Lecture 6 Ideal gas in microcanonical ensemble Entropy

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 1

Ideal gas in microcanonical ensemble Let us consider first the ideal gas in microcanonical ensemble. In the microcanonical ensemble for N noninteracting point particles of mass M confined in the volume V with total energy in E at E we must calculate (6. 1) where the momentum space integral is to evaluated subject to the constraint that (6. 2) 2

by the construction of the ensemble. The accessible volume in momentum space is that of a shell of thickness ( E)(M/2 E)1/2 on a hypersphere of radius (2 ME)1/2. If the result were sensitive to the value of 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 , we may even replace E by entire range from 0 to E. The proof now follows. We write (6. 3) 3

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 v is large enough so that s >>R, Vs is practically the volume V(R) of the whole sphere. If ~1023 as for a macroscopic system, the requirement s>>R/1023 may be satisfied without any practical imprecision in the specification of the energy of the microcanonical ensemble. 4

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 is essentially equal to the volume of the entire hypersphere. In other words, we want to evaluate the volume of the 3 -dimensional sphere of radius R=(2 ME)1/2. We can evaluate the volume V of the hypersphere by the following argument: Consider the integral (6. 7) We may also write (6. 8) 5

where R -1 S denotes the surface area of the -dimensional sphere On comparison of the two results (6. 7) and (6. 8) we find (6. 7) (6. 9) (6. 8) so that the volume of the sphere is (6. 10) V 3=4 R 3/3 and surface are a 3=4 R 2. In the two-dimensional case, we obtain V 2= R 2 and surface are a 2=2 R and in the one-dimensional case V 1=2 R and surface are a 1=2. =2 In the 3 -dimensional case, We can write now that the phase space (6. 11) and using the Stirling approximation to evaluate factorial 6

(6. 12) where in the expression for V we have to put =3 N and R=(2 ME)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! under classical conditions. Taking this factor into account e as the base of natural logarithms (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 7

(6. 15) so that (6. 16) in agreement with the elementary result for the internal energy of a perfect monatomic gas. We can consider (6. 16) as establishing the connection between and T. Further whence (6. 17) (6. 18) Using (6. 16) and S=k , we have the famous Sackur-Tetrode formula for the entropy of an ideal gas: (6. 19) 8

Thermal De Broglie Wavelength We note that (2 Mk. T)1/2 has the character of an average thermal momentum of a molecule. We define (6. 20) as thermal de Broglie wavelength associated with a molecule. Than 9

(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) (6. 17) where p is a pressure and f( ) is a function of the temperature alone. 10

Ideal gas in canonical ensemble. Let us derive the Sackur-Tetrode formula using the canonical ensemble. We’ll start first with partition function. For an ideal gas involving N independent identical spineless particles the partition function can be written as follows (6. 25) Now (6. 26) whence (6. 27) 11

Now the integral may be evaluated using the definite integral (6. 28) giving for each integral (2 Mk. T)1/2. Thus using the definition of the de Broglie wavelength . (6. 29) We can calculate now the free energy using (4. 41) and taking into account that ln. N! Nln N. N (6. 30) whence other properties may be obtained from the relations (6. 31) and 12

(6. 32) which is identically to E. We could also have observed that (6. 33) by the definition of <E>. From (6. 30) and (6. 31) we have the Sackur-Tetrode equation (6. 34) in agreement with (6. 19). 13

Entropy of a system in a canonical ensemble Let Es be the sth energy eigenvalue of a system, and let (6. 35) be the probability according to the canonical ensemble that the system will be found in the state s. Let us show that the entropy may be expressed in the convenient and instructive form (6. 36) We have from the above results whence (6. 37) (6. 38) 14

so that (6. 39) and (6. 40) Now we may write (6. 36) as, using (6. 35) (6. 36) (6. 41) which is identical to (6. 40). 15

Let us consider the significance of the two terms on the right of (6. 41): so that in agreement with the definition of F. 16

Maxwell Velocity Distribution Let us apply the canonical ensemble equally to macroscopic and to atom subsystems. Applied to a single atom of mass M in volume V, we have for (E) (the occupancy probability of a unit volume of phase space at energy E ) from (4. 36) (6. 42) as the probability of finding the atom in the momentum range dpx, dpy, dpz at px, py, pz. We see that (6. 43) is the probability of finding the atom in the velocity range dvx, dvy, dvz at vx, vy, vz. 17

It remains to evaluate e. F/k. T. From (4. 41) we have (4. 41) (6. 44) and for a single atom (N=1) we have from (6. 29) the result (6. 45) (6. 29) Thus the probability (6. 46) This result is known as the Maxwell distribution of velocities. 18

It may be noted that the steps following (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 (6. 41) distribution as (6. 47) and determine the normalization constant C by integration w(v)dv=1. The probability P(v)dv that the atom will have its speed in dv is (6. 48) The probability P(E)d. E that the atom will have its energy in d. E is (6. 49) 19

Principle of equipartition of energy Let us consider one of the variables in the energy, say variable pj. A situation of particular importance occurs when pj is an additive quadratic term in the energy: E=bpj 2+other terms not containing pj (6. 50) We can then calculate the mean energy associated with the variable pj: (6. 51) 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) 20

(6. 53) in general (6. 54) Therefore we have (6. 55) In classical problem the mean energy associated with each variable which contributes a quadratic term to the energy has the value in thermal equilibrium. The result is known as the principle of equipartition of energy; it depends specifically on the quadratic assumption. 21

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 (6. 57) From this we have he Dulong-Petit result that the heat capacity (at high temperature) of one mole of monatomic solid is 3 R, where R is the gas constant. At low temperature the quantum energy h may exceed k. T, and the problem requires special treatment, leading to the Einstein and Debye theories of the heat capacities of solids. 22

Ideal gas in the grand canonical ensemble We have for the grand partition function (6. 58) We have already evaluated the quantity (6. 59) Thus (6. 60) 23

so that (6. 61) (6. 62) or (6. 63) Further (6. 64) which is the ideal gas low. When we calculate (6. 65) we get the Sackur-Tetrode formula as derived previously. 24

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