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Transcript of Chapter 5
Chapter 5Chapter 5
GasesGases
Chapter 5: Gases
5.1 Early Experiments5.2 The gas laws of Boyle, Charles, and Avogadro5.3 The Ideal Gas Law5.4 Gas Stiochiometry5.5 Dalton’s Law of Partial Pressures5.6 The Kinetic molecular Theory of Gases5.7 Effusion and Diffusion5.8 Collisions of Gas Particles with the Container Walls5.9 Intermolecular Collisions5.10 Real Gases5.11 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 H2 2.0
• Nitrogen N2 28.0
• Nitrogen Monoxide NO 30.0
• Oxygen O2 32.0
• Hydrogen Chloride HCL 36.5
• Ozone O3 48.0
• Ammonia NH3 17.0
• Methane CH4 16.0
Substances that are Gases under Normal Conditions
Substance Formula MM(g/mol)
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
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/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 = 33.8 ft
Density of Water
HeightWater HeightMercury
DensityMercury
DensityWater
=
Common Units of Pressure
Unit Atmospheric Pressure Scientific Field Used
Pascal (Pa); 1.01325 x 105 Pa SI unit; physics, kilopascal (kPa) 101.325 kPa Chemistry
Atmosphere (atm) 1 atm Chemistry
Millimeters of mercury 760 mmHg Chemistry, medicine (mmHg) biologyTorr 760 torr Chemistry
Pounds per square inch 14.7 lb/in2 Engineering (psi or lb/in2)
Bar 1.01325 bar Meteorology, Chemistry
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 341.6 mm Hg. Calculate the CO2 pressure intorr, atmospheres, and kilopascals.Plan: The pressure is in mmHg, so we use the conversion factors fromTable 5.2(p.178) to find the pressure in the other units.Solution:
PCO2 (torr) = 341.6 mm Hg x = 341.6 torr 1 torr1 mm Hg
converting from mmHg to torr:
converting from torr to atm:
PCO2( atm) = 341.6 torr x = 0.4495 atm 1 atm760 torr
converting from atm to kPa:
PCO2(kPa) = 0.4495 atm x = 45.54 kPa101.325 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.
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
kV
kP
Figure 5.4: Plotting Boyle’s data from Table 5.1.
Figure 5.5: Plot of PV versus P for several gases.
Applying Boyles Law to Gas ProblemsProblem: 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 thento liters, then we do the pressure change to obtain the final volume!Solution:
V1 (cm3)
V1 (ml)
V1 (L)
V2 (L)
1cm3 = 1 mL
1000mL = 1L
x P1/P2
P1 = 1.23 atm P2 = 3.16 atmV1 = 15.8 cm3 V2 = unknownT and n remain constant
V1 = 15.8 cm3 x x = 0.0158 L 1 mL 1 cm3
1 L1000mL
V2 = V1 x = 0.0158 L x = 0.00615 LP1
P2
1.23 atm3.16 atm
Boyle’s Law - A gas bubble in the ocean!
A bubble of gas is released by the submarine “Alvin” at a depth of6000 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 1.026 g/cm3,and use the mass of Hg in a barometer for comparison!)
Initial Conditions Final Conditions
V 1 = 1.00 cm3
P 1 = ?
V 2 = ?
P 2 = 1.00 atm
Calculation Continued
Pressure at depth = 6 x 103 ft x x x0.3048 m 1 ft
100 cm 1 m
1.026 g SH2O 1 cm3
Pressure at depth = 187,634.88 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:
Pressure = x x x x 10 mm 1 cm
Area1 cm2
1.00 cm3 Hg 13.6 g Hg
187,635 g = 1.00 atm760 mm Hg
Pressure = 182 atm Due to the added atmospheric pressure = 183 atm!
V2 = = = 183 cm3 = 0.183 litersV1 x P1
P2
1.00 cm3 x 183 atm 1.00 atm
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 = 1.4 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 T1 / V1 = T2 / V2 or:
T1
V1
T2=V2
T1 = V1 xT2
V2
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 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 = 192 K
• T2 = 192 K oC = K - 273.15 = 192 - 273 oC = -81oC
1.54 L3.20 L
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 = ( 75 - 32 )5/9 = 23.9o C• K = 23.9o C + 273.15 = 297.0 K• T2 = ( -70 - 32 ) 5/9 = - 56.7o C• K = - 56.7o C + 273.15 = 216.4 K
Antarctic Balloon Problem - II
• V1 / T1 = V2 / T2 V2 = V1 x ( T2 / T1 )
• V2 = 20.0 L x
• V2 = 14.6 L
• The Balloon shrinks from 20 L to 15 L !!!!!!!
• Just by going outside !!!!!
216.4 K297.0 K
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 0.185 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:
T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oCT1 = -17.68 oC + 273.15 K = 255.47 KT2 = 95.6 oC + 273.15 K = 368.8 K
P1 P2
T1 T2
=
P2 = P1 x = ?T2 T1
P2 = 4.28 atm x = 6.18 atm368.8 K255.47 K
Avogadro’s Law - Amount and Volume
The Amount of Gas (Moles) is directly proportional to the volumeof 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 gasn2 = final moles of gas V2 = final volume of gas
n1
V1
n2
V2
= or: n1 = n2 x V1
V2
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 1.143 atm and 28.5 oC is 2.93 m3, what will be the mass of SF6 in a container whose volume is543.9 m3 at 1.143 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 = 146.07 g/mol
V2
V1
n2 = n1 x = 0.0183 mol SF6 x = 3.40 mol SF6
2.67g SF6
146.07g SF6/mol= 0.0183 mol SF6
543.9 m3
2.93 m3
mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = 496 g SF6
Volume - Amount of Gas Relationship
Problem: A balloon contains 1.14 moles(2.298g H2) of Hydrogen and has a volume of 28.75 L. What mass of Hydrogen must be added to the balloon to increase it’s volume to 112.46 Liters. Assume T and P are constant.Plan: Volume and amount of gas are changing with T and P constant, sowe will use Avogadro’s law, and the change of conditions form. Solution:
n1 = 1.14 moles H2
n2 = 1.14 moles + ? moles
V1 = 28.75 LV2 = 112.46 L
T = constantP = constant
n1 n2
V1 V2= n2 = n1 x = 1.14 moles H2 x
V2
V1
n2 = 4.4593 moles = 4.46 moles
112.46 L 28.75 L
mass = moles x molecular massmass = 4.46 moles x 2.016 g / molmass = 8.99 g H2 gasadded mass = 8.99g - 2.30g = 6.69g
Change of Conditions, with no change in the amount of gas !
• = constant Therefore for a change
of conditions :
• T1 T2
P x V
T
P1 x V1=
P2 x V2
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 = 0.00310 L• T1 = 25 oC + 273 = 298 K• T2 = 155 oC + 273 = 428 K
Change of Condition Problem I : continued
•
•
•
•
•
=T1 T2
P1 x V1 P2 x V2
( 0.978 atm) ( 45.9 L) P2 (0.00310 L)
( 298 K) ( 428 K)=
P2 =( 428 K) ( 0.978 atm) ( 45.9 L)
= 2.08 x 104 atm( 298 K) ( 0.00310 L)
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 m3 V2 = ?• T1 = 25 oC + 273.15 T2 = -90 oC +273.15• = 298 K = 183 K • P1 = 1.0 atm P2 = 15 mm Hg
760 mm Hg/ atm
P2= 0.0198 atm
Change of Conditions Problem II: continued
• P1 x V1 P2 x V2
V2 =
• V2 = =
• V2 = 310.148 m3 = 310. m3 or 3.1 times the volume !!!
T1 T2= P1V1T2
T1P2
( 1.0 atm) ( 100 m3) (183 K)
(298 K) (0.198 atm)
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 + 273 = 298 K
• 900 oC + 273 = 1173 K
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/T2 V2 = V1P1T2/ P2T1
• V2 = = 197 L • VCO2 = (197 L C3H8) x (3 L CO2 / 1 L C3H8) =
VCO2 = 590. L CO2
( 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 = 00 C = 273.15 K
Standard Pressure = 1 atmosphere = 760 mm Mercury
At these “standard” conditions, if you have 1.0 mole of a gas it willoccupy a “standard molar volume”.
Standard Molar Volume = 22.414 Liters = 22.4 L
Table 5.2 (P 149) Molar Volumes for Various Gases at 00 and 1 atm
Gas Molar Volume (L)
Oxygen (O2) 22.397
Nitrogen (N2) 22.402
Hydrogen (H2) 22.433
Helium (He) 22.434
Argon (Ar) 22.397
Carbon dioxide (CO2) 22.260
Ammonia (NH3) 22.079
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 = 8.314 J / mol K = 8.314 J mol-1 K-1
• R = 0.08206 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 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
Evaluation of the Ideal Gas Constant, R
Ideal gas Equation PV = nRT R = PVnT
at Standard Temperature and Pressure, the molar volume = 22.4 L P = 1.00 atm (by Definition!) T = 0 oC = 273.15 K (by Definition!) n = 1.00 moles (by Definition!)
R = = 0.08206 (1.00 atm) ( 22.414 L)( 1.00 mole) ( 273.15 K)
L atm mol K
or to three significant figures R = 0.0821 L atm mol K
Values of R in Different Units
R* = 0.08206
R = 62.36
R = 8.314
R = 8.314
Atm x LMol x KTorr x LMol 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
Gas Law: Solving for Pressure
Problem: Calculate the pressure in a container whose Volume is 87.5 Land 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:
nXe = = 38.37014471 mol Xe 5038 g Xe131.3 g Xe / mol
T = 18.8 oC + 273.15 K = 291.95 K
PV = nRT P = nRT V
P = = 10.5108 atm = 10.5 atm(38.37 mol )(0.0821 L atm)(291.95 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 0.150 m3 and the temperature is 36.0 oC.
• n = 375 g N2/ 28.0 g N2 / mol = 13.4 mol N2
• V = 0.150 m3 x 1000 L / m3 = 150 L• T = 36.0 oC + 273.15 = 309.2 K• PV=nRT P= nRT/V
• P = • • P = 2.27 atm
( 13.4 mol) ( 0.08206 L atm/mol K) ( 309.2 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 oF and the pressure is 748 mm Hg?
• P = 748 mm Hg x 1atm / 760 mm Hg= 0.984 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 + 273 = 303 K
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
( 0.984 atm) ( 3.99 x 105 L)
( 0.08206 L atm/mol K) ( 303 K )
Inflated dual air bag
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) 2 Na (s) + 3 N2 (g)
• mol NaN3 = 60.0 g NaN3 / 65.02 g NaN3 / mol =
= 0.9228 mol NaN3
• mol N2= 0.9228 mol NaN3 x 3 mol N2/2 mol NaN3
= 1.38 mol N2
Sodium Azide Calc: - II
• PV = nRT V = nRT/P
• V =
• V = 30.8 liters
( 1.38 mol) (0.08206 L atm / mol K) (294 K)
( 823 mm Hg / 760 mmHg / atm )
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 + 273 = 328 K n = mass / Molar mass = g / M • d = = • d = 0.626 g / L
P x MR x T
( 0.989 atm) ( 17.03 g/mol)
( 0.08206 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 370 C according to the reaction below: CaCO3(s) CaO(s) + CO2 (g)
Solution:
36.8g x = 0.368 mol CaCO3 1 mol CaCO3
100.1g CaCO3
Since each mole of CaCO3 produces one 1 mole of CO2 , 0.368 molesof CO2 at STP, we now have to convert to the conditions stated:
V = =nRT P
(0.368 mol)(0.08206L atm/mol K)(310.15K) (715mmHg/760mm Hg/atm)
V = 9.96 L
Calculation of Molar Mass
• n =
• n = =
•
Mass
Molar Mass
P x VR x T
Mass
Molar Mass
Molar Mass = = MMMass x R x T
P x V
Dumas Method of Molar Mass
Problem: A volatile liquid is placed in a flask whose volume is 590.0 mland allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass of the flask before and after the experiment was 148.375g and 149.457 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 = 0.9684 atm 1 atm760 mm Hg
Molar Mass = = 58.02 g/mol
mass = 149.457g - 148.375g = 1.082 g
(1.082 g)(0.0821 Latm/mol K)(373.2 K)
( 0.9684 atm)(0.590 L)
note: the compound is acetone C3H6O = MM = 58g mol !
Calculation of Molecular Weight of a Gas Natural Gas - Methane
Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 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 = 550.0 Torr x x = 0.724 atm1mm Hg 1 Torr
1.00 atm760 mm Hg
V = 250.0 ml x = 0.250 L 1.00 L1000 ml
T = 25.0 oC + 273.15 K = 298.2 K
n = P VR T
n = = 0.007393 mol (0.0821 L atm/mol K) (298.2 K)
(0.724 atm)(0.250 L)
MM = 0.118 g / 0.007393 mol = 16.0 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.
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
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.
Dalton’s Law of Partial Pressures - Prob #1
• A 2.00 L flask contains 3.00 g of CO2 and 0.10 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 + 273 = 290 K• nCO2 = 3.00 g CO2/ 44.01 g CO2 / mol CO2
• = 0.0682 mol CO2
• PCO2 = nCO2RT/V • • PCO2 = • PCO2 = 0.811 atm
( 0.0682 mol CO2) ( 0.08206 L atm/mol K) ( 290 K)
(2.00 L)
Dalton’s Law Problem - #1 cont.
• nHe = 0.10 g He / 4.003 g He / mol He• = 0.025 mol He• PHe = nHeRT/V
• PHe = • PHe = 0.30 atm
• PTotal = PCO2 + PHe = 0.812 atm + 0.30 atm• PTotal = 1.11 atm
(0.025 mol) ( 0.08206 L atm / mol K) ( 290 K )
( 2.00 L )
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 = 4.46 + 0.74 + 2.15 = 7.35 mol
• XNe = 4.46 mol Ne / 7.35 mol = 0.607
• PNe = XNe PTotal = 0.607 ( 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 = 0.293 (2.00 atm) = 0.586 atm for Xe
Relative Humidity
• 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/ 12.788 mm Hg )x100%
= 51.1 %
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) 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 - 17.5 mm Hg = 752 mm Hg• T = 20oC + 273 = 293 K• P = 752 mm Hg /760 mm Hg /1 atm = 0.989 atm• V = 0.156 L
COLLECTION OVER WATER CONT.-II
• PV = nRT n = PV / RT
• n =
• n = 0.00641 mol
• mass = 0.00641 mol x 2.01 g H2 / mol H2
• mass = 0.0129 g Hydrogen
(0.989 atm) (0.156 L)
(0.0821 L atm/mol K) (293 K)
Chemical Equation Calc - III
Reactants ProductsMolecules
Moles
MassMolecularWeight g/mol
Atoms (Molecules)Avogadro’sNumber
6.02 x 1023
Solutions
Molaritymoles / liter
Gases
PV = nRT
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 Lcontains Ammonia gas at a pressure of 0.776 atm and a temperature of 18.7 oC. The second with a volume of 1.16 L contains HCl gas at a pressure of 0.932 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 molesof 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 + 273.15 = 291.9 K
Gas Law Stoichiometry - II - NH3 + HCl
n = PVRT
nNH3 = = 0.0903 mol NH3(0.776 atm) (2.79 L)
(0.0821 L atm/mol K) (291.9 K)
nHCl = = 0.0451 mol HCl(0.932 atm) (1.16 L)(0.0821 L atm/mol K) (291.9 K)
limiting
reactant
Therefore the product will be 0.0451 mol NH4Cl or 2.28 g NH4Cl
Ammonia remaining = 0.0903 mol - 0.0451 mol = 0.0452 mol NH3
V = 1.16 L + 2.79 L = 3.95 L
PNH3 = = = 0.274 atmnRTV
(0.0452 mol) (0.0821 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, ux, uy, and uz.
Figure 5.12: (c) In the xy plane,ux2 + uy2 = uyx2 by the Pythagorean theorem.
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.
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 = 1/2mu2
• 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!!!
Figure 5.15: A plot of the relative number of O2 molecules that have a given velocity at STP.
Figure 5.16: A plot of the relative number of N2 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 theaverage speeds of the three molecules.Solution: for Hydrogen, H2 = 2.016 g/mol
EK = x T = 1.5 x x 300K = 3 R2 NA
8.314 J/mol K6.022 x 1023 molecules/mol
EK = 6.213 x 10 - 21 J/molecule = 1/2 mu2
m = = 3.348 x 10 - 27 kg/molecule 2.016 x 10 -3 kg/mole6.022 x 1023 molecules/mole
6.213 x 10 - 21 J/molecule = 1.674 x 10 - 27 kg/molecule(u2)
u = 1.926 x 103 m/s = 1,926 m/s
Molecular Mass and Molecular Speeds - II
for methane CH4 = 16.04 g/mole
3 R2 NA
8.314 J/mol K6.022 x 1023 molecules/mole
EK = x T = 1.5 x x 300K =
EK = 6.213 x 10 - 21 J/molecule = 1/2 mu2
m = = 2.664 x 10 - 26 kg/molecule 16.04 x 10 - 3 kg/mole6.022 x 1023 moleules/mole
6.213 x 10 - 21 J/molecule = 1.332 x 10 - 26 kg/molecule (u2)
u = 6.838 x 102 m/s = 683.8 m/s
Molecular Mass and Molecular Speeds - III
for Carbon dioxide CO2 = 44.01 g/mole
u = 4.124 x 102 m/s = 412.4 m/s
6.213 x 10 - 21 J/molecule = 3.654 x 10 - 26 kg/molecule (u2)
m = = 7.308 x 10 - 26 kg/molecule 44.01 x 10 - 3 kg/mole6.022 x 1023 molecules/mole
EK = 6.213 x 10 - 21 J/molecule = 1/2mu2
EK = x T = 1.5 x x 300K =3 R2 NA
8.314 J/mol K6.022 x 1023 molecules/mole
Molecular Mass and Molecular Speeds - IV
Molecule H2 CH4 CO2
MolecularMass (g/mol)
Kinetic Energy (J/molecule)
Velocity (m/s)
2.016 16.04 44.01
6.213 x 10 - 21 6.213 x 10 - 21 6.213 x 10 - 21
1,926 683.8 412.4
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 H2 versus O2 and N2 gases
• Average Molecular weight of air:
• 20% O2 32.0 g/mol x 0.20 = 6.40
• 80% N2 28.0 g/mol x 0.80 = 22.40
28.80 • 28.80 g/mol
• or approximately 29 g/mol
Graham's Law calc.
• RateHydrogen = RateAir x (MMAir / MMHydrogen)1/2
• RateHydrogen = RateAir x ( 29 / 2 )1/2
• RateHydrogen = RateAir x 3.95
• Or RateHydrogen = RateAir x 4 !!!!!!
Fig 5.19 (P 203)
NH3 (g) + HCl(g) = NH4Cl (s)
• HCl = 36.46 g/mol NH3 = 17.03 g/mol
• RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2
• RateNH3 = RateHCl x 1.463
Figure 5.19: (a) demonstration of the relative diffusion rates of NH3 and HCL
molecules through air.
Figure 5.19: (b) When HCL(g) and NH3 (g) meet in the tube, a white ring of
NH4CL(s) forms.
Figure 5.20: Uranium-enrichment converters from the Paducah gaseous
diffusion plant in Kentucky.
Gaseous Diffusion Separation of Uranium 235 / 238
• 235UF6 vs 238UF6
• Separation Factor = S =• • after Two runs S = 1.0086• after approximately 2000 runs
• 235UF6 is > 99% Purity !!!!!
• Y - 12 Plant at Oak Ridge National Lab
(238.05 + (6 x 19))0.5
(235.04 + (6 x 19))0.5
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:Calculate ZA : ZA = A
A = 1.00 x 10-4 m2
N RTV 2M
N PV RT
= =
1.00 atm
0.08206( )L atmK mol
(300. K)
N V
= 4.06 x 10-2 x 6.022 x 1023 x = Mol L
molecules mol
1000 L m3
= 2.44 x 1025 molecules/m3
M = 32.0 g/mol x = 3.2 x 10-2 kg/mol 1 kg1000g
ZA = (1.00 x 10-4 m2) (2.44 x 1025 m-3) x 8.3145 JK mol
(300. K)
2(3.14)( 3.20 x 10-2 kg/mol)ZA = 2.72 x 1023 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 280C and 2.0 atm. Assume that the diameter of an N2 molecule is 290 pm.Solution:
nV
= = = 8.097 x 10-2 mol/L PRT
2.0 atm
( )0.08206 L atmK mol
(301. K)
NV
= (8.097 x 10-2 mol/L)( 6.022 x 1023 molecules/mol)(1000L/m3)
NV
= 4.876 x 1025 molecules/m3 d= 290 pm = 2.90 x 10-10 m
Also, for N2, M = 2.80 x 10-2 kg/mol
Z = 4(4.876 x 1025 m-3)(2.90 x 10-10 m)2 x (8.3145 J K-1mol-1)(301 K)
2.80 x 10-2 kg/molZ = 8.69 x 108 collisions/sec
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:
= = 5.49 x 10-8
= 1
2 (N/V)(d2)
1
2 ( 4.876 x 1025 )()(2.90 x 10-10 )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 batm L2
mol2( ) ( ) Lmol
He 0.034 0.0237 Ne 0.211 0.0171Ar 1.35 0.0322Kr 2.32 0.0398Xe 4.19 0.0511H2 0.244 0.0266N2 1.39 0.0391O2 1.36 0.0318Cl2 6.49 0.0562CO2 3.59 0.0427CH4 2.25 0.0428NH3 4.17 0.0371H2O 5.46 0.0305
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 20.000C at a pressure of 2.000 atm. if the tank is pressureized to a new volume of 1.000 L and a temperature of 150.000C. What is the new pressure using the ideal gas equation, and the Van der Waals equation?Plan: Do the calculations!Solution:
n = = = 1.663 mol PV (2.000 atm)(20.0L)RT (0.08206 Latm/molK)(293.15 K)
P = = = 57.745 atmnRT (1.663 mol) (0.08206 Latm/molK) (423.15 K) V (1.000 L)
P = - = - nRT n2a (1.663 mol) (0.08206 Latm/molK)(423.15 K)(V-nb) V2 (1.00 L) - (1.663 mol)(0.0562)
(1.663 mol)2(6.49) (1.00 L)2
= 63.699 - 17.948 = 45.751 atm
Table 5.4(P173) Atmospheric Composition near Sea Level (dry air)#
Component Mole Fraction
N2 0.78084O2 0.20946Ar 0.00934CO2 0.000345Ne 0.00001818He 0.00000524CH4 0.00000168Kr 0.00000114H2 0.0000005NO 0.0000005Xe 0.000000087
# 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 1990. 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.
Atmospheric Environmental Problems
Rural Ozone - Nitric oxide Chemistry
The Greenhouse Effect
Ozone destruction – By CFC’s
Antarctic Ozone Hole
Rural air pollution reaction between Ozone and Nitric Oxide (NO).
The reaction between Nitric Oxide (NO) and Ozone is very fast andOccurs when ever air containing the two molecules interacts. This is shown by date collected using sensors in a truck that is driving around a housing development near Fairbanks Alaska in 1997.
NO(g) + O3 (g) NO2 (g) + O2 (g)
The sensor used for the NOx is primarely measuring No, but some NO2 is normally reduced in the sensor, so it is called NOx!
Air pollution Measurements near Fairbanks Alaska 1977
Ozone and Nitric Oxide (NOx) near Fairbanks Alaska, 1977
The “Greenhouse effect”
Incoming solar radiationwarms the surface of the
Earth, and the warm surfaceemits infrared radiation to
space, but CO2 & H2O molecules in the atmosphereabsorb the IR radiation and
warm the atmosphere.
The Electromagnetic Spectrum
Earth’s Surface
The “Greenhouse Effect”
The Greenhouse effect
Carbon Dioxide Increase in the Atmosphere at Mauna Loa Observatory, HawaiiYear CO2 Conc. Year CO2 Conc. Year CO2 Conc.
1971 324.62 1983 342.89 1995 360.621972 325.43 1984 343.99 1996 362.36 1973 326.49 1985 345.53 1997 363.471974 329.30 1986 346.73 1998 366.491975 330.85 1987 349.08 1999 368.131976 331.78 1988 351.43 2000 369.131977 333.68 1989 352.89 2001 371.081978 335.28 1990 354.12 2002 373.161979 336.38 1991 355.45 2003 375.791980 338.05 1992 356.20 20041981 339.69 1993 356.92 2005 1982 341.20 1994 358.64 2006
320
330
340
350
360
370
380
1965 1970 1975 1980 1985 1990 1995 2000 2005
Series1
Carbon Dioxide record at the Mauna Loa Observatory in Hawaii 1971-2003
Stratospheric Chemical reactions leading to Ozone Depletion - INormal Ozone production in the stratosphere and its absorption of Ultra violet. Ozone is made when Ultra Violet Light (UVA) is absorbed by molecular oxygen to create two oxygen atoms. If these oxygen atoms collide with molecular oxygen they react to form Ozone. The Ozone formed will absorb Ultra Violet light in theUVB wavelengths, thus absorbing most of the short wavelength UV light, which is also called ionizing radiation, and will destroy biological molecules.
O2 2 O h = 170-190 nm (UVA)
O + O2 O3
O3 O2 + O = h = 200-300 nm (UVB)
h
h
Stratospheric Chemical reactions leading to Ozone Depletion - IIIn 1971 Dr. Harold Johnson A professor at the University of California at Berkeley, calculated that the Nitrogen oxides produced by the Jet engines of the then proposed Super Sonic Transports (SST’s) would destroy much of the Ozone layer that protects us on the Earth’s surface from Ultra Violet light (both UVA ad UVB). The reactions he proposed are:
N2 + O2 2 NO =This symbol indicated heat energy
NO + O3 NO2 + O2 This reaction is very fast! Adding these last two reactions NO2 + O NO + O2 together and canceling out similar reactants gives the catalytic reaction! O3 + O 2 O2 This reaction shows Ozone destruction!The results of this research stopped the US from making an SST!
Chlorofluorocarbons (Freon’s)
Freon-11 CF2Cl2 : Freon-12 CFCl3
Stratospheric Reaction Chemistry:
CF2Cl2 CF2Cl + Cl
Cl + O3 O2 + ClO ClO + O O2 + Cl
O3 + O 2 O2
Catalytic Ozone destruction
.h
..
.
Other Chlorine reactions in the Stratosphere:
H2O + ClNO3 HNO3 + HOCl
HCl + ClNO3 HNO + Cl2
HOCl ClO + H
Cl2 2 Cl
Cl + H HCl (HCl is removed from the atmosphere on ice crystals)
.
.
The absorption spectra for Ozone and Molecular Oxygen. The region that is absorbed by molecular
Oxygen and Ozone are referred to as the UVA (100-240 nm) and the UVB (240-360 nm) region of the electromagnetic
Spectrum and are referred to as ionizing radiation.
UVAUVB
Plot of wavelength verses intensityof in coming solar radiation.
UVA and UVB are known as ionizing radiation and will destroy molecules!
UVA UVBVisible
AntarcticaIce Depth 13,000 ft
Wind Wind
Why Antarctica?
The Antarctic continent is a large Ice covered land mass with An averagealtitude of about 11,000 ft, the highest areas are about 15,000 ft, wind can not blow from the sea over the continent, but descends from the high interior through catabatic flow to the coast. This draws air from high over the continent to make up the air flowing off the continent, bringing stratospheric are to the in land surface. Freon’s and other troposphereic gases are mixed in this way up into the stratosphere over the Antarctic Region, where they are exposed to UVA radiation that ionizes them releasing chlorine to the stratosphere, where it can destroy ozone.
Ozone Levels over Antarctica, The Antarctic Ozone Hole
The Ozone Hole over Antarctica
Ozone and ClO
Antarctic OzoneHole - 1998
Antarctic Ozone Hole 2002 & 2003
South PoleAntarctica
Continental North America
Mauna Loa Observatory