Solutions Manual for Statistical Mechanics | 4th Edition (2021) | Pathria, & Beale | Covers All 16 Chapters

All 16 Chapters Covered




SOLUTIONS

,Chapter 1


1.1. (a) We expand the quantity ln Ω(0) (E1 ) as a Taylor series in the variable
(E1 − Ē1 ) and get

ln Ω(0) (E1 ) ≡ lnΩ1 (E1 ) + ln Ω2 (E2 ) (E2 = E (0) − E1 )
= {ln Ω1 (Ē1 ) + ln Ω2 (Ē2 )}+
 
∂ ln Ω1 (E1 ) ∂ ln Ω2 (E2 ) ∂E2
+ (E1 − Ē1 )+
∂E1 ∂E2 ∂E1 E1 =Ē1
( 2 )
1 ∂ 2 ln Ω1 (E1 ) ∂ 2 ln Ω2 (E2 ) ∂E2

+ (E1 − Ē1 )2 + · · · .
2 ∂E12 ∂E22 ∂E1
E1 =Ē1

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 may
be written as
   
1 ∂β1 ∂B2
2 1 1 1
+ E1 − Ē1 = − + (E1 −Ē1 )2 ,
2 ∂E1 ∂E2 eq. 2 kT12 (Cv )1 kT22 (Cv )2

with T1 = T2 . Ignoring the subsequent terms (which is justified
if the systems involved are large) and taking the exponentials, we
readily see that the function Ω0 (E1 ) is a Gaussian in the variable
(E1 − Ē1 ), with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that
if (Cv )2 >> (Cv )1 — corresponding to system 1 being in thermal con-
tact with a very large reservoir — then the variance becomes simply
kT 2 (Cv )1 , regardless of the nature of the reservoir; cf. eqn. (3.6.3).
(b) If the systems involved are ideal classical gases, then (Cv )1 = 32 N1 k
and (Cv )2 = 32 N2 k; the variance then becomes 32 k 2 T 2 · N1 N2 /(N1 +
N2 ). Again, if N2 >> N1 , we obtain the simplified expression
3 2 2
2 N1 k T ; cf. Problem 3.18.

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

5

,6 CHAPTER 1.

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

Ω2 f ′ (Ω1 Ω2 ) = f ′ (Ω1 ) and Ω1 f ′ (Ω1 Ω2 ) = f ′ (Ω2 ),

so that
Ω1 f ′ (Ω1 ) = Ω2 f ′ (Ω2 ). (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 may be written
as
N −1 N −1 
N 2 v0
  
X jv0 X jv0
C+N ln V + ln 1 − ≃ C+N ln V + − ≃ C+N ln V − .
j=1
V j=1
V 2V

Equation (1.4.2) is then replaced by

N 2 v0
 
P N N N v0
= + = 1 + , i.e.
kT V 2V 2 V 2V
 −1
N v0
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 = 12 N v0 .
A little reflection shows that v0 = (4π/3)σ 3 , with the result that
 3
1 4π 3 4π 1
b= N· σ = 4N · σ .
2 3 3 2

1.5. This problem is essentially 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 6π 2 16π

, 7

Substituting V = L3 and S = 6L2 , we obtain eqns. (1.4.15 and 16).
The expression for T now follows straightforwardly; 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 thermodynamic system is an ex-
tensive quantity, we may write
   
V E V E
S(N, V, E) = Nf , = Nf (v, ε) v = ,ε = .
N N N N
It follows that
       
∂S ∂f −V ∂f −E
N =N f +N · 2 +N · 2 ,
∂N V,E ∂v ε N ∂ε v N
       
∂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. Clearly, the initial temperatures and the initial particle densities of the two
gases (and hence of the mixture) are the same. The entropy of mixing may,
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 by eqn. (1.5.3a). Without loss of
generality, we may keep N1 , N2 and V1 fixed and vary only V2 . The
first and second derivatives of this expression are then given by
   
N1 + N2 N2 N1 + N2 N2
k − and k − + 2 (1a,b)
V1 + V 2 V2 (V1 + V2 )2 V2
respectively. Equating (1a) to zero gives the desired condition, viz.
N1 V2 = N2 V1 , i.e. N1 /V1 = N2 /V2 = n, say. Expression (1b) then
reduces to
 
n n knV1
k − + = > 0.
V1 + V2 V2 V2 (V1 + V2 )
Clearly, (∆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.