5 - 1

Kinetic Molecular Theory (KMT)Kinetic Molecular Theory (KMT)

Gases consist of very small particles (atoms or

molecules) which are separated by largedistances.

Most of the volume occupied by a gas is empty space.

Gas molecules move randomly at very highspeeds in all directions. Pressure results from molecules colliding withthe walls of the container.

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Kinetic Molecular Theory (KMT)Kinetic Molecular Theory (KMT)

There are no intermolecular forces betweenthe gas molecules except when they

collide.

All collisions between molecules are elasticcollisions.

The internal energy of the system remains the same.

The KE of molecules is dependent upon theabsolute temperature.

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STPSTP

Standard temperature and pressure,

T = 273.15K ≈ 273K

P = 1 atm = 760 mm Hg = 76 cm Hg

1.0 mol of any gas = 22.4 L = 22.4 dm3

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The Gaseous PhaseThe Gaseous Phase

In the gaseous state, the molecules havesufficient energy to overcome theintermolecular forces that attract them toeach other.

Each molecule acts independentof the others.

Gases have low densitiesand completely fill the volumethat is available to them.

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Units Used In Measuring GasesUnits Used In Measuring Gases

VolumeVolume mL, L, cm3, dm3, m3

TemperatureTemperature Must use K in gas laws.

PressurePressure atm, torr, kPa, Pa

MolesMoles Amounts measured in moles.

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What Makes a Gas an Ideal Gas?What Makes a Gas an Ideal Gas?

Any ideal gas assumes the:

volume that the gas molecules themselves occupy is neglected.

gas molecules do not attract each other.

Ideal gases are approximated when theintermolecular attractions are very small.

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Low pressures and high temperatures are the

conditions by which a gas approaches theideal.

Low pressures enable gas molecules to spread very far apart.

High temperatures represent a large amount of kinetic energy which overcomes attractive forces.

Small gas molecules such as hydrogen withonly dispersion forces best approximate theideal.

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Deviations from the ideal gas law occur most

at low temperatures and high pressures.

The ideal gas law (IGL) is given by

PV = nRT

where the gas constant R = 0.0821

P is the pressure in atm, V is the volume in L,

n is the number of moles, and T is the Kelvin

or absolute temperature.

L•atmmol•K

,

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Ideal Gas Law ProblemsIdeal Gas Law Problems

What is the pressure exerted by 31.0 moles of

carbon dioxide that is contained in a 110. liter

tank at 28°C?

n = 31.0 moles V = 110. L T = 28°C = 301 K

R = 0.0821

L•atmmol•K

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PV = nRT

31.0 mol × 0.0821 L•atmmol•K=P

× 301 K

110. L

P = 6.96 atm

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A gas with a mass of 3.47 g occupies 282 mL

at 23.0°C and 743 mm Hg. Determine themolecular mass of the gas.

m = 3.47 g V = 282 mL T = 295 K

P = 743 mm Hg R = 0.0821

PV = nRT n =

L•atmmol•K

mMM

MM

=mRTPV

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MM =

3.47 g × 0.0821 L•atmmol•K

× 295 K

743 mm× 1 atm760 mm

× 282 mL× 1 L103 mL

MM = 305 g•mol-1

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Charles’s LawCharles’s Law

Charles’s law says that at constant pressure, the volume of a gas is directly proportional to the absolute temperature (K).

Remember the gas laws require temperatures to be expressed in Kelvin to avoid dividing by zero.

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Two quantities, V and T, are directlyproportional if by whatever factor T

changes,V changes by the same factor.

A graph of V vs T yields a straight linethrough the origin.

The following graph looks to be an exception

but it is not because the temperature isexpressed in °C.

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The graph shows that when the line isextrapolated to -273.15°C, you expect thevolume to be zero.

At 0 K, a gas is predicted to have a zerovolume but all gases either liquefy or

solidifybefore reaching that temperature.

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.

-273.15 °C = 0 K (absolute zero)

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Deriving Charles’s LawDeriving Charles’s Law

Charles’s law can be derived from the idealgas law.

Assume the same gas has existed at twodifferent temperatures while the pressureand the number of moles of gas remainedconstant.

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Applying the IGL at the two differenttemperatures yields:

P1V1 = nRT1 P2V2 = nRT2

Dividing one equation by the other yields:

P1V1 nRT1

P2V2 nRT2

= =>P1V1 nRT1=P2V2 nRT2

V1 V2= T1 T2

(Charles’s Law)

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Charles’s Law ProblemCharles’s Law Problem

A sample of oxygen has a volume of 350 mL at

60.0°C. What will be the volume of oxygen at

standard temperature if the pressure remains

constant?

V1 = 350 mL V2 = ?T1 = 60.0°C = 333 K T2 = 0.0°C =

273 KP1 = P2 P2 = P1

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.

V1 V2= T1 T2

V2 =V1

T1

× T2

V2 =350 mL × 273 K333 K

= 290 mL O2

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Boyle’s LawBoyle’s Law

Boyle’s law says that at a constant temperature for a fixed quantity of gas, the volume of the gas is inversely proportional to the pressure.

Two quantities, P and V, are inverselyproportional because by increasing either variable by a factor causes a decrease in the other variable that was the inverse of the first factor.

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A graph of V vs P yields a hyperbola.

When the pressure is doubled (2 ×), the volumeis halved (1/2).

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A graph of V vs 1/P yields a straight line.

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Deriving Boyle’s LawDeriving Boyle’s Law

Boyle’s law can be derived from the idealgas law.

Assume the same number of moles of gas has

remained at the same temperature while the

pressure and volume have varied.

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Applying the IGL at the two differentpressure yields:

P1V1 = nRT1 P2V2 = nRT2

Dividing one equation by the other yields:

P1V1 nRT1

P2V2 nRT2

= =>P1V1 nRT1=P2V2 nRT2

P1V1 P2 V2= (Boyle’s Law)

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Boyle’s Law ProblemBoyle’s Law Problem

A gas tank holds 2750 L of nitrogen at920 mm Hg. If the temperature remainsconstant, what will be the volume of

nitrogenat standard pressure?

P1 = 920 mm P2 = 760 mm HgV1 = 2750 L V2 = ?T1 = T2 T2 = T1

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.

V2 =P1V1

P2

V2 = 920 mm × 2750 L760 mm

= 3300 L N2

P1V1 = P2V2

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Avogadro’s Hypothesis and LawAvogadro’s Hypothesis and Law

Avogadro’s hypothesis says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Alternatively, 1.00 mole of any gas at STP (0°C, 1.00 atm) occupies a volume of22.4 L.

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Avogadro’s law says that the volume of a gas

maintained at constant temperature andpressure is directly proportional to thenumber of moles of gas.

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Deriving Avogadro’s LawDeriving Avogadro’s Law

Avogadro’s law can be derived from the ideal

gas law.

Assume the pressure and temperatureremain constant.

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Applying the IGL at the two differentvolumes yields:

P1V1 = n1RT1 P2V2 = n2RT2

Dividing one equation by the other yields:

P1V1 n1RT1

P2V2 n2RT2

= =>P1V1 n1RT1=P2V2 n2RT2

(Avogadro’s Law)=n1

V1 V2

n2

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General Gas LawGeneral Gas Law

The general gas law (combined gas law) isused when you are not restricting the IGL toonly two variables (P, V or V, T). In mostinstances, two of the three physicalproperties of a gas are subject to change.

The general or combined gas law can bederived from the ideal gas law.

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Assume the same gas has existed at twodifferent temperatures while only the

numberof moles of gas have remained constant.

Mr. Bean has run out of gas!

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Applying the IGL at two different pressures,temperatures, and volumes yields:

P1V1 = nRT1 P2V2 = nRT2

Dividing one equation by the other yields:

P1V1 nRT1

P2V2 nRT2

= =>P1V1 nRT1=P2V2 nRT2

P1V1 P2 V2= (General Gas Law)T1 T2

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General Gas Law ProblemGeneral Gas Law Problem

What volume will 4.50 L of SO2 at -27°C and600. mm Hg occupy if the temperaturechanges to 40.°C and the pressure to700. mm Hg?

P1 = 600. mm Hg P2 = 700. mmV1 = 4.50 L V2 = ?T1 = -27°C = 246 K T2 = 40.°C =

313 K

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P1V1 P2V2=T1 T2

V2

600. mm× 4.50 L×313 K246 K×700. mm

=

V2 = 4.91 L SO2

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Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

The total pressure of a mixture of gasesequals the sum of the pressures that eachgas would exert if it was by itself.

Each gas of behaves independently of theothers.

PT = P1 + P2 + … Pn

where n = total number of gases

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To determine the pressure of one of thegases present in the mixture, the molefraction is used.

Mole fraction, X, is a dimensionless number(no unit) which expresses the ratio of thenumber of moles of a particular component

tothe total number present in the mixture.

P1 = X1 × PT and X1 = n1

n2

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Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

A 195 cm3 sample of helium is collected over

water at 16°C and 97.5 kPa. What volumewould the dry gas occupy at 117 kPa and28°C?

P1 = 97.5 kPa P2 = 117 kPaV1 = 195 cm3 V2 = ?T1 = 16°C = 289 K T2 = 28°C =

301 K

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PT = PHe + PW

PHe = PT – PW

PHe = 97.5 kPa – 1.81 kPa = 95.7 kPa

PW is the vapor pressure of water at 16°Cwhich can be found in a table of vaporpressures of water.

It is important to note that the total pressure,

97.5 kPa, is the sum of the water vaporpressure and the pressure of helium.

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P1V1 P2V2=T1 T2

V2

97.5 kPa×195 cm3×301 K289 K× 117 kPa

=

V2 =1.70 × 102 cm3 He

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Molar Volume Not At STPMolar Volume Not At STP

What is the density of argon at 37°C and103.7 kPa?

P1 = 101.3 kPa P2 = 103.7 kPa

V1 = 22.4 dm3 V2 = ?T1 = 0°C = 273 K T2 = 28°C

= 310 K

P1V1 P2V2

T1 T2

=

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.

101.3 kPa × 22.4 dm3 × 310 K

273 K × 103.7 kPaV2 =

V2 = 24.8 dm3

D = MV

D =39.95 g

24.6 dm3=1.62 g/dm3

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Gas StoichiometryGas Stoichiometry

What mass of water is produced when625 dm3 of hydrogen gas measured at 28°Cand 97.3 kPa is burnt in the presence ofoxygen as shown in the following reaction.

2H2(g) + O2(g) → 2H2O(g)

P1 = 97.3 kPa P2 = 101.3 kPa

V1 = 625 dm3 V2 = ?T1 = 27°C = 301 K T2 = 0°C = 273

K

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.

=P1V1 P2V2

T1 T2

97.3 kPa × 625 dm3 × 273 K

301 K × 101.3 kPaV2 =

V2 = 544 dm3 H2

STP, standard temperature and pressure, is101.3 kPa and 0°C.

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.

m= 544 dm3 H2 ×1 mol H2

22.4 dm3 H2

×2 mol H2O2 mol H2

×18.02 g H2O

2 mol H2O= 438 g H2O

Remember that Avogadro’s hypothesis statesthat 1 mol of any gas occupies a volume of22.4 L (or 22.4 dm3) at STP.

This is the reason that the volume of thehydrogen had to be determined at 0°C beforedoing the stoichiometry.

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More Gas StoichiometryMore Gas Stoichiometry

If 528 g of carbon disulfide react with oxygen,

what volume of sulfur dioxide will be formed

at 29°C and 98.2 kPa?

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

m = 528 g CS2

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V = 528 g CS2 ×1 mol CS2

76.13 g CS2

×2 mol SO2

1 mol CS2

×22.4 L SO2

1 mol SO2

= 311 L SO2

P1 = 101.3 kPa P2 = 98.2 kPaV1 = 311 L SO2 V2 = ?T1 = 0°C = 273 K T2 = 29°C = 302 K

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P1V1 P2V2

T1 T2

=

101.3 kPa × 311 L × 302 K

273 K × 98.2 kPaV2 =

V2 = 355 L SO2

Starting with 528 g CS2, you determine thevolume of SO2 produced at 0°C, but the problemasked for the volume at 29°C.

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Effusion and DiffusionEffusion and Diffusion

According to Kinetic Molecular Theory (KMT),the average kinetic energy of gas moleculesis given by

KE = ½mv2

Because the KE determines the temperatureof a substance, the molecules of hydrogengas have the same KE as those of Kr at thesame temperature.

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A consequence of this is that the H2 molecules must be moving faster than

thoseof Kr to compensate for their smaller mass.

Two different types of molecular motion arediffusion and effusion.

Diffusion is the spreading out of one substance throughout a second substance. An example of this is opening a bottle of ammonia (NH3).

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Effusion is the escape of gas molecules through a tiny hole into evacuated space.

To quantitatively compare gases, Graham’slaw is used. Graham’s law is given by:

where r1 and r2 are the rates of effusion andthe MM’s are the molar masses.

MM2

MM1

r1

r2=

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Graham’s law states that the rate of effusion

of gas molecules is inversely proportional tothe square root of their masses.

This means that a lighter gas such as H2

effuses quicker than a more massive gas like

Ar.

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Graham’s Law ProblemGraham’s Law Problem

Compare the rates of effusion of hydrogenand krypton.

r1 = H2 r2 = KrMM1 = 2.02 g•mol-1 MM2 = 83.80 g•mol-1

MM2

MM1

r1

r2= = 83.80 g•mol-1

2.02 g•mol-1= 6.44

H2 effuses 6.44 times faster than Kr.

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Real GasesReal Gases

Ideal gases require two conditions:

The gas molecules themselves do not occupy any volume.

There are no intermolecular attractions.

Real gases start to approach these idealconditions at low pressures and hightemperatures.

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Real gases contradict these conditionsbecause:

the volume available to the molecules making up a real gas is less than the volume of the container because the molecules take up some space.

the pressure exerted on the sides of the

container is diminished because the intermolecular attractions result in

fewer collisions with the sides of the

container.

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van der Waal’s Equationvan der Waal’s Equation

×(Pobs+a(n/V)2) (V – nb)= nRT

The first term involving the Pobs is the corrected pressure factor. This correction is necessary because of the intermolecular attractions. As a result of these attractions, the observed pressure will be smaller than predicted from the ideal gas law.

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The second term, V – nb, is the correctedvolume which will be smaller than thecontainer volume because the gas

moleculesthemselves have volume.

There will be less free space for the molecules to move about.

The constants a and b are called empiricalconstants because their value is

determinedby fitting the equation to the experimentalresults.

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The values for a and b are different fordifferent molecules and can be found in atable of physical constants.

5 - 60

Calculate the pressure that H2O will exert at 50°C if 1.00 mol occupies 32.0 L assumingH2O obeys the van der Waals equation. Theconstants for H2O are:

a = 5.46 b =

van der Waals Problemvan der Waals Problem

L2•atmmol2

0.0305 Lmol

T = 50°C = 323K V = 32.0 L n = 1.00 mol

5 - 61

.

(Pobs+a(n/V)2) (V – nb)= nRT×

(Pobs + 5.46 L2•atmmol2

(1.00 mol/32.0 L)2)×

× 32.0 L – 0.0305 Lmol

= 0.0821L•atmmol•K ×323 K=0.829 atm