Mullis 1

Characteristics of GasesCharacteristics of Gases

► Vapor = term for gases of substances Vapor = term for gases of substances that are often liquids/solids under that are often liquids/solids under ordinary conditionsordinary conditions

► Unique gas propertiesUnique gas properties1.1. Highly compressible Highly compressible

2.2. Inverse pressure-volume relationshipInverse pressure-volume relationship

3.3. Form homogeneous mixtures with other Form homogeneous mixtures with other gasesgases

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Pressures of Enclosed Gases Pressures of Enclosed Gases and Manometersand Manometers

► Barometer: Used to measure atmospheric Barometer: Used to measure atmospheric pressure*pressure*

► Manometer: Used to measure pressures of gases Manometer: Used to measure pressures of gases not open to the atmospherenot open to the atmosphere

► Manometer is a bulb of gas attached to a U-tube Manometer is a bulb of gas attached to a U-tube containing Hg.containing Hg. If U-tube is closed, pressure of gas is the difference in If U-tube is closed, pressure of gas is the difference in

height of the liquid.height of the liquid. If U-tube is open, add correction term:If U-tube is open, add correction term: If PIf Pgasgas < P < Patmatm then P then Pgasgas + P + Phh = P = Patmatm

If PIf Pgasgas > P > Patmatm then P then Pgasgas = P = Phh + P + Patmatm

* Alternative unit for atmospheric pressure is 1 bar = 10* Alternative unit for atmospheric pressure is 1 bar = 1055 Pa Pa

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Kinetic Molecular TheoryKinetic Molecular Theory► Number of moleculesNumber of molecules► TempTemp► VolumeVolume► PressurePressure

► Number of dancersNumber of dancers► Beat of musicBeat of music► Size of roomSize of room► Number and force of Number and force of

collisionscollisions

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Kinetic Molecular TheoryKinetic Molecular Theory

►Accounts for behavior of atoms and Accounts for behavior of atoms and moleculesmolecules

►Based on idea that particles are Based on idea that particles are always movingalways moving

►Provides model for an ideal gasProvides model for an ideal gas► Ideal Gas = Imaginary: Fits all Ideal Gas = Imaginary: Fits all

assumptions of the K.M theoryassumptions of the K.M theory►Real gas = Does not fit all these Real gas = Does not fit all these

assumptionsassumptions

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5 assumptions of Kinetic-5 assumptions of Kinetic-molecular Theorymolecular Theory

1.1. Gases = large numbers of tiny particles Gases = large numbers of tiny particles that are that are far apartfar apart..

2.2. Collisions between gas particles and Collisions between gas particles and container walls are container walls are elastic collisionselastic collisions (no net (no net loss in kinetic energy).loss in kinetic energy).

3.3. Gas particles are always Gas particles are always movingmoving rapidly rapidly and randomly.and randomly.

4.4. There are There are no forcesno forces of repulsion or of repulsion or attraction attraction betweenbetween gas particles. gas particles.

5.5. The average kinetic energy of gas particles The average kinetic energy of gas particles depends on temperaturedepends on temperature..

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Physical Properties of GassesPhysical Properties of Gasses

►Gases have no definite shape or Gases have no definite shape or volume – they take shape of volume – they take shape of container.container.

►Gas particles glide rapidly past each Gas particles glide rapidly past each other (fluid).other (fluid).

►Gases have low density.Gases have low density.►Gases are easily compressed.Gases are easily compressed.►Gas molecules spread out and mix Gas molecules spread out and mix

easilyeasily

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►Diffusion = mixing of 2 substances due Diffusion = mixing of 2 substances due to random motion.to random motion.

►Effusion = Gas particles pass through Effusion = Gas particles pass through

a tiny a tiny ooppeenniinng……..…g……..…

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Real GasesReal Gases►Real gases occupy space and exert Real gases occupy space and exert

attractive forces on each other.attractive forces on each other.►The K-M theory is more likely to hold The K-M theory is more likely to hold

true for particles which have little true for particles which have little attraction for each other.attraction for each other.

►Particles of NParticles of N22 and H and H22 are nonpolar are nonpolar diatomic molecules and closely diatomic molecules and closely approximate ideal gas behavior.approximate ideal gas behavior.

►More polar molecules = less likely to More polar molecules = less likely to behave like an ideal gas. Examples of behave like an ideal gas. Examples of polar gas molecules are HCl, ammonia polar gas molecules are HCl, ammonia and water.and water.

Mullis 9

Gas BehaviorGas Behavior

► Particles in a gas are very far apart.Particles in a gas are very far apart.► Each gas particle is largely unaffected by its Each gas particle is largely unaffected by its

neighbors.neighbors.► Gases behave similarly at different pressures Gases behave similarly at different pressures

and temperatures according to gas laws.and temperatures according to gas laws.► To identify a gas that is “most” ideal, choose To identify a gas that is “most” ideal, choose

one that is light, nonpolar and a noble gas if one that is light, nonpolar and a noble gas if possible.possible.

► Ex: Which gas is most likely to DEVIATE Ex: Which gas is most likely to DEVIATE from the kinetic molecular theory, or is the from the kinetic molecular theory, or is the “least” ideal: N“least” ideal: N22, O, O22, He, Kr, or SO, He, Kr, or SO22? ?

Answer: sulfur dioxide due to relative polarity Answer: sulfur dioxide due to relative polarity and mass.and mass.

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Boyle’s LawBoyle’s Law►Pressure goes up if volume goes Pressure goes up if volume goes

down.down.►Volume goes down if pressure goes Volume goes down if pressure goes

up.up.►The more pressure increases, the The more pressure increases, the

smaller the change in volume.smaller the change in volume.

Boyle's Law

0

2

4

6

8

10

12

14

0 20 40 60 80 100 120 140

Pressure (N/cm2)

Vo

lum

e (c

m3 )

P , V

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Boyle’s lawBoyle’s law►Pressure is the force created by Pressure is the force created by

particles striking the walls of a particles striking the walls of a container.container.

►At constant temperature, molecules At constant temperature, molecules strike the sides of container more often strike the sides of container more often if space is smaller.if space is smaller.

VV11PP11 = V = V22PP22

►Squeeze a balloon: If reduce volume Squeeze a balloon: If reduce volume enough, balloon will pop because enough, balloon will pop because pressure inside is higher than the walls pressure inside is higher than the walls of balloon can tolerate.of balloon can tolerate.

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Charles’ LawCharles’ Law

►As temperature goes up, volume goes As temperature goes up, volume goes up.up.

►Assumes constant pressure.Assumes constant pressure.

VV11 = V = V22

TT11 T T22

Charles' Law

0

5

10

15

20

25

0 5 10 15 20 25

Temperature (deg C)

Vo

lum

e (c

m3)

T , V

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Charles’ lawCharles’ law

►As temperature goes up volume goes up.As temperature goes up volume goes up.►Adding heat energy causes particles to Adding heat energy causes particles to

move faster.move faster.►Faster-moving molecules strike walls of Faster-moving molecules strike walls of

container more often. The container container more often. The container expands if walls are flexible.expands if walls are flexible.

► If you cool gas in a container, it will If you cool gas in a container, it will shrink.shrink.

►Air-filled, sealed bag placed in freezer Air-filled, sealed bag placed in freezer will shrink.will shrink.

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Gay-Lussac’s LawGay-Lussac’s Law►As temperature increases, pressure As temperature increases, pressure

increases.increases.►Assumes volume is held constant.Assumes volume is held constant.

PP11 = P = P22

TT11 T T22

►A can of spray paint will explode near A can of spray paint will explode near a heat source.a heat source.

►Example is a pressure cooker.Example is a pressure cooker.

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Combined Gas LawCombined Gas Law

► In real life, more than one variable may In real life, more than one variable may change. If have more than one condition change. If have more than one condition changing, use the combined formula.changing, use the combined formula.

► In solving problems, use the combined In solving problems, use the combined gas law if you know more than 3 gas law if you know more than 3 variables.variables.

VV11PP11 = V = V22PP22

TT11 T T22

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Using Gas LawsUsing Gas Laws►Convert temperatures to Kelvin!Convert temperatures to Kelvin!►Ensure volumes and/or pressures are in Ensure volumes and/or pressures are in

the same units on both sides of equation.the same units on both sides of equation.►STP = 0STP = 0°° C and 1 atm. C and 1 atm.►Use proper equation to solve for desired Use proper equation to solve for desired

value using given information.value using given information.

VV11PP11 = V = V22PP2 2 VV11 = V = V2 2 PP11 = P = P2 2 VV11PP11 = = VV22PP22

TT11 T T2 2 TT11 T T2 2 TT11 TT22

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Gay Lussac’s law of combining Gay Lussac’s law of combining volumes of gasesvolumes of gases

►When gases combine, they combine in When gases combine, they combine in simple whole number ratios.simple whole number ratios.

►These simple numbers are the These simple numbers are the coefficients of the balanced equation.coefficients of the balanced equation.

NN22 + 3H + 3H22 2NH2NH33

►3 volumes of hydrogen will produce 2 3 volumes of hydrogen will produce 2 volumes of ammoniavolumes of ammonia

Mullis 18

Avogadro’s Law and Molar Volume of Avogadro’s Law and Molar Volume of GasesGases

►Equal volumes of gases (at the same temp Equal volumes of gases (at the same temp and pressure) contain an equal number of and pressure) contain an equal number of molecules.molecules.

In the equation for ammonia formation, In the equation for ammonia formation,

1 volume N1 volume N22 = 1 molecule N = 1 molecule N22 = 1 mole N = 1 mole N22 ►One mole of any gas will occupy the same One mole of any gas will occupy the same

volume as one mole of any other gasvolume as one mole of any other gas►Standard molar volume of a gas Standard molar volume of a gas is the is the

volume occupied by one mole of a gas at volume occupied by one mole of a gas at STP.STP.Standard molar volume of a gas Standard molar volume of a gas is is 22.4 22.4 L.L.

Mullis 19

Sample molar volume Sample molar volume problemproblem

►A chemical reaction produces 98.0 mL of A chemical reaction produces 98.0 mL of sulfur dioxide gas at STP. What was the sulfur dioxide gas at STP. What was the mass, in grams, of the gas produced?mass, in grams, of the gas produced?

***Turn mL to L first! ***Turn mL to L first! (This way, you can can use 22.4 (This way, you can can use 22.4 L)L)

98 mL 1 L98 mL 1 L 1 mol SO 1 mol SO22 64.07g SO 64.07g SO22 = 0.280g = 0.280g SOSO22

1000 mL1000 mL 22.4 L 22.4 L 1 mol SO 1 mol SO22

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Sample molar volume problem 2Sample molar volume problem 2What is the volume of 77.0 g of nitrogen What is the volume of 77.0 g of nitrogen

dioxide gas at STP?dioxide gas at STP?

77.0 g NO77.0 g NO2 2 1 mol NO1 mol NO22 22.4 L 22.4 L = 37.5 = 37.5 L NOL NO22

46.01g NO46.01g NO2 2 1 mol NO1 mol NO22

Mullis 21

Ideal Gas LawIdeal Gas Law► Mathematical relationship for PVT and Mathematical relationship for PVT and

nnumber of moles of gasumber of moles of gas

PV = PV = nnRTRT nn = number of moles = number of molesR = ideal gas constantR = ideal gas constantP = pressureP = pressureV = volume in LV = volume in LT = Temperature in KT = Temperature in K

R = 0.0821 if pressure is in atmR = 0.0821 if pressure is in atmR = 8.314 if pressure is in kPaR = 8.314 if pressure is in kPaR = 62.4 if pressure is in mm HgR = 62.4 if pressure is in mm Hg

Mullis 22

Sample Ideal Gas Law Sample Ideal Gas Law ProblemProblem

► What pressure in atm will 1.36 kg of NWhat pressure in atm will 1.36 kg of N22O gas exert O gas exert when it is compressed in a 25.0 L cylinder and is when it is compressed in a 25.0 L cylinder and is stored in an outdoor shed where the temperature stored in an outdoor shed where the temperature can reach 59can reach 59°°C in summer?C in summer?

V = 25.0 LV = 25.0 L T = 59+273 = 332 K T = 59+273 = 332 K P = ?P = ?R = 0.0821R = 0.0821L-atmL-atm n = 1.36 kg n = 1.36 kg converted to molesconverted to moles

mol-K mol-K

► 1.36 kg N1.36 kg N22O 1000 g O 1000 g 1 mol N1 mol N22OO = 30.90 mol N = 30.90 mol N22OO 1 kg1 kg 44.02 g N44.02 g N22OO

► PV = nRTPV = nRT► P = P = 30.90 mol x 0.0821 L-atm x 332 K30.90 mol x 0.0821 L-atm x 332 K = 33.7 = 33.7

atmatm25.0 L mol-K25.0 L mol-K

Mullis 23

Volume-Volume CalculationsVolume-Volume Calculations►Volume ratios for gases are expressed the Volume ratios for gases are expressed the

same way as mole ratios we used in other same way as mole ratios we used in other stoichiometry problems.stoichiometry problems.

NN22 + 3H + 3H22 2NH2NH33

Volume ratios are:Volume ratios are:

2 volumes NH2 volumes NH33 3 volumes H3 volumes H2 2 2 volumes 2 volumes NHNH3 3

3 volumes H3 volumes H2 2 1 volume N1 volume N2 2 1 volume N1 volume N22

Mullis 24

Sample Volume-Volume ProblemSample Volume-Volume Problem

►How many liters of oxygen are needed How many liters of oxygen are needed to burn 100 L of carbon monoxide?to burn 100 L of carbon monoxide?

2CO + O2CO + O22 2CO2CO22

100 L CO 1 volume O100 L CO 1 volume O22 = 50 L O = 50 L O22

2 volume CO2 volume CO

Mullis 25

Sample Volume-Volume Problem 2Sample Volume-Volume Problem 2Ethanol burns according to the equation below. Ethanol burns according to the equation below.

At 2.26 atm and 40At 2.26 atm and 40° C, 55.8 mL of oxygen are ° C, 55.8 mL of oxygen are used. What volume of COused. What volume of CO22 is produced when is produced when measured at STP?measured at STP?

CC22HH55OH + 3OOH + 3O22 2CO2CO2 2 + 3H+ 3H22OONumber moles oxygen under these conditions Number moles oxygen under these conditions

is?is?PV = nRT: PV = nRT: 2.26 2.26 atmatm(.0558(.0558LL)) = n = 0.0049 mol = n = 0.0049 mol

OO22

((0.0821 0.0821 L-atmL-atm)(313K)(313K)) mol-Kmol-K

0.0049 mol O0.0049 mol O22 2 mol CO 2 mol CO2 2 22.4 L =0.073 L CO 22.4 L =0.073 L CO 22

3 mol O3 mol O2 2 1 mol CO1 mol CO22

Mullis 26

Gas Densities and Molar Gas Densities and Molar MassMass

►Need units of mass over volume for density Need units of mass over volume for density (d)(d)

►Let Let MM = molar mass (g/mol, or mass/mol) = molar mass (g/mol, or mass/mol)PV = nRTPV = nRT

MMPV = PV = MMnRTnRTMMP/RT = nP/RT = nMM/V/V

MMP/RT = mol(mass/mol)/VP/RT = mol(mass/mol)/VMMP/RT = densityP/RT = density

MM = = dRTdRT PP

Mullis 27

Sample Problem: Density Sample Problem: Density 1.00 mole of gas occupies 27.0 L with a density of 1.41 g/L at a 1.00 mole of gas occupies 27.0 L with a density of 1.41 g/L at a

particular temperature and pressure. What is its molecular weight particular temperature and pressure. What is its molecular weight and what is its density at STP?and what is its density at STP?

M.W. =M.W. = 1.41 g 1.41 g |27.0 L |27.0 L = = 38.1 38.1 g___g___ LL |1.0 mol|1.0 mol molmol

MM = = dRTdRT d= d= M M P P = 38.1 = 38.1 g (1 atm)______________ g (1 atm)______________ = = 1.70 1.70 g/Lg/L

PP RTRT mol (0.0821 mol (0.0821 L-atm )(273K)L-atm )(273K) ( mol-K )( mol-K )

OR…AT OR…AT STPSTP: : 38.1 g | 1 mol38.1 g | 1 mol = = 1.70 g/L1.70 g/L mol | 22.4 Lmol | 22.4 L

Mullis 28

Example: Molecular WeightExample: Molecular Weight

A 0.371 g sample of a pure gaseous A 0.371 g sample of a pure gaseous compound occupies 310. mL at 100. compound occupies 310. mL at 100. º C º C and 750. torr. What is this compound’s and 750. torr. What is this compound’s molecular weight?molecular weight?

n=n=PVPV = = (750 torr)(.360L)(750 torr)(.360L) = 0.0116 mole = 0.0116 mole

RTRT 62.4 62.4 L-torr(373 K)L-torr(373 K)

mole-Kmole-K

MW = MW = x gx g_= _= 0.371 g 0.371 g = 32.0 g/mol = 32.0 g/mol

molmol 0.0116 mol0.0116 mol

Mullis 29

Partial PressuresPartial Pressures► Gas molecules are far apart, so assume Gas molecules are far apart, so assume

they behave independently.they behave independently.► Dalton:Dalton: Total pressure of a mixture of Total pressure of a mixture of

gases is sum of the pressures that each gases is sum of the pressures that each exerts if it is present alone.exerts if it is present alone.

PPtt = P = P1 1 + P+ P22 + P + P3 3 + …. + P+ …. + Pnn

PPtt = (n = (n11 + n + n22 + n + n33 +…)RT/V = n +…)RT/V = nii RT/V RT/V

► Let nLet nii = number of moles of gas 1 exerting = number of moles of gas 1 exerting partial pressure Ppartial pressure P11::

PP11 = X = X11PP11 where X where X11 is the mole fraction is the mole fraction (n(n11/n/ntt))

Mullis 30

Collecting Gases Over WaterCollecting Gases Over Water

► It is common to synthesize gases and collect It is common to synthesize gases and collect them by displacing a volume of water.them by displacing a volume of water.

►To calculate the amount of a gas produced, To calculate the amount of a gas produced, correct for the partial pressure of water:correct for the partial pressure of water:

►PPtotaltotal = P = Pgasgas + P + Pwaterwater

►The vapor pressure of water varies with The vapor pressure of water varies with temperature. Use a reference table to find.temperature. Use a reference table to find.

Mullis 31

Kinetic energyKinetic energy► The absolute temperature of a gas is a measure The absolute temperature of a gas is a measure

of the of the average*average* kinetic energy. kinetic energy.► As temperature increases, the average kinetic As temperature increases, the average kinetic

energy of the gas molecules increases.energy of the gas molecules increases.► As kinetic energy increases, the velocity of the As kinetic energy increases, the velocity of the

gas molecules increases.gas molecules increases.► Root-mean square (rms) speed of a gas Root-mean square (rms) speed of a gas

molecule is molecule is u.u.► Average kinetic energy, Average kinetic energy, εε ,is related to rms ,is related to rms

speed:speed:εε = ½ = ½ mu mu 22 where where m m = mass of molecule= mass of molecule

*Average is of the energies of individual gas molecules.*Average is of the energies of individual gas molecules.

Mullis 32

Maxwell-Boltzmann Maxwell-Boltzmann DistributionDistribution

► Shows molecular speed Shows molecular speed vs. fraction of vs. fraction of molecules at a given molecules at a given speedspeed

► No molecules at zero No molecules at zero energy energy

► Few molecules at high Few molecules at high energy energy

► No maximum energy No maximum energy value value (graph is slightly (graph is slightly misleading: curves approach misleading: curves approach zero as velocity increases)zero as velocity increases)

► At higher At higher temperatures, many temperatures, many more molecules are more molecules are moving at higher moving at higher speeds than at lower speeds than at lower temperatures (but you temperatures (but you already guessed that)already guessed that)

Just for fun: Link to mathematical details: Just for fun: Link to mathematical details: http://http://user.mc.net/~buckeroo/MXDF.htmluser.mc.net/~buckeroo/MXDF.html

Source: http://www.tannerm.com/maxwell_boltzmann.htm

Mullis 33

Molecular Effusion and Molecular Effusion and DiffusionDiffusion

►Kinetic energy Kinetic energy εε = ½ = ½ mu mu 22

►u u = = 3RT 3RT Lower molar mass M, higher rms Lower molar mass M, higher rms speed speed uu

M

Lighter gases have higher speeds Lighter gases have higher speeds than heavier ones, so diffusion than heavier ones, so diffusion and effusion are faster for lighter and effusion are faster for lighter gases.gases.

Mullis 34

Graham’s Law of EffusionGraham’s Law of Effusion

►To quantify effusion rate for two gases To quantify effusion rate for two gases with molar masses with molar masses MM11 and and MM22::

rr11 = = MM22

rr22 MM11

►Only those molecules that hit the small Only those molecules that hit the small hole will escape thru it.hole will escape thru it.

►Higher speed, more likely to hit hole, so Higher speed, more likely to hit hole, so

rr11/r/r2 2 = = uu11//uu22

Mullis 35

Sample Problem: Molecular Sample Problem: Molecular SpeedSpeed

Find the root-mean square speed of Find the root-mean square speed of hydrogen molecules in m/s at 20hydrogen molecules in m/s at 20ºº C. C.

1 J = 1 kg-m1 J = 1 kg-m22/s/s2 2 R = 8.314 J/mol-KR = 8.314 J/mol-K

R = 8.314 kg-mR = 8.314 kg-m22/mol-K-s/mol-K-s2 2

uu22 = = 3RT 3RT = = 3(8.314 kg-m3(8.314 kg-m22/mol-K-s/mol-K-s22)293K)293K

MM 2.0162.016 g g ||1 kg___ 1 kg___

mol |1000gmol |1000guu22= 3.62 x 10= 3.62 x 1066 m m22/s/s22

u u = 1.90 x 10= 1.90 x 1033 m/s m/s

Mullis 36

Example: Using Graham’s Example: Using Graham’s LawLaw

An unknown gas composed of homonuclear An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is diatomic molecules effuses at a rate that is only 0.355 times that of Oonly 0.355 times that of O2 2 at the same at the same temperature. What is the unknown gas?temperature. What is the unknown gas?

rrx x = = MMO2 O2 0.355 0.355 = = 32.0 g/mol32.0 g/mol

rrO2O2 MMxx 11 MMxx

Square both sides:Square both sides: 0.3550.35522 = = 32.0 g/mol32.0 g/mol

MMxx

MMx x = = 32.0 g/mol 32.0 g/mol = 254 g/mol = 254 g/mol Each atom is 127 g, Each atom is 127 g,

0.3550.35522 so gas is I so gas is I22

Mullis 37

The van der Waals equationThe van der Waals equation► Add 2 terms to the ideal-gas equation Add 2 terms to the ideal-gas equation

to correct for to correct for 1.1. The volume of molecules (V-nb)The volume of molecules (V-nb)2.2. Molecular attractions (nMolecular attractions (n22a/Va/V22))Where a and b are empirical constants.Where a and b are empirical constants.P + P + nn22aa (V – nb) = nRT (V – nb) = nRT VV22

► The effect of these forces—If a striking The effect of these forces—If a striking gas molecule is attracted to its gas molecule is attracted to its neighbors, its impact on the wall of its neighbors, its impact on the wall of its container is lessened.container is lessened.