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1 Lecture 6 Ideal gas in microcanonical ensemble. Entropy. Sackur-Tetrode formula. De Broglie wavelength. Chemical potential. Ideal gas in canonical ensemble.

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Presentation on theme: "1 Lecture 6 Ideal gas in microcanonical ensemble. Entropy. Sackur-Tetrode formula. De Broglie wavelength. Chemical potential. Ideal gas in canonical ensemble."— Presentation transcript:

1 1 Lecture 6 Ideal gas in microcanonical ensemble. Entropy. Sackur-Tetrode formula. De Broglie wavelength. Chemical potential. Ideal gas in canonical ensemble. Entropy of a system in a canonical ensemble. Free Energy. Maxwell Velocity Distribution. Principle of equipartition of energy. Heat capacity. Ideal gas in the grand canonical ensemble

2 2 Ideal gas in microcanonical ensemble Let us consider first the ideal gas in microcanonical ensemble. In the microcanonical ensemble for N non- interacting point particles of mass MM MM confined in the volume V with total energy in EE EE at EE EE we must calculate (6.1) where the momentum space integral is to evaluated subject to the constraint that (6.2)

3 3 by the construction of the ensemble. The accessible volume in momentum space is that of a shell of thickness (  E)(M/2E) 1/2 (  E)(M/2E) 1/2 on a hypersphere of radius (2ME) 1/2 (2ME) 1/2. If the result were sensitive to the value of  E  E employed, we would have difficulty in deciding on a value  E. Fortunately we can prove that for a system of large numbers of particles the value of ln   is not sensitive to the value of EE EE, we may even replace  E  E by entire range from to EE.EE. The proof now follows. We write (6.3)

4 4 -dimensional sphere R hypersphere for the volume of a -dimensional sphere of radius R. The volume of a shell of thickness s at the surface of this hypersphere is (6.4) or, by the definition of the exponential function, (6.5) Therefore if vv vv is large enough so that >>R, s >>R, VsVs VsVs is practically the volume V(R) V(R) of the whole sphere. If ~10 23 ~10 23 as for a macroscopic system, the requirement s>>R/10 23 s>>R/10 23 may be satisfied without any practical imprecision in the specification of the energy of the microcanonical ensemble.

5 5 We can replace now the constraint (6.2) by the relaxed condition (6.6) because for any reasonable (not too thin) shell the volume of the shell shell is essentially equal equal to the volume of the entire hypersphere hypersphere. In other words, we want to evaluate the volume of the 3 -dimensional sphere of radius radius R=(2ME) 1/2 R=(2ME) 1/2. We can evaluate the volume V V of the hypersphere by the following argument: Consider the integral (6.7) We may also write (6.8)

6 6 where R -1 S R -1 S denotes the surface area of the -dimensional -dimensional sphere sphere. On comparison of the two results (6.7) and (6.8) we find so that the volume of the sphere is In the 3-dimensional case, V 3 =4  R 3 /3 V 3 =4  R 3 /3 and surface are a 3 =4  R 2 a 3 =4  R 2. In the two-dimensional case, we obtain V2=R2V2=R2 V2=R2V2=R2 and surface are a 2 =2  R 2 =2  R and in the one-dimensional case V 1 =2R V 1 =2R and surface are a 1 =2 1 =2. We can write now that the phase space (6.9) (6.10) (6.11) and using the Stirling approximation to evaluate factorial (6.7) (6.8)

7 7 where in the expression for V we have to put =3N and R=(2ME) 1/2. It turns out that if the N particles are identical we must not count as different conditions, which differ only by interchange of identical particles in phase space. We have to overestimate the volume of phase space by a factor which is N! N! under classical conditions. Taking this factor into account ee ee as the base of natural logarithms (6.12) (6.13) To complete the expression for the entropy of ideal gas we need to introduce the unit of volume in phase space, so that (6.14) From (3.13) we have

8 8 so that in agreement with the elementary result for the internal energy of a perfect monatomic gas. gas. We can consider (6.16) as establishing the connection between  and T. Further (6.16) (6.17) whence (6.18) Using (6.16) and S=k  S=k , we have the famous Sackur-Tetrode formula for the entropy of an ideal gas: (6.19) (6.15)

9 9 Thermal De Broglie Wavelength We note that (2  MkT) 1/2 (2  MkT) 1/2 has the character of an average thermal momentum of a molecule. We define (6.20) as the thermal de Broglie wavelength associated with a molecule. Than

10 10 (6.21) showing that the entropy is determined essentially by the ratio of the volume per particle to the volume 3 associated with the de Broglie wavelength. The chemical potential of a perfect gas will be as following (6.22) or (6.23) or using (6.17) we can write that (6.24) where p is a pressure and f(  ) f(  ) is a function of the temperature alone. (6.17)

11 11 Ideal gas in canonical ensemble. Let us derive the Sackur-Tetrode formula formula using the canonical ensemble. We’ll start first with partition function function. For an ideal gas involving N independent identical spineless particles particles the partition function can be written as follows Now (6.25) (6.26) whence (6.27)

12 12 Now the integral may be evaluated using the definite integral giving for each integral (2  MkT) 1/2 (2  MkT) 1/2. Thus using the definition of the de Broglie wavelength. (6.28) (6.29) We can calculate now the free energy energy using (4.41) and taking into account that lnN!  Nln lnN!  Nln NN.NN. (6.30) whence other properties may be obtained from the relations (6.31) and

13 13 in agreement with (6.19). From (6.30) and (6.31) we have the Sackur-Tetrode equation by the definition of. which is identically to E. We could also have observed that (6.32) (6.33) (6.34)

14 14 Entropy of a system in a canonical ensemble Let Es Es be the sth energy eigenvalue of a system, and let be the probability according to the canonical ensemble that the system will be found in the state s. s. Let us show that the entropy may be expressed in the convenient and instructive form We have from the above results whence (6.35) (6.36) (6.37) (6.38)

15 15 so that and Now we may write (6.36) as, using (6.35) (6.39) (6.40) (6.41) (6.35) (6.36) which is identical to (6.40).

16 16 Let us consider the significance of the two terms on the right of (6.41): in agreement with the definition of F.F.F.F. so that

17 17 Maxwell Velocity Distribution Let us apply the canonical ensemble equally to macroscopic and to atom subsystems. Applied to a single atom of mass MM MM in volume V V, we have for  (E)  (E) the (the occupancy probability of a unit volume of phase space at energy EE EE ) from (4.36) (6.42) as the probability of finding the atom in the momentum range dp x, dp y, dp y, dp z dp z at p x, p x, p y, p y, p z p z. We see that (6.43) is the probability of finding the atom in the velocity range dv x, dv x, dv y, dv y, dv z at v x, v x, v y, v y, vzvz vzvz. (4.36)

18 18 It remains to evaluate e F/kT e F/kT. From (4.41) we have (6.44) and for a single atom ( N=1 ) we have from (6.29) the result (6.45) Thus the probability (6.46) (6.29) (4.41) This result is known as the Maxwell distribution of velocities.

19 19 C w(v)dv=1. and determine the normalization constant C by integration  w(v)dv=1. P(v)dv dv The probability P(v)dv that the atom will have its speed in dv is (6.47) (6.48) P(E)dEdE The probability P(E)dE that the atom will have its energy in dE is (6.49) It may be noted that the steps following (6.41) (6.41) are devoted to normalizing the distribution to a single atom. It would be just as easy to do this by writing the result of the canonical distribution as

20 20 Principle of equipartition of energy Let us consider one of the variables in the energy, say variable pjpj.pjpj. A situation of particular importance occurs when pjpj pjpj is an additive quadratic term in the energy: p j : We can then calculate the mean energy associated with the variable p j : (6.51) E=bp j 2 +other terms not containing p j (6.50) as the other terms in the exponent may be canceled in the numerator and denominator. To evaluate (6.51) we require the definite integrals (6.52)

21 21 (6.53) in general (6.54) (6.55) Therefore we have In classical problem the mean energy associated with each variable which contributes a quadratic quadratic term to the energy has the value in thermal equilibrium. The result is known as the principle of equipartition of energy energy; it depends specifically on the quadratic assumption.

22 22 For a free atom the Hamiltonian is the total of three quadratic terms. The thermal energy is therefore (6.56) for N atoms, and the heat capacity at constant volume is From this we have he Dulong-Petit Dulong-Petit result that the heat capacity (at high temperature) of one mole of monatomic solid is 3R, where R is the gas constant. At low temperature the quantum energy h h may exceed kT, and the problem requires special treatment, leading to the Einstein and Debye theories of the heat capacities of solids. (6.57)

23 23 Ideal gas in the grand canonical ensemble We have for the grand partition function We have already evaluated the quantity (6.58) (6.59) Thus (6.60)

24 24 (6.62) or (6.63) Further (6.64) which is the ideal gas low. When we calculate we get the Sackur-Tetrode formula formula as derived previously. (6.65) (6.61) so that

25 25 Problems (Maxwell distribution) 1)Show that the root mean square velocity of a Maxwellian gas at constant volume and temperature is (3kT/M) 1/2 (2) Show that the most probable speed of a Maxwellian gas at constant volume and temperature is (2kT/M) 1/2 (3) Show that the mean speed of a Maxwellian gas at constant volume and temperature is (8kT/  M) 1/2.


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