Problems
we must deal with in the next period of time: Urban Pollution – Photochemical Smog Acid Rain Greenhouse Effect Stratospheric Ozone Destruction

139 Figure 5.28: Concentration (in molecules per million molecules of “air”) of some smog components versus time of day.

140 Figure 5.29: Our various modes of transportation produce large amounts of nitrogen oxides, which facilitate the formation of photochemical smog.

141 This photo was taken in 1990. Recent renovation has since replaced the deteriorating marble.

142 Figure 5.31: Diagram of the process for scrubbing sulfur dioxide from stack gases in power plants.

143


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