Lec3

3 LectureProf. Dr. Mamdouh A. Abdel Rahim

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Lecture 03Contents

1. The Kinetic Molecular Theory of Gases

2. The Meaning of Temperature

3. Root Mean Square Velocity

4. The Mean Free Path5. Diffusion and Effusion6. Real Gases and Van der Waals equation


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The Kinetic Molecular Theory of GasesFour postulates could describe the behavior of molecules in a gas: 1. Gases consist of large numbers of particles,

the volume of the particle can be neglected. 2. The particles are in constant

motion. The pressure of the gas is due to the collision of the particles with the wall of the container.

3. No force of interaction exists between particles.4. The average kinetic energy of a molecule is

directly proportional to the Kelvin temperature. KEavg α T

4.


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The Kinetic Molecular Theory of Gasesz

x

y

lFor a large number of molecules moving in random directions:

P V =13

n NA m u2

where n = number of moles,NA = Avogadro’s numberm = the mass of moleculesu = the velocity of molecules

= the average velocity of molecules= the average square velocity

uu 2


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The Kinetic Molecular Theory of Gases

PV =13

nNA m u2 2

= 3n2 KEavg

P V =3n2 KEavg α

Since the average kinetic energy of a molecule is directly proportional to the Kelvin temperature,

T P V =nR T

= 3n NA m u2( )

21

2= 3

n NA KE

P Vn


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The Meaning of TemperatureThe exact relationship between temperature and average kinetic energy is:

P V =nR T P V =

3n2 KEavg

KEavg = 23 R T

The Kelvin temperature is an index of the random motions of the particles of the gas.

KEavg = 23 R T


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Root Mean Square VelocityThe square root of the average speed is called the root mean square velocity.

Urms = u2

From the equations:

KEavg = NA ( )21 m u2 KEavg = 2

3 R T

u2 = NA

3 R Tm


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Root Mean Square Velocity

u2 = Urms =NA

3 R Tm

u2 =NA

3 R Tm

m represents the mass in kilograms of a single gas particle.

Urms =M

3 R T meter s-1

In this case R = 8.314 J K-1 mol-1M is the molecular weight of the gas.

Urms =M

3 R T


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Example:Calculate the root mean square velocity for the atoms in a sample of helium gas at 25C.

Solution:M for helium = 4 g mol-1 = 4 × 10-3 Kg mol-1

T = 25 + 273 = 298 KR = 8.314 J K-1 mol-1

=

Urms = 1.8610 6 m2s-2

Urms= M3 8.314 298

4 × 10-3K-1mol-1 K

Kg mol-1Kg m2s-2

=1.3610 3 m s-1


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Root Mean Square Velocity

At the same temperature,

compare Urms H2 with Urms O2

Urms H2Urms O2

=MH2

3 R T

MO2

3 R T=

MO2MH2

322= = 4

Hydrogen molecules move 4 times faster than oxygen molecules.


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The Mean Free PathIt is the average distance a particle travels between collisions in a particular gas sample.

Molecular velocity (m/s)

Rel

ativ

e nu

mbe

r of

mol

ecul

es w

ith

give

n ve

loci

tyO2gas

0 400 8000


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The Mean Free Path


Rel

ativ

e nu

mbe

r of

m

olec

ules

with

giv

en

velo

city

273 K

1273 K

2273 K

O2 gas

N2 gas

00 1000 2000


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Diffusion and EffusionDiffusion and EffusionDiffusion: Diffusion:

When gas molecules escape from their container through tiny holes in the container, the process is said to be

When gas molecules escape from their container through tiny holes in the container, the process is said to be Effusion Effusion

Is the spread of one gas throughout a space or throughout a second gas. Is the spread of one gas throughout a space or throughout a second gas. The movement of gas particles during mixing with another gas occurs from a higher to a lower concentration area.

The movement of gas particles during mixing with another gas occurs from a higher to a lower concentration area.


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found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.

found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.

Diffusion and EffusionDiffusion and EffusionThomas Graham Thomas Graham

For gases 1 and 2: For gases 1 and 2:

11 M1M1

11 andand 22 M2M2

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where is the rate of effusion and M is the atomic or molecular weight of the gas particles.where is the rate of effusion and M is the atomic or molecular weight of the gas particles.

Thomas Graham Thomas Graham (1805(1805--1869)1869)


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Diffusion and EffusionDiffusion and Effusionand

M1

constant1 =M2

constant2 =

12

= orM1

M2 12

=M1

M2

This is called Graham’s law of effusion.

12

=M1

M2


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Diffusion and EffusionDiffusion and EffusionComparing the molecular velocities of nitrogen, hydrogen and helium gases:

Molecular mass of nitrogen = 28Molecular mass of hydrogen = 2

Atomic mass of helium = 4


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Diffusion and EffusionDiffusion and Effusion


Rel

ativ

e nu

mbe

r of

m

olec

ules

with

giv

en

velo

city

N2 gas

He gas

H2 gasUrms

UrmsUrms


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Cotton witted with HCl(aq)

Cotton witted with NH3(aq)

White ring of NH4Cl(s)

450 m/s660 m/s


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HCl36.5 g/mol

NH3

17 g/mol

NH4Cl


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Diffusion and EffusionDiffusion and EffusionNH3 (g)NH3 (g) ++ HCl (g)HCl (g) NH4Cl (s)NH4Cl (s)

distance traveled by NH3 distance traveled by NH3

distance traveled by HCldistance traveled by HCl

==

Urms for NH3 Urms for NH3

Urms for HClUrms for HCl

MNH3MNH3

MHClMHCl 36.536.5

1717

==

1.51.5

Urms for NH3 Urms for NH3

Urms for HClUrms for HCl ====


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- What is the rate of effusion for H2 if 15.00 ml of CO2 takes 4.55 sec to effuse out of a container?

- What is the rate of effusion for H2 if 15.00 ml of CO2 takes 4.55 sec to effuse out of a container?

- What is the molar mass of gas X if it effuses 0.876 times as rapidly as N2(g)?

- What is the molar mass of gas X if it effuses 0.876 times as rapidly as N2(g)?

Problems:


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Real GasesReal Gases

1.01.0PVPVRTRT

P (atm)P (atm)00 200200 400400 600600 800800 10001000

2.02.0

00

H2

N2

CH4

CO2

PV = nRTPV = nRT

n = n = PVPVRTRT = 1.0= 1.0 Repulsive Forces

Attractive Forces

Ideal Gas


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Van Van derder Waals equationWaals equationThe deviation from ideal behavior has been explained by the fact that:The deviation from ideal behavior has been explained by the fact that:- molecules do occupy space. - molecules do occupy space. - there is a slight forces of attraction- there is a slight forces of attraction (Van derWaals forces) exist between molecules.

(Van derWaals forces) exist between molecules.

The ideal gas equation is:The ideal gas equation is:

PVPV == nnRR TTThe first step: the actual volume is the volume of the container V minus a correction factor for the volume of molecules nb.

The first step: the actual volume is the volume of the container V minus a correction factor for the volume of molecules nb.


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Van Van derder Waals equationWaals equation

This leads to: P = nR T(V – nb)The second step: consider the attraction forces exist between molecules.

Pobs = P – correction factorNo attraction forces exist between molecules.

The ideal behaviour


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There is attraction forces exist between molecules.

The real behaviourThe Pobs decreases due to the attraction forces.


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Van Van derder Waals equationWaals equationTo determine the value of the correction factor:To determine the value of the correction factor:- The factor depends on the concentration of

the gas molecules.- The factor depends on the concentration of

the gas molecules.The concentration of the gas molecules is:Moles of the gas per literThe concentration of the gas molecules is:Moles of the gas per liter

(n / V)(n / V)- The number of interacting pairs of particles

depends on the square of concentration, - The number of interacting pairs of particles

depends on the square of concentration,

(n / V)2(n / V)2


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Van Van derder Waals equationWaals equationIn a gas sample containing N particles, there is N – 1 partners available for each particle.

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3

45

6

7

8

9

10For N particles, there are N(N – 1)/2pairs.

N(N–1)/2 = 10(10-1)/2= 10(9)/2= 90/2 = 45

45


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Van Van derder Waals equationWaals equationFor N particles, there are N(N – 1)/2 pairs.For large N, the value N(N – 1)/2 could be: N2 / 2Taking into consideration the attraction forces between gas molecules, the pressure is corrected in the form:

Pobs = P – correction factorcorrection factor α (concentration)2

(n / V)2

= aα

(n / V)2


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Pobs = P – a nV

( )2

Therefor:

P = Pobs + a nV

( )2

Pobs = P – correction factor


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= nR TPobs + a nV( )2 (V – nb)

Pressure correction Volume correction

Van der Waals equation

= nR TPobs + a nV( )2 (V – nb)

Pideal Videal


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Thank you

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