Presentation on theme: "Chapter 5 Gases. Chapter 5: Gases 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5."— Presentation transcript:
Chapter 5 Gases
Chapter 5: Gases 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5 Daltons Law of Partial Pressures 5.6 The Kinetic molecular Theory of Gases 5.7 Effusion and Diffusion 5.8 Collisions of Gas Particles with the Container Walls 5.9 Intermolecular Collisions 5.10 Real Gases 5.11 Chemistry in the Atmosphere
Hurricanes, such as this one off the coast of Florida, are evidence of the powerful forces present in the earth's atmosphere.
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.
Helium He 4.0 Neon Ne 20.2 Argon Ar 39.9 Hydrogen H Nitrogen N Nitrogen Monoxide NO 30.0 Oxygen O Hydrogen Chloride HCL 36.5 Ozone O Ammonia NH Methane CH Substances that are Gases under Normal Conditions Substance Formula MM(g/mol)
Some Important Industrial Gases Name - Formula Origin and use Methane (CH 4 ) Natural deposits; domestic fuel Ammonia (NH 3 ) From N 2 + H 2 ; fertilizers, explosives Chlorine (Cl 2 ) Electrolysis of seawater; bleaching and disinfecting Oxygen (O 2 ) Liquified air; steelmaking Ethylene (C 2 H 4 ) High-temperature decomposition of natural gas; plastics
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! Pressure = Force Area
Figure 5.1: A torricellian barometer.
Construct a Barometer using Water! Density of water = 1.00 g/cm 3 Density of Mercury = 13.6 g/cm 3 Height of water column = H w H w = Height of Hg x Density of Mercury H w = 760 mm Hg x 13.6/1.00 = 1.03 x 10 4 mm H w = 10.3 m = __________ ft Density of Water Height Water Height Mercury Density Mercury Density Water =
Common Units of Pressure Unit Atmospheric Pressure Scientific Field Used Pascal (Pa); x 10 5 Pa SI unit; physics, kilopascal (kPa) kPa chemistry Atmosphere (atm) 1 atm Chemistry Millimeters of mercury 760 mmHg Chemistry, medicine (mmHg) biology Torr 760 torr Chemistry Pounds per square inch 14.7 lb/in 2 Engineering (psl or lb/in 2 ) Bar bar Meteorology, chemistry
Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from the decomposition of Lime stone (CaCO 3 ) in a closed end manometer, the height of the mercury is mm Hg. Calculate the CO 2 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: P CO2 (torr) = mm Hg x = torr 1 torr 1 mm Hg converting from mmHg to torr: converting from torr to atm: P CO2 ( atm) = torr x = atm 1 atm 760 torr converting from atm to kPa: P CO2 (kPa) = atm x = _____________ kPa kPa 1 atm
Figure 5.2: A simple manometer, a device for measuring the pressure of a gas in a container.
Figure 5.3: A J-tube similar to the one used by Boyle.
Boyles 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 ! P 1 V 1 = k P 2 V 2 = k k = k Then : P 1 V 1 = P 2 V 2 kVkV kPkP
Figure 5.4: Plotting Boyles data from Table 5.1.
Figure 5.5: Plot of PV versus P for several gases.
Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm 3, what will be the volume if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm 3 to ml and then to liters, then we do the pressure change to obtain the final volume! Solution: V 1 (cm 3 ) V 1 (ml) V 1 (L) V 2 (L) 1cm 3 = 1 mL 1000mL = 1L x P 1 /P 2 P 1 = 1.23 atm P 2 = 3.16 atm V 1 = 15.8 cm 3 V 2 = unknown T and n remain constant V 1 = 15.8 cm 3 x x = L 1 mL 1 cm 3 1 L 1000mL V 2 = V 1 x = L x = ________ L P1P2P1P atm 3.16 atm
Boyles 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 cm 3, 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/cm 3, and use the mass of Hg in a barometer for comparison!) Initial ConditionsFinal Conditions V 1 = 1.00 cm 3 P 1 = ? V 2 = ? P 2 = 1.00 atm
Calculation Continued Pressure at depth = 6 x 10 3 ft x x x m 1 ft 100 cm 1 m g SH 2 O 1 cm 3 Pressure at depth = 187, g pressure from SH 2 O For a Mercury Barometer: 760 mm Hg = 1.00 atm, assume that the cross-section of the barometer column is 1 cm 2. The mass of Mercury in a barometer is: Pressure = x x x x 10 mm 1 cm Area 1 cm cm 3 Hg 13.6 g Hg 187,635 g = 1.00 atm 760 mm Hg Pressure = _____ atm Due to the added atmospheric pressure = ___ atm! V 2 = = = ____ cm 3 = liters V 1 x P 1 P cm 3 x ____ atm 1.00 atm
Boyles 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? P 1 = 1.0 atm P 2 = 0.40 atm V 1 = 0.55 L V 2 = ? V 2 = V 1 x P 1 /P 2 = (0.55 L) x (1.0 atm / 0.40 atm) V 2 = __________ L
Figure 5.6: PV plot verses P for 1 mole of ammonia
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 T 1 / V 1 = T 2 / V 2 or: T1T1 V1V1 T2T2 = V2V2 T 1 = V 1 x T2T2 V2V2 Temperatures must be expressed in Kelvin to avoid negative values!
Figure 5.7: Plots of V versus T (c) for several gases.
Figure 5.8: Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature.
Charles Law Problem A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 o C. Calculate the temperature ( o C) at which the gas will occupy 1.54 L if the pressure remains constant. V 1 = 3.20 L T 1 = 125 o C = 398 K V 2 = 1.54 L T 2 = ? T 2 = T 1 x ( V 2 / V 1 ) T 2 = 398 K x =______K T 2 = ___ K o C = K = ______ o C = ______ o C 1.54 L 3.20 L
Charles Law Problem - I A balloon in Antarctica in a building is at room temperature ( 75 o F ) and has a volume of 20.0 L. What will be its volume outside where the temperature is -70 o F ? V 1 = 20.0 L V 2 = ? T 1 = 75 o F T 2 = -70 o F o C = ( o F - 32 ) 5/9 T 1 = ( )5/9 = 23.9 o C K = 23.9 o C = __________ K T 2 = ( ) 5/9 = o C K = o C = ___________ K
Antarctic Balloon Problem - II V 1 / T 1 = V 2 / T 2 V 2 = V 1 x ( T 2 / T 1 ) V 2 = 20.0 L x V 2 = __________ L The Balloon shrinks from 20 L to 15 L !!!!!!! Just by going outside !!!!! K K
Applying the Temperature - Pressure Relationship (Amontons Law) Problem: A copper tank is compressed to a pressure of 4.28 atm at a temperature of o F. What will be the pressure if the temperature is raised to 95.6 o C? 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: T 1 = (0.185 o F o F)x 5/9 = o C T 1 = o C K = K T 2 = 95.6 o C K = K P 1 P 2 T 1 T 2 =P 2 = P 1 x = ? T2T1T2T1 P 2 = 4.28 atm x = _______ atm K K
Avogadros 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. n 1 = initial moles of gas V 1 = initial volume of gas n 2 = final moles of gas V 2 = final volume of gas n 1 V 1 n 2 V 2 =or: n 1 = n 2 x V1V2V1V2
Avogadros 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 SF 6 at atm and 28.5 o C is 2.93 m 3, what will be the mass of SF 6 in a container whose volume is m 3 at atm and 28.5 o C? Plan: Since the temperature and pressure are the same it is a V - n problem, so we can use Avogadros Law to calculate the moles of the gas, then use the molecular mass to calculate the mass of the gas. Solution: Molar mass SF 6 = g/mol V2V1V2V1 n 2 = n 1 x = mol SF 6 x = 3.40 mol SF g SF g SF 6 /mol = mol SF m m 3 mass SF 6 = 3.40 mol SF 6 x g SF 6 / mol = __________ g SF 6
Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles(2.298g H 2 ) of Hydrogen and has a volume of L. What mass of Hydrogen must be added to the balloon to increase its 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 Avogadros law, and the change of conditions form. Solution: n 1 = 1.14 moles H 2 n 2 = 1.14 moles + ? moles V 1 = L V 2 = L T = constant P = constant n 1 n 2 V 1 V 2 =n 2 = n 1 x = 1.14 moles H 2 x V2V1V2V1 n 2 = moles = 4.46 moles L L mass = moles x molecular mass mass = 4.46 moles x g / mol mass = ___________ g H 2 gas added mass = 8.99g g = 6.69g
Change of Conditions, with no change in the amount of gas ! = constant Therefore for a change of conditions : T 1 T 2 P x V T P 1 x V 1 = P 2 x V 2
Change of Conditions :Problem -I A gas sample in the laboratory has a volume of 45.9 L at 25 o C and a pressure of 743 mm Hg. If the temperature is increased to 155 o C by pumping (compressing) the gas to a new volume of 3.10 ml what is the pressure? P 1 = 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm P 2 = ? V 1 = 45.9 L V 2 = 3.10 ml = L T 1 = 25 o C = 298 K T 2 = 155 o C = 428 K
Change of Condition Problem I : continued = T1T1 T2T2 P 1 x V 1 P 2 x V 2 ( atm) ( 45.9 L) P 2 ( L) ( 298 K)( 428 K) = P 2 = ( 428 K) ( atm) ( 45.9 L) = ________ atm ( 298 K) ( L)
Change of Conditions : Problem II A weather balloon is released at the surface of the earth. If the volume was 100 m 3 at the surface ( T = 25 o C, P = 1 atm ) what will its volume be at its peak altitude of 90,000 ft where the temperature is - 90 o C and the pressure is 15 mm Hg ? Initial Conditions Final Conditions V 1 = 100 m 3 V 2 = ? T 1 = 25 o C T 2 = -90 o C = 298 K = 183 K P 1 = 1.0 atm P 2 = 15 mm Hg 760 mm Hg/ atm P 2 = ___________ atm
Change of Conditions Problem II: continued P 1 x V 1 P 2 x V 2 V 2 = V 2 = = V 2 = m 3 = __________ m 3 or 3.1 times the volume !!! T1T1 T2T2 = P1V1T2P1V1T2 T1P2T1P2 ( 1.0 atm) ( 100 m 3 ) (183 K) (298 K) (0.198 atm)
Change of Conditions :Problem III How many liters of CO 2 are formed at 1.00 atm and 900 o C if 5.00 L of Propane at 10.0 atm, and 25 o C is burned in excess air? C 3 H 8 (g) + 5 O 2 (g) = 3 CO 2 (g) + 4 H 2 O (g) 25 o C = 298 K 900 o C = 1173 K
Change of Conditions Problem III:continued V 1 = 5.00 L V 2 = ? P 1 = 10.0 atm P 2 = 1.00 atm T 1 = 298K T 2 = 1173 K P 1 V 1 /T 1 = P 2 V 2 /T 2 V 2 = V 1 P 1 T 2 / P 2 T 1 V 2 = = 197 L V CO2 = (197 L C 3 H 8 ) x (3 L CO 2 / 1 L C 3 H 8 ) = V CO2 = _________ L CO 2 ( 5.00 L) (10.00 atm) (1173 K) ( 1.00 atm) ( 298 K)
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 = 0 0 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
Table 5.2 (P 149) Molar Volumes for Various Gases at 0 0 and 1 atm Gas Molar Volume (L) Oxygen (O 2 ) Nitrogen (N 2 ) Hydrogen (H 2 ) Helium (He) Argon (Ar) Carbon dioxide (CO 2 ) Ammonia (NH 3 )
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
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 Boyles Law –for a fixed amount at constant volume P / T = nR / V = constant Amontons 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 Avogadros Law
Evaluation of the Ideal Gas Constant, R Ideal gas Equation PV = nRT R = PV nT at Standard Temperature and Pressure, the molar volume = 22.4 L P = 1.00 atm (by Definition!) T = 0 o C = K (by Definition!) n = 1.00 moles (by Definition!) R = = (1.00 atm) ( L) ( 1.00 mole) ( K) L atm mol K or to three significant figures R = L atm mol K
Values of R in Different Units R * = R = R = Atm x L Mol x K Torr x L Mol x K kPa x dm 3 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 m 2 /s 2
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 o C. Plan: Convert all information into the units required, and substitute into the Ideal Gas equation ( PV=nRT ). Solution: n Xe = = mol Xe 5038 g Xe g Xe / mol T = 18.8 o C K = K PV = nRT P = nRT V P = = _______ atm = atm (38.37 mol )( L atm)( K) 87.5 L(mol K)
Ideal Gas Calculation - Nitrogen Calculate the pressure in a container holding 375 g of Nitrogen gas. The volume of the container is m 3 and the temperature is 36.0 o C. n = 375 g N 2 / 28.0 g N 2 / mol = 13.4 mol N 2 V = m 3 x 1000 L / m 3 = 150 L T = 36.0 o C = K PV=nRT P= nRT/V P = P = atm ( 13.4 mol) ( L atm/mol K) ( K) 150 L
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 o F and the pressure is 748 mm Hg? P = 748 mm Hg x 1atm / 760 mm Hg= atm V = 1.41 x 10 4 ft 3 x (12 in/1 ft) 3 x(2.54 cm/1 in) 3 x x (1ml/1 cm 3 ) x ( 1L / 1000 cm 3 ) =3.99 x 10 5 L T = o C =(86-32)5/9 = 30 o C 30 o C = 303 K
Mass of Air in a Hot Air Balloon - Part II PV = nRT n = PV / RT n = = 1.58 x 10 4 mol mass = 1.58 x 10 4 mol air x 29 g air/mol air = 4.58 x 10 5 g Air = 458 Kg Air ( atm) ( 3.99 x 10 5 L) ( L atm/mol K) ( 303 K )
Inflated dual air bag
Sodium Azide Decomposition-I Sodium Azide (NaN 3 ) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 o C and 823 mm Hg by the decomposition of 60.0 g of NaN 3. 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) mol NaN 3 = 60.0 g NaN 3 / g NaN 3 / mol = = mol NaN 3 mol N 2 = mol NaN 3 x 3 mol N 2 /2 mol NaN 3 = ___________mol N 2
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 )
Ammonia Density Problem Calculate the Density of ammonia gas (NH 3 ) in grams per liter at 752 mm Hg and 55 o C. Density = mass per unit volume = g / L P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm T = 55 o C = 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)
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 37 0 C according to the reaction below: CaCO 3(s) CaO (s) + CO 2 (g) Solution: 36.8g x = mol CaCO 3 1 mol CaCO g CaCO 3 Since each mole of CaCO 3 produces one 1 mole of CO 2, moles of CO 2 at STP, we now have to convert to the conditions stated: V = = nRT P (0.368 mol)( L atm/mol K)(310.15K) (715mmHg/760mm Hg/atm) V = __________ L
Calculation of Molar Mass n = n = = Mass Molar Mass P x V R x T Mass Molar Mass Molar Mass = = MM Mass x R x T P x V
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 o C 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: Pressure = 736 mm Hg x = atm 1 atm 760 mm Hg Molar Mass = = g/mol mass = g g = g (1.082 g)( Latm/mol K)(373.2 K) ( atm)(0.590 L) note: the compound is acetone C 3 H 6 O = MM = 58g mol !
Calculation of Molecular Weight of a Gas Natural Gas - Methane Problem: A sample of natural gas is collected at 25.0 o C 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: P = Torr x x = atm 1mm Hg 1 Torr 1.00 atm 760 mm Hg V = ml x = L 1.00 L 1000 ml T = 25.0 o C K = K n = P V R T n = = mol ( L atm/mol K) (298.2 K) (0.724 atm)(0.250 L) MM = g / mol = ____________ g/mol
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.
Figure 5.9: Partial pressure of each gas in a mixture of gases depends on the number of moles of that gas.
Daltons 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
Daltons Law of Partial Pressure - II Pressure exerted by an ideal gas mixture is determined by the total number of moles: P=(n total 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 = (n 1 RT)/V + (n 2 RT)/V + (n 3 RT)/V +... the total pressure is the sum of the partial pressures.
Daltons Law of Partial Pressures - Prob #1 A 2.00 L flask contains 3.00 g of CO 2 and 0.10 g of Helium at a temperature of 17.0 o C. What are the Partial Pressures of each gas, and the total Pressure? T = 17 o C = 290 K n CO2 = 3.00 g CO 2 / g CO 2 / mol CO 2 = mol CO 2 P CO2 = n CO2 RT/V P CO2 = P CO2 = _____________ atm ( mol CO 2 ) ( L atm/mol K) ( 290 K) (2.00 L)
Daltons Law Problem - #1 cont. n He = 0.10 g He / g He / mol He = mol He P He = n He RT/V P He = P He = 0.30 atm P Total = P CO2 + P He = atm atm P Total = 1.11 atm (0.025 mol) ( L atm / mol K) ( 290 K ) ( 2.00 L )
Daltons 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 X Ne = 4.46 mol Ne / 7.35 mol = P Ne = X Ne P Total = ( 2.00 atm) = 1.21 atm for Ne X Ar = 0.74 mol Ar / 7.35 mol = 0.10 P Ar = X Ar P Total = 0.10 (2.00 atm) = 0.20 atm for Ar X Xe = 2.15 mol Xe / 7.35 mol = P Xe = X Xe P Total = (2.00 atm) = atm for Xe
Relative Humidity Rel Hum = x 100% Example : the partial pressure of water at 15 o C 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
Figure 5.10: Production of oxygen by thermal decomposition of KCIO3.
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.
Collection of Hydrogen gas over Water - Vapor pressure - I 2 HCl (aq) + Zn (s) ZnCl 2 (aq) + H 2 (g) Calculate the mass of Hydrogen gas collected over water if 156 ml of gas is collected at 20 o C and 769 mm Hg. P Total = P H2 + P H2O P H2 = P Total - P H2O P H2 = 769 mm Hg mm Hg = 752 mm Hg T = 20 o C = 273 = 293 K P = 752 mm Hg /760 mm Hg /1 atm = atm V =___________ L
COLLECTION OVER WATER CONT.-II PV = nRT n = PV / RT n = n = mol mass = mol x 2.01 g H 2 / mol H 2 mass = ______________ g Hydrogen (0.989 atm) (0.156 L) ( L atm/mol K) (293 K)
Chemical Equation Calc - III ReactantsProducts Molecules Moles Mass Molecular Weight g/mol Atoms (Molecules) Avogadros Number 6.02 x Solutions Molarity moles / liter Gases PV = nRT
Gas Law Stoichiometry - I - NH 3 + 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 o C. The second with a volume of 1.16 L contains HCl gas at a pressure of atm and a temperature of 18.7 o C. 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: NH 3 (g) + HCl (g) NH 4 Cl (s) T NH3 = 18.7 o C = K
Gas Law Stoichiometry - II - NH 3 + HCl n = PV RT n NH3 = = mol NH 3 (0.776 atm) (2.79 L) ( L atm/mol K) (291.9 K) n HCl = = mol HCl (0.932 atm) (1.16 L) ( L atm/mol K) (291.9 K) limiting reactant Therefore the product will be mol NH 4 Cl or 2.28 g NH 4 Cl Ammonia remaining = mol mol = mol NH 3 V = 1.16 L L = 3.95 L P NH3 = = = ____ atm nRT V ( mol) ( L atm/mol K) (291.9 K) (3.95 L)
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.
Figure 5.11: An ideal gas particle in a cube wholse sides are of length L (in meters).
Figure 5.12: (a) The Cartesian coordinate axes.
Figure 5.12: (b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, u x, u y, and u z.
Figure 5.12: (c) In the xy plane, u x 2 + u y 2 = u yx 2 by the Pythagorean theorem.
Figure 5.13: (a) Only the x component of the gas particles 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.
Figure 5.14: Path of one particle in a gas.
A balloon filled with air at room temperature.
The balloon is dipped into liquid nitrogen at 77 K.
The balloon collapses as the molecules inside slow down because of the decreased temperature.
Velocity and Energy Kinetic Energy = 1/2mu 2 Average Kinetic Energy (KE avg ) –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 =Avogadros 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!!!
Figure 5.15: A plot of the relative number of O 2 molecules that have a given velocity at STP.
Figure 5.16: A plot of the relative number of N 2 molecules that have a given velocity at three temperatures.
Figure 5.17: A velocity distribution for nitrogen gas at 273 K.
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, H 2 = g/mol E K = x T = 1.5 x x 300K = 3 R 2 N A J/mol K x molecules/mol E K = x J/molecule = 1/2 mu 2 m = = x kg/molecule x kg/mole x molecules/mole x J/molecule = x kg/molecule(u 2 ) u = _______________ m/s = ___________ m/s
Molecular Mass and Molecular Speeds - II for methane CH 4 = g/mole 3 R 2 N A J/mol K x molecules/mole E K = x T = 1.5 x x 300K = E K = x J/molecule = 1/2 mu 2 m = = x kg/molecule x kg/mole x moleules/mole x J/molecule = x kg/molecule (u 2 ) u = ___________________ m/s = __________ m/s
Molecular Mass and Molecular Speeds - III for Carbon dioxide CO 2 = g/mole u = ___________________ m/s = _________ m/s x J/molecule = x kg/molecule (u 2 ) m = = x kg/molecule x kg/mole x molecules/mole E K = x J/molecule = 1/2mu 2 E K = x T = 1.5 x x 300K = 3 R 2 N A J/mol K x molecules/mole
Molecular Mass and Molecular Speeds - IV Molecule H 2 CH 4 CO 2 Molecular Mass (g/mol) Kinetic Energy (J/molecule) Velocity (m/s) x x x ,
Important Point ! At a given temperature, all gases have the same molecular Kinetic energy distributions. or The same average molecular Kinetic Energy!
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.
Figure 5.18: The effusion of a gas into an evacuated chamber.
Relative Diffusion of H 2 versus O 2 and N 2 gases Average Molecular weight of air: 20% O g/mol x 0.20 = % N g/mol x 0.80 = g/mol or approximately 29 g/mol
Graham's Law calc. Rate Hydrogen = Rate Air x (MM Air / MM Hydrogen ) 1/2 Rate Hydrogen = Rate Air x ( 29 / 2 ) 1/2 Rate Hydrogen = Rate Air x 3.95 Or Rate Hydrogen = Rate Air x 4 !!!!!!
Figure 5.19: (a) demonstration of the relative diffusion rates of NH 3 and HCL molecules through air.
Figure 5.19: (b) When HCL(g) and NH 3 (g) meet in the tube, a white ring of NH 4 CL(s) forms.
Figure 5.20: Uranium-enrichment converters from the Paducah gaseous diffusion plant in Kentucky.
Gaseous Diffusion Separation of Uranium 235 / UF 6 vs 238 UF 6 Separation Factor = S = after Two runs S = after approximately 2000 runs 235 UF 6 is > 99% Purity !!!!! Y - 12 Plant at Oak Ridge National Lab ( (6 x 19)) 0.5 ( (6 x 19)) 0.5
Example 5.9 (P167) Calculate the impact rate on a 1.00-cm 2 section of a vessel containing oxygen gas at a pressure of 1.00 atm and 27 0 C. Solution: Calculate Z A : Z A = A A = 1.00 x m 2 N RT V 2 M N P V RT = 1.00 atm ( ) L atm K mol (300. K) NVNV = 4.06 x x x x = Mol L molecules mol 1000 L m 3 = 2.44 x molecules/m 3 M = 32.0 g/mol x = 3.2 x kg/mol 1 kg 1000g Z A = (1.00 x m 2 ) (2.44 x m -3 ) x J K mol (300. K) 2(3.14)( 3.20 x kg/mol) Z A = 2.72 x collisions/s
Figure 5.21: The cylinder swept out by a gas particle of diameter d.
Like Example 5.10(P 169) Problem: Calculate the collision frequency for a Nitrogen molecule in a sample of pure nitrogen gas at 28 0 C and 2.0 atm. Assume that the diameter of an N 2 molecule is 290 pm. Solution: nVnV = = = x mol/L P RT 2.0 atm ( ) L atm K mol (301. K) NVNV = (8.097 x mol/L)( x molecules/mol)(1000L/m 3 ) NVNV = x molecules/m 3 d= 290 pm = 2.90 x m Also, for N 2, M = 2.80 x kg/mol Z = 4(4.876 x m -3 )(2.90 x m) 2 x ( J K -1 mol -1 )(301 K) 2.80 x kg/mol Z = collisions/sec
Like Example 5.11 (P171) Problem: Calculate the mean free path in a sample of nitrogen gas at 28 0 C and 2.00 atm. Solution: Using data from the previous example, we have: = = 5.49 x = 1 2 (N/V)( d 2 ) 1 2 ( x )( )(2.90 x ) 2 Note that a nitrogen molecule only travels 0.5 nano meters before it collides with another nitrogen atom.
Figure 5.22: Plots of PV/nRT versus P for several gases (200 K).
Figure 5.23: Plots of PV/nRT versus P for nitrogen gas at three temperatures.
Figure 5.24: (a) gas at low concentration - relatively few interactions between particles (b) gas at high concentration - many more interactions between particles.
Figure 5.25: Illustration of pairwise interactions among gas particles.
Table 5.3 (P172) Values of Van der Waals Constants for some Common Gases Gas a b atm L 2 mol 2 ( ) L mol He Ne Ar Kr Xe H N O Cl CO CH NH H 2 O
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).
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: n = = = mol PV (2.000 atm)(20.0L) RT ( Latm/molK)( K) P = = = atm nRT (1.663 mol) ( Latm/molK) ( K) V (1.000 L) P = - = - nRT n 2 a (1.663 mol) ( Latm/molK)( K) (V-nb) V 2 (1.00 L) - (1.663 mol)(0.0562) (1.663 mol) 2 (6.49) (1.00 L) 2 = = atm
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.
Figure 5.27: The variation of temperature and pressure with altitude.
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
Figure 5.28: Concentration (in molecules per million molecules of air) of some smog components versus time of day.
Figure 5.29: Our various modes of transportation produce large amounts of nitrogen oxides, which facilitate the formation of photochemical smog.
This photo was taken in Recent renovation has since replaced the deteriorating marble.
Figure 5.31: Diagram of the process for scrubbing sulfur dioxide from stack gases in power plants.