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Thermodynamics and Kinetics of Solids 33 ________________________________________________________________________________________________________________________ III. Statistical Thermodynamics 5. Statistical Treatment of Thermodynamics ..., er-1 (r-states). The number of particles of the energy state ei is Ni. 5.1. Statistics and Phenomenological Thermodynamics. - Calculation of the energetic state of each atomic or molecular constituent by making use of mechanics/quantum mechanics. Because of the large number of species: Consideration of probabilities, i.e. statistics. Determination of the partition function (e.g. velocity). In view of the very large number only one distribution among all possible distributions is most probable (Fig. 5.1: comparision of the relative probability for throwing a certain number of spots with 1, 2, 3 and many dices). Number of microstates for the realization of a macrostate (total number of spots) The total number of particles is N: W = N! (5.1) and under consideration of the exchange of dices without new configurations W= N! N 0 !N 1 !N2 !K (5.2) Determination of the distribution function for the following model: - N particles - Particles may be distinguished. - Particles are independent of each other (no mutual influence). - Each particle has one of the energetic states e0, e1, e2, r- 1  Ni = N (5.3) i= 0 The total energy of the system is E (fixed value) r-1  N ie i = E (5.4) i= 0 Of interest is the most probable distribution function, i.e. the distribution for which W= N! N 0 !N 1 !K N r-1 ! (5.5) has a maximum. We consider lnW (which has a maximum at the same value as W): r-1 lnW = ln N!-  ln Ni ! (5.6) i =0 Considering Stirling’s formula (ln n! = n ln n - n for large numbers of n) we have r-1 lnW = N lnN - N -  r-1 N i lnN i + i =0  Ni (5.7) i= 0 Considering eq. (6.3) this results in r-1 lnW = N lnN -  Ni ln Ni (5.8) i =0 A maximum with regard to Ni is obtained from the derivation with regard to Ni under consideration of eqs. (5.3) and (5.4): d ln W = -d  N i lnN i = -  dN i -  lnN idN i = 0 (5.9) Fig. 5.1. Comparison of the relative probabilities for throwing a certain number of spots when using 2, 3 and many dices. 01.08.97 34 Thermodynamics and Kinetics of Solids ________________________________________________________________________________________________________________________  dN i = 0, sin ce dN = 0 (5.10)  e i dNi = 0, sin ce dE = 0 (5.11) Application of Lagrange’s multiplier method, i.e. multiplication of eqs. (5.10) and (5.11) by l and m (constant, but not fixed values) and adding them to eq. (5.9):  dNi +  lnN idN i + l  dN i + m e idNi = 0 (5.12) (5.18)  - ei e -e / kT  e -e / kT i U=N (5.19) i The denominator is named partition function Z Z=  e- e / kT i The counter of eq. (5.19) is kT2 (5.20) dZ . dT Accordingly we have or  dNi (1+ lnN i + l + me i ) = 0 (5.13) Since the dN i may be arbitrarily chosen, the quantities in parenthesis have to disappear: 1 + ln N i + l + me i = 0 -(1+l) -mei e (5.15) According to eq. (5.3) we have N= 1 dZ dln Z = NkT2 Z dT dT (5.21) By knowlegde of Z the inner energy may be determined and from that all other thermodynamic functions. (5.14) or Ni = e U = NkT 2  Ni = e- (1+l )  e -me i (5.16) Entropy The entropy provides information about the direction of irreversible processes and is a criterium for the presence of equilibrium. From a statistic point of view there is a transition to the most probable macrostate. The probability plays accordingly the same role as the entropy. Therefore, a functional relationship is assumed: S = S (W ) (5.22) and by making use of eq. (5.15) follows Ne -me i Ni = e -mei (5.17)  m has to have the inverse dimension of energy. Under consideration of the average oscillation energy of a particle, e = kT , we obtain Ni = N e Derivation of this relationship: 2 independent systems of the same type of particles (1 and 2) are combined isothermally to the total system (1, 2) In this case, the entropies of the individual systems (S1,2 = S1 + S2) are added up, while statistic weights are being multiplied (W1,2 = W1 · W2): S1, 2 = S ( W1, 2 ) = S( W1 ⋅ W 2 ) = S( W1 ) + S ( W2 ) (5.23) -e i / kT  e- e / kT i (5.18) (more strict derivation of this equation by considering quantum mechanics and going to classical mechanics). This equation may be only fulfilled if S = k* ln W (5.24) k* will be later identified as Boltzmann’s constant k. Equation (5.18) represents the Boltzmann’s distribution (Boltzmann’s e-relation). 5.2. The Various Statistics. Inner Energy from Boltzmann’s distribution: Because of E ≡ U in eq. (5.4) we have according eq. Boltzmann’s statistics: - particles which build up the system are independent of each other and distinguishable - any number of particles may occupy the same state. 01.08.97 Thermodynamics and Kinetics of Solids 35 ________________________________________________________________________________________________________________________ - location (3-dim. space) - velocity or momentum (3-dim. momentum space) Both are combined to the 6-dimensional phase space. Quantum mechanical treatment of a particle in a cube with the length a of each edge. Possible energies of the particle: e= Fig. 5.2. Possibilities of realizing the total number 7 and 8 spots with 2 distinguishable dices (B) and undistinguishable dices (BE) as well as prohibition of the same number of spots. Quantum statistics: It is impossible to determine simultaneously exactly location and momentum of a particle. The possibility to distinguish of particles is therefore questionable. Fig. 5.2.: Possibilities to throw dices with a total number of 7 and 8 spots with 2 distinguishable dices (B), not distinguishable dices (BE) and prohibition of the same number of spots (FD). Particles for which the sum of numbers of electrons, protons and neutrons is even (H2, D+, D2, N2, 4He, photons) have an integer spin. Particles for which the sum of the numbers of electrons, protons and neutrons is odd (e-, H+, 3He, NH +4 , NO) have a half-numbered spin. A system which consists of many particles is described in the case of an integer-numbered spin by a symmetric and in the case of a half-numbered spin by an antisymmetric eigenfunction. h2 n 2x + n 2y + n2z 8m a 2 ( ) (5.25) By considering the relationship between energy and momentum e= 1 p 2x + p 2y + p2z 2m ( ) (5.26) eq. (5.25) results in the following possible components of the momentum: h n 2a x h py = ny 2a h pz = nz 2a px = (5.27) nx, ny and nz are the integer quantum numbers. By using a cartesian coordinate system with the unit h/2a of the axis px, py and pz, the states of the particle in the cube are represented by the lattice points with integer numbered coordinate values (Fig. 5.3.). In the first case (integer-numbered spin) any number of particles may be present in the same energy state, while in the latter case (half-numbered spin) each energy state may be only occupied by 1 particle (Pauli’s law). The non-distinguishability results in the following quantum statistics: - Bose-Einstein statistics in the case of an integernumbered spin. - Fermi-Dirac-Statistics in the case of a half-numbered spin (Pauli’s law). 5.3. Momentum- and Phase space Fig. 5.3. States of a particle in a cube with the length of a of each edge in the momentum space. Classical description of the state of a particle: 01.08.97 36 Thermodynamics and Kinetics of Solids ________________________________________________________________________________________________________________________ Volume of each cell: h3/8a3. Since eq. (5.26) holds for positive and negative momenta, all 8 octants of the momentum space have to be taken into consideration. Accordingly, a state of a species corresponds to a cell of the volume h3/a3 in the full 3-dimensional momentum space. By considering the phase space, i.e. adding the physical space to the momentum space, a state of the particle corresponds in the 6-dimensional phase space toa cell of the volume h3, since the species will occupy the volume a3. 2 2 ÊÁ a 8me ˆ˜ Ëh ¯ + ny ÊÁ a 8 me ˆ˜ Ëh ¯ 2 + n2z ÊÁ a 8me ˆ˜ Ëh ¯ 2 ( 12 8me 2 ) ( 12 p 2y 8me 2 + ) ( 12 p2z 8me ) 2 =1 (5.29) This equation is the surface of a bowl with a radius 3 1 4 1 2 2 8me and the volume 3 p 8 ( 8me ) . Æ The number of cells with energies < e is: N(e) = 4 1 3 h 3 8 2 V 32 32 p (8m e ) 2 / 3 = p 3m e 3 8 a 3 h (6.30) where (V = a3). The number of states with energies between e and e + de is dN ( e ) = D( e ) de = dN ( e ) de de N (e ) = 1.03 ⋅10 28 which is a continuum in a first approach. If there is not only 1 helium atom in the volume of 1 L, but 3 · 1022 atoms under atmospheric pressure, the number of states is N (e ) = 3.3⋅10 5 5.4. DistributionFfunctions Calculation of the number of quantum states with translational energies < e: Feeding eq. (5.27) into eq. (5.28) results in: + ) = 1 (5.28) There exist as many different quantum states that belong to the energy e as integer solutions nx, ny, nz are possible. All those integer numbers nx, ny, nz, that result in a value < 1 for the left hand side of eq. (5.28 belong to quantum states with energy values < e. p 2x ( Even under such conditions, the number of quantum states and accordingly the number of cells in the phase space is still much larger than the number of species. Eq. (5.25) may be rewritten in the following way: n2x of translation of a helium atom (m = 6.7 · 10-27 kg) in a volume of 1 l at 300 K is under considering the -21 translation of energy e = 32 kT 6.2 ⋅10 J : (5.31) D(e) is the density of states. Differentiation of eq. (5.30) results in V 3 1 (5.32) dN ( e ) = D( e ) de = 4 2 p 3 m 2 e 2 de h The order of magnitude of the number of quantum states Since the discrete energy levels are very close to each other, we do not consider the occupation of the individual levels but the occupation of the total number of energy values between ei and ei + dei. The number of energy levels between ei and e i + de i: Ai. These are occupied by Ni species. For the determination of the distribution function it has to be calculated by how many microstates a macrostate may be built up. That macrostate which may be generated by the largest number of microstates is the most probable one and characteristic for the system. Bose-Einstein-Statistics. The species may not be distinguished and any number of species may occupy one state. The energy levels are: I, II, ..., Ai. Fig. 5.4. Shows the distribution of 2 species over 3 cells. Dots are used in that figure to indicate that the species may not be distinguished. In general: Distribution of Ni species over Ai cells: Number of the possibilities of distributions: Fig. 5.4. Illustration or the derivation of Bose-EinsteinsStatistics. Two not distinguishable species (dots) are distributed over three cells (I, II, III) of a group of energy levels. 01.08.97 Thermodynamics and Kinetics of Solids 37 ________________________________________________________________________________________________________________________ A i ⋅ (A i +1)L( Ai + Ni - 1) 1⋅ 2LNi d ln W =  i Expansion of this expression by (Ai - 1)! results in the following number of microstates - i ( N i + Ai - 1) ! N i !( A i - 1) ! i Ni dN i -  ln NidN i = 0 Ni (5.38) or The same holds for all energy intervals. Since each distribution within one group may be combined with any distribution in another group, the number of different microstates is W= P Ni + Ai dNi +  ln( N i + A i )dN i Ni + Ai i ( N i + A i -1 )! N i !( A i - 1)! ÊA ˆ  lnÁË N ii +1˜¯ dNi = 0 (5.39) i with the boundary conditions dN = (5.33)  dN i = 0 (5.40)  e idN i = 0 (5.41) i and dE = Conditions that have to be fulfilled: i i) The total number of species is constant N= Application of Lagrange’s multiplier method:  Ni (5.34) È ÊA  dN i ÍlnÁ N i i Î Ë i i ˘ ˆ + 1˜ + a + be i ˙ = 0 ¯ ˚ (5.42) ii) The total energy of the system is constant E=  This results in Ni e i (5.35) i That macrostate is the most stable one for which W or ln W takes up a maximum under the boundary conditions eqs. (5.34) and (5.35). Equation (5.33) results in ln W =  ln( (N i + Ai )!) -  ln( N i !) -  ln(A i !) i i (5.36) i Considering Stirling’s formula (ln n! = n ln n - n for large numbers n): ln W =  ( N i + A i ) ln( N i + A i ) -  ( N i + A i ) i - = i i (5.37) i i Ni 1 = -a -be i Ai e -1 (5.44) or Determination of a and b: Making use of eq. (5.37a), the expression S = k ln W may be written as: È Â ÍÎ Ni ln i  ( Ni + A i ) ln( Ni + Ai ) -  N i lnN i -  A i lnA i i (5.43) S=k i  N i lnN i +  Ni -  A i lnA i +  Ai i Ê Ai ˆ ln Á + 1˜ + a + be i = 0 N Ë i ¯ i Ni + Ai N + Ai ˘ + Ai ln i Ni Ai ˙˚ (5.45) and according to eq. (5.44) S=k È 1 ˘  ÍÎ Ni ( lnB - be i ) - Ai ln ÊÁË1 - B ebe ˆ˜¯ ˙˚ i (5.46) i (5.37a) Maximum: 01.08.97 38 Thermodynamics and Kinetics of Solids ________________________________________________________________________________________________________________________ 1 be i e <<1 (confirmation later!), the B following approximation holds under considering of ln (1-x) = -x: with B = e-a. If È Ai ˘ S = k Í N i ( ln B - bei ) +  -be ˙ i Î i ˚ i Be and because of eq. (5.44) for Be -bei È ∂ lnB ˘ S = k N Í lnB - b + 1˙ ∂b Î ˚  For the inner energie U of the system of N particles holds U = Ne = N (5.48) e may be expressed by the distribution function (5.44): E = Ne =  N i e i =  i -be i Be -1 (5.50)  Ni =  i i Ê ∂U ˆ ∂ 2 ln B Á ˜ =N Ë ∂b ¯ v ∂b 2 (5.57) Ê ∂S ˆ ∂2 ln B Á ˜ = -kNb Ë ∂b ¯ v ∂b 2 (5.58) 1 Ê ∂U ˆ ˜ =T= Æ Á Ë ∂S ¯ v kb or b = - Under considering of N= Be Ai -be i -1 (5.51) 1 kT N= i e= i  Ai e be i = ∂  A e be i ∂b i i = ∂ È Íln ∂b Î i ˘  A ie be ˙˚ i (5.52)  A ie be i - lnB (5.60) • 1 V 3 e 2 de N = 4 2p 3 m 2 e h Be kT - 1 o i Ú According to eq. (5.51) results in the limiting case Be -bei >> 1 ln N = ln Ai Bee i / kT -1 Changing from summation to integration: The number of states Ai has to be expressed as a function of e: Ai, i.e. the number of energy levels between ei and ei + dei, is identical with d N(e) in eq. (5.32):  A ie be i i (5.59) Accordingly, eq. (5.51) may be written in the following way e becomes in the case Be -bei >> 1 :  Ai e i e be i (5.56) determine b: (5.49) Aiei ∂ ln B ∂b Ê ∂U ˆ ˜ = T , eqs. (5.55) and (5.56) allow to Because of Á Ë ∂S ¯ v With E = Ne ( e : average energy of each particle) eq. (5.48) results in S = kN[ lnB - be + 1] (5.55) (5.47) >> 1 : È ˘ Í ˙ S = k ÍlnB Ni - b Ni e i + N ˙ 123 Í ˙ i 24 1 4 3 N ÍÎ ˙˚ E  Accordingly, the following expression holds for the entropy (5.49) Be e kT >>1 : (5.53) i (5.61) N = 4 2p V 32 1 • 12 - e kT m e e de B Ú0 h3 N = 4 2p 2 V 32 1 3 ( kT ) 2 2 u2 e -u du 3m h B 0 (5.62) and eq. (5.52) may be rewritten: or ∂ ∂ ( ln N + ln B) = ln B e= ∂b ∂b B is a function of b. (5.54) • Ú (5.63) 01.08.97 Thermodynamics and Kinetics of Solids 39 ________________________________________________________________________________________________________________________ 1 The integral has the value pÆ. 4 This results in 3 (2p mk T ) 2 V B= h3 N (5.64) (B ≡ e-a) Application of the same procedure as for the BoseEinstein and Fermi-Dirac-Statistics. The cells may be occupied without limitations; all species may be distinguished from each other. Number of possibilities to distribute Ni species over Ai states: A i Ni Fermi-Dirac-Statistics Again the species may not be distinguished, but in addition Pauli’s law holds, i.e. each quantum state may only be occupied by one species. Number of microstates: A i ( A i - 1)( Ai - 2 )L (A i - N i +1) 1⋅ 2⋅L ⋅ N i W= N! i A N 0 A N1 LA N i L N 0 !N 1 !L N i !L 0 1 = N! Ai ! N i !( A i - N i )! A Ni ’ N ii ! (5.68) i With the same boundary conditions and the same procedure as before, this results in For all energy intervals holds i N! N 0 !N1 !L N i !L Number of possibilities to realize the distribution Expansion by (Ai - Ni)! results in the number of micro states W=’ Number of possibilities to distribute N species in groups of N0, N1, N2, ... species each with the same propertieson A0, A1, A2, …energy levels: Ai ! N i !( A i - N i )! (5.65) Analogously results as above under the same boundary conditions Ni 1 = -a -be i Ai e Also for the Boltzmann-Statistics holds b=- Ni 1 = -a -be i Ai e +1 (5.69) (5.66) 1 kT (5.70) 3 B= For b=- 1 kT and in the limiting case for B ⋅ e -be i >> 1 B = e -a the same value holds as in the case of Bose-Einstein’s statistics. (2p m kT ) 3 h 2 V N (5.71) (5.67) Comparison of the Statistics Table 5.1. B-values for the H2-molecule and conducting electrons in sodium at different temperatures and pressures. Boltzmann-Statistics 01.08.97 40 Thermodynamics and Kinetics of Solids ________________________________________________________________________________________________________________________ Difference in the distribution functions: "1" in the denominator. -a e i If e e kT >> 1 , the quantum statistics result in the Boltzmann statistics. z:=  gi e - For e e >>1 the right hand side of the distribution functions becomes very small, i.e. Ni / Ai (occupation probability) becomes very small. The number of quantum states is very much larger than the number of species (holds, e.g., for a gas under normal conditions). ei Since e i ≥ 0 , e kT takes up values between 1 and • . For B e ei kT kT >>1 , B has to be sufficiently large: large mass, high temperature, high dissolution. B-values for H2 and e-: Table 1. 1. row: The same density as at 273 K and p = 1.013 bar is assumed at all temperatures. 2. + 3. row: p = constant For H2, the condition B>>1 is fulfilled except for extremely low temperatures and high pressures. Electrons: small mass, large concentration (in the case of metals) Æ Fermi-Dirac-Statistics. 5.5. Partition Function and Thermodynamic Potential Making use of the different statistics, thermodynamic quantities are derived. According to Boltzmann’s statistics, the ratio of the number of species Ni with the energy ei relative to the total number of species N is ei Ni g i e - kT = ei N gi e - kT (5.74) ("molecular partition function") results from eq. 5.73, as may be easily shown by substitution, e = kT2 ei kT i When is that the case? -a ei ∂z / ∂T ∂ lnz = kT2 z ∂T (5.75) The average energy may be determined from the differentiation of the partition function with regard to the temperature. When we do not consider the occupation probability of an energy state ei by a single species and not the average energy of this single species, but a large number (n moles) of species, then the energy (= inner energy) is analogously for the entire system Ê ∂ ln Z ˆ U = kT2 Á ˜ Ë ∂T ¯ v (5.76) Z: "system partition function" Eq. 5.76 allows to relate the statistical treatment to phenomenological thermodynamics by knowledge of the partition function: i) Heat capacity ∂ Ê 2 ∂ ln Z ˆ ∂ Ê ∂ ln Z ˆ Ê ∂U ˆ Cv = Á ˜ = Á kT ˜ = -k = Á Ë ∂T ¯ v ∂T Ë ∂T ¯ v ∂T Ë ∂ (1 / T) ˜¯ v Ê ∂[ ∂ lnZ / ∂(1/ T )] ˆ Ê ∂[ ∂ lnZ / ∂(1/ T )] ˆ k = - kÁ = - 2 T 2Á ˜ ˜ = ∂T T ∂T Ë ¯v Ë ¯v (5.72)  i = gi: degree of degeneration of the i-th state (statistical weight). This results in the average energy e of one species:  N ie i  e i g i e e= i = i N  gi e ei ei k Ê ∂[ ∂ lnZ / ∂(1/ T )] ˆ k Ê ∂ 2 ln Z ˆ =- 2 Á ˜ 2Á T Ë ∂(1/ T ) T Ë ∂(1/ T )2 ˜¯ v ¯v ii) Entropy dS = Cv dT T (5.78) kT kT (5.73) Integration: T i S - S0 = With the abbreviation (5.77) Cv dT T 0 Ú (5.79) Making use of eq. 5.77, this results in 01.08.97 Thermodynamics and Kinetics of Solids 41 ________________________________________________________________________________________________________________________ Ê ∂ lnZ ˆ p⋅ V = kT Á ˜ Ë ∂ lnV ¯ T T 1 ∂ Ê 2 ∂ lnZ ˆ S - S0 = Á kT ˜ dT = T ∂T Ë ∂T ¯ v 0 Ú T = 1È v) Enthalpy Ê ∂2 lnZ ˆ Ê ∂ lnZ ˆ ˘ ˜ ˙ dT = 2 ˜ + 2kT Á ∂T ¯ v Ë ∂T ¯ v ˙˚ Ú T ÍÍÎkT 2 ÁË 0 T T Ê ∂2 ln Z ˆ Ê ∂ lnZ ˆ = kTÁ dT + 2k Á ˜ dT 2 ˜ ∂T Ë ∂T ¯ v Ë ¯v 0 0 Ú (5.89) Ú Ê ∂ lnZ ˆ Ê ∂ lnZ ˆ H = U + pV = kT 2 Á ˜ + kT Á ˜ Ë ∂T ¯ v Ë ∂ lnV ¯ T (5.80) Partial integration Æ ÈÊ ∂ ln Z ˆ Ê ∂ ln Z ˆ ˘ H = kT ÍÁ ˜ +Á ˜ ˙ ÎË ∂T ¯ v Ë ∂ lnV ¯ T ˚ 2 (5.90) vi) Gibbs Energy T T Ú Ú Ê ∂ ln Z ˆ Ê ∂ ln Z ˆ Ê ∂ ln Z ˆ S - S 0 = kTÁ ˜ - kÁ ˜ dT + 2 k Á ˜ dT = Ë ∂T ¯ v 0 Ë ∂T ¯ v Ë ∂T ¯ v 0 T Ê ∂ ln Z ˆ = kTÁ ˜ + k lnZ Ë ∂T ¯ v 0 (5.81) Ê ∂ lnZ ˆ G = F + p⋅ V = -kT lnZ + kT Á ˜ Ë ∂ lnV ¯ T È Ê ∂ ln Z ˆ ˘ G = -kT Íln Z - Á ˜ Ë ∂ ln V ¯ T ˙˚ Î (5.91) Making use of eq. 5.76, this results in U S - S 0 = + k lnZ - ( k ⋅ lnZ )T =0 T Æ S 0 = ( k ln Z) T =0 (temperature independent) (5.82) (5.83) Relationship between the entropy S and the statistical weight W because the entropy adds up while the probability is multiplied when several systems are combined, the assumption is made S ~ lnW or S = k* ln W (5.92) Accordingly, we have U S = + k lnZ T It has been (5.84) r-1 lnW = N lnN -  Ni ln Ni oder ÈÊ ∂ lnZ ˆ ˘ S = k ÍÁ ˜ + lnZ ˙ ÎË ∂ lnT ¯ v ˚ (5.93) 0 (5.85) Making use of the partition function (without degeneration) ei N i e - kT = N z iii) Free Energy F = U - TS (5.86) In view of eq. (5.84) this results in ÊU ˆ F = U - T Á + k ln Z˜ = -kT ln Z ËT ¯ (5.87) (5.94) this results in È S = k* ÍN lnN - N ÍÎ Â i e- ei Ê e- ei kT ˆ ˘ ln Á N ˜˙ z z ¯ ˙˚ Ë kT (5.95) ei ei ei e - kT e - kT e - kT e i ˘ ˙ = k Nln N - N  ln N + N  ln z + N z z z kT ˙˚ ÍÎ i i i È *Í iv) p·V Ê ∂F ˆ Ê ∂ lnZ ˆ p = -Á ˜ = kT Á ˜ Ë ∂V ¯ T Ë ∂V ¯ T (5.88) È ei N ei e= k Nln N - N ln N + N ln z +  kT i z ÍÎ *Í kT ˘ ˙ ˙˚ 01.08.97 42 Thermodynamics and Kinetics of Solids ________________________________________________________________________________________________________________________ Making use of eqs. (5.73) and (5.74) this results in Ne ˘ U˘ È È S = k* ÍN lnz + = k* ÍN lnz + ˙ Î Î kT ˚ kT ˙˚ (5.96) Comparison with eq. (5.84) results in k* = k (5.97) and Z = zN (5.98) Relationship between the molecular partition function and the system partition function. According to eq. 5.98, the system partition function (without degeneration) may be written as Z =Âe i i - Ne kT =Âe - EkTi (5.99) i Ei: energy eigen value of the i-th quantum state of the macro system of N species. Determination of Z from z: i) Boltzmann: The system consists of N not interacting distinguishable species (their exchange provides a new state). Example: Crystal of N species, which may be distinguished from each other because of the localization at specific lattice sites. Exchange of 2 such species provides a new state E i = N ei ; Z = zN (5.100) All species of the system are equal to each other and have the same energy eigen values. ii) B o s e - E i n s t e i n : The system consists of N not interacting and not distinguishable species (their exchange provides no new state) Example: Ideal Gas of free molecules. The exchange of 2 species provides not a new state. Under the assumption that the number of states is much larger than the number of species, it may be assumed that each quantum state is only occupied by 1 species. 01.08.97