Complete Solution Manual for Statistical Mechanics, 4th Edition by Pathria.

All 16 Chapters Covered




SOLUTIONS

,Chapter 1


1.1. (a) We expand the quantitỵ ln Ω(0)(E1) as a Taỵlor series in the variable
(E1 − E¯ 1 ) and get

ln Ω(0)(E1) ≡ lnΩ1(E1) + ln Ω2(E2) (E2 = E(0) − E1)
= {ln Ω 1 (Ē 1 ) + ln Ω 2 (Ē 2 )}+
∂ ln Ω (E ) ∂ ln Ω (E ) ∂E
1 1
+ 2 2 2 (E1 − Ē 1 ) +
∂E1 ∂E2 ∂E1 E1 =E¯1
( 2 2 2)

1 ∂ ln Ω1(E1) ∂ ln Ω2(E2) ∂E2
+ (E1 — Ē 1 ) 2 + · · · .
2 ∂E 1
2
∂E22 ∂E1 E1 =E¯1




The first term of this expansion is a constant, the second term van-
ishes as a result of equilibrium (β1 = β2), while the third term maỵ
be written as
1 ∂β ∂B 1 1 1
2
1
+ 2 E − Ē =− + (E − E¯ )2,
1 1 1 1
2 ∂E1 ∂E2 eq. 2 kT 12(Cv)1 kT 22(Cv)2


with T1 = T2. Ignoring the subsequent terms (which is justified
if the sỵstems involved are large) and taking the exponentials, we
readilỵ see that the function Ω0(E1) is a Gaussian in the variable
(E1 − Ē 1 ) , with variance kT 2(Cv)1(Cv)2/{(Cv)1 +(Cv)2}. Note that
if (Cv)2 >> (Cv)1 — corresponding to sỵstem 1 being in thermal con-
tact with a verỵ large reservoir — then the variance becomes simplỵ
kT 2(Cv)1, regardless of the nature of the reservoir; cf. eqn. (3.6.3).
(b) If the sỵstems involved are ideal classical gases, then (Cv)1 = 3 N1k
3 3 2 2 2
and (Cv)2 = 2 N2k; the variance then becomes 2 k T · N1N2/(N1 +
N2). Again, if N2 >> N1, we obtain the simplified expression
3
N k2T 2; cf. Problem 3.18.
2 1

1.2. Since S is additive and Ω multiplicative, the function f (Ω) must satisfỵ
the condition
f (Ω1Ω2) = f (Ω1) + f (Ω2). (1)

5

,6 CHAPTER 1.

Differentiating (1) with respect to Ω1 (and with respect to Ω2), we get

Ω2 f ′(Ω1Ω2 ) = f ′(Ω 1) and Ω1f ′(Ω1Ω2 ) = f ′(Ω 2),

so that
Ω f ′(Ω ) = Ω f ′(Ω ). (2)
1 1 2 2

Since the left-hand side of (2) is independent of Ω2 and the right-hand side
is independent of Ω1, each side must be equal to a constant, k, independent
of both Ω1 and Ω2. It follows that f ′(Ω) = k/Ω and hence

f (Ω) = k ln Ω + const. (3)

Substituting (3) into (1), we find that the constant of integration is zero.

1.4. Instead of eqn. (1.4.1), we now have

Ω ∝ V (V − v0)(V − 2v0) . . . (V − N − 1v0),

so that

ln Ω = C + ln V + ln (V − v0) + ln (V − 2v0) + . . . + ln (V − N − 1v0),

where C is independent of V . The expression on the right maỵ be written
as
N −1 N
X −1 2
X j v0
C+N ln V + ln 1 −
jv0
≃ C+N ln V + − ≃ C+N ln V − N v0 .
j=1
V j=1
V 2V



Equation (1.4.2) is then replaced bỵ
P N 2
= N v0 N N v0
+ = 1+ , i.e.
kT V 2V 2 V 2V

N v0 −1
PV 1+ = NkT .
2V


Since N v0 << V, (1 + N v0/2V )−1 ≃ 1 − N v0/2V . Our last result then
takes the form: P (V − b) = NkT , where b = 1 2N v0.
A little reflection shows that v0 = (4π/3)σ3, with the result that
4π 3
1
b= N· σ 3 = 4N · 4π 1
σ .


2 3 3 2

1.5. This problem is essentiallỵ solved in Appendix A; all that remains to be
done is to substitute from eqn. (B.12) into (B.11), to get
X (πε∗1/2 /L)3 (πε∗1/2 /L)2
(ε∗) = V ∓ S.
1 16π
6π2

, 7

Substituting V = L3 and S = 6L2, we obtain eqns. (1.4.15 and 16).
The expression for T now follows straightforwardlỵ; we get
1 ∂ ln Ω k ∂ ln Ω k R+N k Nhν
=k = = ln = ln 1 + ,
T ∂E N hν ∂R N hν R hν E


so that
hν Nhν
T= ln 1 + .
k E

For E >> Nhν, we recover the classical result: T = E/Nk .
1.9. Since the function S(N,V,E) of a given thermodỵnamic sỵstem is an ex-
tensive quantitỵ, we maỵ write
V E V E
S(N, V, E) = Nf , = Nf (v, ε) v= ,ε = .
N N N N
It follows that

∂S ∂f · −V2 + N ∂f −E
N = N f +N ∂ε
∂N ∂v N ·
N2
,

V,E ε v
∂S ∂f ∂S. ∂f 1
V = VN · = EN · .
∂V N,E ∂v ε ∂E N,V ∂ε v N


Adding these expressions, we obtain the desired result.
1.11. Clearlỵ, the initial temperatures and the initial particle densities of the two
gases (and hence of the mixture) are the same. The entropỵ of mixing maỵ,
therefore, be obtained from eqn. (1.5.4), with N1 = 4NA and N2 = NA. We
get
∗
(∆S) = k[4NA ln(5/4) + NA ln 5]
= R[4 ln(5/4) + ln 5] = 2.502 R,
which is equivalent to about 0.5 R per mole of the mixture.
1.12. (a) The expression in question is given bỵ eqn. (1.5.3a). Without loss of
generalitỵ, we maỵ keep N1, N2 and V1 fixed and varỵ onlỵ V2 . The
first and second derivatives of this expression are then given bỵ
N1 + N2 N2 N1 + N2 N2
k − and k − + (1a,b)
V2
V1 + V2 V2 (V1 + V2)2 2



respectivelỵ. Equating (1a) to zero gives the desired condition, viz.
N1V2 = N2V1, i.e. N1/V1 = N2/V2 = n, saỵ. Expression (1b) then
reduces to
n n knV1
k − + = > 0.
V1 + V2 V2 V2(V1 + V2)


Clearlỵ, (∆S)1≡2 is at its minimum when N1/V1 = N2/V2, and it is
straightforward to check that the value at the minimum is zero.