§7. Boltzmann Statistics

1. §7. Boltzmann Statistics §7.1 The statistics expression of thermodynamics quantities 7.1.1 The statistics expression of internal energy Introduce a new function Z1 called partition function Then With the formula (7.1.1) and (7.1.3) it gets

2. 7.1.2 Generalized work Generalized force performed by environment follows that For instance ,

3. In infinitesimal quasi-static process the work performed by environment follows that The total differential of is given by Remark! (1) The first term denotes the work performed by environment ; (2) The second term denotes the heat absorbing from environment.

4. 7.1.3 The statistics expression of entropy Based on the first law of thermodynamics , with integral factor 1/T for dQ, it gets that With the formulas (7.1.4) and (7.1.6) it gets The total differential of lnZ1 follows that Therefore dQ gets another integral factor β

5. Let It will prove that k is Boltzmann constant. Comparing the formula (7.1.10) and (7.1.11) and with the formula (7.1.12) one gets Here the integral constant is chose to be zero.

6. With logarithm calculation for the formula (7.1.3) it gets Based on Boltzmann distribution , it gets therefore Comparing with the formula (6.6.4) it gets This is called Boltzmann relation.

7. Remark! • The more the microstate number is, the bigger the entropy is. (2) For Bose and Fermi systems satisfying the classical limited condition, the entropy is given by

8. 7.1.4 The statistics expression of free energy F=U-TS This formulaisapplied to local systems. or This formulaisapplied to Bose and Fermi system.

9. 7.1.5 The classical statistics expression When Δωlis small enough it follows The internal energy, equation of state and entropy are the same with above results as long as partition function expression (7.1.18) is applied.

10. §7.2 Theequation of state of the ideal gas The energy of monatomic molecule In the scope of dxdydzdpxdpydpz, the probable microstate number is given by Therefore partition function follows that

11. With logarithm calculation for the formula (7.2.3) it gets here is the volume of the ideal gas. Remark! (1) The Boltzmann constant is obtained from comparing the formula with pV= nRT. (2) For biatomic molecule, multi-atomic molecule, although the energies include translational energy, rotational energy and vibrational energy, the formula (7.2.5) suits for every case.

12. (3) The results are the same by using classical statistics theory. (4) The other expression of classical limited condition is that the average space between molecules is more bigger then de Broglie wave length. If ε=3kT/2, thus ,

13. §7.3 Maxwell Velocity Distribution Law 7.3.1 Maxwell velocity distribution The classical expression of Boltzmann distribution No outfield, The state number of translation of molecule mass center follows that In the scope of V and dpx dpy dpz , the molecule number follows that

14. The parameter α is given by In the scope of V and dpx dpy dpz , the molecule number follows that

15. Let , denoting the molecule number of the unit of volume, thus the molecule number within the velocity scope of dvx dvy dvz is given by This formula is known very well and called Maxwell velocity distribution law.

16. The volume element of spherical polar coordinates substitutes dvx dvy dvz , and integral for variables θ and φ, thus the molecule number in unit of volume and the velocity scope of dv is given by

17. 7.3.2 Three characteristic velocity • The most probable velocity（vm) μ is mole mass. (2) Mean velocity ( )

18. (3) Square mean root(vs) 7.3.3 Application of Maxwell velocity distribution law Calculate the molecule number of collision in unit of time at unit area. Solution dA is a area element, dΓdAdt denotes the molecule number of collision in dt at dA. dΓdAdt=

19. namely

20. §7.4 Energy Equipartition Theorem 7.4.1 Energy equipartition theorem For a classical system which is in equilibrium state with temperature T the average value of every square term of a particle energy equals . εpandεq denote the particle kinetic energy and potential energy respactivily. here pi is momentum, ai is a positive coefficient.

21. The first term equals zero, it follows that

22. Potential energy can be denoted as square terms bi is a positive coefficient. Similarly it gets that 7.4.2 Application of energy equipartition theorem (1) Monatomic molecule gas According to energy equipartition theorem the mean energy is

23. The internal energy of monatomic molecule of the ideal gas The heat capacity as constant volume with The heat capacity as constant pressure with

24. (2) Biatomic molecule gas The first term: translational energy; M=m1+m2 The second term: rotational energy encircling center of mass, I=μr2 moment of inertia, The third term: relative movement energy of two atoms, relative movement kinetic energy, u(r) is the interaction energy of two atoms. For rigidity biatomic molecule

25. The internal energy and heat capacity quantities follows as , ,

26. (3) The solid The atomic vibration in the solid is considered as harmonic oscillation of independence each other.The energy of one degree of freedom is The internal energy of the solid is U=3NkT The heat capacity as constant volume with This result is agreement with the experimental result of Dulong-Petit.

27. §7.5 The Internal Energy And Heat Capacity of the Ideal Gas 7.5.1 The basic expression of internal energy and heat capacity et, ev and er denote translational energy, vibrational energy and rotational energy of biatomic molecule ideal gas. The total partition function can be written as the product of translational partition function , vibrational partition function , rotational partition function .

28. The internal energy of biatomic molecule ideal gas is The heat capacity as constant volume with The translational partition function has been given by

29. 7.5.2 Why is the contribution of vibration degree of freedom to heat capacity nearly zero in the case of normal temperature. The relative vibration can be considered as linear harmonic oscillation vibrational partition function Based on it gets

30. Introduce vibration characteristic temperature θv Theformulas (7.5.8) and (7.5.9) follows as

31. θv ~103K, normal temperature T<< θv , therefore Uv and can beapproximately The formula (7.5.9’) indicates that the contribution of vibration degree of freedom to heat capacity is nearly zero in the case of normal temperature. Energy level interval , transition energy is very big, oscillator can not be excitated to high energy level and freeze in ground state.

32. 7.5.3 Why is not the heat capacity of hydrogen agreement with experiment? (1) Heteronuclear (CO, NO , HCl) Rotational energy level and rotational partition function are Introduce vibration characteristic temperature θr

33. In the case of normal temperature, , can be considered to be a continuous variable. Thus integral substitutes calculation sum. Let Therefore it gets

34. (2) The question about H2 Ortho-hydrogen state: spin parallel , odd number for l, probability is . Parahydrogen state: spin reverse parallel , even number for l, probability is . denote the rotation partition functions of Ortho-hydrogen and parahydrogen respectively.

35. H2is atthe stateof high l. Similarly it gets Because the moment of inertia I of hydrogen is small, so the vibration characteristic temperature θr is big. In the case of low temperature (92K), energy equipartition theorem is not applicable. 7.5.4 Why does not the contribution of electron to the heat capacity of gas be taken into account. The difference between excitation state energy and ground state energy for a electron is 1~10eV, namely 10-19~10-18J, corresponding temperature 104~105 K. It is too high to exciting a electron to excitation state.

36. 7.5.5 Calculation thermodynamics quantities by using classical partition function. The energy of different core diatomic molecule is

37. We obtain that With the formulas (7.5.21)-(7.5.23) , we have

38. §7.6 The Entropy of the Ideal Gas 7.6.1 The entropy of the ideal gas with For monatomic molecule ideal gas we have

39. 7.6.2 The chemical potential of the monatomic molecule ideal gas According to the formulas (7.1.16’) and (7.6.4) , it gets For the ideal gas , μ<0

40. §7.7 The Einstein Theory of Solid Heat Capacity 3N oscillators : oscillator energy level is

41. Introduce vibration characteristic temperature θE Dots denote experimental result; Solid line denotes Einstein theory result. θE =1320K

42. Discussion: (1) When T >> θE, CV =3Nk (7.7.7) This formulais agreement with energy equipartition theorem. The effect of quantum is neglected and The classical statistics is applied. (2) When T << θE, The difference between excitation state energy and ground state energy is much big so that 3N oscillators are in ground state.

43. §7.8 Paramagnetism Solid An especially interesting application of classical statistics ( Boltzmann statistics) is the paramagnetic behavior of substances. If a homogeneous field points in z-direction, the total angular momentum of a magnetic ion is 1/2. The magnetic momentum is Two possible energy are –μB and μB . The partition function of this system is

44. Generalized force performed by environment Magnetization M is

45. Discussion: • High T or weak field The relation (7.8.4) is known as Curie’s law. (2) Low T or strong field M=Nμ (7.8.5)

46. The internal energy of this system is This is potential energy in outside field. The entropy of this system is Discussion: • High T or weak field

47. Then The microstate number is (2) Low T or strong field Then The microstate number is 1, namely all magnetic momentum points in the direction of H.

48. §7.9 The state of Negative Temperature If S decreases with increasing U,thus T isnegative. The example of a paramagnetic system with j=1/2(two-level system, nuclear spin system ) allows us to discuss a possible extension of the notion of temperature. Each of the N particles of the system shall be able to assume two possible energies, . Let the number of nuclear magnetic momentum in the level + ε be N+, and that in - ε be N-. Of course we have N=N+ + N- (7.9.2)

49. The total energy of system is Equations (7.9.2) and (7.9.3) can be solved for N+ and N- , Equation (7.9.4) immediately allows for the calculation of the entropy of the system. Where the formula lnm!=m (lnm-1) is used for N+ , N- >>1.

50. Equation (7.9.6) yields for the temperature Discussion: (1) As long as E<0, we have T >0, as usual. (2) When E>0, we have T <0. (3) When E= -Nε, all magnetic momentums point the direction of B. Ω=1, S=0.