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 = __________ 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 (psl 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 = _____________ 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 = ________ 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 = _____ atm Due to the added atmospheric pressure = ___ atm!

V2 = = = ____ cm3 = litersV1 x P1

P2

1.00 cm3 x ____ 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 = __________ 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 =______K

• T2 = ___ K oC = K - 273.15 = ______ - 273 oC = ______oC

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 = __________ K• T2 = ( -70 - 32 ) 5/9 = - 56.7o C• K = - 56.7o C + 273.15 = ___________ K

Antarctic Balloon Problem - II

• V1 / T1 = V2 / T2 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 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 = _______ 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 = __________ 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 = ___________ 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)

= ________ 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= ___________ atm

Change of Conditions Problem II: continued

• P1 x V1 P2 x V2

V2 =

• V2 = =

• V2 = 310.148 m3 = __________ 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 = _________ 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 = = _______ atm = 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 = 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

= ___________mol N2

Sodium Azide Calc: - II

• PV = nRT V = nRT/P

• V =

• V = __________ 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 = __________ 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 = __________ 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 = ____________ 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 = _____________ 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%

= ___________ %

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 =___________ 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 = ______________ 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 = = = ____ 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 = _______________ m/s = ___________ 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 = ___________________ m/s = __________ m/s

Molecular Mass and Molecular Speeds - III

for Carbon dioxide CO2 = 44.01 g/mole

u = ___________________ m/s = _________ 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 !!!!!!

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 = 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 = = = 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 = 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.