9. Kinetic Theory Thermodynamics – empirical description of matter based on experimental observations (macroscopic) All phenomena described in thermodynamics can be explained by considering the behaviours of

the molecules at microscopic scale This can be done by 2 approaches :

(1) Consider motion of molecules based on law of mechanics - Kinetic theory

(2) Consider the probability of the behaviour of group of large number of molecules - Statistical thermodynamic From kinetic theory → equation of state, internal energy, specific heat Statistical thermodynamic → entropy At microscopic scale, some "classical" law of physics may not be obeyed ⇒ Quantum Mechanical approach

9.1 Basic assumptions of kinetic theory (1) A macroscopic volume of a gas contains a very large number of molecules 1 mole : 6.03 × 1023 molecules (Avogradro number) (2) The distance between any two molecules in the gas is much larger than their individual size (3) Each molecule will not experience any force exerted by other molecules except when a

collision occurs. (4) The molecules are assumed to be hard spheres and the collisions between the molecules are

perfectly elastic → conservation of momentum and energy (5) The molecules are distributed uniformly inside the volume of the gas

- number density of the molecules VNn = and

VNn

∆∆

= for any element ∆V

(6) The directions of the motion of molecules are isotropic (all directions are equally probable)

Assumption (6) may lead us to the following : If N : total number of possible direction Each of these direction can be represented by a point on the surface of a sphere with surface area of 4 π r2

the number of directions between (r, θ , φ ) and (r, θ+dθ , φ+dφ ) is given by

Nr

AN ⎟⎟⎠

⎞⎜⎜⎝

⎛

π

∆=∆ 24

(through ∆A)

where φ∆θ∆θ=θ∆φ∆θ=∆ sinsin 2rrrA

and tA

N∆∆

∆=Φ is the flux of number of directions passing through ∆A

Hence the number of molecules having velocity with direction between (r, θ , φ ) and (r, θ+dθ ,

φ+dφ ) is given by

( ) ( )φ∆θ∆θ⎟⎠⎞

⎜⎝⎛π

=φ∆θ∆θ⎟⎟⎠

⎞⎜⎜⎝

⎛

π=∆ sin

4sin

42

2Nr

r

NN

The number density of molecules having velocity with direction between (r, θ , φ ) and (r, θ+dθ , φ+dφ ) is given by

( )φ∆θ∆θ⎟⎠⎞

⎜⎝⎛π

=∆ θφ sin4nn

If the number density of molecules having velocity between v and v+dv is ∆nv The number density of molecules having velocity with direction between (r, θ , φ ) and (r, θ+dθ ,

φ+dφ ) and with magnitude between v dan v+dv is given by

( )φθθπθφ ∆∆⎟⎠⎞

⎜⎝⎛ ∆=∆ sin

4v

vnn

Consider in a time ∆t, distance travelled by particle with velocity v is r = v ∆t , ⇒ ( )( )tvAV ∆θ∆=∆ cos

∆V The number of molecules having velocity with direction between (r, θ , φ ) and (r, θ+dθ , φ+dφ ) and

with magnitude between v and v+dv is given by :

( )( )( )tvAn

VnN vvv ∆θ∆φ∆θ∆θ⎟

⎠⎞

⎜⎝⎛

π∆

=∆∆=∆ θφθφ cossin4

Hence, the flux of molecules passing through ∆A :

( )φ∆θ∆θθ⎟⎠

⎞⎜⎝

⎛π

∆=∆Φθφ cossin

4v

nvv

Total flux of molecules :

( )∑∫ ∫ ∫ ∆=Φ=Φπ

=φ

π

=θθφ

vallv

vallv nvd

412

0

2/

0

Define N

vNvNvNv L+∆+∆+∆= 332211

Hence, ( )∑ ∆=v

vNvN

v1 or ( )∑ ∆=

vvnv

nv

1

⇒ vn41

=Φ which is the molecular flux

9.2 The equation of state for an ideal gas Consider the collision of molecules at the wall of the container :

In the direction normal to the plane of the wall, ∆(momentum) = mvcosθ - (-mvcosθ ) = 2mvcosθ Consider a unit area at the plane of the wall, it will be collided by ∆Φθ v molecules per unit time ∆Φθ v ∆(momentum) = rate of momentum transfer per unit area = pressure

Hence, ( )θ⎟⎠⎞

⎜⎝⎛ θ∆θθ∆=∆ θ cos2cossin

21 mvnvP vv

θ∆θθ∆= 22 cossinvnmv For molecules with velocity v and for all possible directions,

( )vvv nmvdPP ∆==∆ ∫π

=θθ

22/

031

For all possible velocities, ( )∑∫ ∆==∞

−∞= vv

vv nvmdPP 2

31

Define mean square velocity, ( )

nnv

NNv

v vv ∑∑ ∆=

∆=

222

⇒ 231 vnmP =

or 231 vNmPV =

But 221

23 vmkT = →

mkTv 32 =

Hence, NkTPV = or PV = nRT where AN

Rk =

which is the equation of state for ideal gas.

9.3 Principle of equipartition of energy For a system with 2 components at thermodynamic equilibrium, we may write

211111 3

1vmNkTNVP == for component 1

222222 3

1vmNkTNVP == for component 2

Also, 2112

123 vmkT =

2222

123 vmkT =

The experssion 221 vm is the average kinetic energy for particles with mass m moving with

velocity v

Consider Cartesian coordinate (x, y, z), v : (vx , vy , vz )

We can write : 221

21

xvmkT =

221

21

yvmkT =

221

21

zvmkT =

This means : the kinetic energy of the particles is divided into 3 parts according to its 3 degree of

freedom

Each degree of freedom ≡ kT21

For monoatomic gas, 3 degree of freedom (linear)

kTE23

=

For molecular gas, 3 degree of freedom for linear motion 2 degree of freedom for rotational motion 2 degree of freedom for vibrational motion

kTE27

=

In general, we can write kTfE2

=

where f is the degree of freedom. Each degree of freedom represents an equal partition of the

kinetic energy, which is kT21

Other types of degree of freedon include excitation, ionization etc which involve transfer of

energy into the internal energy of the particles

9.4 Specific heat capacity Consider 2 states a and b. We can write Ua - Ub = W (adiabatic)

Write nRTfNkTfU22

== , or RTfu2

=

RfTuc

vv 2

=⎟⎠⎞

⎜⎝⎛∂∂

=

Rf

Rcc vP 22+

=+=

⇒ f

fcc

v

P 2+==γ

For solid system, since the molecules are closely position, it is required to consider the effect of potential energy asssociated with its motion in addition to kinetic energy

Hence each degree of freedom, the energy partition is kT ⎟⎠⎞

⎜⎝⎛ + kTkT

21

21

This means = 3 RT fRTu = and Rcv 3=

9.5 van der Waals equation of state Real gas effect – 2 possible corrections to ideal gas assumption Correction 1: Inside volume V, if the space occupied by all the other particles is b, then the space that can be

occupied by the particle is (V – b). Molecule – hard sphere of diameter d

Closest distance between 2 molecules is d

Space occupied by 2 molecules can be taken to be 3

34 dπ

Hence, the total space occupied by all N molecules is

nbdnNdN A =×=× 33

32

34

21 ππ ⇒ 3

32 dNb A π×=

Correction 2 : Effect of intermolecular force Force of attraction between molecules (gravitational force) Inside the gas, the forces experienced by a molecule from all direction are the same ⇒ there is no net force in any direction However, this is not true for molecules near to the boundary (such as the wall of the container) These molecules will collide at the wall to produce the effect of “pressure” But because of the attractive force due to the rest of the molecules on these molecules, the

pressure produced by these collisions will be reduced

The reduction in the pressure is proportional to 222

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

vN

VnN

VN AA

or, we can write 2

2

va

vN

P A =⎟⎠⎞

⎜⎝⎛=∆ α , 2

ANa α=

Hence, the equation of state of the gas can be written as

2va

bvRTP −−

= atau ( ) RTbvvaP =−⎟⎠⎞

⎜⎝⎛ + 2

which is the van der Waals equation ot state

9.6 Collision between particles

VNn = : number of particles per unit volume

d : diameter of particle Number of particles that may collide with a particle after moving

through unit distance : 2dnπ πd 2 – is used as a measure of the probability of collision ( Unit - area : cm2 ) nπd 2 – number of collision per unit distance 1/(nπd 2) – distance between two consecutive collision (mean free path) Write σ = πd 2, which is called the cross-section of collision Mean free path (mfp), λ = 1/(nσ) The probability for a particle to travel distance x without any collision: λ/x

x eP −=

If the target particle being collided is also moving, σ

λn1

43

=

For particles with velocities following M-B distribution, σ

λn1

21

=

For the case of electrons colliding atoms or ions, σ

λn4

=

9.7 Viscosity Consider gas or liquid in between 2 plates The top plate is moving to the right by force F Imagine that the gas or liquid is divided into thin sheet The sheet next to the top plate will be moving almost at the same velocity as the top plate, but the sheets further down will be moving with lower velocities The bottom plate expereiences the same force F but in opposite direction Particles are moving randomly (thermal) ⇒ possibility of particles jumping from one sheet to another This results in the transfer of momentum between sheets, hence force exerted on each other

If dydu is the variation of velocity in the y-direction,

dydu

AF η= , η is the coefficient of viscosity, A is the surface area of the plate

Consider element ∆A in the SS plane,

y = l cosθ l : mean free path Molecular flux at ∆A :

( )φθθθπθφ ∆∆⎟⎠⎞

⎜⎝⎛ ∆=∆Φ cossin

4v

nvv

Flux in the θ-direction : θθθπ

φθφθ ∆=Φ=∆Φ ∫ ∫

=

cossin212

0

nvdvall

v

Total flux : nv41

=Φ

The average distance travelled by particles after their last collision reaching the SS plane from above

or below the SS plane is given by

lnv

dnlvdly

32

cossincos

41

2

0

221

2

0 ==Φ

Φ=

∫∫ππ

θ θθθθ

This means that on the average, particles make their last collision at distance of 2/3 l from the SS plane

before they finally reach the plane

Velocity: dyduluu o 3

2+=

Momentum ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

dydulummu o 3

2

Hence, the total momentum per unit time per unit area carried by the particles passing through surface SS from above or below the plane is

⎟⎟⎠

⎞⎜⎜⎝

⎛+↓=

dyduluvnmG o 3

241r

and for the upwards direction }from below to above

⎟⎟⎠

⎞⎜⎜⎝

⎛−↑=

dyduluvnmG o 3

241r

Net momentum transferred per unit time per unit area :

dydu

AF

dydulvnmGGG η=≡↑=−↓=

31rrr

Hence, σ

η vmlvnm31

31

==

9.8 Thermal conductivity Now consider the plate on top to be not moving, but at temperature much higher than the

bottom plate

⇒ temperature gradient dydT ( K m-1 )

Heat flow per unit cross-sectional area per unit time (heat flux) :

dydTH λ−= ( J m-2 s-1 )

λ is the heat conductivity ( J m-1 s-1 K-1 )

Actually, the heat flow can be expressed as

dydTlcvnH v

*

31

−= A

vv N

cc =* ⇒

σλ

**

31

31 v

vcv

lcvn ==

9.9 Diffusion Particles drift caused by density gradient

n = N/V , gradient : dydn

Particle flux passing through unit cross-sectional area in unit time

dydnD−=Γ

D is the diffusion coefficient

It can be shown that σnvlvD

31

31

==