Statistical Mechanics - Pathria Homework 1 4d5r5s

Statistical Mechanics - Homework Assignment 1 Alejandro G´omez Espinosa∗ February 5, 2013

Pathria 1.4 In a classical gas of hard spheres (of diameter D), the spatial distribution of the particles is no longer uncorrelated. Roughly speaking, the presence of n particles in the system leaves only a volume (V − nv0 ) available for the (n + 1)th particle; clearly, v0 would be proportional to D3 . Assuming that N v0 << V , determine the dependence of Ω(N, V, E) on V (compare to equation (1.4.1)) and show that, as a result of this, V in the ideal-gas law (1.4.3) gets replaced by (V − b), where b is four times the actual volume occupied by the particles. To determine the dependence of Ω in this case, the relation must reflect that each time that we introduce a particle in the volume V we will end up with less space. This scenario can be represented by N Y Ω(N, E, V ) = (V − (n − 1)v0 ) (1) n=1

Now, let us use the eq. 1.4.2 to derive the dependences on V: P T

= k

∂ ln Ω ∂V

∂ = k ln ∂V = k

N Y

! (V − (n − 1)v0 )

n=1

N ∂ X ln(V − (n − 1)v0 ) ∂V n=1

=

≈ = =

N X

N 1 k X 1 k = (n−1)v0 V − (n − 1)v0 V n=1 n=1 1 − V N k X (n − 1)v0 k N (N + 1)v0 N v0 1+ = N+ − V V V 2V V n=1 −1 (N − 1)v0 kN (N − 1)v0 kN 1+ ≈ 1− V 2V V 2V −1 (N − 1)v0 kN V − 2

0 where the term (N −1)v is clearly the volume of the particles added to the system. Then, taken into 2 the condition that v0 ∝ D3 we can replace in the ideal-gas law the term V with (V − 4b). ∗

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Pathria 1.7 Study the statistical mechanics of an extreme relativisitic gas characterized by the singleparticle energy states hc 2 (2) ε(nx , ny , nz ) = (n + n2y + n2z )1/2 2L x instead of (1.4.5), along the lines followed in Section 1.4. Show that the ratio /CV in this case is 4/3, instead of 5/3. Using the procedure following in Pathria (eq. 1.4.5) with (2), we can define: 2Lε hc 4V 2/3 ε2 4L2 ε2 = h2 c2 h2 c2

(nx , ny , nz )1/2 = (nx , ny , nz ) =

where L3 = V . Extending this argument for Ω(N, E, V ), gives the relation: 3N X

n2r =

r=1

4V 2/3 E 2 h2 c2

(3)

From this expresion, we can conclude that the combination V 2/3 E 2 will enter into the expression for Ω. Therefore S(N, V, E) = S(N, V 2/3 E 2 ), which defines a reversible adiabatic process when: V 2/3 E 2 = const

⇒

E = const V −1/3

(4)

1 const V −4/3 3

(5)

Then, using Eq. 1.4.10: P =−

∂E ∂V

= N,S

resulting in the adiabatic relation: P V 4/3 = const Finally, comparing to Eq. 1.4.30 for adiabatic process: P V γ , where γ = result: 4 = Cv 3

2

(6) Cv ,

we found the final (7)

Pathria 1.8 Consider a system of quasiparticles whose energy eigenvalues are given by ε(n) = nhν;

n = 0, 1, 2, ...

(8)

Obtain an asymptotic expression for the number Ω of this system for a given number N of the quasiparticles and a given total energy E. Determine the temperature T of the system as a function of E/N and hν, and examine the situation for which E/(N hν) >> 1. For a system of quasiparticles with energy (8), the total energy of the system must be the sum of the particles in the system i: ∞ X E= inhν (9) i=0

In addition, since the energy of the eigenvalues of the quasiparticles is quantized, we know that the energy must be equal to E = phν, where p is the momentum of the particle. Comparing (10) with P this relation is easy to see that p = in. Next we have to estimate the number of states in our system. Since we know now the momentum of the quanta, we have to distribute this momentum into N particles. Then, the number of state is approximated to the combination of P + N states with P combinations. Hence, (P + N )! P !N ! ln Ω = ln(P + N )! − ln P ! − ln N ! Ω ≈

≈ (P + N ) ln(P + N ) − (P + N ) − P ln P + P − N ln N + N P +N P +N = P ln + N ln P N Replacing P = E/hν, we found the relation for the entropy: S = k ln Ω kE N hν E = ln 1 + + kN ln +1 hν E N hν Following by the temperature: 1 ∂S = T ∂E N 1 k N hν kE NEhν 2 N hν = ln 1 + − + kN E hν E hν 1 + NEhν N hν + 1 k N hν kN kN = ln 1 + − + hν E E + N hν E + N hν hν T = k ln 1 + NEhν Finally, in the case that approximated by:

N hν E

>> 1, ln 1 +

N hν E

T ≈

3



≈

E kN

N hν E .

Therefore, the temperature could be (10)

Pathria 1.13 If the two gases considered in the mixing process of Section 1.5 were initially at different temperatures, say T1 and T2 , what would the entropy of mixing be in that case? Would the contribution arising from this cause depend on whether the two gases were different or identical? In the case of two ideal gases with different temperature, eq. 1.5.1 becomes: 3 2πmi kTi Si = Ni k ln Vi + Ni k 1 + ln ; i = 1, 2 2 h2

(11)

After the mixing, the total entropy would be: ST =

2 X i=1

3 2πmi kT Ni k ln V + Ni k 1 + ln 2 h2

(12)

where V = V1 + V2 and T is the final temperature of the system. Therefore, using (11) and (12), the entropy of mixing is given by (∆S) = ST −

2 X

Si

i=1

3 3 3 3 = N1 k ln V + N2 k ln V − N1 k ln V1 − N2 k ln V2 + N1 k + N2 k − N1 k − N2 k + 2 2 2 2 3 2πkm1 T 3 2πkm2 T 3 2πkm1 T1 3 2πkm2 T2 + N1 k ln + N2 k ln − N1 k ln − N2 k ln 2 h2 2 h2 2 h2 2 h2 3 V T V T = N1 k ln + N2 k ln + N1 k ln + N2 k ln V1 V2 2 T1 T2 " " 3/2 # 3/2 # V1 + V2 T V1 + V2 T = N1 k ln + ln + ln + N2 k ln V1 T1 V2 T2 Since the increase in the entropy in this case depends upon the temperature, there will be no distinction whether the two gases were different or identical. This is because the temperature measures the energy of a system, and since the final temperature after the mixing does not depend upon the particles are the same, the contribution cannot be atributed to the difference of the gases.

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