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CHAPTER 10

GASES

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CONTENT

10.1 Characteristics of Gases

10.2 Pressure

10.3 The Gas Laws

10.4 The Ideal-Gas Equation

10.5 Further Applications of The Ideal Gas Equation

10.6 Gas Mixture and Partial Pressures

10.7 Kinetic-Molecular Theory10.8 Molecular Effusion and Diffusion

10.9 Real Gases: Deviations from Ideal Behaviour

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Learning Outcomes

Able to apply the Ideal Gas Lawincalculations involving gaseous system

Able to calculate partial and total pressurein a mixture of gases (with or withoutreaction)

Able to differentiate effusion and diffusion

Able to apply kinetic molecular theoryinproblem solving

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Bonds

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10.1 Characteristics of Gases

Expand spontaneously to fill its container.

Volume of gas = Volume of the container.Highly compressible.

Form homogeneous mixtures with other gases.

Occupy about 0.1% of the volume of theircontainers.

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10.2 Pressure

Measured properties of a gas:

Temperature (T), Volume (V), Pressure (P)

Pressure, P is the force F acting on an objectper unit area, A.

P = F/A

Our atmosphere exerts a downwardforce/pressure on Earths surface because ofgravity.

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10.2.1 Atmospheric Pressureand The Barometer

The acceleration produced by earths gravity is9.8 ms-2.

A column of air 1m2

in cross section has a massof roughly 10,000 kg.

The force exerted by this column is:

F = ma where, a = 9.8 m/s2

= (10,000 kg) (9.8 m/s2)

= 1 105kg m/s2

= 1 105N

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Cont: 10.2.1 AtmosphericPressure and The Barometer

The pressure exerted by the column is the forcedivided by its cross-sectional area, A:

P = F/A

= (1 105N)/(1m2)= 1 105 N/m2

= 1 105 Pa

= 1 102kPa

SI unit for P = N/m2 (also called Pascal, Pa)1Pa = 1 N/m2

1 bar = 105Pa

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Therefore, the atmospheric pressure at sea level

is about 100kPa.The actual atmospheric pressure at any location

depends on weather conditions and on altitude.

Atmospheric pressure can be measured by amercury barometer.

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Bonds

HGFGJDF

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Standard atmospheric pressure is the pressurerequired to support a column of mercury 760 mm

in height.Units: 1 atm = 760 mmHg

= 760 torr

= 1.01325 105Pa

SI units: 1 atm = 1.01325 105Pa

Other common units 1 atm = 14.70 psi

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Bonds

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Example 1

Convert 0.378 atm to: a) torr; b) pascal

(1atm = 760 torrs = 101325 Pa)

torratm

torratm 271

1

760357.0

Paatm

Paatm 36173

1

101325357.0

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10.2.2 Pressure of EnclosedGases and Manometers

Closed-tube manometer

measures pressure below atmospheric pressure.

difference in height of mercury level equals topressure of enclosed gas.

Open-tube manometer

measures pressure near atmospheric pressure.

difference in height of mercury level relates topressure of enclosed gas.

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Bonds

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Example 2

A vessel connected to an open-end mercurymanometer is filled with gas to a pressure of

0.835 atm. The atmospheric pressure is 755torr.

a) In which arm of the manometer will the level ofmercury be higher?

b) What is the height difference between the 2 armsof the manometer?

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Example 2 (Answer)

a) Since the atmospheric pressure is greaterthan the enclosed gas, the level attached tothe gas will be higher.

b) P gas= 0.835 atm 760 torr = 635 torr1atm

P gas + Ph = Patm635 torr + Ph = 755

Ph = (755 - 635 ) torrs= 120 torr = 120 mm Hg

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10.3 The Gas Law

4 variables are needed to define the physicalcondition or state of a gas:

1. Temperature, T

2. Pressure, P

3. Volume,V

4. Number of moles, n

Gas Laws:The equations that relate T, P, Vand n.

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10.3.1 Boyles Law: ThePressure-Volume Relationship

Molecular Explanation of Boyles Law

A certain mass of gas enclosed in a container ofvolume,V exerts a pressure, P.

Gaseous particles hitting the walls of the

container-gives P.

If T is constant - the number of collisions with the

wall remains constant.

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Cont: 10.3.1 Boyles Law: ThePressure-Volume Relationship

If the volume of the sample is reduced by

half,V/2, there will be twiceas many molecules

per unit volume.

As a result, twice as many molecules strike the

walls in a given period of time. So that the

average force exerted on the walls is doubled.

Hence, when the volume is halved, thepressureof the gas is doubled.

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Bonds

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The volumeof a fixed quantity of gasmaintained at constant temperature is

inverselyproportional to the pressure.(n,T constant)

PV = k = constant

PV

1

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If V1and V2are the different volumes of a fixedmass of gas at P1and P2, then:

P1 V1 = P2 V2

V

P

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Example 3

A sample of gas has a volume of 54 ml at

pressure 452 mmHg. What will the

volume be if the pressure is changed to

649 mmHg while the temperature is kept

constant?

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Example 3 (Answer)

Vol/ ml Pressure/mmHg

Initial(1) 54 452Final(2) ? 649

P1V1 = P2V2(452mmHg)(54ml) = (649mmHg)V2

V2 = 37.6 ml

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10.3.2 Charles Law: TheTemperature-Volume Relationship

Molecular Explanation of Charles LawRaising the temperature of a gas increases the

average speed of its molecules.

The molecules collide with the walls morefrequently and do so with greater impact.Therefore they drive back the walls, and the

gas occupies a greater volume than it did

initially.At constant volume, pressure will increase.Particles will hit the walls of a container more

frequently.

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Bonds

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Cont: 10.3.2 Charles Law: TheTemperature-Volume Relationship

At constant pressure, volume will expand equallyfor equal rises in temperature.

Volume

0K 273K 373K temp./KExtrapolating - volume versus temp. at constant

pressure line, the line intercept at -273C on thex-axis.

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This -273C (more accurately -273.15C) wastermed absolute zero or O K (Kelvin scale).

Theoretically gases would have a zero volume at273C.

In practice they all liquefy above thistemperature.

0 K is equal to -273.15 C

T(K) = T(C) + 273.15

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General definition of Charles law:

The volumeof a fixed amount of gas maintainedat constant pressure is directly proportionalto its

absolute temperature. V T (abs.) (n, Pconstant)

V/T= k = constant

If V1and V2are the different volumes of a fixedmass at absolute temperature T1andT2, then :

V1/T1 = V2 /T2

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Bonds

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Example 4

A sample of gas has a volume of 364 ml at a

temperature of 25 C. What will the volume be

if the temperature is changed to 100.0 C while

the pressure is kept constant?

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Example 4 (Answer)

Vol/ml Temp/C Temp/K

Initial(1) 364 25.0 25 + 273.15 = 298.15

Final (2) ? 100.0 100 + 273.15 =373.15

mlTT

VV

T

V

T

V

6.45521

12

2

2

1

1

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10.3.3 Avogadros Law: TheQuantity-Volume Relationship

Avogadros hypothesis:Equal volume of gases atthe same temperature andpressure contain equalnumbers of molecules.

Ar N2 H2

Volume 22.4L 22.4L 22.4L

Pressure 1atm 1atm 1atm

Temperature 00

C 00

C 00

CMass of gas 39.95g 28.01g 2.02g

No. of molec. 6.02 x 1023 6.02 x 1023 6.02 x 1023

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Cont: 10.3.3 Avogadros Law: TheQuantity-Volume Relationship

Avogadros law: For a gas at constant temperature

and pressure, the volumeof a gas is directly

proportionalto the number of molesof the gas.

Mathematically:

V n (T,Pconstant)

V/n= k = constant

Note: This relationship is obeyed closely bygases at low pressure.

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Bonds

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Bonds

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Example 5

Suppose we have a 12.2 L sample containing

0.5 mol oxygen gas (O2) at a pressure of 1

atm. and a temperature of 25C. If all this O2were converted to ozone (O3) at the same

temperature and pressure, what would be the

volume of ozone?

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Example 5 (Answer)

3 O2(g) 2 O33 mol of O2 gives 2 mol of O3

0.5 mol of O2 = 0.5 mol O2 2 mol O3 = 0.33 mol O3

3 mol O2

Lmolmol

Lnn

VV

n

V

n

V

05.833.05.0

2.122

1

12

2

2

1

1

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10.4 The Ideal-Gas Equation

Boyles law: V1/P (constant n,T)

Charles law : VT (constant n,P)

Avogadros law : Vn(constant P,T)

Combine these relationships

Let the proportionality constant be R:

Rearrange: PV = nRT The ideal-gas equation

P

nTV

P

RnTV

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Cont: 10.4 The Ideal-GasEquation

An ideal gas is a hypothetical gas whosepressure, volume and temperature behaviour iscompletely described by the ideal-gas equation.

Rin the ideal-gas equation is called the gasconstant.

In working problems with PV= nRT, the units ofP,V, n and T must agree with the units of gasconstant.R= 8.314 J/mol.K

R= 0.0821 L.atm/mol.K

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Example 6

Determine volume of 1 mole gasxat STP.

Standard temperature and pressure(STP):

P= 1atm

n= 1 mol

R = 0.0821 L-atm/mol K

T(K) = T(C) + 273.15

= 0C + 273.15

= 273.15K

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Example 6 (Answer)

The volume occupied by 1 mol of ideal gas atSTP, 22.4L is known as the molar volume ofan ideal gas at STP.

LV

atm

KmolKatmLmolV

P

nRTV

4.22

0.1

15.273/.0821.00.1

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The ideal-gas equation does not always accuratelydescribe real gases.The measured volume,Vfor givenP, nand Tmight differ from the volume calculated

from PV = nRT. The standard conditions for gas behaviour (where Ris

calculated based on 0 C and 1 atm) are not the sameas the standard conditions in thermodynamics (25C

and 1 atm).

Note:Very small volume difference is noticed in

calculation involving real and ideal gas.

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From Ideal-Gas Equation : PV = nRT

1) ( constantn)

2) (constant n& P)

3) (constantn &

V)

4) (constant n& T)

2

22

1

11

T

VP

T

VP

2

2

1

1

T

V

T

V

2

2

1

1

TP

TP

2211 VPVP

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Example 7

A gas occupies a volume of 136 cm3at 210Cand 102 kNm-2.

What would its volume be at STP?{use PV/T = constant}

STP: T= 273K, P= 1 atm = 101.3kNm-2

V1= 136 cm3 V2= ?T1= (273 + 21)K T2= 273K

P1= 102 kNm-2 P2 = 101.3 kNm-2

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Example 7 (Answer)

3

2

2

32

2

2

2

1

11

2

2

22

1

11

127

3.101

273

294

136102

cmV

kNm

K

K

cmkNmV

P

T

T

VPV

T

VP

T

VP

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Example 8

What volume would be occupied by 100g of oxygen

(O2) at 18 C and 105 kNm-2?

n = 100 g = 3.125 mol T = 18 + 273 = 291 K

32 g/mol R = 8.31 Nm/K.mol

3

23

0720.0

10105

291./31.8125.3

mV

Nm

KKmolNmmolV

P

nRT

V

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Example 9

A metal cylinder holds 50.0 L of oxygen at 18.5atm and 21C. What volume will the gas occupy ifthe pressure is reduced to 1.00atm and

temperature is maintained at 21C.FromPV= nRT

When T, nare constant P1V1 =P2V2V2 =P1V1

P2V2 = (18.5 atm) (50.0L) = 925 L

(1.00 atm)

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Exercise 10.1

The gas pressure in an aerosol can is 1.5

atm at 25 C. Assuming that the gas inside

obey the ideal- gas equation. What would

the pressure be if the can was heated to

450 C?

Use: ; Answer: 3.6 atm

P

1

1

2

2

T

P

T

P

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Exercise 10.2

An inflated balloon has a volume of 6.0 L at sea

level (1.0 atm) and is allowed to ascend in altitude

until the pressure is 0.45 atm. During ascent the

temperature of the gas falls from 22C to 21

C.

Calculate the volume of the balloon at its final

altitude.

Use: ; Answer: 13.3 L

1

11

2

22

TVP

TVP

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10.5 Further Applications ofThe Ideal-Gas Equations

10.5.1 Determination of Gas Densities and Molar Masses

Density = mass/volume Unit : grams/liter (g/L)

PV = nRT

Rearrange,

Molar mass, M = grams/mol

n/V has units of mol/Liter Multiply both sides with M, thus

Density,

RTP

Vn

RT

PM

V

nM

RT

PMd

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Example 10

What is the density of CCl4vapour at 714 torr and

125 oC?

Molar mass of CCl4, M = 12.0 + 4(35.5) = 154.0 g/mol

Pressure: 760 torr = 1 atm

714 torr : 714 torr 1 atm = 0.939 atm

760 torr

Lgddensity

KKmolatmL

molgatm

RT

PMddensity

/43.4,

398./.0821.0

/154939.0,

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Example 11

A 2.78 g sample of gas occupies a volume

of 4.24 Liter at 23.6 C and 755.1 mmHg.

What is the molecular mass of the gas?

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Example 11 (Answer)

760 mmHg - 1 atm 755.1 mmHg - 0.994 atm

mass = 2.78 g, mol, n = 0.173 mol

molar mass, M = mass/mol

M = 2.78 g/ 0.173 mol = 16.1 g mol-1

moln

KKmolatmL

Latm

RT

PV

n

173.0

6.296./.0821.0

24.4994.0

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Example 12

The industrial synthesis of nitric acid involves the

reaction of nitrogen dioxide gas with water

3NO2(g) + H2O (l) 2HNO3 (aq) + NO(g)

How many moles of nitric acid can be prepared

using 450 L of NO2at a pressure of 5.00 atm and

a temperature of 295 K?

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Example 12 (Answer)

VNO2= 450L, T= 295K, P= 5.00 atm

3NO2(g) + H2O (l) 2HNO3 (aq) + NO(g)

3 mol of NO2will produce 2 mol of HNO3

92.9 mol NO2will produce=

moln

KKmolatmLLatm

RTPVn

9.92

295./.0821.045000.5

mol

molmol

mol

9.61

23

9.92

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Exercise 10.3

A large flask is evacuated and found toweigh 134.567g. It is then filled to apressure of 735 torr at 310C with a gas of

unknown molar mass and then reweighed; itsmass is 137.456g. The flask is then filledwith water and again weighted: its mass now1067.9g. What is the molar mass of theunknown gas?{use M=dRT/P, answer 79.7 g/mol}(the density of water at 310C is 0.997g/cm3)

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Exercise 10.4

The density of a gas measured at 1.50 atm

and 27C and found to be 1.95 g/L.

Calculate the molar mass of this gas.

{use M=dRT/P, answer 32.0 g/mol}

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10.6 Gas Mixtures and PartialPressures

Dalton:

In gas mixtures each gas acts independently ofthe other gases present

OR

Total pressure of a mixture of gases equals thesum of the pressures that each would exert if it

were present alone.Mathematically

PTotal= P1+ P2+ P3+ ...

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Bonds

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Cont: 10.6 Gas Mixtures andPartial Pressures

E.g. : a mixtureof gases G1, G2and G3in onecontainer.

G1: P1, n1 G2: P2 , n2 G3: P3, n3Total pressure, Pt= P1 + P2 + P3Each gases obeys the ideal gas equation.P1= n1(RT/V) ; P2= n2(RT/V) ; P3= n3(RT/V)

Pt=n1(RT/V) + n2(RT/V) + n3(RT/V)Pt=(n1+n2 +n3) (RT/V)

= ntotal (RT/V )

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Example 13

Mixture of helium and oxygen are used in

scuba diving tanks to help prevent the bends.

For particular dive, 46 L He at 25 C and 1.0

atm and 12 L O2at 25 C and 1.0 atm were

pumped into a tank with a volume of 5.0 L.

Calculate the partial pressure of each gas and

the total pressure in the tank at 25 C.

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Example 13 (Answer)

PHe= 1.0 atm; VHe= 46 L ; R= 0.08206 L.atm/K.mol

THe= 25 +273 = 298K

PO2= 1.0 atm ; VO2= 25 L ; TO2 = 25 +273 = 298K

From n = PV/RT

mol

KKmolatmL

Latmn

molKKmolatmL

Latmn

O

He

49.0298./.08206.0

120.1

9.1298./.08206.0

460.1

2

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Example 13 (Answer)

Total tank volume, V= 5.0 L; T= 298 K

From P = nRT/V

Total Pressure,Pt=PHe+PO2= 9.3 + 2.4 = 11.7 atm

atmL

KKmolatmLmolP

atmL

KKmolatmLmolP

O

He

4.20.5

298./.08206.049.0

3.90.5

298./.08206.09.1

2

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10.6.1 Mole Fractions andPartial Pressure

Mole fraction,Xis the ratio of number of moles ofa given component in a mixture to the totalnumber of moles in the mixture. X1= n1/nT

From ideal gas equation PV = nRT:

i.e for each component in the mixture

RTVPn

RTVPn 2211 ,

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Cont: 10.6.1 Mole Fractionsand Partial Pressure

Mole fraction represented in terms of pressure:

...3211

11

RTVPRTVPRTVP

RTVP

n

nX

TOTAL

TOTALP

P

PPP

P

PPPRT

V

RTVP

1

321

1

321

1

....

....

TOTAL

i

TOTAL

ii

PP

nnX

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Example 14

A 2.0 L tank containing oxygen at a pressure

of 100 kPa is connected to a 0.1 L tank

containing helium at a pressure of 3.00 MPa

and the gases are allowed to mix. What is the

final pressure assuming that the temperature is

held constant?

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Example 14 (Answer)

Volume of He increases from 0.1 L to 2.1 L

Volume of oxygen increases from 2.0 L to 2.1 L

Dalton: The total pressure is the sum of partial pressure.

O2 ; V1= 2.0 L V2= 2.1 L

P1= 100 kPa P2= ?

From P1V1 = P2V2

kPaL

LkPaP 95

1.2

0.21002

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Example 14 (Answer)

From P1V1 = P2V2

He ; V1= 0.1 L V2= 2.1 L

P1

= 3.00 MPa P2

= ?

kPaL

LkPaP 143

1.2

1.030002

kPaPPP HeOt 238143952

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Example 15

A mixture of 8.00 g of CH4and 9.00 g of

C2H6is stored at a total pressure of 500

kPa. What is the partial pressure of each

component present?

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Example 15 (Answer)

P1=X1Pt

Mof CH4= 16.0 g

Mof C2H6= 30.0g

No. of mole of each component:

nCH4

= 8.00g/16.00g mol-1= 0.50 mol

nC2H6= 9.00g/30.00g mol-1= 0.30 mol

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Example 15 (Answer)

nt= nCH4 + nC2H6

= 0.50 + 0.30

= 0.80 mol

Mole fraction: (mol/total mol)XCH4= 0.50/0.80 = 0.625

XC2H6= 0.30/0.80 = 0.375

Partial Pressure:

P1=X1Pt

PCH4= 0.625 500kPa = 312 kPa

PC2H6= 0.375 500kPa = 188 kPa

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Exercise 10.5

A gaseous mixture made from 6.00 g O2and 9.00 g

CH4is placed in a 15.0 L vessel at 00C. What is the

partial pressure of each gas and what is the totalpressure in the vessel?

Note: use Pi= ni(RT/V )

{Answer: PO2=0.281 atm, PCH4=0.841 atm, Pt=1.122 atm}

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Exercise 10.6

A synthetic atmosphere composed of 1.5 molpercent CO2, 18.0 mol percent O2and 80.5 molpercent Ar.

a) Calculate the partial pressure of O2in themixture if the total pressure of theatmosphere is to be 745 torr. {Ans.=134 torr}

b) If this atmosphere is to be held in a 120 Lspace at 295 K, how many moles of O2areneeded? {use ni=Pi(V/RT), ans.=0.874 mol}

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10.7 Kinetic Molecular Theory

(1822-1888) The kinetic-molecular theory(theory of moving molecules) was developed:

Assumptions:

1) Gases consist of large numbers of molecules in

constant random motion.

2) Volume of individual molecules is negligiblecompared to volume of container (total volume).

C t 10 7 Ki ti M l l

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Cont: 10.7 Kinetic MolecularTheory

3) Attractive and repulsive (intermolecular) forcesbetween gas molecules are negligible.

4) The collisions between molecules are perfectlyelastic.The average kinetic energy does notchange as the temperature is constant.

5) Average kinetic energy of the molecules isproportional to the absolute temperature.

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v

d

C t 10 7 Ki ti M l l

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Cont: 10.7 Kinetic MolecularTheory

The average kinetic energy, is related to rootmean square speed, u and mass

= 1/2 mu2

u= root-mean-square (rms) speed, the speed ofa molecule possessing average kinetic energy.

A gas composed of light particles such as He willhave the same as one composed of muchheavier particles such as Xe - provided that bothare at the same temperature.

10 8 M l l Eff i d

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10.8 Molecular Effusion andDiffusion

The mass,m of the lighter gas is smaller than theheavier gas. Therefore, the lighter gas must havea higher rms speed, u.

An equation that expresses this fact:

;M=molar mass

* Less massive the gas molecules, the higher therms speed, u.

M

u

M

RTu

13

Cont 10 8 Molec la Eff sion

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Cont: 10.8 Molecular Effusionand Diffusion

Consequences of the dependence of molecular speedson mass:

1. Effusion: the escape of gas molecules through atiny hole into an evacuated space.

2. Diffusion: the spread of one substancethroughout a space or throughout a secondsubstance. E.g. molecules of a perfume diffusethroughout a room.

10 8 1 Grahams Law of

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10.8.1 Grahams Law ofEffusion

The effusion rate of a gas is inversely proportionalto the square root of its molar mass.

Consider 2 gases at same Tand Pin containers with

identical pinholes (different containers).Gas 1 : Molar mass, M1- rate of effusion r1

Gas 2 : Molar mass, M2- rate of effusion r2

The relative rate of effusion:

1

2

2

1

M

M

r

r

Cont: 10 8 1 Grahams Law of

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Cont: 10.8.1 Grahams Law ofEffusion

Lighter gas effuses more rapidly.

Only those molecules which hit the small hole willescape through it.

Thus, the higher the rmsspeed the greaterpossibility of a gas molecule hitting the hole and

effuse.

Cont: 10 8 1 Grahams Law of

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Cont: 10.8.1 Grahams Law ofEffusion

Rate of effusionis directly proportionalto the rmsspeed of the molecules.

Example:

Rubber and various plastic have tiny openingsthrough which gas molecules can pass.

1

2

2

1

2

1

2

1

/3/3MM

MRTMRT

uu

rr

10 8 2 Diffusion and Mean Free

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10.8.2 Diffusion and Mean FreePath

Diffusion(like effusion) is faster for light gasmolecules.

Molecular collisionsmake diffusion more complicated

than effusion.Average distance of a gas molecule between collisions

is called mean free path.

The higher the density of a gas, the smaller the meanfree path. The more molecules are in a given volume,the shorter the average distance traveled betweencollisions.

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v

d

10 9 Real Gases: Deviations

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10.9 Real Gases: Deviationsfrom Ideal Behaviour

Ideal gas- the molecules are assumed to occupyno spaceand no attractionsfor one another.

Real gas- Molecules have finite volumesand theyattractone another.

At lower pressures (usually below 10 atm), the

deviation from ideal behaviour is negligible.

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v

d

Cont: 10 9 Real Gases:

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Con t: 10.9 Real Gases:Deviations from Ideal Behaviour

As pressure increases (for PidealV/RT = n = 1):The free space which the molecules can move

becomes a smaller fraction of the container volume

(due to finitevolumes of the molecules).In addition, the attractive forcesbetween

molecules becomes significant (at short distances),the impact with the wall of the container is

lessened (Pat

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90

v

d

Cont: 10 9 Real Gases:

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Con t: 10.9 Real Gases:Deviations from Ideal Behaviour

Temperature determines how effective attractiveforces between gas molecules are.

At low temperature: gases deviate from ideality:the average kinetic energy decreases,intermolecular attractions remain constant.

At high temperature: gas molecules are far apart,thus the finite volumes of the moleculespredominate.

10 9 1 The Van der Waals

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10.9.1 The Van der WaalsEquation

Ideal gas equation - cannot be used to predict thepressure-volume properties of gases at highpressure.

The ideal-gas equation predicts that the pressureof a gas i.e.

This expression has to be corrected for the 2effects:

i) the finite volume occupied by the gas molecules.ii) the attractive forces between gas molecules.

V

nRTP

Cont: 10 9 1 The Van der Waals

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Con t: 10.9.1 The Van der WaalsEquation

Van der Waals introduced 2 constants, aand bto makethese corrections:

nb: accounts for the correction of the finite volume ofgas molecules.The Van der Waals constant b is a measure of theactual volume occupied by a mole of gas molecules -units of L/mol.

The pressure is decreased by the factor n2a/V 2:accounts for the attractive forces.The Van der Waals constant a - units L2-atm/mol2.

2

2

V

an

nbV

nRTP

Cont: 10 9 1 The Van der Waals

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Con t: 10.9.1 The Van der WaalsEquation

Van der Waals equation:

aand b- different for different gases.

Values of both aand bgenerally increase with anincrease in mass of the molecule and complexity of its

structure.Larger, more massive molecules, have larger volumes

and greater intermolecular attractive forces.

nRTnbVV

anP

2

2

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Example 16

If 1.00 mol of an ideal gas were confined to

22.41 L at 0.0 C, it would exert pressure of

1.00 atm. Use the Van der Waals equationand the constants in Table 10.3 to estimate

the pressure exerted by 1.00 mol of Cl2(g)

in 22.41 L at 0.0 C.

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Example 16 (Answer)

n = 1.00 mol,R = 0.08206L.atm/mol.K,

T = 273.15K, V = 22.41L,a= 6.49L2atm/mol2, and b = 0.0562L/mol

nRTnbVV

anP

2

2

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Example 16 (Answer)

atmP

atmatmP

L

molatmLmol

molLmolL

KKmolatmLmolP

990.0

013.0003.1

41.22

/.49.600.1

/0562.000.141.22

15.273./.08206.000.1

2

222

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Example 17

A sample of KClO3is partially decomposed,

producing O2gas that is collected over water. The

volume of gas collected is 0.250 l at 26C and 765

torr total pressure.

2KClO3(s) 2KCl (s) + 3O2(g)

a) How many moles of O2are collected?

b) How many grams of KClO3were decomposed?c) When dry, what volume would the collected O2

gas occupy at the same temperature and

pressure?

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Example 17 (Answer)


Vtotal inside

= 0.250L

Ttotal inside= 26C (273.15 + 26) = 299.15K

Ptotal inside= 765 torr

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Example 17 (Answer)

a) PO2 = (765 - 25) torr. = 740 torr

moln

KKmolatmL

Ltorratmtorrn

RT

VPn

O

O

O

O

31091.9

15.299./.0821.0

250.0760/1740

2

2

2

2

l ( )

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Example 17 (Answer)


b) 2 mols KClO3 3 mols of O2

Molar mass of KClO3 = 122.6 g/mol

gram KClO3 g

molKClO

gKClO

molO

molKClOmolO

811.0

1

6.122

3

21091.9

3

3

2

323

l ( )

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Example 17 (Answer)

c)Use Boyles law:

V1= 0.250L ; P2= 765 torr.(assumed as water partial

pressure replaced by O2)

P1= 740 torr (from O2 - water vapour)

V2= (assumed dry O2without water vapour)

2

11

2P

PVV

LV

torr

torrLV

242.0

765

740250.0

2

2

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END of CHAPTER 10

chapter 10 jan13

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