Chapter09 Kinetic Theory of Gases - Weebly · 1 9 Kinetic Theory of Gases By Liew Sau Poh 2 Content 9.1 Ideal gas equation 9.2 Pressure of a gas 9.3 Molecular kinetic energy 9.4 The

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9 Kinetic Theory of Gases

By Liew Sau Poh

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Content

9.1 Ideal gas equation 9.2 Pressure of a gas 9.3 Molecular kinetic energy 9.4 The r.m.s. speed of molecules 9.5 Degrees of freedom and law of

equipartition of energy 9.6 Internal energy of an ideal gas

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Objectives

(a) use the equation of ideal gas, pV = nRT (b) state the assumptions of the kinetic theory of

an ideal gas (c) derive and use the equation for the pressure

exerted by an ideal gas, p = 1/3 <c2> (d) state and use the relationship between the

Boltzmann constant and molar gas constant k = R / NA

(e) derive and use the expression for the mean translational kinetic energy of a molecule, ½ m<c2> = 3/2 kT

(f) calculate the r.m.s. speed of gas molecules 4

Objectives g) sketch the molecular speed distribution graph

and explain the shape of the graph (description of the experiment is not required)

h) predict the variation of molecular speed distribution with temperature

i) define the degrees of freedom of a gas molecule

j) identify the number of degrees of freedom of a monatomic, diatomic or polyatomic molecule at room temperature

k) explain the variation in the number of degrees of freedom of a diatomic molecule ranging from very low to very high temperatures

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Objectives

l) state and apply the law of equipartition of energy

m) distinguish between an ideal gas and a real gas n) explain the concept of internal energy of an

ideal gas o) derive and use the relationship between the

internal energy and the number of degrees of freedom.

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9.1 Ideal gas equation

7


Gases at low pressures are found to obey the ideal gas law:

at constant temperature is inversely

if T = constant

nRTPV

VP 1

8


Equation above also can be write as

P1V1 = P2V2

Where P1 = initial pressure P2 = final pressure V1 = initial volume V2 = final volume

VP 1

9

P

V 0 T1

T2

10

P

1/ V 0

T1

T2

11

PV

P 0

T1

T2

12

PV

V 0

T1

T2

13

States : constant pressure is directly proportional to its

V T if P = constant, thus Where V1 = initial absolute volume, T1 = initial absolute temperature, V2 = final volume, T2 = final temperature

constantTV

2

2

1

1

TV

TV

14

V

T/ oC 0 -273.15

V

T/ K 0

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Gay-

States : constant volume is directly proportional to its

P T if V = constant Equation above also can be write as P/T = constant or P1/T1 = P2/T2 where

re temperatuabsolute final:2Tre temperatuabsolute initial:1T pressure initial:1P

pressure final:2P16

Graphs of Gay-

P

T/ oC 0 -273.15

P

T/ K 0

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Equation of State for an Ideal Gas

An ideal gas is defined as a perfect gas which

Gay-

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Consider an ideal gas in a container changes its pressure, volume and temperature as shown in figure below.

1st stage1P1V1T

2P'V

1T

2P2V2T

2nd stage

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1st stage, temperature is kept at T1

112 ' VPVP2

11'PVPV

1st stage1P1V1T

2P'V

1T

2P2V2T

2nd stage

20


2nd stage, pressure is kept constant at P2 ,

2

2

1

'TV

TV

2

12'T

TVV

1st stage1P1V1T

2P'V

1T

2P2V2T

2nd stage

21


Thus Or Hence,

For n mole of an ideal gas, the equation of state is written as

2

22

1

11TVP

TVP

constantT

PV

RT

PVm

nRTPVm

22


Where n : the number of mole gas where where

nRTPV

Mmn

gas theof mass:m

12310026 constant sAvogadro': -A molx.N

ANNn molecules ofnumber :N

massmolecular :M

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If the Boltzmann constant, k is defined as then the equation of state becomes Where k = R/NA

PV = nRT = (N/NA)RT = N(R/NA)T = NkT

123 1038.1 KJxNRk

A

NkTPV

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Real Gases

Assumptions of real gases by van der Waals: The volume of the molecules may not be negligible in relation to the volume V occupied by the gas. The attractive forces between the molecules may not negligible. Therefore the equation of state for an ideal gas has to be modified i.e.

nRTPV

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Real Gases

van der Waals equation of state

The constants a and b are empirical constants, different for different gases. Where a volume of 1 mole of the gas molecules & depends on the attractive intermolecules forces nb total volume of the molecules

nRTnbVVanP )(2

2

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Graphs representing the real gases

P

T 0

T1

T2

T3

T4

T1 T2 T3 >T4 >T3

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Graphs comparing the real-ideal gases

moln /

11 molKJT

PV

Ideal gas8.31

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9.2 Pressure of a gas

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The macroscopic behaviour of an ideal gas can be describe by using the equation of state but the microscopic behaviour only can be describe by kinetic theory of gases.

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Kinetic Theory of Gases Assumptions

The main assumptions of the kinetic theory of gases are: a) All gases are made up of identical atoms

or molecules. b) All atoms or molecules move randomly

and haphazardly. c) The volume of the atoms or molecules is

negligible when compared with the volume occupied by the gas.

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Kinetic Theory of Gases Assumptions

d) The intermolecular forces are negligible except during collisions.

e) Inter-atomic or molecular collisions are elastic.

f) The duration of a collision is negligible compared with the time spent travelling between collisions.

g) Atoms and molecules move with constant velocity between collisions. Gravity has no effect on molecular motion. 32


Pressure of a gas, P is defined as: How to get this?

231 cP

where gasby pressure :Pgas theofdensity :

molecules gas theof velocity square mean :2c

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Consider the molecules are contained in a cubic box with the side length d as shown in figure (a).

Assume the velocity of a molecule v The velocity,v can be

resolved into components of vx, vy and vz.

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If a molecule of mass, m collides with wall A hence it will bounce off in opposite direction with velocity, -vx because of elastic collision as shown in figure (b).

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Therefore the change in linear momentum for x-component :

xxx mvmvP

xx mvP 2

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By assuming the molecule move from wall A to B and bounce back to wall A without collides with other molecules, the time taken for that movement is

xvdt 2

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From the definition of the impulse,

111 xxx PtFJ

tPF x

x1

1

1

11

22

x

xx

vd

mvF

211 xx v

dmF

where Fx1 is the average force of one molecule.

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For N molecules of ideal gas in the cubic box, 22

221 ....... xNxxx v

dmv

dmv

dmF

222

21 ....... xNxxx vvv

dmF

The mean square of vx is

Nvvvv xNxx

x

222

212 .......

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so the x-component for total force exerted on the wall of the cubic box:

2xx v

dNmF

The velocity v is resolved into vx, vy and vz, hence 2222

zyx vvvv then

2222zyx vvvv

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Since the velocities of the molecules in the ideal gas are assumed to be random, there is no preference to one direction or another. Hence

222zyx vvv

22 3 xvv

3

22 vvx

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By substituting the relationship above in the equation for total force, Fx hence the total force exerted on the wall in all direction is given by

dvmNF

2

3

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From the definition of pressure,

AFP where 2dA and

dvmNF

2

3

3

2

31

dvNmP 2

31 v

VNmP

because Vd3 Then

2

31 vNmPV

where box. in the gas theof mass:Nm

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Since the density of the gas, is given by

VNm hence equation (15.1)

can be written as

2

31 vP

where gasby pressure :Pgas theofdensity :

molecules gas theof velocity squaremean :2v

231 cPOR

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9.3 Molecular Kinetic Energy

45


Rearrange equation 2

31 v

VNmP into 2

21

32 vm

VNP

This equation shows that P increases ( ) When

increases VN

and increases21 2vm

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into

2

21

32 vm

VNPRearrange eq.

2

21

32 vmNPV

From the equation of state in terms of Boltzmann constant, k : NkTPVBy equating eq. 2

21

32 vmNPV

with eq. NkTPV

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thus kTvm23

21 2 and

trKvm 2

21

kTK tr 23

where

energy kinetic onal translatiaverage : trK

moleculea of

re temperatuabsolute :T

48


For N molecules of ideal gas in the cubic box, the total average (mean) translational kinetic energy, E is given by

trNKE kTNE23

nRTNkTE23

23

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9.4 The root mean square (R.m.s.) speed of molecules

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Distribution of molecular speeds

Not all the molecules have the same speed Apparatus for studying molecular speed distribution (1831-1879)

Distribution of molecular speeds

Gas molecules constantly collide elastically with each other and with the wall of the container Under the collision, kinetic energy transfer from one molecule to another, hence the kinetic energy of one molecule increases while the other one decreases Molecules move in difference speed as the speed changes after each collision The distribution of molecular speed is known as Maxwell distribution

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Maxwell Distribution

distribution law: P(v)dv fraction of molecules with speeds in the range from v to v + dv

James Clerk Maxwell(1831-1879)

RTMv

evRT

MvP 222/3 2

24)(

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P(v)dv fraction of molecules with speeds in the range from v to v + dv Maxwell Distribution

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Number of molecules, n (v)

Speed, v 0 V0 Vm Vrms

T1

V0 < Vm < Vrms V0 = Most probable

speed Vm = mean speed Vrms = root mean square

speed

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P(v)dv fraction of molecules with speeds in the range from v to v + dv

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The distribution of speeds for nitrogen gas molecules at three different temperatures

urms = 3RT

Maxwell distribution

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Number of molecules, n (v)

Speed, v 0

T2 > T1 T1

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The distribution of speeds of three different gases at the same temperature

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9.4 The r.m.s. speed of molecules

Because of kTvm23

21 2

thus mkTv 32

Then 2vvrms m

kTvrms3

or

where (speed) velocity squaremean root : rmsv gas molecule a of mass:m

gas of massmolecular relative:Mre temperatuabsolute :T 60

9.4 The r.m.s. speed of molecules

Since therefore the equation of root mean square velocity also can be written as

2

31 vP thus

Pv 32

Pvrms3

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9.4 Distribution of molecular speeds

Distribution function is normalized to 1: Average speed: Root mean square speed:

1)(0

dvvP

MRTdvvvPvavg

8)(0

MRTdvvPvvrms

3)(0

2

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9.4 Distribution of molecular speeds

Most probable speed:

0PvdvdP

MRTvP

2

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

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9.5 Degrees of freedom and

law of equipartition of energy

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9.5 Degrees of freedom

Definition is defined as the number of independent ways in which an atom or molecule can absorb or release or store energy

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Example

Monatomic gas (e.g. He, Neon, Argon)

The number of degrees of freedom is 3 i.e. three direction of translational motion where contribute translational kinetic energy.

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Example

Diatomic gas (e.g. H2, O2, N2) The number of degrees of freedom is: Translational kinetic energy = 3 Rotational kinetic energy = 2/5

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Example

Polyatomic gas (e.g. H2O, CO2, NH3) The number of degrees of freedom is Translational kinetic energy 3 Rotational kinetic energy = 3/6

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Example: degrees of freedom

kT23 kT25 kTkT 326

Molecule Eg.

Degrees of Freedom (f) Average kinetic

energy per molecule,<K> Translat

ional Rotational

Total

Monatomic He 3 0 3

Diatomic He 3 2 5

Polyatomic H2O 3 3 6

kT23

kT25

kTkT 326

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Degrees of freedom depend on the absolute temperature of the gases. For example : Diatomic gas (H2)

H H

vibration

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Hydrogen gas have the vibrational kinetic energy as shown in figure above where contribute 2 degrees of freedom which correspond to the kinetic energy and the potential energy associated with vibrations along the bond between the atoms

At 250 K f = 3 At temperature (250 750 K) f = 3 At temperature >750 K f = 3

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9.5

energy of every degrees of freedom of a molecule is or kT

21

RT21

RTfkTfK22

where

gas molecule a ofenergy kinetic average : K

freedom of degrees:f re temperatuabsolute:T

constantBoltzmann : k constant gasmolar : R72

9.6 Internal Energy of An Ideal Gas

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Definition is defined as the sum of total kinetic energy and total potential energy of the gas molecules. But in ideal gas, intermolecular forces are assumed to be negligible hence the potential of the molecules can be neglected. Thus for N molecules, ..EKNU

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NkTfU2

and

ANRk

nRTfU2

where gas theofenergy internal:U

..EKNU

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Example

A quantity of plasma is composed of hydrogen ions (protons) and electrons in thermal equilibrium. Both the protons and electrons are assumed to behave like molecules of an ideal gas. The r.m.s. speed of an electron in the plasma is estimated to be 3 x 106 m s-1. a. Determine the r.m.s. speed of the hydrogen ions. b. Estimate the temperature of the plasma. (Given mass of electron = 9.11 x 10-31 kg, mass of hydrogen ion = 1.67 x 10-27 kg, Boltzmann constant, k = 1.38 x 10-23 J K-1)

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Solution

(vrms)e=3 x 106 m s-1

a. By using the equation of vrms, electron : hydrogen ion :

eerms m

kTv 3)(

HHrms m

kTv 3)(

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Solution

then eq. (1) divided by eq. (2), thus

e

H

Hrms

erms

mm

vv

)()(

14 1001.7 smxv Hrms

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Solution

b. The temperature of the plasma is given by

eerms m

kTv 3)(

kmvT eerms

3)()( 2

KxT 51098.1

Summary

Kinetic Theory of Gases

Ideal Gas Equation PV = nRT = (N/NA)RT

Molecular Kinetic Energy ½ mc2 = 3/2 kT

R.M.S. Speed Molecular speed distribution & graph

Pressure of a Gas P = 1/3 <c2>

Degree of Freedom & Law of Equipartition of Energy

K.E. Molecule = f/2 kT

Internal Energy Interal energy of n moles, U = f/2 nRT

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End of Topic

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Chapter09 Kinetic Theory of Gases - Weebly · 1 9 Kinetic Theory of Gases By Liew Sau Poh 2 Content 9.1 Ideal gas equation 9.2 Pressure of a gas 9.3 Molecular kinetic energy 9.4 The