Stat 3

Statistical Physics

36

Chapter 3 Statistical equilibrium and Maxwell-Boltzman Distribution Law 3.1.

Statistical equilibrium

Consider that a thermodynamically isolated system consist of N particles. The energy states available to the particles are E1 , E2 , E3 etc. These energy states may be quantized or may be continuous and are due to vibrational and rotational energy of particles. Suppose that at any given instant of time, n1 particles are in state of energy E1, n2 particles are in state energy and E2 so on. The total number of particles in the system is,

N = n1 + n2 + n3 + ... = ∑ ni

( 3.1)

i

Here I = 1, 2, 3 etc. The total energy of the system,

U = n1 E 1 +n 2 E 2 + n3 E3 + ... = ∑ni Ei i

( 3.2 )

equation ( 3.2 ) refers to the energy of a system in which the particles are non-interacting. Here, the energy of each particle depends only on the co-ordinates of the particle system. For isolated system, the total energy U is constant. For isolated system,

U = ∑ni Ei = constant i

(

3.3 ) Consider a gas having N molecules at certain temperature and pressure. Its volume, temperature and pressure are kept constant i.e. the system is isolated. The total energy of this system remains constant., But, the molecules of the gas collide with each other and also with the walls of the container. Consequently, the numbers of molecules change from one energy state to the other energy state. It means that the values of n1 , n2 , n3 etc. continuously change. It can be reasonably assumed that for each macroscopic state of a system of particles, there is a particular most favoured distribution. When this distibution or partition is reached, the system attains statistical equilibrium. For isolated system, the values n1 , n2 , n3 etc. vary only near the values corresponding to the most probable distribution. Hence the basic problem in statistical thermodynamics is to obtain the most probable distribution law for a given composition of the system. Physics Department, Padjadjaran University I Made Joni

Statistical Physics

37

In practise there are three most probable distributions law are used. They are :

3.2

(i)

Maxwell – Boltzman distribution laws

(ii)

Ferrini – Dirac distribution laws

(iii)

Bosch – Einstein distribution laws

Probability Theorems in Statistical Thermodynamics The following are the important probability theorems commonly used in statistical thermodynamics:

1.

The number of ways in which N distinguishable particles can be arranged in order is equal to N!

2.

The number of different ways in which n particles can be selected from N distinguishable particles irrespective of the order of selection is equal to N! ( N −n) !

3.

n!

The number of different ways in which n

indistinguishable particles can be arranged in g

distinguishable state with not more than one particle in each state is equal to g ! ( g − n) !

3.3

n!

The Maxwell – Boltzman distribution laws

Consider a system that contains a large number of particles that are identical and distinguishable. The identical particles refer to the particles having the same structure. The distinction between the particles is due to their energy state at a given instant. As shown in Fig. 3.1. 4 particles are in the energy state E 1, 2 in state E2, zero in state E3 and so on. It is assumed that all the energy states are accessible to each particle. Consequently, it can be assumed that the probability of any particular partition is proportional to the number of different ways in which the particles can be distributed in the existing available energy states so as to produce the desire partition. n5=1

E5 E4 E3 E2

l g

h

k

e f E1 Physics Department, Padjadjaran University a b d c I Made Joni

n4=3 n3=0 n2=2 n1=4

Statistical Physics

38

Fig. 3.1. To obtain the required distribution as shown in Fig. 3.1., the particle a in state E 1 can be selected in N ways. The second particle b in state E1 can be selected in (N-1) ways and so on. Therefore, the total number of ways in which the first four particles in state E1 can be selected is given by, N ( N −1)( N − 2)( N − 3) =

N! ( N − 4)!

Moreover, the four particles in state E1 can be arrange in 4! different orders (a,b,c,d is different from c,d,a,b and so on). There are 24 ways. But, it is immaterial for these particles to be arrange in any particular order in state, E1, because they are identical. Thus, the total number of distinguishable different ways is N! 4!( N −4)!

In general, if the first state consist of n1 particles, the distinguishable different ways, for arranging n1 particles in state E1 are P1 =

N! n1 !( N − n1 )!

For the second state E2, only (N-n1) particles are available and n2 particles are in state E2. The number of different ways is

P2 =

( N − n1 )! n 2 !( N − n1 − n 2 )!

If the process is continued for all available state, the total number of distinguishable ways is obtained by multiplying P1, P2, P3 etc. P = P1 xP2 xP3 x ......

   ( N − n1 )!  x ....... N! P= x   n1!( N − n1 )!   n 2 !( N − n1 − n2 ) 

=

N! n1 ! n 2 ! n3 ! ...

As shown in Fig. 3.1. the distinguishable ways are P =

N! 4!3!2!1!

Physics Department, Padjadjaran University I Made Joni

( 3.4 )

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39

It has been assumed so far that all the available states have the same probability of occupation by the particles. However, it may happen that the states have different intrinsic probabilities say gi.. For example, a particular energy state may be favourable with more different angular momentum states than the rest and hence it is more likely to be occupied. Taking into account this instrisic probability factor, the value of P will be different. If gi is the probability of locating a particle in a certain energy state E i, then probability of locating 2 particles in the same state is gix gi = gi2. For ni particles the probability is gi ni. Hence the total probability for a given distribution n

P=

n

n

N ! g1 1 g 2 2 g 3 3 .... n1! n2 !n3 ! ....

( 3.5 )

If all particles are further assumed to be indistinguishable. i.e, in state E1 and particle in state E4 can’t be distinguished, then N! permutations among the particles themselves and occupying the different state result in the same distribution. The probability in this case is given by n n n 1  N ! g1 1 g 2 2 g 3 3 ....  P=   N !  n1!n2 !n3 ! ... 

N

P=∏ i

n

gi i ni !

( 3.6 )

The most probable distribution can be obtained by evaluating the maximum value of ln P in Eq. ( 3.6 ). This condition should also satisfy the two conditions that Σ ni = N and

Σ niEi = U

According to Stirling’s approximation

ln x!= x ln x − x From equation ( 3.6 ), applying Stirling’s approximation, we get ln P = ∑ni ln g i − ln ni !

Physics Department, Padjadjaran University I Made Joni

Statistical Physics

40

ln P = ∑ ni ln g i − ( ni ln ni − ni )

= ∑ n i −∑ ( ni ln ni − ni ln g i ) n = ∑ ni − ∑ ni ln i  gi n = N − ∑ ni ln i  gi

  

   ( 3.7 )

Differentiating Equation ( 3.7 ) n d ( ln P ) = −∑d ( ni ) ln  i  gi

 dn i  − ∑ni ni 

But Σ dni = 0 (number of part constant)

 n d ( ln P ) = ∑ ln  i i   gi

 dn i 

To obtained maximum value P, that means d(ln P) = 0

  ni     dni = 0   i 

∑ ln g

( 3.8 )

But Σ dni = 0

( 3.9 )

Σ Eidni = 0

( 3.10 )

Multiplying Eq.( 3.9 ) by x α and Eq.(3.10) by β and adding to Eq. ( 3.8 ) we get

  ni   dni + α ∑ dni +β ∑ Ei dni = 0 i     n  ∑  ln gi + α + βEi dni = 0 i  

∑ ln g

The equilibrium distribution is possible if Physics Department, Padjadjaran University I Made Joni

Statistical Physics

41

ln

ni + α + βEi = 0 gi

ni = e −α − βEi gi

Or

( 3.11 )

ni = g i e −α − βEi This gives the maximum probability distribution, where α ,β are parameter that depend up on the physical properties of the system. Total number of particles is

N = n1 + n2 + n3 + ... N = g1i e −α−βE 2 + g 2 e −α−βE2 + ...

(

N = e −α g1e −βE1 + g 2 e −βE2 + ...

)

  N = e −α  ∑g i e −βEi   i  Take

Z = ∑ g i e − βEi i

( 3.12 )

Here, Z is called the partition function. N = e −α Z e −α =

N Z

Substituting this value in equation n i = e −α g i e −βEi ni =

N g i e −βEi Z

( 3.13 ) The Equation (3.13 ) refers to Maxwell-Boltzman distribution law

3.4.

Maxwell_Boltzman distribution in term of temperature

The total of an isolated system is given by

Physics Department, Padjadjaran University I Made Joni

Statistical Physics

42

U = ∑ ni E i i

= n1 E1 + n2 E2 + n3 E3 + ... = g1e −α −βE1 E1 + g 2 e −α −βE 2 E2 + g 3e −α −βE3 E3 + ...

(

= e −α g1e −βE1 E1 + g 2 e −βE 2 E2 + g 3e −βE 3 E3 + ...

U =

)

N  ∑g i e −βE i Ei  Z  i 

(

3.14 )

here Z = ∑g i e −βEi i

dZ d   =  ∑g i e −βEi  = −∑g i E i e −βEi dβ dβ  i i  d   g i E i e −βEi = −  ∑g i e −βEi  ∑ d β i  i 

We substitute in Eq. ( 3.14 ) N d    ∑g i e −βEi  − Z  dβ  i  N dZ U =− Z dβ d ( ln Z ) U = −N dβ U =

The average energy of a particle is E ave =

U d ( ln Z ) =− N dβ

( 3.15 )

This shows that for a given system, the total energy U, partition function Z, and the average energy of the particle Eave depend on parameter β . Therefore , β may be taken to characterize the internal energy of the system and has the units per joule. If T is the temperature in degrees kelvin, then it is more convinient to represent β as

β=

1 kT

Physics Department, Padjadjaran University I Made Joni

( 3.16 )

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43

Here kT has the unit of energy i.e. joule and k is the Boltzmann’s constant k = 13805x10-23 j/K Maxwell-Boltzman distribution by term of temperature of the system N g i e −( Ei / KT ) Z d ( ln Z ) U = −N dβ ni =

here

β=

1 1 , andd β = dT KT KT 2

U = −NKT E ave = KT

2

d ln Z dT d ( ln Z ) dT

2

( 3.17 )

The Equation (3.17 ) gives the relation between the average energy of the particle and it’s temperature under equilibrium position. Hence, the temperature of a system in statistic equilibrium is the physical quantity related to the average energy of particle of the system.

3.5.

Application of M_B distribution to an ideal gas Firstly, let us discuss the number of energy levels in a small energy dE for a particle in a very

large potential box in one dimension and the schrodinger’s wave equation is

d2 ψ + k 2ψ = 0 ; dx2

k2 =

2m E 2

( 3.18 )

Particle is moving between x = 0 and x = a and the wave function is

ψ( x ) Ae iKx + Be −iKx Inserting the boundary condition at x=0 ψ ( x = 0) = A + B = 0 ⇒ B = − A ψ ( x ) = Ae − iKx − Ae − iKx ψ ( x ) = A( e − iKx − e − iKx ) ψ ( x ) = 2iA sin Kx = 0

And second BC at x = a gives ψ ( x = a ) = C sin Ka = 0

In this case c ≠ 0 , because we would then have no solution or wave function

Physics Department, Padjadjaran University I Made Joni

( 3.19 )

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44

sin ka = 0 ka = nπ k =

nπ a

The momentum of particle in x direction is

P = kx Px = 

n xπ a

( 3.20 )

Similarly for the 3 dimension, we have Px = nx Py = n y

π a

π

a π Pz = nz a

Therefor the kinetic energy is then

E=

2 2 2 ny ny  1 2 1 2 1 2 2  nx  mv = P = π  + + 2 2   a2 2 2m 2m a a  

=

(

1 π 2 2 2 2 2 nx + n y + n z 2 2m a

(Taken a = b = c)

)

Let us introduce the coordinate kz , ky, and kz a certain representative space Fig 3.2; each point, of coordinate kz = nz , ky =ny , and kz =nz represent an energy level, and to each point there corresponds a cell of unit volume in this representative space. Let us defined k 2 = nx2 + ny2 + nz2, and say that the number of points having positive integral coordinates and lying on the surface of a sphere of radius k give the different state associated with the energy

1 π 2 2 2 E= k 2 2m a

; where k2 = nx2 + ny2 + ny2

(

3.21 ) So we finally have the relation

k=

2ma 2 E π 2 2

Physics Department, Padjadjaran University I Made Joni

( 3.22 )

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45

kz

k kx ky

Fig 3.2. To find the number of stae N(E) with energy between zero and E, we must find the volume of an actant of space of radius k, since only positive values of nx ,ny and nz are allowed. Thus remembering that  = h / 2 π , we obtained

14  N ( E ) =  πk 3  83  1 1 4  2ma 2 E  2   = π  8 3  π 2 2    

π

( 3.23 )

3

3

 2mE  2 = a  2 2 6 π   1 1 8π = 2 V 2m 3 2 E 2 3 3

(

)

Where V = a3 is the volume of the potential box. The number of states with energy between E and E+dE is obtained by differentiating the above expression, this yields :

(

4π V 2m 3 dN ( E ) = h3

)

1

2

1

(

E 2 dE

3.24 ) it is convenient to write

dN ( E ) = g ( E ) dE

(

dN 4πV 2m 3 g( E) = = dE h3

)

1

2

E

1

2

( 3.25 )

For simplicity we shall now consider ideal gas composed of monoatomic molecules, all molecules energy is translational kinetic Physics Department, Padjadjaran University I Made Joni

Statistical Physics

Ei =

46

1 2 Pi 2m

Inserting this energy into the partition function

Z = ∑ gi e

−

Ei

KT

( 3.26 )

i

But we note that the kinetic energy of an ideal gas accupying a rather large volume may be considered as not being quantized but as having a continuous spectrum. Therefore we must rewrite Eq. ( 3.26 ), replacing the sum by an integral in the form ∞

Z = ∫ g ( E )e

−E

KT

dE

0

where g(E) d(E) replaces gi represent the number of molecular states in the energy range between E and E+dE.

g( E )d ( E) =

4πV 2m 3 3 h

(

4πV Z = 3 2m 3 h

(

~

) ∫E 1

2

1

)

2

1

e

2

1

E 2 dE

−E

( 3.27 ) KT

dE

0

Let the integration part of above equation be I ~

I = ∫E

1

2

e

−E

~

KT

0

=

1 2

dE = ∫ E

1

2

e −αE dE

0

βπ( KT

)3

Which gives the partition function of an ideal monoatomic gas as function of temperature and the volume of the gas.

Z=

4πV 2m 3 3 h

(

)

1

V ( 2πmKT ) Z= h3

2

3

1 3 π ( KT ) 2

2

Then, by taking the natural logarithm of each side, we have

Physics Department, Padjadjaran University I Made Joni

( 3.28 )

Statistical Physics

47

V ( 2πmKT ) 3 2  ln Z = ln   h3   3 ln Z = C + ln KT 2 where C is includes all the remaining constant quantities. Substituting this value, we obtained the average energy of the molecules as

Eave = kT 2

∂ ( ln Z ) ∂T

∂  3  C + ln kT  ∂T  2  3 1 = kT 2 2T = kT 2

The average energy is then Eave =

3 kT 2

( 3.29 )

The total energy of gas composed of N molecules is then U = NE ave =

3 NkT 2

( 3.30 )

We conclude that the internal energy of an ideal monoatomic gas dipends only on its temperature. This is a direct result of our definition of an ideal gas and of temperature. We do not expect the same relation to hold for real gases or other substances, since their internal energy is partly potential and thus depends on the separation of the molecules ; that is, on the volume of the substance.

3.6.

Heat capacity of an ideal gas

Heat capacity of an ideal monoatomic gas We defined the capacity of a substance at constant volume and at constant pressure as,

1  ∂V  1   = CV n  ∂T V n 1  ∂V  1 CP =   = CP n  ∂T  P n CV =

Physics Department, Padjadjaran University I Made Joni

( 3.31 )

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48

where H=U+PV is the enthalpy of the substance. Let us first consider an ideal monoatomic gas. The internal energy of such a gas U =

3 nRT 2

by using Eq. ( 3.31 ) CV =

3 R = R.4715 jmol 2

−1

K −1 = 2.9807 cal .mol

−1

K −1

( 3.32 )

equation of state of an ideal gas PV = nRT H = U + PV = CP =

3 5 nRT + nRT = nRT 2 2

5 R = 20 .7858 j.mol 2

−1 0

. c −1 = 4.9678 cal .mol

−1 0

. c −1

(

3.33 ) Therefore all ideal monoatomic gas is have the same heat capacity, independent of the structure of the atoms. We may note cP - cV = R. So that cP is longer then cV by the amount R. 3 R 2 5 cP = R 2 5 γ = 3 cV =

( 3.34 )

This relation is just for an monoatomic gas. These result suggest that at very low temperature the rotational energy is “FROZEN” and when energy is added to the gas the only result is to increase the translational energy of the molecules. At room temperature, however intermolecular collisions, cause frequent changes in both the rotational and translational energy of a molecule.

3.7.

Heat capacity of an ideal polyatomic gas When the ideal gas is not monoatomic, we must take structure of the molecules into account when

we calculate the heat capacities. The energy of a polyatomic molecules is Emolecule = Etr + Erot + Evib

Physics Department, Padjadjaran University I Made Joni

( 3.35 )

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49

We shall ignore the electronic energy of the molecules because it seldom participate in thermal excitation of the gas. Electronic excitation requires an energy of the orde at 1 eV at least. Such energy is about 40 times greater that the average thermal kinetic energy, at room temperature (298 K) and therefore a temperature of the order of 104K is required to produced a substantial number of molecules is of the order 10-4eV, and therefore molecules can easily be carried to excited rotational level, even at temperatures that are compared with room temperatures vibrational energy are in the range of 10 -3eV to 10-1eV (at room temperature may be found in a few low – laying excited vibrational state). The rotational kinetic energy of diatomic molecules.

Er =

=

L2 2I

 2 l ( l + 1) 2I

( 3.36 )

where l = 0, 1, 2, 3, … and I = µ oro2; ro = separation of the particle is equilibrium. Using Maxwell_Boltzman statistics, gi is intrinsic probability, in this case = the angular momentum = 2l+1 (it may have 2l+1 different orientation, all with a same energy)

n rot =

E N − i g i e KT Z rot

n rot =

Er N ( 2l + 1) e − KT Z rot

− N = ( 2l + 1) e Z rot

h2 l ( l +1) 2 IKT

Q

−l ( l +1) r N T = ( 2l + 1) e Z rot

where Qr =

h2 2 IK

It is called “the characteristic temperature of rotation” The rotational partition function

Z rot = ∑ g i e

−

Ei

KT

i

Z rot = ∑ ( 2l + 1) e

− l ( l +1) Qr

T

l

Rotational energy Physics Department, Padjadjaran University I Made Joni

( 3.37 )

Statistical Physics

50

U rot = KNT

2

d ( ln Z rot dT

1

Z rot = ∫ ( 2l + 1) e

−l ( l +1)

Qr

)

T

0

The rotation of the molecules around its center of mass as is the molecules were rigid body, because their mass is so small, we can neglect the moment of inertia due to the electron. Angular momentum around Zo axis is also zero and the total angular momentum L of the molecules is perpendicular to the molecular axis. So, we substitute 2l+1 = 2l and l(l+1) = l2. 1

Z rot = ∫ 2le

l2

Qr

T

dl

0

we know that ~

∫ xe

−αx 2

dx =

0

1 2α

The rotational partition function become

  1 Z rot = 2  2 Qr T 

  T =  Qr 

Thus in Zrot = ln T – ln Qr The internal energy is U rot = kNT

2

= kNT

2

d ( ln Z rot ) dT d ( ln T − ln Qr ) dT

= kNT

; kN = nR

( 3.38 )

= nRT

The total energy of T >> Qr is U = U trans + U rot =

3 5 nRT + nRT = nRT 2 2

( 3.39 )

Heat capacity diatomic molecules cV =

3.8

1  ∂V  5   = R n  ∂T V 2

The laws of thermodynamics from the point of view of statistical mechanics.

Physics Department, Padjadjaran University I Made Joni

(3.40 )

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51

First law of Thermodynamics This law is simply a statement system

Exchange energy through work or heat or both

surroundings We defined change in internal energy ∆U = Q −W dU = dQ − dW dU = Tds − PdV

( 3.41 )

So far we have kept the number of particle N fixed. If we also allow exchange number of particle dU = Tds − Pdv + µdN

( 3.42 )

where µ = chemical potential, measured the increase in internal energy on adding particle.

Second law of Thermodynamics In statistical equilibrium, we obtained the equilibrium partition of the system. Which depend on the properties of the components of the system and corresponds to the must probable distribution of the molecules of the system among the different available energy state under such condition P ( or ln P) is maximum Ln (P)  maximum ( in equilibrium ) d(lnP) =0 Ln (P)  lower probability then max ( non equilibrium, non isolated ) To describe this trend toward statistical equilibrium by involving toward partition of maximum probability, entropy S has been invented

S = k ln P

( 3.43 )

If an isolated system is not in equilibrium it will naturally evolve in the direction in which its energy increase. Since these are the processes that carry the system toward state of maximum probability or statistical equilibrium. ds ≥ 0

( 3.44 )

Then in equality holds when isolated system is not initially in equilibrium and the process are irreversible we may state the second law of Thermodynamic. The most probable process that may occur in an isolated system are those in when the entropy either increase or remains constant. Physics Department, Padjadjaran University I Made Joni

Statistical Physics

52

If a system is not isolated its entropy must then also change. But the total amount of all changes at entropy made by all amount system involved in the process must be agreement with ( ds ≥ 0 ). Entropy of system in statistical equilibrium and which obey M_B statistic N

P=∏

n

n

n

n

g i i g1 1 g 2 2 g 3 3 ... = ni ! n1!n2 ! n3!...

S = k ln P

( 3.45 )

where ln P = n1 ln g + n 2 ln g 2 + n3 ln g 3 + ..... − ln n1!− ln n 2 !− ln n3 !−... ln n! = n ln n − n ln P = n1 ln g1 + n 2 ln g 2 + n3 ln g 3 + ... − ( n1 ln n1 − n1 + n 2 ln n 2 − n2 + ... )

= n1 ln g 1 + n 2 ln g 2 + n3 ln g 3 + ... − n1 ln n1 − n 2 ln n 2 − n3 ln n3 − ... + n1 + n 2 + n3 + ... = n1 ln

g1 g + n 2 ln 2 + ... + ∑ ni n1 n2

= −n1 ln

n1 n − n 2 ln 2 − ... + N g1 g2 N

ln P = −∑ ni ln i

ni +N gi

The entropy then become

S = k ln P N

= KN − K ∑ ni ln i

ni gi

we have

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53

N −E g i e KT Z ni N −E = e KT gi Z

ni =

n ln i  gi

 N −E  = ln e KT + ln Z  =−

E Z − ln KT N

The entropy of the equilibrium system is N Z  E S = KN − K ∑ ni  − − ln  N  KT i N Ei Z = K ∑ ni + K ∑ ni ln + K ∑ ni KT N i i N

N E  N Z  = K  ∑ ni i + ∑ ni ln + ∑ ni  KT N i  i 

S=

1 T

∑n E i

i

i

+ K ( ∑ni ) ln

Z + K ∑n i N

we know U = ∑ni E i i

and N = ∑ni i

therefore, S =

U Z + KN ln + KN T N

where kN ln

Z Z   + kN = k  N ln + N  N N   = k [ N ( ln Z − ln N ) + N ]

= k ( N ln Z − N ln N + N ) = k [ N ln Z − ( N ln N − N ) ] = k ( N ln Z − ln N !)

Z ZN  = k  N ln  = k ln N!  N!  Physics Department, Padjadjaran University I Made Joni

Statistical Physics

54

And we get the entropy in M_B statistics is S =

U ZN + k ln T N!

( 3.46 )

Entropy of an ideal gas in statistical equilibrium, we know for an ideal gas U =

3 NkT 2

Partition function :

Z

( 2πNkT ) =V

3

2

h3

U ZN + k ln T N!  V ( 2πNkT ) 3 2 / h3 U S = + k ln  T N!   S=

{(

) }

[ ( {(

) })

N

   

]

3 U + k N ln V ( 2kT ) 2 / h 3 − ln N ! T   V ( 2 N kT) 3 2  U  = + k  N ln  − ( N ln N − N )  T h3    

S=

U V ( 2 NkT) S = + kN ln T h3

3

2

− kN ln N + kN

3 NkT V ( 2 NkT) = + N + kN ln 2 T h3 5 V ( 2 NkT) S = kN + kN ln 2 h3 N

3

3

2

− kN ln N

2

(

3.47 ) This result is know as Sackur_Terroda equation. It can be written in alternative form 3 S = kN ln VT 2  + S0  

Where

Physics Department, Padjadjaran University I Made Joni

( 3.48 )

Statistical Physics

V =

55

V N 3

S0 =

5  2πmkT  kN + kN ln V   3 2  h 

2

Physics Department, Padjadjaran University I Made Joni

Statistical Physics

Physics Department, Padjadjaran University I Made Joni

56

Statistical Physics

Physics Department, Padjadjaran University I Made Joni

57