CONTENTS
Preface.
i
CHAPTER 1:
Introduction to Statistical Methods
1
CHAPTER 2:
Statistical Description of Systems of Particles
11
CHAPTER 3:
Statistical Thermodynamics
l6
CHAPTER !):
Macroscopic Parameters and their Measurement
18
CHAPTER 5:
Simple Applications of Macroscopic Thermodynamics
20
CHAPTER 6:
Basic Methods and Results of Statistical Mechanics
32
CHAPTER 7:
Simple Applications of Statistical Mechanics
4l
CHAPTER 8:
Equilibrium between Phases and Chemical Species
5^
CHAPTER 9:
Quantum Statistics of Ideal Gases.
6k
CHAPTER 10: Systems of Interacting Particles
8^
CHAPTER 11: Magnetism and Low Temperatures
90
CHAPTER 12: Elementary Theory of Transport Processes
93
CHAPTER 13: Transport Theory using the Relaxation -Time Approximation
99
CHAPTER l *-: Near-Exact Formulation of Transport Theory
107
CHAPTER 15: Irreversible Processes and Fluctuations
117
1
PREFACE
This manual contains solutions to the problems in Fundamentals of Statistical and Thermal
Physics , by F. Reif.
The problems have been solved using only the ideas explicitly presented
in this text and in the way a student encountering this material for the first time would
probably approach them.
Certain topics which have implications far beyond those called for in
the statement of the problems are not developed further here.
numerous treatments of these subjects.
The reader can refer to the
Except when new symbols are defined, the notation
conforms to that of the text.
It is a pleasure to thank Dr. Reif for the help and encouragement he freely gave as this
work was progressing, but he has not read all of this material and is in no way responsible for
its shortcomings.
Sincere thanks are also due to Miss Beverly West for patiently typing the
entire manuscript.
I would greatly appreciate your cal ling errors or omissions to my attention.
R. F. Knacke
CHAPTER 1
Introduction to Statistical Methods
1.1
There are 6*6*6 = 216 ways to roll three dice.
Throw
The throws giving a sum less than or equal to 6
1,1,1
1,1,2
1A,3
1,1,2*
1,2,2
1,2,3
2,2,2
1
3
3
3
3
6
1
No. of
Permutations
Since there are a total of 20 permutations, the probability is
^
1.2
Probability of obtaining one ace = (probability of an ace for one of the dice) x (probabil-
(a)
ity that the other dice do not show an ace) X (number of permutations) =
5
=
= (|)
(b)
. 1*02
.
The probability of obtaining at least one ace is one minus the probability of obtaining
none, or
^
1 - (§)
(c)
= *667
By the same reasoning as in (a) we have
1
2
£-
(£) (f)
4
g,
57ir
=
-
040
1*3
The probability of a particular sequence of digits such that five are greater than
less than 5 is
1 5 i 5
(— .
(-)
v
2 y '2 )
5 and.
five are
Then multiplying by the number of pe imitations gives the probability
irrespective of order
5151
~
(W(h'
'2'
2 46
1.2*.
(a)
To return to the origin, the drunk, must take the same number of steps to the left as to the
right.
Thus the probability is
where
N is even.
(b)
The drunk cannot return to the lamp post in an odd number of steps.
1
1.5
r
«
(a)
(|)
(b)
(Probability of being shot on the
trial) = (probability of surviving H-l trials) X
I-J
(probability of shooting oneself on the N
N_1
th
trial) =
5
1
(g-)
(g-)
6
(c)
~
-v
m = nr = 0 since these are odd moments.
m
and
i? = (2n - N)
2
ve find m
2
4
Using n
= (2n
u)
-
2
= (p
S
r
N
(p+<l)
2
2 4nU + N
= U n
2
= l6 i? - 32 Nn 3 + 2k N
n
2
-
8N3 n + N*
2 2N.
= N and i? = 3U
The probability of n successes out of N trials is given by the sum
W(n) =
2
2
E
E
2
E w.w.
...
...
1 J
m=l
i=l j=l
w
m
with the restriction that the sum is taken only over terms involving w^ n times.
2
Then
V(n) =
•
E
i=l
2
2
w.
1
E
j=l
w, ...
J
E v
m
~
If we sum over all i ... m, each sum contributes
.
m=l
(wi+v2 ) and
W'(n) = (w
x +
U
V
'
(n)
^
1
2
n
n
7 /v* r-r w.,1 w.^
n:(w-n):
2
i
E
=
w )N
n=0
—
by the binomial theorem .
Applying the restriction that w^ must occur n times we have
>
NJ
= nl(N-n)i
'
V
Ve consider the relative motion of the two drunks.
n „ N-n
1
V
2
With each simultaneous step, they have a
Probability of l/b of decreasing their separation, l/k of increasing it, and l/2 of maintaining
it by takin g steps in the same direction.
and n , respectively.
Let the number of times each case occurs be n^, n 5 ,
Then the probability of a particular combination in N steps is
n
n
n
2
l
3
t
B(a n n 3>
- 11
( )
(|)
i a
n:ln ; (|)
5
V
W°3
3
2
Then the probability that they meet after N steps irrespective of
The drunks meet if n^ = n^.
the number of steps n^ which leave their separation unchanged is
n
N
P = S
^
n::nfc'
12 3
„
n =0
3
n
l
^
where we have inserted a parameter x which cancels if n^n^.
sum over n^ n^ n^ and choose the term in which x cancels.
1
p’ =
x +
1
N
i
+
2N
,
/
= (|)
|)
(
x ^
®
21T
/
+ x
hMtt
2
We then perfoim the unrestricted
By the binomial expansion,
)
2U 2N
Expansion yields
2
<*
1/2 >° (^-1/2
)
Since x must cancel we choose the terms where n = 2U-n or n = H
(
P .
Thus
(a)
In (l-p)
00
InS}T
r-\
(c)
W
w
N_n =
~ -Up
-p (U-n)
=
•
•
•
(
-
(|)
2
n
N " n+i) “
K,
l
(mf
«U
thus (l-p)
^
if n
11
Nl
_n ,, -_ % U-n
X
U
n -Up
15
e
(1 p )
= niTu-nVi p
ST P
n
51
n
-X
^T e=e
\
y
S
n=0
»
—
2
a
-X
.n -X
-
n=0
n
\
X
y
h-T=ee=l
.
v
"A.
o
X
-XX = X
—
= X
ni
y
A
^X
nI
-X
e
2
=
2
n=0
Z
(B)
^
nI
4
-o
?
/
\2
- n~ = X
= n
(An)
(a)
«
The mean number of misprints per page is 1 .
..
-1
n
= —7 e
W(n)
v
'
=
ni
W( 0) = e”
i
—
= .37
3
ni
n:
e
2
- h * X
If
N-n
« e”^
2
P = 1 -
(*)
-1
E ^-r=.08
n
*
n=0
1.12
(a)
Dividing the time interval t into small intervals At we have again the "binomial distribution
for n successes in N = t/At trials
W(n > =
In the limit At -> 0 , W(n)
’
^
n:
P
nX( lJ-n )~T
N-n
X* X
—
as in problem 1.9 where X = S
n«
nJ
e
f
the mean number of disintegra-
"t
tions in the interval of time.
(b)
W(n) =
e'
n
4
W(n)
0
1
2
3
4
5
6
7
8
.019
.076
.148
.203
.203
.158
.105
.061
.003
1.13
We divide the plate into areas of size b
2
2
.
Since b
is much less than the area of the plate,
the probability of an atom hitting a particular element is much less than one.
Clearly n
«N
so we may use the Poisson distribution
W(n) fj
e-
n
6
o
Hil .003
1
1
3
.086
—
.162
1.14
We use the Gaussian approximation to the binomial distribution.
v2
(215-200)*
1
W(n) =
2(400/1+)
= .013
1.15
The probability that a line is in use at any instant is l/30 .
We want II lines such that the
probability that N+l or more lines are occupied is less than .01 when there are 2000 trials
during the hour.
IT
.oi = i -
n=0‘/^r
where
-
n =
2000
~^
= 66.67,
2a
Z
a
15
a =
= 8.02
The sum may he approximated by an integral
2
.99 =
exp [-(n-n) /2a
%/&?
-§
p(l
•
99 =
7=^ e
V27T
-
1
~ M/
4(-n-gj/CT
The lower limit may he replaced hy
2
]
dn
a
2y2
dy
'
after a change of integration variable.
with negligible error and the integral found frcm the
-°o
tables of error functions.
*
ni
=
!Z l
We find
2.33
N = 66.67 + (2.33) (8.02)
- .5
* 85
1.16
(a)
The probability for N molecules in V is given by the binomial distribution
N I
=
N_N
N
^
NI(N
-N)i
K
'
o
o
v
o
o
Thus
2
(N-fl )
0>)
2
jj
(c)
(d)
V
If
If V -» V
« VQ
2
_
-
IT
H
N
1 -
2
2
H
2
(AMP
,
o
o
..
2
M
1
M
—9
IT
Q,
1.17
Since 0
« — « 1 and since Mq is large, we may use a Gaussian distribution,
o
P(tf)dN =
1
—
2
2
exp [-(N-N) /2 AN
-
i
2
(2?r AM )
5
]
dM
1.18
N
U
§.
R = Z r. = -t Z —
—
-a
_
(S.’s are unit vectors)
i
p2
^
„ 2
s.
r-r = tr Z
i
p2
v*
§.• A
§. =#+rZ
Z '
"
v
Z
p2
.
+ -t
1
„.p2
X
i /
,
J
i
o
Z cos e
/ 0
2
Hence R
The second term is 0 "because the directions are random.
2
= Eft .
1.19
Z v.1 ; hence the mean square is
The total voltage is V =
i
-
where v. = vp and v.
2
x
i
—=
-s
N
v
= Z v.
1
1
N
+ Z
N _
/
.
1
Z v.X v.
J
.
/ 0
2
= v p.
2
2
P = V /r = (u
^/R) p 2 ri+(l-p)/lTp]
Hence
1.20
(a)
= ME, and since the intensity
The antennas add in phase so that the total amplitude is E
t
is proportional to E^
(h)
2
2
,
1^
= H !.
To find the mean intensity, we must calculate the mean square amplitude.
Since amplitudes
may he added vectorially
E
2
2
= E
= E, *E,
Z
2
S
2
+ E Z
where the S^'s are -unit vectors.
2
= HE
—g
A
n
Since a
2
N
= Z
i
or
I
^
N
+ Z
a
n.
i
.
HI
t =
N _
/
.
i
Zan. an.
.
i
/ 0
0
= 0, the second term on the right is 0 and
i
i
(A
l
1
2 .2
= H2 (a
)
1
0.2
2
= H2 103 a
)
i_
"but
*S,
The phases are random so the second term is 0 and
E
1.21
Z S.
i/O
i
(A
2
2
)
=
A„
Equating these expressions we find N = 10°.
6
=
Ha c
1.22
N _
x =Es.,
(a)
_
= 'L,
but since
(x-x)
= E
i
N
x = E
(s
—
i
nL
=
-L
II
N
+ E
E
i
/ j
(s.--t)(s,- £)
1
0
— 0**0
Since
(a)
(s
-£) = £ - -t = 0, the second term is 0 and (x-x)^ = S a
i
(x-x )
2
IT
= E (s. -£)
i
2
u
+ E
i
ii
E (s.-l)(s.-l)
j/i
To find the dispersion (s^-E )
s
.
N _
II
x = E s^ = E^=
The mean step length is -t.
(b)
o
= Na
(s.~V)=-t--t = C>
,
1
J
,
we note that the probability that the step length is between
and s + ds in the range -t-btO'L + b is
•
4 —®— T
7—^2 - r Ul° 'v1 2 ds i
)
Hence
*
2
‘
,
,
—\ 2
(X-X)
\1(9)
2
b
= E
i
2
-r- =
3
Nb
2
3
H
d9 =
27T a
w(9) de =
(a)
—
2
sin © d9
sin 9 d9
We find the probability that the proton is in 9 and 9 + d9 and thus the probability for the
resulting field.
sin 9 d9
W(9) d9 =
From problem 1.2!
2_ ,\
u
Since b = >— (3 cos 0-1) we have
,
db
6u cos
—
= -=
a3
W(b)db = 2W(0)d0 =
0
sin 9
a3
a 3 a/T db
6a/
7
|ia\
(b)
If the spin is anti -parallel to the field,
W(b)db =
Then if either orientation is possible, we add the probabilities and renormalize
W(b)db =
1.26
0“
,W
{
.
48 '
S
Q may be evaluated by contour integration.
V( s ) =
iks
a«
dS
TT
For k > 0, the integral is evaluated on the path
-kb
e
residue (ib) = ^-r
For k < 0
8
Q(k) = e
-9
*
Thus Q(k) = e
-kb
C/W -fj
,,
dk
e
-ikx -IT
Q
'-Leo
«-tr
dk e
-ikx
e
Hkb
r
+ i.
dk e
ar Jr,
,
-ikx
-Mkb
-Nkb
dk e
cos k x
r
* Jn
-i
,
This integral may he found from tables.
dx
2 2
x + N *)
does not become Gaussian because the moments s n diverge.
1
TTb
(P(x) dx = -
2
7r
1.27
From equation (1.10.6) and (1.10.8) we see that
<?(*) =
b X **
e_iix
Qi(k) =
where
(k)
s.
-**V k)
ds e
iks
(i)
W.(s)
V-CO
Q.(k) = 1 + i s.k k
| s*
We expand (^(k).
^(k) = i Sik -
In
Q± (k) = exp
(P(x) = h:
f dk
thus from (l)
|
(As .)
(A Si )
s.k |
[i
exp[i(Z s -x)k
-
±
i
“-<*>
2
2
2
k
k
2
2
| Z
—
W2k
2
(As
i
This integral -may be evaluated by the methods of section 1.11.
<P(x) =
]
exp [-(x-h) /2ct
k
2
]
We find
2
]
a
N
N _
where
2
i)
H=Zs.
c =
Z (As
i
.
1
1.28
The probability that the total displacement lies in the range r to r + dr after IT steps is
(p(v) d 3 r =
J 2' f W(s x W(s 2
)
)
...
W(s^) d 3
(d3 l)
where the integral is evaluated with the restriction
IT
r<
Z
i=l
s.
< r + dr
1
9
^
...
d3
^
We remove the restriction on the limits by introducing a delta function
3
3
d3 s. .. «d s„
CP (e) d r = O[ ... 0f W^) ... W(s^) S(r-Z1 s i )d r
In three dimensions
5(r
"
Z _1 )
=
s
5-
i
(2tt)
3
-N
ik* [Z s. -r]
X
^
o
-j
-
-1
L
-CO
'
d^k
"
/
-U
e
ik* [Z s^-r]
thus
or
W( -1 ) -*’ W(
<?(*)
=
Ptr)
-^5 /_
<p (r) =
I'C'f
—
^
1
% ^3 X d3 - e
d3
)
i3i
e
3
d k e
— Q^(k)
where
“ £l
ir-- d3
%
i—
•••£ dV<%)
Q(k) =
d
3
s_
*—
e"^*— W(s)
1.29
For displacements of uniform length hut of random direction W(s) =
(
s~
%
)
,
in spherical
s
coordinates. W(^) is a properly normalized probability density since
f
Jn
0
1
s
f
Jf0 u
Jn
Jn
0
2
sin 0 ds d0 dcp
W 2
= kn
B
^
Jfn
= 1
We find the Fourier transform Q(k)
fff ^ e^w(s] rrr
27T p7T
>
<Kk) =
p“
=
J J J
sin 0 ds d0 dcp
s
J Q J Qj Q
sin 0 d0
mar r* —sr~
Jn
’
ikL cos
0
%
s
2
£
iks cos 0
sin kt
-TT
.3,
From
d k e
(27T)
S(s-t)
3
-ik*r -N/, \
Q (k)
we have, for N = 3
—~r
(P(r) =
(27r)
d
1
2ir p7r
/
^0
p®
II
k“ sin 0 dk d0
dcp
e
-ikr cos 0
(kt)
The angular integrations are trivial and
?w
sinJ kt sin kr
sk />
To evalua te the integral, we use the identities
*3
sin kE, sin kr
5
k
sin3ki
1
0
,1-cos
——2 kEU
= —=
sin kr sin kc (
2
k
.
.
.
10
.
3
CHAPTER 2
II
2.3
(a)
Tiie
probability of displacement x is the probability that cp assumes the required value in
the expression x = A cos (wt + cp)
r(<p)dcp =
w(
P(x)dx = 2v(cp)
and
dtp
=
^
7
vhere the factor of 2 is introduced because two values of cp give the same x.
dcp
Since
(b)
.
dx
P(x)dx =
we have
rj
p
VA - x
A sin (ut + cp)
dx
We ta kp the ratio of the occupied volume, 8v, in phase space when the oscillator is in the
range x to x + dx and E to E + 8E, to the total accessible
volume, v.
The area, A , of the ellipse is up^x^.
g
2 2
2
P_.+ m
*
E _
-
a
Since
we have
'
= if 5E
mw
a-
*(*)* - V
For any x, p
2
™
Ae = 7r
Hence
*
03
~
27r
o-r»
533
^ = 8E
,
where we have used (l).
Thus
» ,
mu
mu
dx
*
T - w V2mE
j
dx
-
-
—
g
mux
12 2
From (l) and x = A cos (ut + <p), it follows that E = — u mA .
r(=0a*
UJ
i 7=§=r
2.k
(a)
The number of ways that N spins can be arranged such that n^ are parallel and n
g
parallel to the field, is
r
HI
'V
and
12
“W - f (“i> ffe
are anti-
=
Since
-(n^) nH = -(a^-H) nH,
- ^),
n
(N +
2 = |
^)
2 hE
v
2
2nH
y *
'2
2pH'
Using Stirling's approximation. In n! = n In n-n, ve have
(h)
Ni
In
l
^ T + VB) ^ {1 ' T*
-
&
-n
^T#3^yi
^
a(E) - it* - fe)
(c)
^ = | (N
M_
I*
„/ E \
Thus
ve have
(
*>
§<»
-
,
(uln^)
a t>out tlie maximum, n^
1
In f(n ) = In f^) +
§ B2
1
(
In f(n
ti
B
= el. - n
1
i.
=
2
^
,
I
f^) = f^) e
-
1)
1)
c.
“i
i
From (1)
&
'
We expand In f (n^) = In
where
*
2
|b
1
2
2
21
t]
|
1
(2)
f (n^) is evaluated by noticing that the integral of f(n^) over all n^ must equal the total number
of permutations of spin directions , 2^.
B
‘I
f
2
I"
dij
= 2
11
*
*<=l)
n^ is found frcm the maximum condition
d In f(n^)
= 0 = -In n^ + In (N-r^) , or
dn.
2
and
N
d
In f(n^)
_
-
dni
1_ _
IV
2
“l
IV
“l
1
k
N
»v
N -°l
\ -B/2
Substituting these expressions into (2), ve have
2
13
= |
“M * r <“i> §s
Since
and
>
n
i
(s -
I
s
);
it follows that
,N
a(E) =
e *P
y-rr
f" 4 (of^)
N v 2pE /
3
J
2ioH
g
2.5
^f
(a)
cA _ SB
Since second derivatives are equal,
Sc
ciy
f ^ f = f (B) - f (A) =0
f is exact,
If
(b)
B.g
dy
A.j£
ox
is an exact differential.
If
Adx + Bdy
=
A = B
if
2.6
2
2
(a)
not exact
(b)
dy +
J
O
d Xf
ST 6y
2
J
(x -2) dx = i,
J"
O
d Xf
/
dy ox
'
2
(x -l)dx +
2dx =
J
2
^ J (x -x)dx + J' x dx =
,
M
u>
exact
(c)
.2,2
(2,2)
(a)
0.
f =
x +
1,1
e"
for all paths
= 1
s
xl
(i.i)
2.7
The particle in a state with energy E does work,
(a)
XX
to L
+ dL
.
x
dE—
= -
dL
.
x
n
2 2
The energy levels are given by
ft
E =
(
n
P -’
_
r\
—
p
,
j
= l
**
,
%K
g
mV
by symmetry
*
W-z
1
* '
2
2
n
V*
y
5E
z
2
2
2J2
_
^7t
h
K "VA
'
/ JC_
v L 2
X
s n
(fx
where
V = Lx L
y
L
XX
Since c^W = F dL ,
JE_ + _2_ +
2
2
.2
L
L
L
x
If L. . = L
when the length of the box is changed
This work is done at the expense of the energy, i.e., c^W = -dE.
it follows that F
(b)
li W,
z
= n , and using the expression for E we have
nx2 = :2
n
z
y
2.8
p = a V
10
6
,
2
dynes/m
a may "be found by evaluating this expression at a point on the curve.
o
O
and lO^cm, a = 32.
For the adiabatic process,
AE = Eg - E =
32V
a
-J
-5 /3
dV = -36 =-3600 Joules
Since energy is a state function, this is the energy change for all paths.
08
dV = 22,400 joules
Q = AE + W = -36OO + 22,400 = 11,800 joules
P = 32^- 4jr (V-l)
W
=J
-
32
Q = 11,550
-
2k (v-l)J = 11,550 joules
3600 = 7,950
joules
r.8
W =
J
(1) dV = 700 joules
Q = 700 - 3600 = -2900 joules
15
In units of
CHAPTER 3
Statistical Thermodynamic
3-1
(a)
Thus, in the larger
In eauilibrium, the densities on "both sides of the box are the same.
volume, the mean number of Ne molecules is 750 and the mean number of He molecules is 75 •
1000
(1)
100
,
p - (?)
- 10
(?)
-185
3-2
From problem 2.b- (b)
(
In fl(E) - -
a)
2
(H -
^
In
(1 -
2
where we have neglected the small term
In
"
2
^
(H +
In
1+ ;
^
6E
^-
.
x - -(1
1
k
2
Consequently
E = -NpH tanh
(b)
T < 0
(c)
2^
——
H^H'
^
E > 0
if
M = n(n^-n ) = n(2n^-Il) where n^ is the number of spins a.lllgned parallel to H.
2
^ = i(N —
771
T7*
1
Since
-
M - u(H - jig - S) - -
from problem 2.k (a).
M = Nu tanh
|
it
3-3
*2
(a)
In 0 (e) =
for system A, we have from 2.k (c)
In 2
(1)
2 2
2 i H lT
ttN
t
2
where we have neglected the term
3 = §=r In fl(E) =
2 2
*
3’ =
u H H
In equilibrium.
3 = 3’
4’
2 2
H H’
Hence
u
(b)
Since energy is conserved,
Substituting from (a) we find
(c)
In
,2
N*
E + E* = bNpH + b’H’ioH
+ b'K'ji'H}
E =
Q - E - Muz . MgO-VV* -
2
V
p !! + h
,2
H*
a
ri
free (2)
HIT + li'TT
16
(2)
SE
2uH
P(E) = J1 (e) fl'(E -E)
0
(d)
where
E
is the total energy, E^ = hWpTT + b'N'ii’H*
q
2
From
E
P(E) = exp
(l)
1
exp
2 2
2
L 2 sL , H2ir
tt' J
2^2 N
L
This expression may he revrritten in terns of the most probable energy E, equation (2).
P(E)dS = C exp pCg rlL
L
M
? ,g
2 2
g
2 = u u H
where
dE
J
2a
t
4
The constant C is determined by the normalization requirement
L
P(E)dE = -=y-
Thus
exp
(e)
(A*E)
2
=
2
2a
M
~ =
dE, where a is given above
2
-1
2 2 2
= u u» S
2
2
,
n N+n N*
,
(3)
11
NfbuHW + TaVU'H)
liN +11
»N
N' » N
A*E = qE Vn"
From equation (3)>
(f)
dE = 1
2a
(eJV
a^T a
(E-E)
C exp
2
,
N*
Nu^g
__ ,,
4’
N’
A*S
hence
jiL
n bi/iT
E
3^
The entropy change of the heat reservoir is As' =
system,
AS + AS* = AS
a^d for the entropy c hange of the entire
"
Q
-
^7 ^ 0. , by the second law.
AS a
Hence
3.5
(a)
N
In section 2.5 it is shown that n(E)= V x(e), where V is the volume and X (e) is independent
of volume.
Then for two non-interacting species with total energy E
n(E) = C ^(E)
0 >)
N +N2
fi
(E -E) = CV
2 o
l
X
X (E) X (E)
1
2
In n(E) = (M^+Ng) In V + In C + In X (e) Xg(E)
1
N
+N )
(
1
l 2
or pv = (l^+N^kT
In 0(E) =
-P =
,
-IS
f §v
|
—
v
17
3-6
2 atmospheres.
CHAPTER b
Macroscopic Parameters and their Measurement
k.l
AS^ = me
(a)
J
^ = 4l80 In
Since
(h)
Sres
ASres =
“
= -
Aot =
A
+
1310 Joules
=
A
=
Tf (373-273)
-H20 joules
=
- 190 JouleS
^res
A
A
5= mc(50) = 2.09 X 10^ joules.
The heat given off "by either reservoir is
Since the
entropy change of the water is the same as in (a) we have
AS = 1310 (c)
2,09
= 102 joule s/°K
There will "be no entropy change if the system is brought to its final temperature by inter-
action with a succession of heat reservoirs differing infinitesimally in temperature.
k.2
(a)
The final temperature is found by relating the heats, Q, exchanged in the system.
required for melting the ice is m^-t where m^ is the mass of the ice.
The heat
Since for the other compo-
nents of the system, Q = mcAT, where m is mass and c is the specific heat, we have
+
ml +
i
^(melted ice)
^(V
+
273 ) +
^(original water)
+
^(calorimeter)
^
Vw(V293) + mc C Tf‘293) = 0
C(
(30) (333) + (30)(^.l8)(T f -2T3)+(200)(4.l8)(T f -293)+(750)(.Ja8)(T f -293)
T
283°K
f =
m.-t
(b)
The chang e in entropy when the ice melts is
For the other components
Tf
AS = me
r
/
T
dT
f
—
= me In —
J-
i
£
Hence
AS=^ + mw cln||| + m cln||| + mc
i
18
c
c
In §|l = 1.6 JoulesA
=0
(c)
The work required is the amount of heat it would take to raise the water to this temperature.
w =
Vw <V*i> + mc c c
(
750 ) (. 418 ) +
(
230 ) 4 . 18 ) (293-283) = 12,750 Joules
4.3
2 Q = cdT+pdV
r
by the first law
y dv
= c dT +
Since pV = BT
T,
AS
-
-r
C
T
V
f
—f + Rln—
V
r
dT ^
+
T
f
Rdv
V
Jv
T
cln
=
independent of process
i
1
4.4
The spins become randomly oriented in the limit of high temperatures.
Since each spin has 2
available states, for If spins we have
S(T -»«) = k In n = k In 2^ = Nk In 2
(l)
rp
But since
S(T)-S(0) =
S(T ~ m)
we have
dr
J
'4* ci
(
where we have set S(o) = 0 by the third law.
C
11
f
f
' c (1 -ln 2)
(2)
i
From (l) and ( 2 )
Hk In 2
1 ~ 1 - In 2
Hk In 2 = c^(l-ln 2)
= 2.27 Hk
4.5
f
The entropy at temperature T is found by evaluating S(t) - S(o) =
dT.
Labeling the
Jn
undiluted and diluted systems by u and d, we have
S
T
u< l>
- S (0)
u
S (Ti) - S (0)
fl
d
S (0) = S (o) by the third law.
d
u
c
=/^ l(
C
=
**1
T
2 T
Since there are only
case as in the other, we must have S^T^) = .7 S^(T^) .
— = .7 (1-ln
C
1
19
-
2
2
1)
f=
c
l(
l-ln 2)
fl
= C
T
2
as many magnetic atoms in the diluted
Hence
2) = .214
CHAPTER 5
Simple Applications of Macroscopic Thermodynamics
(a)
pV
7 =
constant for an adiabatic quasistatic process.
7
v ET.V
vR T.V 7
—
1
-
V
(b)
Substituting pV = v KT ve have
=
—
V
i
1-7
V_
f-f
or
T
f
= T
f
(
k
i
—
)
;
The entropy change of an ideal gas in a process taking it from T^V^ to
was found in
V
T
—
+ vR In —
problem it.3
AS = vCy In
AS = 0 for an adiabatic quasistatic process.
Hence
In^
A
° =
R//c
V
T
V
^
(
v
i
)
i
Since R/c„ = c -clVc^ = 7-1, and since In 1 = 0, it follows that
*
P Y V
1 --y
T
f
= T
Vf
i
(
k
—
V±
)
'
(a)
W =
(b)
Since the energy of an ideal gas is only a function of the temperature and since pV = KT,
p d7 = area enclosed by curve
AS =
=
^
31^ joules
P V
c c
A„
3 „ (—
AT
R ,
=
|
"P
v
‘A An
g-)
= ^ (6-2) X 10^ ergs = 600 joules
(c)
Q = AE + W where AE is calculated in (b) since the energy change is independent of path.
Again the work is the area under the curve.
Q = 600 +
1+00
+
1157 joules
_
_ SE _ 5 R
®V " cff " 2
W = / p dV = area under curve = 1300 joules
(c)
To find the heat, Q, we first find the energy
change in going from A to C.
course, is independent of the path.
AE = c^AT =
| R(T c -Ta ) = | (P CV C - PAVA )
20
= 1500 joules
This, of
Then since
Q = AE + W
Q = 1500 + 1300 = 2800 joules
(d)
From problem 4.3
V
AS ' C
+ R In
In
V
c
A
2
12 X 10 /R
x 103
+ R In 3
= 2.84 R = 23.6 joules/°K
2
6 X 10 /R
103
= f-R In
5.4
(a)
The total system, i.e. , water and gases, does not exchange energy with the environment, so
its energy remains the same after the expansion.
change is given hy the heat it absorbs,
Since the water cannot do work, its energy
AE = Q = C^AT, where
is the heat capacity.
The
energy of the two ideal gases is, of course, dependent only on temperature, AE = GAT, where C is
the heat capacity.
*
“(vater)
Sl<
“(He)
+
' 0
“(A)
rt> + CHe<T rt> + CA< T rt> ’ 0
T
no temperature change.
Thus
T
(b)
Let -t be the distance from the left and v the number of moles.
T^,
f =
V
P
He
V
V
He He = (5)(30) _ ,
P V
"tl 50 T 3
"
A A
A
Hence, in equilibrium -t = 3(80-t), -t = 60 cm.
V
(c)
Since there is no temperature change, the entropy change is vR In
AS = A£>
a +
AS
—f
.
|ln|2 + Rln|^ = 3.24 joules/°K.
He =
5-5
(a)
The temperature decreases.
(b)
The entropy of the gas increases since the process is irreversible.
(c)
The system is isolated, Q = 0, and from the first law
The gas does work at the expense of its internal energy.
AE = -W = -
(V -V
f
Q)
The internal energy of the gas is only a function of temperature , AE = vc^(t^,-T q )
Thus
In equilibrium p =
vCy(T
f
^o
)
= - 2£ (V -V )
f Q
^ and since pV^ = vRT^ we have
,
21
(1)
VAST,
V
=
=
f
mg
T
f = ± '+
V 1
r
1
Substitution in (l) yields
R/^) [
(
T
0
+
5.6
mx = pA - mg - P A
The equation of motion is
(
q
1)
Since the process is approximately adiabatic
pV^
constant = (p + 5£)
o
A
=
v/
o
AV '
** = (P
and
+ ?)
Q
~^
-
The displacements from the equilibrium position,
V
dinate change x =
V
^
+ T) and expand about
X7
V
/
mg - P„A
(2 )
(Ax)'
O
—
\'
.
—A
are small so we can introduce the
,
.
o
(t + ’l)
coor-
o
We keep only the first and second terns and substitute into (2)
mq
This is the equation for harmonic motion and the frequency may be determined by inspection.
1
Hence
,
V =
2^
(P 0
+ T?)
A' V m
o
-
%p v p m V.
P A
Q
+ mg A
5-7
The volume element of atmosphere shown must be in
z+dz
equilibrium under the forces (pressure) x (area) and
z
gravity.
Then if n is the number of particles per un i t
volume, m the mass per particle, and g the acceleration
of gravity.
p(z+dz)A-p(z)A = -n(Adz)mg
dz
dz = 3J,
22
dz,7
where
= m
w
N,
Since p = nil.
(
1)
(2)
(3)
Consider the element shown.
to move a distance
g
Ax is its initial width.
and the right
a distance |4Af.
The wave disturbance causes the left face
Let P
q
and p be the equilibrium pressure
and density, and p = P-P is the departure from the equilibrium pressure.
o
K
S
“
1
AV
V
P-P
or
23
Q
_1_
AAx
AA|_
p
we have neglected the pressure difference Ap across the slab,
where
from the equation of motion.
->2
m —-§• = PA - (P-*AP) A = -APA
dt
£
Ax
AP
Since
ap
we have
SE
Then
=
AAx
= “ P
Sc
A
St
aA 2
a?
StJ
Sc
2
is.
«
3
St
2
2
2
SjN a E
Sx/ ^2
at
1 a p
2
_
^x
hut since the order of differentiation is Immaterial,
2
2
1
a p
2
K P
a p
2
at
ax
S
5.9
(a)
For an adiabatic process pV^ = const.
tt
av
dp
(t)
Substituting
v
S
yp
u = (p k )^
s
p =
^
and
1
7P
i av
V dp
.
=
(
p =
^
>
^e find
u =
V *)
1
(c)
(d)
u « T^ , independent of pressure
u = 35^ meters/ sec.
5.10
Substituting the given values into the equation
C
V
2 2
= VT a
we find
1
cl.
= 2k.
joules mole “deg
,
7 = l.lt
5.11
When the solid, of dimensions xyz, expands an increment dV, we have
2k
The displacement is found
V+dV = (x+dx) (y+dy) (z+dz)
To first order
V+dV = V + yzdx + xzdy + xydz
_ldV_ldx
“ "
V dT
a = 3
Therefore
ldz
ldjr
y dT
x dT
dT
z
°L
5.12
For the adiabatic, quasistatic process
33 -
Dstog
(11 ‘f
(M)
® + (l) *=
°
T
to
and the Maxwell relation
_/dv\
\Sv
\ /
we have
,
"D
c
1 /oV
.
aVT
AT =
Thus
—
pc
Ad
where
O'
where
c
= ?
^
= -£•
pv
P
5.13
From the first law
c^ = T
* T
l
T
Substituting the Maxwell relation
fLX
= T
(i
v
1^1
si H/
and the definition
-T (^)
2
ov = a
vT - vT
a = - — fvs=\
.S we have
g
5,1k
TdS = dE
(a)
(b)
From
(l)
-
(1)
we may read off the Maxwell relation
Ss
5l
oF
(
Sr.
2)
2
F = aT (L-L
Since
l)T (c)
FdL
S (L ,T)
o
-
s(L ,T
o
q
)
=
C
f T -i
r
J
o
t
_ ~
dT * =
JT
-2aT
(3)
o)
HI
_
^
o
25
dT* = b(T-T
0)
S(L,T) - S(L ,T) =
q
J
In tlie quasistatic adiabatic process
S(T ,L ) + t(T -T )
o o
f o
m
Hence
-
aT (L -L )
± f o
Thus if L^> L
a
we have T
= S(T ,L ) +
o q
-aT (L -L )
f f Q
t-a(L -L Y
q
i.e. the temperature increases,
f >
From the first law
/>c\
^
Substituting (2) and
2
q
±
i
T
(s) t
we find
(3)
vSirl
= “2aT (L - L
o}
S(L,T )-S(L ,T o )
(?)
o
o
j.
pT
S(L,T) - S(L,T
0)
-/
u_
C ax
dT*
= b(T-T
Thus
-/
S(L,T) = S(T ,L
2
(T-T
o
o
q)
dE
-
+ b(T-T )
-
)'
dT*
T*
-a(L-L )
o)
hT* - aT'(L-L
P
rT
*
-fr-
q)
aT(L-L )
£
Q
o
5.15
(a)
(h)
•d Q
= TdS =
Hence
2a £ dx
as-f
T
From (l)
dx +
1 fo£\
dT =
T
(
T
,
+
1
t VSF;
Equating coefficients, we have
d
:
2
0)
2
t-a(L -L Y
_ m
f
= -aT(L-L
= 0
f
(e)
o
S(L,T) = S(L ,T ) + b(T-T ) - aT(L-L )
q 0
0
Q
Hence
(d)
-2 aT(L*-L )dL'
(^)
_ -
<1
26
+ 2ffi
—
_ - 2ot dx
«3T
,
1)
2
From (l), ve can read off the Maxwell relation
/§(-2g£)\
SS\
Since a = o
q
W
do
(¥)
-off.
W
T
= 2lea + 2la
T
- 2ic£T = 2la
0
°
If the film is stretched at constant temperature
E(x) - E(o) = 2^o
(0 -> x) = -
By the first lav
Hence
~F
J'
H W”
3
dv +
.
/SS N
1 /SE\
\
v
4V +
.
T
1 /"dE\
T (Sf )
Equating c oefficients, ve have
II -*(g)
From (l), ve may read off the Maxwell relation
Ss'N
df.
fbY'
{-m.
= zfT
If one mole is produced
-
AE = zfT
zfV
By equation (5-8.12)
I)T
Substituting
p = n Iff
[1
1
(i)v -
+ B„(t) n] ve find
= P +
^
-2 <j£x
T
,
T <vV
x
J 2o& dx' =
+
-f
T
as
SsA
dx = ~
q
2
-3S " P = z *?
27
—
® + zfV d”
.
5.18
i)T ® + (I)v 41
dE =
(a)
from (5.8.12)
@x-v-(i) =<i;
(h)
(c)
From the first lav, dE = TdS
pdV.
Hence
0 ' T
/ds\
(sf)
-
p
(,Sv)
* t
For a van der Waals gas
dTN
T
From
^ =
21
v a
v a
v
r 2 r&\
X V,
w=
We
r
v
2
X
(p + ^j(v-b) = RT
&
(v
T
XSSX
7
0
1
*
c
T
O
Solving, ve find
6a
-h)3
(.
v
'
c
T
SL -0
.
2
-b)
c
W4“
a=jRtv.b=o
c c
3
,
(h)
/'oS\
E
E
vRT
(a)
“*„
5
'
=4
Substituting a and h in (l) yields p
RT
—
-
.
c
P’ +
V
^)( "t
’
To find the inversion curve, ve must have
'8T\
,WH
V
/T roV\
" c; (v <Sr>
'
-
p
1
dv = .avf
2
°V
V
a
^V2
_
jj
V
l-
V
T
From the van der Waals equation
dp
RdT
v-b
s
>
(v-b)
•<r
thus
On eliminating v and putting the equation in terns of the dimensionless variables of problem
5.19# it follows that
p
p« = 9 _ 12 {spP'
5.21
Substituting these expressions into ( l) we find
29
-
V?
)
(b)
r
Frcm (2)
T
dT
T
JT
r* a'd-d*
4 C
J
=
f
*
o 1 +
"
T = T
o
”
15
a'dd
f
exp
V.
uo 1 + —- rcTT
,
V
5.22
The system may be represented by the diagram
^2
<l n
W
(a)
By the second lav
g
AS =
*2
l
^
= q^“W , it follows that
For TnBYiTmTTT' heat, the equality holds and since
*1
T
W
Tl -T o
i
°i
O’)
^ = 11.9
5-23
We have the system
W = q -q
(a)
(b)
1
C^-Tj) + C(T 2 *T f ) = C^-KT,,
2 =
r
J Ti
(c)
Tf
1
/
It follows that
T
so +
T
f)
(
f
2
f
4,
l
T
r
®- clJ1 !s_
5 0
T
s
T
W = C^-fTg
T1 2
1
The maximum amount of work will be obtained when
(1)
2T
AS § 0
3y the second lav.
Thus
-
-
2T
f)
30
=
T^ = ^ T^T^
.
1)
To freeze an additional mass m of vater at T 0 , heat mL must he removed from the ice-water mixture
resulting in an entropy change AS, = - jjp
The heat rejected to the "body of heat capacity C
T
f f dT = C In ^f
increases its temperature to T„ with an entropy change AS = C L
.
By the
T
u "o
o
Tf
mL
S 0. For minimum temperature increase and thus minimum
+ C In
second law AS^ + ASg = 1
.
—
—
—
^o
“o
heat rejection the equality holds and it follows that
mL/T C
o
Tf = T q e
mL/T C
The heat rejected is Q = C(T^-T
o)
= CT
°
(e
q
-1)
The prohahility that the system occupies a state with energy E is proportional to the number of
states at that energy, fi(s).
By the general definition of entropy, S = k In J1 (e), we have for the
prohahility
= e sA
P « n(E) _
To raise the weight a height x, an amount of energy, mgx, in the form of heat, Q, must he given
off hy the vater.
The temperature change is found from
Q = CAT = -mgx
AT = -mgx/c
For a mole of liquid and x = 1 cm, AT is on the order of
where C is the heat capacity of water.
10
^
°K.
Hence we may approximate the water as a heat reservoir and find for the entropy when
the weight is at height x
5 = s
where S
q
is the entropy at height x = 0.
o-ro = s oThus (l) becomes
-mgxAr„
P(x) = Ae
s
Here we have absorbed the term e
usual normalization condition.
oA into the
constant A which is found to he kT /mg by the
Then the probability that the weight is raised to a height L or
more is
r-\
L
w
oo
hr
-TV i
pWax -/ -£
T
Substituting L = 1 cm and the given values yields
GW
10
31
-mgxAT
-mgiAr
dx = e
5.26
la the processes a -> h and c -> d no heat is absorbed, so by the first law W = - AE, and since
a? = ry\T , where C is the heat capacity, ve have
V* * -<VEa) * -*V
"c^d
' -< E E
*
d- =)
y—
However, in an adiabatic expansion TV'
Hence
<~d -
The volume is constant in process b-»c
d^e>
= const.
Vd, *
w
'T
^
^
1-
T
1T
( iy -S b C <v^
)
"V T = (1_ <^>
)
-1
o
so no work is performed.
= (Eg-E^) = vGy. (T
W
11
*
)
=
s—
+ Vc
,
c
^)
7-1
/i
j
— cl
_
aX
f c\
vv
s.
CHAPTER 6
Basic Methods and Results of Statistical Mechanics
6.1
(a)
The probability that the system is in a state with energy E is proportional to the Boltzman
factor e V* ®.
1
Hence the ratio of the probability of being in the first excited state to the
probability of being in the ground state is
-3»o/2KT
=
"
e
(b)
e
-flw/2hT
-fiu/KT
By the definition of mean value
3^-3fiu/2M
+ 2
-3h^/2iT~
+ e
~ftu
fiu
2
6.2
The mean energy per particle is
_
e = -
Hence for N particles.
He -dHAr
.-&fcr
_
7a>r
+
E = -NpH tanh
32
Hi
gr
!^
1 + 3e
" ’-hu/kr
+
g
6.3
(a)
The probability that a spin is parallel to the field is
o nH/kT
P =
x
j-nS/ST +
nH/K
e
-2|iE/kT
1 “ e
which yields
2nH
k ln[p/l-P]
For P = .75 and S = 30,CC0 gauss.
T = 3.66°K
(h)
Fran (l), with 4 = 1.4l X 10
-2
^
(
1)
"
T = 5*57 X 10
°K
ergs/gauss,
6.4
The power absorbed is proportional to the difference in the number of nuclei in the two levels.
This is.
-MH/kT
jiH/KF
n+ - n - -
U e ~
N e'
e
nH/kT + e -v£/XZ ' e nH/kT
where IT is the total number of nuclei.
Since pH
e
“ViH/KT
« kT, we may expand the exponentials and keep
only the first two terms.
(i.g -i.jS)
Hence ,
n
+
- n_
« N
HuH
m
1 + h5 + 1 _h5
kT
kr
-1
Thus the absorbed power is proportional to T
6.5
p(z+dz)A
z
The volume element of atmosphere shown must be in
equilibrium under the forces of the pressure and
gravity.
mn(z)Adzg
Then if m is the mass per particle, A
the area, and g the acceleration of gravity, we
have
p(z+dz) A
-
p(z)A =
dz
dz A = - mn(z) A dzg
Substituting the equation of state, p = hKC leads to
Thus
n ( z ) = n(0)e'
33
^r
mez/
trp(z)A
+ dz
6.6
E
(a)
As T approaches 0 the system tends to the low energy state, while in the limit of high temperatures all states "became equally probable.
~ iT.
limit when
(b)
The energy goes from the low to the high temperature
The specific heat is, Cy = (^J)
,
i.e., the slope of E.
and
T -* 0, E -> N€ 1#
T —* c0 .
E -> N
To find the temperature at which E changes from the low to the high temperature limit, we
evaluate T where
-(« -€ )/kT
r
—
e +e
H
i
E = N
€l + |
2
2
€
1
=
N
-(€ -6 j/KT
1 + e
which yields
£
_€
2
KT
1
= In 3
- 1
or
3k
(e
2~e l) 88 351
The heat capacity is
.
N
se
°V =^T
r
=
<V
M?
-» 0
and
as
T -» 0, T -» 00
€
»2
e
i>
-(« " € )Ap
2 1
“( € 2~ €
l)>^^l
* x
2
J
Jl+e
.
6.7
per mole is
The mean energy
(a)
where
0
Oe"
E = N.
^ 2^^
+
2 N
ee_eAT
A
1 + 2e- /^
(
£
1)
1 + to*/**
is Avogadro ' s number.
N.
We find the molar entropy from the partition function Z. Clearly, Z =
-e /kT
i
, the sum being over states of a single nucleus,
5 = L e
(h)
i
5=1 + e" 6
Thus
S = k (in Z + PE) = N k In (l + 2e'
and
(c)
^
£
+ e -eAr
e
^) +
—
N.
ee
-e/kT
T(l+2e^
(2)
As T -+ 0, the nucleii go to the ground state, and by the general definition of entropy
Ul
In 1 = 0
A
S = N,k In !J =
A
as T -» °°, all states become equally probable,
SI
-* 3
and
S = N k In 3
A
It is easily verified that egression (2) approaches these limits.
(d)
where
From (l) we have
aA A 6Al
~
c
= 31
oT
For large T, e
2
€
€ Kr
3sT
(l+2e" /
A^
q.
and
)
c
« T
6.8
The polarization, P, is defined as
N el where
e-
XM
e-
N is the number of negative ions per unit volume,
e is the electronic charge, and -t is the vector
from the negative to the positive ion.
Then the
o
l+-
35
o
th
P is
•eszt x component of
P
vhere x
x =
cos 9
|p|
=
Ne
x
= Ne
-prj-
l-t-l
0f Q
The positive ion can have energy + —
nd 9 are shown in the diagram.
O
or
,
and since there are two lattice sites for each energy we have
efa
eia.
2
P
x
= Ne
(f)*
P
3
/+
—
e£a
2
e
—
(~|) e
Nea
e*a
P
+
,
.
tanh
P
^
e£a
e
6.9
(a)
The electric field is found from Gauss ’s Theorem
/
E
•
J a
n da = %q
A
where q is the enclosed charge and n is the vector normal to the surface,
a.
Ea = E 2nr L = %q
By symmetry,
KS) = II £
The electric potential is given by
E-dl =
U(r) =
-Jr
-
|a In
o
Since U(r) = -V at r = R , we can evaluate q.
q
Thus
(b)
E(r) £ ,
J In §o
U(l)
i
V 5“
R
In
The energy of an electron at position r is qU(r) = -eU(r) where e is the electronic
charge.
Since the probability of being at r is proportional to the Boltzman factor, we have
for the density
p(r) = p(0) exp
|^-
^A
r
-eV/tT ln(R/r
0 )J
= p( °)
)
{jr
6.10
(a)
2
The centrifugal force mw r yields the potential energy
-
mco
2 2
r
Since the probability that
a molecule is at r is proportional to the Boltzmann factor, the density is
2 2
p(r) = p(0) exp [mw r /2hT]
(b)
(
1)
Substituting p = N m > the molecular weight, into (l) and evaluating this expression at r^
A
6.11
The mean separation, x, is found from the probability that the separation is between x and x+dx,
h£eI
_
hT
P(x)dx = e
dx
12
61
©
U(x) = U
Where
" 2
©
It is easily verified that the minimum of the potential is at x = a.
Since departures from this
position are small, we may expand U(x) about a to obtain
—
36 U
-U
U(x)
o
+
252 U
-
—
a
- (x-a)
=
a
where we have kept only the first three terms.
-
(x-a)
3
It follows that
36 UO (x-a)*
2
'*
252 UO (x-a) 3
'
P(x)dx = C exp
2
a !!
We have absorbed the irrelevent constant term,
dx
3
a !:!
e
into the normalization constant C.
,
This constant is evaluated from the usual requirement
f*
P(x)dx = C
36 U
exp
/
J0
a
The predomi nent factor in the integrand is
(x-a)
o
2
2
-]
-
p
exp
252 U (x-a) 3 1
o
a kr
exp -36 U
(x-a^/a^krj
Q
expanded in a Taylor’s series as in Appendix A. 6.
dx = 1
3
kT
so the second factor may be
Furthermore, the region of integration can be
extended to -® since the exponential is negligably small except near a.
Thus
1 = C
r
exp
V
36
252 u
x ~a ) 1
1 +
2
a !!
a
where we neglect all but the first two terms.
(x-a)
3
3>
dx
(
1)
kT
The first integral is evaluated in Appendix A.b
while the second is 0 since the integrand is an odd function.
BEy definition,
0 . S
a
x =
f
x P(x) dx,
x =
—
/
(
7r
kT/
and to the same approximation as in (l) we have
/U0 \5
a l 7rkT
V
\
J
r
36 u
L
a
exp
/
(x-a )
kT
c
1 +
K
3
252 U (x-a) \
—r
a^kT
/
n
/
The first term in the integrand is a Gaussian times x and the integral is just a.
may be evaluated by making the substitution | = x-a
3?
The second
Uo \ 2
252 U
X = a + —
a\^7rkTy
;
h
S
3
J
U
36
exp
a kr
°v
d| +
exp
a|
/
G6 uO
e 2
ia?
ae
Noting that the second integral is 0 since the integrand is an odd function, and evaluating the
we find
first by Appendix A.
a =
Thus
1 +
dx
dT
aU
1
x
kT N
.15
x = a
U
.15
ale
__
,
+ (.15)akT *
15 k
since U
U
» kT
o
6.12
(a)
Let the dimensions of the box be x, y, z.
Then the volume and area are
V = xyz
A = xy + 2yz + 2xz
(
1)
(
2)
Eliminating x, we have
A
V = yz
From the conditions for an extremum,
SV
oV
-
^ ^=
2yz
+
2z
y
0, it follows that
(A-2yz) (y+2z) - 2yz (y+2z) - y (A-2yz) = 0
(A-2yz) (y+2z) - 2yz (y+2z) -2z (A-2yz) = 0
.
Subtracting these equations yields y = 2z.
and from (l) we find x
2
A2
.
Then substitution in (2) gives
y =
,
v3
—
At .
=
From equation (l), we have
dV = yz dx + xz dy + xy dz
(3)
The equation of constraint is
g(x,y,z) = xy + 2yz + 2xz
and
Adding
-
A = 0
kdg = k(y+2 z) dx + \(x+2 z) dy + k( 2 x + 2y) dz = 0
(3 )
w
(5)
and (4) and noting that the differentials are now independent, we have
yz + \(y+2 z) = 0
xz + k(x+2 z) = 0
(
xy + \( 2x+2y) = 0
We multiply the first equation by x, the second by y, and the third by z, and add.
3 xyz + \( 2xy + kxz
Substituting (k) yields
x.
= - 2
S£
2A
38
+ hyz) = 0
a
—
r=q ,
2 \/ 3
V3
(b)
z =
6)
and substituting X into equations (6) gives
1 -
1 1 -
g + 2z) = 0
g (x + 2z) = 0
§ (2x+ 2y) = 0
(7
The first two eauations show that x = y and the third yields x =
first and second equations gives
1
But since E P
^
^
= E P
S.
y =
,
&
a 1/2
,
z =
—
k
2V3
a/3
^In P
+ S„ = -k E P
2
Then substitution in the
l/2
^ -kZP ^ln
we may write
+ S
_l
x =
a1/2
A
cL
= -k E E ?
r
r s
= -k E E P
T*
S
r
2
^P ^ In P ^
^P 2 ^ln Pr ^ - kEEPsr
s
^
^
s
s
^Ps ^ In Pr
r
+
l—
-kEEP ^P ^In P ^P ^
=
= -k E E P
In P
rs
„ s„ rs
r
since P
r
^p
( 2)
= p
s
rs
From the definition, we have
S 1 + S, = -k E P
x
d
_ r
Substituting P
r
^
= E P
and
rs
P
s
^
s
S,
1
^ln Pr ^-k E P s ^In P
s
m (x) we have
= E P
rs
r
v
+ S 0 = -k E E ?
rs
2
r s
-k E E P
r s
'
InP r ^-kEEP rs In ? s
r s
,x
In ? (1)' B
? (2)
r
s
'
rs
For the total system, the entropy is
S = -k E E P
r s
Then from (2) and
rs
In P
rs
(3)
'12' = -k E E Prs in Prs + k E E Prs In Pr ^P s
S-(S +S
)
= -k E E p
r s
in
p
'
r
39
'P
(
'
s
(2)
^
(fc)
In x § x-1,
or
Since -In x £ -x + 1
we have
(2 )
In P
^-w (D -1
P (Dp
S---'
-S -
*
rs
rs
P
Fran
S-(S +S
(!<),
1
= k Z Z P
2)
r s
In
rs
(Dp (2)
r
(Dp (2)
(
s
skZZp.J
_ _ rs
rs
r s
P
\
\
1
rs
/
But since
Z Z P^^P,^
s
r
2'
= 1, it follows that
r s
S-( s
Thus S S S
1 +
5
i* 2 )
S
P^^P^^-l
k Z Z
= 0
r s
Sg.
6.15
(a)
\
j
Since
S = -k Z P
In P
r
S-S
0
r
and
= k Z
r
P
S
= -k Z P
o
rrrr
In P
In P
Then using Z P
r
^
.
(D+ P ( 0 )pn P (°)
r
(D
r
We show that the sum of the last
e"%
P (0) r
Since
we have
2n P
^
where we have added and, subtracted the term k Z P In P
r
r
r
two terms in (l) is zero.
J,
rr
(D _ P
In P
+ P
^
^ln Pr
r
'
= -gE
-
r
In Z
= 1, we have
-ZP
lnp(°)=pZPE
Z Pr ^In Pr
Similarly
+ Z ?
In Z = S3 + In Z
Z P r ^3r - Z Pr ^ln Z = -?3 - In Z
-p
r
r
r
Adding, we obtain
^In ?r
-Z? r InP^+Z
r
r
r
=
r
Thus (l) becomes
S-S
(0)
(b)
Since In
?
-P
r
In P
+ P
P
S
Q
= k Z P
r
r
In
0)
P
(
r
1, equation (2) yields
r
F
Consequently S
In P
0)
„ (0)
"
r
rrrr
= k Z
Q
(
(
§ k Z P
r
°
r
( °)
—
p
r
S S.
UO
-
1
= kZ
r
p <°>.
_ r
p‘
r
= 0
2)
CHAPTER 7
Simple Applications of Statistical Mechanics
7.1
(a)
We label the positions and momenta such that r.
i.
There are N. molecules of species i.
and p.
.
j
molecule of type
Then the classical partition function for the mixture
of ideal gases is
d3
3
exp
,th
refer to the
.
2
(Slrl
2m^
+ ** ,+2>-M
^k
2
^ll'*-
d3
|
)
h
o
^
^Mkd3£ll*-- d3EMk
3M
. . .
h
*
o
The integrations over r yield the volume, V, while the p integrals are identical.
Since there
are N^+N^. . .N^ integrations, we have
„(v- +Vr r - Sr ai-|Ci+ - +V
3P
z
,
=
htl
The term in brackets is independent of volume, consequently
In Z' = (N^+. ..+W ) In V + In (constant)
k
and
>
~slv^ z
'
<V— +V k
'
pV = (N +...+N )kT = (v +...-*V ) RT
]
k
1
k
For the
(b)
i
td
gas,
p..V =
(i)
v^RT, hence by (l)
p = 2 p.
7-2
3y the eauipartition theorem the average value of the kinetic energy of a particle is
(a)
€ =
|
(b)
kT.
The average potential energy is
/
mgz e
-Smgz
a
Then
5
”
R mgZ
dz
dz
3
^
*'^ to /n .-ngz B ** • "Se
8
On carrying out the differentiation, we find
u = kT +
.
^gL/iT
2_
_ e
1-1
—^
B 1
5
7.3
(a)
Before the partition is removed, we have on the left, pV = vRT.
1S
-
-n
Pf
(h)
After removal the pressure
RT _ 2g_
(l+h)v
1+b
The initial and final entropies of the system for different gases are
S,
^
= vR |!n
In
vR
+
+ vR
A
— + ^2 In T+o-2
In —
N. v
1|
_
A
In T+a
+ |
2
1
N. v
|
in T+a
|
+ vR
in
J
ASH
+
2
N v
3
2
a
Here one adds the entropies of the gases in the left and right compartments for
while S
f
is
the entropy of two different gases in volume (l+b)v.
Kiiil . u JL .
Then
AS = S
(c)
In the case of identical gases,
compartments.
f
-S. =
i
vR
V
vR In
is again the sum of the entropies of the left and right
is the entropy of 2v moles in a volume (l+b)v.
—
V
In + ^ In T+a
o
2
N^v
= vR
S
Thus
V
V
^ w.
m 2N^v + 2
2vR
f =
AS = vR
R
In
(
vR
In
bV
R
+ — In T + a
N V
°
2
A
T +
.1.X.1.W
1+b ) v
B v
2» »
A
vR In
B v
a
a
Sb
7.5
(a)
The system is isolated so its total energy is constant, and since the energy of an ideal gas
depends only on temperature, we have
AS
X
+ A3 2 = CyCT^) + Cy (T f -T 2 ) = 0
T
or
T
f
=
1
+ T
2
The total volume is found from the equation of state
HE
2
—im
——i + V——
V
V =
Thus the final pressure is
( V]L+v
p^
=
2)
V
RT
f =
"
2
(v +v
x
2
52
2)
T
l
+T2
U v 1Ti/Pi)+( v 2T 27p 2
)
(b)
Using
^
~
•=
y
,
ve have for the initial and final entropies of different gases
kT
s
S
v R
f = x
k
V
,
11
+ 7 In T. + a.
= V R
1
i
In
v
v 2^2\
3
^ 1^2
%
'
2
2
111
,1+
/
A
2
P-L
l^l
pi
V T P
Thus
(c)
AS = S -S
f
v R
i = x
2 2
l\
+ VgR
In
+ v R
„
T 4T
1 2
3
—
““d — — a^
2
11
V T
k
/
1 1
T
v T
2 2\
+
+ 3
Hr
l"
,
2
)
In
+v R
-h
+ In T„
+ c_
X
2
2
P2
2
1
1+
v T iP
2\
1
v T P
2 2 1
3
+
°a
T ±^o
2
In -i|
2
211
'
For identical gases
kTi
= v,R
In
1
=
AS = v-jR
(Vi^v 2 )
l
1
Px
(Vj+v^R
kT
^
+ —
In T.+a
1 o
2
In
2
In
..
^
+ ^
In T„ + a
2
P2
2
o
V T
V T
T 1^2
1 1
2 2\
3
+
+ a
In
+ %
2
^
o
pl
p2
/
V
—)
+ VgR
2^
2
2/
T-jP,
,
(VV
+ v_R
W
In
v T P \
1 1 2
V
2
Vl
)
/
T
^
42ln ±i2T 2
3
2
7-5
We take the zero of potential energy so that if a segment is oriented parallel to the vertical
it contributes energy Wa and if antiparallel it contributes -Wa to the total energy of the rubber
band (Thus if the rubber band were fully extended, the total energy would be -RWa).
segments are non-interacting,
-Wap
= N
-Wap
- 77.
.^WaS
'
WaS
+ e
= Wa tanh
kr
7.6
Since the total energy is additive,
V Mp.) (V
+ u
C
-n)
=
F
2
+ u
(V°-n)
where U is the energy of interaction, the eauipartition theorem still applies and
€ =
|
7-7
If the gas is ideal, its mean energy per particle is
2
T3
and the mean energy per mole becomes 2 = W^kT
2
p
Since the
7*8
For clas sical harmonic motion the mean energy is
*-£ t- ?
2
5/ equipartition, each of these terms contributes ^ hr.
Hence
°=
p
x
« T
7-9
(a)
The mean elongation is found by equating the gravitational and restoring forces
ax = Mg,
(b)
thus
a
By the equipartition theorem
1
—2
v
/
p a (x-x)
or
(
x =
|
kT
(x-x)'
(x-x)
c)
KT
a
_
2 2
= X
m
2
or
a
.a,
M =
7-10
(a)
Let the restoring force be -ax.
Then the mean energy of N particles is
—
1 »2
E = N (r mx
By equipartition
)
E=If(|kT+|kl)=NkT
=g = Hk
c
(b)
1
p
+^ax
If the restoring force is -ox3, the mean energy per particle is
-
e =
1-21
T
mx +^ax
The kinetic energy term contributes — kT by equipartition.
found from
4
T
exp
ax
exp
-,
Bax
-rr
47
Bax
T
ax
dx
dx
44
The mean of the potential energy is
T
ax
r -p(«» te
la
T*
.ur
=
e
^
-03
e
-<^
A p'lA
fly
^ “03
iA x.
where we have made the substitution* y = p 7
T
T
coc
Thus
¥
(-
_
Hence
i In p + In
s
-oy
A dy}
=
K
¥
—
e =
-2
E
1
1
1
1
S
imx
+Jcoc = |kT+£kT=|kT
Then the total energy is E = He and
m
=|| = t
C
7-11
nfito
IT
Since the probability of excitation is proportional to e
where n is an integer, it is
clear that the two parallel components are negligibly excited at 300
» 300 k.
where
« 300 kj and since 300° is in the classical
The perpendicular component is excited since
region, the equipartition theorem holds and we have for the mean energy per mole
mi
a (|
E = N
Thus
0 *
§
2
+ | mUj
V
*
2
2
x ) = H
A
kT
5
7.12
(a)
k
Aa
o
and therefore the pressure is Ap = k Aa/a
o
Hence
(b)
The force necessary to decrease the length of a side by Aa is
Consider a cube of side a.
*
The Einstein temperature is 0
O
Q
The change in volume is AV = -a^Aa.
.
2
a Aa
.
AV
~ " 1 f
V Up.
k
Since
ftco/k.
E =
K
x
s
«*.._£
E
k i m
to
a_
Aa/a
(
K
where m is mass, we have from
=
„
= t
f_
k Vmn
(2)
The length a is found by considering that the volume per atom is n/pU where
A
pi
is the atomic
weight, p the density, and N is Avogadro's number
A
1/3
Thus
Substitution of m = pi/N
a =
~)
(~J
1/3
=
(
23
(8.9)(6xl0 )
and the given values yields
0
E
» 150 K
^5
1)
)
Q
= 2.3 X 10
cm
(l)
7-13
By equation 7.8.14,
Ve have 2^ = NQg (i Q J Bj(tj).
Si(t) = 2 (coth tj
= 2
-
j
=
coth |)
cosh n
sinh 7]
2
2
cosh (n/2) + sinh Ct/2)
2 sinh(r)/2) cosh (q/2)
IliT^fcosh (n/2)
N S u
0
\
Thus
=
o
— "becomes
^ (tj) for J
,
.
'
"
_
cosh (n/2)
sinh (t]/2j
cosh (b/2)
sinh (t)/2J
'
f {^2] = tanh (n/2)
'
= co
s ^oH
2
7.14
The energy of the magnetic moment in the field E is
2 =
-ji*H = -|iH cos 9
where 9 is the angle "between the magnetic moment and the direction of the field or 2 -axis, Then
the probability that the magnetic moment lies in the range 9 to 0 + d0 is proportional to the
Boltzmann factor and the solid angle 2jt sin 9 dQ.
P(e)d 0 «
Thus
sin
q
'7T
cos 9
and
Vz
«Z
cos 0)
sin g
IJ
*
r
e
^ cos 9 sxn 9 dS
BliH
Kh
.
.
n
N
or
,,
“z
=
o
S
T ¥
0 5_
E eg
W
.
f
e
COS0
B(iH
N
sinf5
de
=
BpH
..
f Ip
111
—FF
-3uH
“
o
BuH _
^ 2 sinh
pEgcosh 3pE -sinh guE
puE
hb sinh p|iH
“z =
V
coth PuH
-
7-15
We have
«z
R
o
8 1*0 J
46
U)
B
J^
lere By (7-8. I*)-)
If T]
(J
J
« 1 and J » 1 in such a way that
=
Bj(t))
+ §) coth (J + hn
»
Jt]
1, B
J coth jT
J
"
l
- | coth
|
r,
Becomes
(f])
= coth J,1
\ (|)
'
where we have used
coth x =
e
x
+ e
,
x -
where
Q
+-.. - ! •? + (1
'
-
x +
.
e
Let pTTp = Jtj = Jga Hp
-x
jj.
e
(1 + X + ...j
-x
-
Xl “ X + ...)
w 1
for small 3
X
= gh J is By (7*8.2) the classical magnetic moment.
Q
Becomes
*Z
V
=
coth ppH
Then (l)
Pul
7.16
E = -4
The energy of a magnetic moment in a field is
(a)
•
H.
For spin — atoms
fc
<VHi>
uH P
2
dH P
1
n(Z_)
v
2.'
n^)
(*)
n(Z )/n(Z
2
(c)
If
1)
uH
> 1
« hT ve have
~
and
^
+_e
_ e
~ pH-B
-fjH 7 P
~
+ e
e
^
Hg >
since
= exp
r-(Z
^(zj
(1 + fiHjp +
| p
2 + u
“
2
n(Z
+ i lA
(l + pHgB
2)
2)
( z^T
+ n
2
2
2
^
(h _h )
2 i
fc
V) + (l - hEgP + ^ U^p 2 ?
2 2
?
)
+ (1 - nE^ +
|
)
u^V)
2 ?
Hg P
2_
“ 1 + 2
1 +
2.2
1
1 +
p
l
H
"
<et>
1 - - v(
2
—
hT
-)'
! )
7.17
To find the fraction,
of molecules with x component of velocity Between -v and v. we must
integrate the distribution Between these limits, i.e..
47
i
|
= =
r
/
n J
v
(“v
m
® v x')dvx =
~ g(v
^27rkT
/2kT
/2KT)
e
;
dv
m
Making the change of variable, y = ,^£L
5
=
f
|
we have.
3" (y
y 2
f e"( / = 2 erf VF
dy =
)
J-y/2
VFr
Then from tables
v ,
-/irr
^0
= .8^3.
7.18
In problem 5 *9 we found that the velocity of sound is
- =
where 7 = C
and 4 is the atomic weight.
Since 4 = N^m,we have
u = v(^);
N.m
A
The most probable speed is v = (2kT/m) 2 .
2
=
(
'
2
*
S
m ‘)
Thus
4
u =
(f
v
For helium, 7 = 1.66 so that u = .91 v and the fraction of molecules with speeds less than u is
}
3
.9iv
.9l(2M/») 2
%
F(v)dv =
0
(ra/kT) 2 v,
Making the change of variable y =
5
=
2
<SB>
T
a
2
8
2
m
mv
2
dv
we have
%
^
y
2
e
2
-y
^ '/2
dy
fl
'0
(27T)
-y /2
This integral may be evaluated by noticing that integration by parts of f e J '
dy yields
0
£
y2 2
y2 2
2
-y
J //2 .
+
e' / dy = ye“ /
dy
y e
/
J
0
0
/
Since the integral on the right is the required one, we have.
I
=
%.1
2
91^2
- I
1
$
r .91V2
-J2 r .91^ 2 "y2// 2
2
2
=
e
dy
%
(2)
y
(2)
J
J
*
_
P
72/
/ 2 dy - .91V2 exp|-(.91 2)2/2)
-1
e
The integral on the right may be evaluated from tables of error functions.
h8
Thus we find |
= .37*
7-19
v
(a)
v
(b)
v
'
2
(v v
(e)
v
'
'•X''}
2
x
x =
=
0
symmetry
"by
—
by eauipartition
m
= (v ^ + v
x
y
2
v
x
(v^v)=v'
v =0
y'
X
x
y
5
(d)
'
'
2
+ v
w
x
^ ^
= (i+b
2
x'
v
2
dv =
J
^
ve
l.
=
)
3£
m
(^)
v
m1
dv
—
and since v = (—
,
EL
71
^
w
dA)
have v
,
2
=
)
,
1.2J
= 2e/m and dv = de/ V2me.
On substituting these
relations into the speed distribution F(v) dv, it follows that
_ 3
F(e)de = 2?m
The
4a
ICT
~
1
e2 e
= 0
e_
1
jl
_ e
_
i
2
(7 tKT)
most probable energy is given by the condition
M de
or from problem 7.20b
M =0
2
e
1ST
e =
|
1
v = (2kT/m) 2
~
The most probable speed is
1 ~2
nrv
= kT
Hence
V = V
(b)
o
(l +
—
)
c
= V
since v
o
The rms frequency shift is given by
(Av)
ms
=
(v_v )
_
2
2
=
_
—x
v
Since v^ = 0 and v^
_
- v
/
/
2
‘
y2
_
v
2v
+
o
= kT/m by ecuipartition, we have
v
2
l.
12
w ve
Since the energy is e =
^
2
2
2
+ b v
y
mv
% (^)
f
v
y
x
o /o
By appendix A. 4 ve find (1/v) = (^-)
VliCl
(b)
y
v ) = 0
z
v f'
(
)
(a)
2
+bv)
=vx2 +2bvx v
y'
(v
(e)
/.
~
\
me
=v o 2 1+ kT
^
x
=0
/ .
\
r 2 +—
W
=
(Av)
o
rms
(c)
5 V
2
-
-
2
V
o
il
2
o
/iT'
o
The intensity distribution is proportional to the distribution of the x component of
Since v
velocity.
X
=
—
(v-v
'o
v
we have
),3
'
o
2
,
>2 me (v-v
qJ
r
—
exp
l(v)dv = I
L
where I
(a)
r
o
is the intensity at v=v
2kT v
dv
J
o
.
o
The number of molecules which leave the source slit per second is
A = r nvA =
4
o
molecules/sec
= 1.1 X 10
1
<J>
where A is the area of the silt.
(b)
7 . 11. 7) > the number of molecules
Approximating the slit as a point source, we have by
(
with speed in the range between v and v + dv which emerge into solid angle d2 is
A$ (v) d^v = A[f(v)v cos
0] [v
2
dv dfi]
cos 0 = 1 for molecules arriving at the detector slit; hence the total number which reach the
detector is
« r-
&
A fl p
^
d s
7T
where we have used the results of appendix A,k and p
(c)
11
= nkT.
Letting L = Im be the distance
is approximately A/L
between source and detector, the solid angle
given values into (l) then gives 1.7 X 10
g
V 2nmKF
.
Substitution of the
molecules/sec arriving at the detector slit.
In the steady state, the number of molecules entering the detection chamber per second is
equal to the number being emitted.
Prom
A,2Pq
~
we find
(l)
-
A P.,
.
S
= 0 A
d
=-—4r-
7rL y27nnlfi?
i/Stt
_
P d _ PS
A
mkT
2.4 X 10
,.2
inm
of Hg.
The rate of change of the number of particles in the container is given by
f=
- cpA
= nvA
J
50
Since the pressure is proportional, to the number of particles, we have
dp
=
dt
vA
"
WP
vA
p(t) = p( 0 ) exp
where p(0) is the original pressure.
tw!|
Hence the time required for the pressure to decrease to
p(t)/p( 0 ) = l/e is
t
As in problem 7*2^, we have
-12
dn
vA
= "
dt
5v p
W In —P
,
4-
t
where p(0) is the original pressure.
The mean speed, v, is for nitrogen at 300°K,
v =
5
JPf = .^75 x 10 cm/sec
to Vil . If . 07 sec
_i
10
(.^75 x lCT)(l)
....
(a)
Since the rate of effusion is
= p/(27r m kT) 2 and the concentration is proportional to the
<p
,
pressure, it follows that
v
c
(b)
(5T
= f2
^
l*
V
For the uranium separation, we have
A.
c
'
'233 // C '23Q
= C
235
//C
=
238 ^235 + o[l 9 ^
(c
C
(1*003)
o^/
238 ;
23
V o-3ft)
The rate of change of the number of particles inside the container is
thus defining A.
dU
1
dt
t
-
nvA =
NA
kT
AN
77
m
Vm
Since pressure is proportional to the number of particles, we find after
integrating
At
/
p/p o - exp
For Helium gas p/p
/8
l/2 at t = 1 hour.
Q =
b ®J
Substituting we have
1^2
A
51
where m^
From (l) and (2) it follows that the ratio of the
is the mass of the helium molecule.
Ne and He concentrations in terns of the initial concentrations is
vV
x
<
(n^/n^)
0
1/vv')
^
<v/V> 0 exp [-^ 2 (V"He/v’
*
Since
•
[-
*]
= 1, we have at t = 1 hour.
(
^Ke/^He
1_
(
^
Ie^
~ 2
1_
AwSfe
A
= 2
^
7.28
(a)
The rate of change of the number of particles in the left side of the box is
®L (t)
dt
= (no. entering/sec) -(no. escaping/sec)
1
.
Since the rate of effusion is
^ nvA, we have
dN (t)
x
t
N
[ 2 ^)-V ]
= W/2
dt
where N is the total number of particles.
Ni(-t)
x
§ ^^(t)
From the equation of state,
^
V
= P (t)
=
P l^ 0 ^ + P 2^ 0 ^
„
N =
and
V
kT
(1)
dp, (t)
Thus
Integrating,we find
(b)
2V
([fl^
+ P
^(0) + p 2 (0j^
p^(t) =
“
2^
A(0) - p2 (0)\
-
f
t
The initial entropy is
S.
+ | In T + a^) + B2 (0)k
. ^(0)* (in
In final equilibrium,
in
^15)
+
f
^ T + °o)
= N =
2
2
S
Then
=
dt
f
=
m (In 5 + | In T + co
)
AS = S^-S^ and using (l), we find
Pn (0)
AS = -±
V
21
+ Po(®)
—
-£
f
„
(p^o)
m
y
1
In
2kJ
p 1 (o)+p 2 (o)
p l(°) v
KT
sr In
2
T
P>)
2P
2Po(0)
3?i(o)
Pi ( 0) + p
2
(0)
52
+ p
2 ( °)
111
i^ToJ + Pp(o)y
P 2^°) Y - ~ kT
1
111
2
T
(a)
Inside the container
(h)
The velocity distribution, 0(v), of the molecules which have effused into the vacuum
= 0 by symmetry.
d\ = f(v) v^ d\
3>(v)
where f(v) is the Maxell distribution.
dv
'
2
dv
/
dv
/
,
v
$ (v)
dv
e
2kT
z
v
2
z
=
v
z
dV
'
xJ
<0
dv
dv
/
$ (v)
-<o
dv
e
PVT
z
v
z
The second equality follows because the integrations over v^ and v^_ are identical in the numera-
tor and denomi n ator and therefore cancel.
The integrals over v
z
are tabulated in apnendix A A.
ZE
1 /2kT\
2
(a)
m
(2
U/
In time dt, molecules with velocity component v
which are in a cylinder of height v
z
with base at the aperture can escape from the container.
z
dt
Molecules of higher speed can escape
from a larger cylinder than can molecules of lower speed, and a proportionally larger share of
the fast molecules effuse.
Thus the mean kinetic energy of particles in the beam is greater
than in the container.
'3
d v
2
1
£q = g mv =
1
g
mv
2
/
\
$ (v;
d^v 0 (v)
o
2
J
Since <&(v) = f(v) v cos 9 and d v = v dv sin 9 d0 dcp, we have
2 n't
P7T
P 7T/£
P 00 rV
To^
r®
/
(v
/
/
jo^ O
2
mv ) e
—
/
/
i
i
J o
r°° r 71/ 2 of
/
.
dv sin 6 d9 dcp)
5 e
1 i f v
r
?Vp
v cos
9
my2
2
(v^ dv sin e d0
2kT
"
dcp)
e
pm v cos 9
.
dv
= 2kT
TV
e
2kT
,
dv
^o
where the integrals over v are evaluated by appendix A A.
53
Since the mean kinetic energy in the
the enclosure is e. = 7: kT, it follows that e = ^ e.
o
2
i
i
3
7.31
When a molecule is scattered, it commutes momentum Ap
= 2mv
z
z
Ap / dfc.
z
The force is
to the disk.
To find the total force, we must multiply Ap /dt hy the number of molecules with
z
velocity in the range (v, dv} which strike the disk in time dt and then integrate over all
velocities.
Clearly only molecules in the cone of half -angle 0
disk.
F =
/
A$
“ rS
2
o r *, 2
dt
(v)
= sin ~ r/l will strike the
AP Z
1
dJv
q
dt
(v av sin 0 d0 dtp) (A f(v)
-X
2m v cos
dt
v cos 0 dt)(-
ou o
0.
(2mA) (2rr) (-
j
v
cos^0)
f (v)dv
u o
o
The integral can he expressed in terms of the mean square speed inside the container, i.e..
v^f (v)dv = j£-
mx.
Thus
1
2
„
F = = nm v A
,
.
(-
3
1
1
3„\ 0 = —
nm v2 AA
w cos 0)
3
o3
Rx
3,-1
=)
(sin
- cos
L
'
—
1
2
_
—
n mv = nkT
= p
.
But hy eauipartition we have
1 - cos 3
F = pA
Hence
,
1
.
[sm
,
-1 R\
—
CHAPTER 8
Equilibrium between Phases and Chemical Species
8.1
By equation (8.3.8), we have the proportionality
CP(v,T) dV dT « exp
-G
(V,T)/kT
dV dT
where V and T are the volume and temperature of the portion of substance of mass M, and T
q
minimum at V = V, and T = T yields
2
b g_
Bg
G q (V,T) = Gn (V,T) + (sr)
(T-T) + («2) (V-V) +
v
| (
BV
bt
2
+
I
o G
£)
(
^2
1)
(T-T)
2
BT
2
(
o G_
+
is
Expanding the Gibbs Free Energy G (V,t) about the
the temperature of the surrounding medium.
Bg
q
(v-v) (t-t) + ...
(aTS?)
BV BT 7
5^
1)
0X
>
where the derivatives are evaluated at T = I and V = V.
the first derivatives in (l) are 0.
By
Since this is 0 at T = T, we have T = T
(8.1;. 7 )
and at T = T
From
(2)
o
(Se/oT)^. is the heat capacity, C
.
q
o G\
3
Since ve are expanding about a minimum.
T
T
3
'
= T,'
*
<V
2
.
SC..
0\
V\
& - fH-g
,
T
we also find the mixed derivative
T
<sr>„
V
T
(fv>„
SCT
,
n - rKsrl - 0
-
at T = T.
By the equation preceding (8.1;.13)
cG
(^)
cV
(M«
-
By definition,
k = -
1
SV
— (2=)
v
,
= (T-T
y
'
o )(^)
'cV-'
0
-(f)
u
T
thus
5
)
(?(V,T)dV dT
°=
exp
-
T = T , V = V
at
(f T
)
Substitution of these results into (l) yields
.
V k
+
)
(T-T
~o
= Mc^ where
and
—
)
o
Q
= “
= -
-
G (V,T) = G (V,T
and since V = M/p
P + po
p
1
(
\svy
Sp T
f
2
0)
+
(V-V )
2
2Vk
c^. is the specific heat per gram, it follows that
rs (IJr/ asrw T ' w/°o
<
'
where we have absorbed the constant term G (V, T
o
o
)
2
>
®®
into the normalization factor.
Performing the
usual normalization, we find
/cl p
\2
<?(v,T) dv dT =
(a)
For the solid. In p = 23.03
-
Me
exp
"^
375
T and for the lia-uid In P = 19* ^9
p
(T_T
p
o)
^o*
triple point, the pressures of the solid and liquid are equal.
23.03
=19.^9 - 306 3/T
- 375
55
p
dvdT
- 3 C63 /T.
At the
T = 691/3. 5h = 195 °K
hence
—
+ constant where
RT
y
(h)
Since In p =
-
•£,
is the latent heat, we have
= 3
^sublimation
^vaporization
~
"^fusion
^
R = 31 ’ 220
3 3063 R = 25
j^es/nole
^8° oules/mole
j
^vaporization
^sublimation
= 31,220 - 25 } b&Q = 5740 joules/mole.
The latent heat of sublima tion is equal to the sum of the latent heats of melting and vapori= £,
zation, i.e.,
Since -t = TAV (dp/dT), the Clausius -Clapeyron equation, where v is
+
the volume per mole, it follows that
=
f—
'dT'
7
T
1
.
- v
TTv
J
sol'
o gas
v
s
Ip
\ o
_
-» 0
as
= f
'dT'
s
2.
p
K
L_
v
-
(v
fv
k
'‘lie
lia*
o
A
+ T fv
v
v
)
sol 'dT'
sol'
dT ;
m
and AS
^
^
AV
dT
(a)
Q/L moles/sec evaporate and are swept out "by the pump.
T -» 0
"by
J±L
L
m
dr _
_
dT
t(v
'
v
vj
L
gas
- v. .
.
.
dT
0
as
T -* 0.
Then "by the equation of state
from the Clausius -Clapeyron equation
, )
liquid'
L
_
~ T v
gas
RT/p we find
&s =
r 5m dp
T
L / m dT
1
*m
^ po
Hence
A dT'
n
) f—®’)
the volume per mole of the gas is much greater than that
liquid
Then using v_
lia*
V TT
or
RT
We find the temperature at the pressure p
of the liquid.
v
the third law, it follows that
V
£
where we have neglected
-
gas
s.
Since
(h)
o
T
m
= T
L /I
R T
o'
(
56
T
o
"
o
T
-
1 >
T
m
8.6
The intensity, I, is given by
—
1
nv =
I =
£
p
V2tt m kT
Differentiating -with respect to T and dividing by I,we have
d£
1 dl
1_
= 1
“
I dT
2T
p dT
(1)
Since p is the vapor pressure, the Clausius -Clapeyron equation yields dp/dT.
dp _
L
dT ~ T(v
gas
^
Substituting into (l) and using V
L
T v
ZJ
liquid'
v~.
gas
= HT/p gives
g as
1 dl _
I dT
L
1_
^2 “ 2T
8.7
Differentiating the molar latent heat yields
41
-
dT
<#)
cip'
- (i>
§
dT
(i)
'ST'
3?
Now -L = t(s ~s^) where (s^-s^) is the molar entropy difference between the phases.
2
8s
’
Thus
<i>
But (os/8p)
T
= -(8v/8t)
,
8s
"
^5p"^
T
J
a Maxwell relation, and since the coefficient of expansion is
a = -l/v (cv/ST)^ ,it follows that
(
ll ‘ ^ [“2V2 ' Vl
(2 )
Differentiating at constant p gives
CS_
#
T(8s/oT)
-
<V
§r
(
is the specific heat at constant pressure.
<#>
CS-
+ 1
S 1>
-&+(<
2
)
Thus
(3)
)
*1
Substituting (2) and (3) into (l) yields
dT
= -T(a v - o v
1 1)
2 2
dp _
-t
dT ~ T(v -Vi)
2
(a V
Since
dt
dT
§+%+
,
v
x
P2
px
'
.
I
T
57
p2
‘
2 2
- c
(c
¥l^
v -v
2
pi
1
)
3.8
The bar moves "because the higher pressure underneath it lowers the melting point.
The water is
then forced up along the sides to the top of the bar and refreezes under the lower pressure.
To melt a mass dm of ice, heat -Idn must be conducted through the bar.
it per unit time is
The heat passing through
AfT1
51
f
= - K (be)
where AT is the temperature drop across the bar and
the thermal conductivity.
k is
the position in the vertical direction, we have dm = p^(bc dx), and equating the
Letting x be
rate at which
heat is conducted to the rate it is absorbed in the melting process we have
f=l P (bC g
-K (be)
(1)
.
AT is given by the Claus ius-Clapeyron equation
dr __ Ap _
at ~ AT
t(v
l
- v.
ice'7
water
where v is the volume per gram or l/p.
Then since Ap = 2 mg/bc.
-6 be
'p
p.
Substitution in (l) yields
dx _ 2 mg kT
dt ~ .
p2
aoc 'Op.
/
1 \
I
(
‘p.
"
pw
i
‘
8.9
From the Clausius -Clapeyron relation, dp/dT = L/TAv we find
r dT _ f AV
dp
L
J
T'J
We know L(cp) and AV(cp), and we can find dp/dcp from the curve of vapor pressure vs. emf.
Thus
273.16
AVfaO
dl*
T'
L^cp*)
,dpv
Me?'
dep*
where 273 «l6 °K is the temperature of the reference junction of the thermocouple.
Then
T = 273.16 exp
AV
L
8.10
We divide the substance into small volume elements.
Tie total
entropy is just the sum of the
entropies of each of these, or
S=ZS. (E± ,N.)
i
where E_. and
denote the energy and number of particles in the i^ volume.
58
If we consider
0,u
an interchange of energy "between the i.th
and
volumes, we have since the entropy is stations
3 S.
cS
dS = 0 = vtF dS. +
diii\
"but
dS
2.
0
since dE. = -dE.« ve have
i
a’
cS.
oS.
oEi
^
or
1
.
= T.
Similarly, if we interchange particles
^
cs.
as.
dS = 0 = v^r1 dN. +
oil.
a
oH.
Ss.
1
implies
SnT
dli.
j
as.
=
i
__2
oiT
or
^i
=ti
j
j
8.11
(a)
The chemical potential can "be found from the free energy by n =
^
oN
.
We consider a small
volume element at height z over which the potential energy may he considered constant.
It is
clear that the addition of a constant term to the energy does not change the entropy of the
substance in the small volume element.
particle function times the factor e
S = k
Thus
In
—
Explicitly, the partition function will he the free
7
npr z
.
e
Z„
free
-pmgz^
+ 3
l
E
free
+ ***'
= k
+ 3E„
In Z_
free
free
Hence the free energy is, letting N he the number of particles in Ay
F = S
=
ip
ana
p =
5
^ = ! hr + mgz
-
kT
-
llkT
^
In
TS
+ Ifingz
-
— + ^ In T +
vp
HkT
iln
f+|mT+
c
-s
a
where we have put AV/u = kr/p.
(h)
Since p is independent of z.
KT an
dp
—
= mg + —
az
p az
2£=
or
t=-
Integration yields
p(z) = p(C) e
n
-
^dz
kT
mgz
IT
8.12
The relative concentration remains the same.
n
The dissociation is described by
C0 ^H-O
n
2- = X (T)
cc
X
n
2
Generally, when the number of particles is conserved, equilibrium is independent of vc
59
c
8.13
The intensity of the I atoms in the "beam is proportional to their concentration in. in the oven.
Since dissociation proceeds according to
I
I + I
2 =
We have
a
l'\ ’
«
which gives
*l\ ' TS^ * VT
by the equation of state,
The total pressure is p = p
into (l)
-
P
>
Pi = Kp Pi
+ p
X
which yields on substitution
2
2
P z = (-1 +*Jl + 4-iy?' )/2K
p
or
Thus, if the total pressure is doubled, the ratio of final to initial intensities is
(p
Of
-1 +yi+8K~p
I) f
-1 +Vl + lt-K p
°i
p
Since K is small we may expand the square roots to obtain
P
4>
-1 + 1 + 4KpP
f
~ -1 +
i
1 + 2Kp*n
For chemical equilibrium we have
-1
“2
=
—
where n is the number of molecules per unit volume.
12b
b
b.
!?/«
i
m
• •
2
K
JT
)
Since n^ = p^./kT
+b + • • • +"b
1 *
a
-"k
follows that
"b_
= (SE)
Kn (T)
= KjT)
3.15
3y the first law aO, = dE + p dV.
Letting d(pv) = V dp + p dY we have
d Q = dE + d(pv)-v dp
= d(S + pV) - Vdp
= dH - V dp
where the last follows from the definition of enthalpy, H = S + pV.
Then if p is constant
d Q = dH
.e., in a constant pressure process, the heat absorbed is equal to the change in enthalpy.
a chemical reaction, Q = LE = Z bJn since enthalpy is an extensive quantity.
60
In
8.l6
~3y
the equation preceding (8.10.25)
Zb.1 In £. * = Zb. in n.
1
i
1
i
(
1)
Substitution of n^ = p.j/fcT into (l) yields
Zb.i In p.i = Z (b. In t.
s
.
D
or
P
D
1
+ b. In hr)'
x
D
•"Pm ”=V T
p2
In K^(T) = Z (b
where
'
»
x
i
;
n.
1
In
d
+ b
«
£
)
In kT)
b.
to K.W - s (\
Thus
ffl
5
to Jj' + -X)
e.
By (8.10.30)
-=•
In
£ .
CLL
1
*
h.
where e. is the mean energy per molecule.
=
1
J5.J.
Then multiplying
~
T
p.V.
vs have
^ 1 = TW
i
b.h.
p.V.
-1
h **,<*)
*1
.
if).
where In 1 is the enthalpy per molecule.
Multiplying and dividing by Avogadro’s number gives
^ ^
h * VT = Ei ET 2 ~ ET 2
>
where lu is the enthalpy per mole.
8.17
We let | be the fraction of R^O molecules which are dissociated.
Since the pressure is pro-
portional to the number of molecules, we have
2p„
s
=
PH 0+pH
p
p
(
PH„0+P Hp
1)
The total pressure is the sum of the partial pressures.
P = Ptt
+ ^0
t
Pn
2
2
H
Substituting (l) and
(2)
into
+ P•
%0
(2 )
2
pO
^(T)
=
-2—Z
pa,o
we find
V
.3
T) =
TTIv
(i-s)
(3)
(2+6)
61
8.18
In problem 8 .l6 it was shown that
AH = RT
2
^ In Kp
(T)
Using Kp(T) from problem 8.17 would, give AH for the dissociation of 2 moles of HgO since the
reaction is 2HpO
2Hg + 0^.
Hence for the dissociation of 1 mole
h^
** =
where K^(t) is given by equation
V
^ In
|
is found by plotting In
of this curve at 1700°.
111
1,
we have
~
2
(2+0
(1-0
«
f
pg
Pi'
T> -
m ~ ~ dl
Hence
Since
in 8 . 17 .
( 3)
~
" 2
dl
111
5
vs. temperature using the given values and measuring the slope
£
We find
In 5 = .CC6 h.
Thus at 1700°K
2
AH = | (1.99)(1700) (.OC60 = 56,000 cal/mole
8.19
(a)
N
The partition function is Z =
where
l\l •
£
is the partition function for a single molecule.
For a liquid we assume a constant interaction potential -tj and that the molecule is free to move
in a volume N
v
3
3
’a „ a r
=
Thus
!,
where the integration is over Ngv
Q
and all p.
’
s
thus defining
=
Vo
“3 f d3P e P ~
e
=
\^
V
£ ’
No
= nTT = nTT
Hence
(b)
/2m
h
v
e
o
%
^
(1)
For an ideal gas, the partition function is
z
s
d
M
N
6
The Helmholz free energy is
N
F
= -KT hi Z
= -hi (in
=
-3sT
(N
Vs
62
[V]
In V
g
S -
In IT
J
InN +N
t’-N
b
g
g
where ve have used Sterling’s approximation In J ! » u InN
g
g
g
SF
=
Then
=
v
-lsr
Sir
&
(c)
g
5*
“
-
w
g
^ Ng
)
For the liquid, ve have by (l)
IT
*£,= -1
(In
&T)
v t’
o _
L 'V
V
=
e
1To
(H£ In v^' +
-IdT
X - In No
pT)
+ Ng)
sf
^
Therefore
(d)
=
£ = -kT (ln v
^
0 S'
+ (*l + 1)
When the liquid is in equilibrium with the vapor, the chemical potentials are equal or
V f
-10? (in
-
’
o
In N ) = -KT (in v £’ + prj + l)
O
&
li.
Thus
N
- v
o
7 4-1
oP
“ !
6
but since p/kT = N /V ve have
g g
P _ e
ev
IT
(e)
(2)
We calculate the molar entropy difference from the Clausius -Clapeyron relation.
dp
dT
As
~
As
~ v
Av
g
where in the second equality ve have assumed that the molar volume of the liquid is much less
than that of the gas.
= ST ve have
Since pv
g
1
p
Then using
(2)
dp
dT
As
~
BT
in this relation ve find
As = R (1 + -r)
v.
The molar heat of evaporation is
L = TAs = HT (1 + gj)
»
(f)
From (3)
§ = \T\ ^
M
m 1+^
1^ -= *
*
TaMng the logarithm of (2) , ve find
+ 3_ = !n
ha?
where v
— ln-S
=
is the volume per molecule of the vapor.
63
(3)
q
» iT
-k~ = ln -£
Hence
when v
(*)
^b
gO
and v
are evaluated at T, .
D
In (4) we can take v to "be the volume per mole instead of the volume per molecule since ve
(b)
have the ratio v /v
S
To estimate the order of magnitude of L/RT^ we let v^ = 22.4 liters/mole
b
I
.
°
-
and v
q
=»
^
7ir ^ 11^
o
-3
= ^ it (.5 X 10 ®)
(6
L
Thus
i
2
x 10
^)
= 10 ^ liters/mole
In « 10
10 _J
mmmm
B
V
V
0
g
.83 cal/gm°K
vt,
experimental
1.45 cal/gm°K
water
.018 liters/mole
31
nitrogen
.035
6.4
•37
.64
benzene
.087
29
.16
.27
liters/mole
These results are remarkably good in view of the very simple theory considered here.
CHAPTER 9
Quantum Statistics of Ideal Gases
9-1
configuration
no. of states
0
MB
e
3e
1
1
1
1
1
1
--
2
1
1
X
2
1
1
X
2
1
1
XX
XX
X
X
X
FD
—
—
XX
X
BE
= 1 + e’2
(a)
00
z
(<0
^FD
be
=
2
1 + e‘
e
® + e"6® + Se'® - 2,- 3® + Se' *®
1
® + s'6® + e‘® + e- 3® +
-® + ,-3»B
«.
.-"WB
9.2
(a)
For an FD gas, we have
-a-£E
In Z = aN + Z In (l + e
r
r
)
_ =
n^
and
e
-a-£E
Hence
S = k [In Z + gE] = k
—— x
aN + Z In (l + e
r
64
~
)
+ gi
r
+ 1
"but
N = Z n
since
r
r
and
E = Z
_
_ nr e ’,
r
r
we find
~
-a-pe
_
Z nr'(a+ p€ r )+ Z In (1 + e
S = k
r
From
n
=
r
——
a+Pe
e
r
)
(
it follows that
r
+ 1
-a- P €
Substituting (2) and (3) in
In (l-n ) - In n
r
r
r =
5
r
1-E
r
(3)
nj + Z la (1 + -H=-)
sr
la
-
V
r
= -k Z
(2)
we find
(l)
2 nr (lu(l-nr )
S = k
Jn^
r_
In n^ + (1-n^) In (l-n^)J
For the EE case
-a-pe
r
In Z = aN - Z lu (l-e
n
and
)
v.
=
r
—a+Pa
cc+Bi
e
r
The calculation of S proceeds in exactly the same way as in (a).
S = -k Z
n
r
(c)
1)
r
a + pe
(h)
'
r
In n
r
(l+n
-
-*•
We quote the result.
In (l+n
r)
(5)
r)
From (^) and (5) we 6ee that in the classical limit where n^
S
-l
«1
-k Z n In n
r
r r
for both FD and BE statistics.
9*3
The partition function in
N/
£
/HI
£
wher
Z
=
exp
K K K
-
6h“ r 2
2
2x
=
+ K +K ‘
(k
v
x
2m
)
y
z
(
1)
x y z
The sum is over all positive
k k k
x y z
when
we use the standing wave solution, equation (9.9.22).
The density of state is now greater than in the traveling wave case, and by (9.9*25)
p
If we assume that successive terms in
K
£
d3 K= X-d3 K
~
3
~
7T
differ little from each other, we can approximate (l)
with integrals.
65
„
L
Thus
k
X
3^ 2
_
-
Pm
e
4-^ —
L
x
2m
e
/
(
'
=0
x die
,
\
)
x'
7r
3
Hence
5
=
(27T
m IT ) 2
u-3
which is identical to
(9 »
10 . 7 )»
9.k
(a)
The chemical potential can he found from the Helmholtz free energy, F = -kT In Z.
-N
F = -kT In
« -kT
[N In
|jy
-
£
H In N + N]
where we have used Sterling's approximation , In HI = H In H-H and
V
£
= -=r
m kT)\2
a)
Then
(b)
/
(27r
The partition function for a single molecule of a 2 dimensional gas with binding energy -€ q
00
IS
Be
{;*
=
Z
exp
t
Thus
where A is the area.
*o f
=
e
£
In equilibrium
jj.
°iT
~
= p. * and by (l) and
(
2)
V
N
(2tt
pe
m kT)'
A_
o
(27T
m kT)
h3
Letting n* = N'/A and since p/lsT = H/V, we find
-1/2
n' = ph
3
(
2nm k^T' )
9*5
(a)
e
As
e
1
By equation
( 9 . 9 .IL)
we have
if L = L = L = L =
z
x
y
66
—
2*
sir,
v (v K 2^
+K ‘)
x
2m
y
Jar)
|
—
.*•< §
L.ill.
N
N'
or
exp
where we put A = V5
II = -kT In
p' = -kT In £7
N'
Hence
(c)
£ '
Ir^S^o
* *
.
(2 )
dK dK
„
2
x y
/
= —
3
(c)
v
. SL*!
m
‘
3
ii '/ ,
2rr
r'
&
dV
—m-2—2 (nx2 + n 2 + n 2
x
17
=
(V)
;
e
4'%
r
\
(
)
'
z
z
y
1)
'
—v
(2)
The mean pressure is p = Z n^
is the mean number of particles in state r.
where
y —
n
2
P “
3V
£
r
r
2
r “ 3
E
T~
The expression for p is valid for a system in which the energy of a quantum state is given
by (l).
For photons,
1
/
e
=
r
2
2
2
= ilc[2nV 3 [n +n +n )'
V
A x
z
y
cfiK
ne
„ =
Pr
(e)
x
\3
—
(d.)
y
/
f
T\
rt
1
i
J.
_
3
e
T
—
.
3114
V
p “
IE
V
3
The mean force on the wall is found by multiplying the change of momentum per collision,
2mv, by the number of particles which collide with the wall per unit time,
^ nvA.
Hence the
pressure as
—
2
i
_
rinrv
p = - (2mv) (g- nvA) =
^
'
Letting v
2
« v^ we have
TD
3
since N(= mv
(a)
_
)
= S.
3
(Ap)
^
3
2
=
=
— ^ £r n e
.
9V
V
(E
2
=
p
r
=
Since 2 = - rq In Z and (AE)
k
By problem 9-5^ P
.
9
—
E
V
= 2
)
o
2
= 2 n
P
(b)
(
Equation (3) is the exact result as shown in section 7.1^.
The dispersion is (Ap)^ = p“ - p
Hence
2n
N /I
—
— 221V '
V *2
1
2
2
= — nmv = —
2
2
-
3
2
)
=
(^E)
2
—^ In Z, it follows that
o
oS
67
y ——g— , and it remains to find p
by
(2)
in 9-5
(Ap)
tut by
(2)
.
=
\^
9v
83^
2
Sp
53
In Z
2
53
9V'
in 9-5, E = | V p.
Thus
(c)
2
(Ap)"
_
3V
2kT
3V
2
op
5T
v
Substituting p = — kT, we find
(AP )
2
/P
2
M
(I
=
-2
)
N
^(Akt)
2_
3N
W
ST
*9-7
The probability that a state of energy e is occupied is proportional to the Boltzmann factor
-e/kT
°.
Since the vibrational levels have a spacing large compared to kT , we see that
e
-e/kT
0
On the other hand,
is small and we may neglect this contribution to the specific heat.
e
the spacing between rotational levels is small so that these states are appreciably excited.
For a dumbbell molecule the rotational energy should be
e
2
1
2
1
+
A
A w
2 2 2
r = g l“l
where A is the moment of inertia and w the angular frequency.
the axis of the molecule because the moment of inertia is
may use the equipartion theorem yielding
= kT
We have neglected rotation about
small.
At room temperature, we
Adding the energy of the translational
.
motion gives for the total energy
6 = e
t
+ e
r = 2
M o + M o = I Mo
Thus the molar specific heat is
0
I
V
2HD = H
D
”A It
o
I
R
9-8
The reaction proceeds according to
2 +
2
and the law of mass action becomes
\
n
^
i>2
-
= K
n
“HD
=
h
2
\~
(
1)
’HD
From the results of section 9*12, we can immediately write down the quantities
notation of that section.
68
In the
P
V
CH = ^3
[27r(2m)
2
KP]^
2A
H
KT
e
h
2
=
2
PfiWg
|
2¥l
^
*D„
I
C
-
*2
2
2
[ar(2M) krr
^
h
e
I
ps
%
3
^ ^
5
m
^-\
h
a'
2
3
*
ii
where we have assumed that molecules are predominantly in the lowest vibrational state and have
treated the rotational degrees of freedom classically.
The electronic ground state energies e
are approximately equal for the three species since
1
this quantity is altered only slightly by the addition of a neutral particle to the neucleus
(to form D) .
Thus
HD
H„
S imi larly, the moments of inertia are of the form
A _
.
1
2
4
^2
* _ 2 _ 1
2
R
R
o
2 nijHflig o
where R , the equilibrium distance between nuclei, depends primarily on electromagnetic forces.
It is then approximately the same for the three species, and we have
Aft ilg a nD - 2 =
l
^2
2 n 2'W
= 1
¥
(m^ ) 2
/
mM
Finally, the frequency for two masses coupled by a spring of constant
k
is
ic(m^-m,
Again,
.
m, m„
1 £
k
must be the same for the three molecules, and
k (m+M) \ 2
HD
V
mM
/2k\ 2
'
M
'
~
111
T
s
= n^
,
2
^
“o
2
n
HUM'
U
o
into (l) and using the results (2), (3)
K
Since n^
W
/2M
~~
£
2m
hn
/2k\ 2
Substituting the
,
n * I $3 mT
and (2) yields
_l
x
m2 +M2 _
exp
[2(mHM)]
we have
69
3)
2
'N-:
'“HD
= K
n*HD
^HD =
N
or
1
1+2
JP
JF
2
iiio
exp
nM
1
^
I
IN
i
m2 + MfL
.
1
2kT
9-9
In a thermally isolated quasistatic process, there is no entropy change.
Hence hy the first law,
TdS = 0 = dE+pdV
Substituting E = Vu and p = i u where u(t) is the energy density, we have
0 = udY+Vdu+^udV
r^k dV'
'J
3~
v
du
1
r
k
/
u
dT
—
*
since u « T
k
Integrating, we find
.
— In 8 = 4 In
—
3
-*•-»
k
-
-*•
\J rp
i
T.
=
X
or
—
2
9.10
(a)
By the fundamental thermodynamic relation
T dS = dE + p dV
u, we have
Then substituting E = Vu and p =
^
_
—
—
k
udV + Vdu
TdS=udV + V du_ + 1i _u dV = —
But since u is only a function of T, du =
^ dT, and it follows that
dS =
|
We also have
|
dS = (If)
dV+^gdT
(
dY + (t§)
(2)
dT
1)
V
T
Equating coefficients in (l) and (2) yields
(b)
/oS-v
k u
o?
' 3 T
j
,
8X1(1
,oS\
V
dT
“ T
^
2
Since (c S/oV cl) = (o 2 s/gT oV), it follows from
70
(3)
du
dT
that
(3)
4u
du
~
dT
T
Thus
In T
k
*/?=/#
or
-
= In u + C where C is the constant of integration.
Finally u « T
k
•
9.11
The energy of the "black "body radiation in the dielectric is
S = Vu = V 215
(c'h)
where c' is the velocity of light in the material,
vacuum by c
1
= c/n
'
,
o'
n
o
3
c' is related
to the velocity of light in
being the index of refraction.
— ^1— no =
3
E = V
Thus
«
2
(cH )3
2
3
Here a is the Stefan Boltzmann constant, a = 7r^k^/60 c fi .
16 a n 3 VT 3
Hence
-
<r
<§
>
-
(
1)
Substituting v, the volume per mole, for V in (l) gives the heat capacity per mole, c^*.
classical lattice heat capacity per mole is cy = 3R.
c*
C
The ratio of these two quantities is
16 a n 3 vT
3
3RC
V
3
For an order of magnitude calculation, we can let v = 10 cm /mole and n
find Cy'/Cy ~ 10
The
= 1.5.
At 300°K, we
^3
.
9.12
(a)
The total rate of electromagnetic radiation or power is
p = (p(krr r
2
= 47rr 2 oT'
)
1l
=
=
(b)
2
(%)(10 )(5.7X10
_5
6
)(10 )
= 7.2X10
ergs/sec
1
7.2 X lO ^ watts
Since the total energy/sec crossing the area of a sphere of radius r* = 1km must be the
same as that crossing the sphere of radius r = 10 cm, we have
<p'(47T r'
2
or
22
(?'
=(?(fr)
r
=
2
)
= <?(% r
2
)
2
6
5
= (5-7 x io“ )(id
(fr-)
r
71
2
)\^)
10^
o
1 *]
= 5.7 X 10
(c)
P
il
= 5.7 X 10' vatts/cm
ergs/sec cm
The power spectrum has a maximum at
max
he
= 3
KT \
max
kT
'
„
Thus
,
x K) 27
x VjP) . bj X 10- 6
cm - b 7 A
)j?
6
16
(1.4
x
io"
)(io )
(3)
9.13
(a)
The power given off "by the sun is
O
Jl
P = a T
(% R
q
)
Half the earth's surface intercepts radiation from the sun at any time.
Hence the rower that
the earth absorbs is
P
1
= a T
4
(4ttR
0
2 )(^L)
2
knL
The pover that the earth emits is
P
ll
P
= a T
e
(km
)
In equilibrium, the power incident upon the earth is equal to the power it radiates, or P^= P
Thus
oT
k
k
..
O. .On-*2.
2
(knR )
k
= tfT
Wl
g
o.
2
(% r )
R_
T = T.
(
2L
1)
-.1/2
.10
(b)
From Eq.
7 X IQ'
T = (5500)
(l)
= 270°K
13
2(1.5*10
)
9.14
We consider the liquid in equilibrium with its vapor (at vapor pressure p).
p/V2rr m KT'
flux of molecules incident on the liquid is
,
By (7.I2.I3), the
and this must also be the flux
Since jf cannot depend on the conditions outside the
of particles escaping from the liquid.
liquid, this is the escaping flux at all pressures
-V
ThUS
~
/—
r
; f
V27T m kT
"
-
The mass of a water molecule is m = 18/6.02* IQ23 = 2.99*10 23 .
) I
p = 23.8 mm Hg = 3*17*10
p
dynes/cm
we have
3*17*10
X
(2ttX2.99
Thus for water at 25°C where
X 10
.22
- 23
X1. 38X10
72
_1
^ 298)2
= 1.14 x 10‘
molecules
sec -cm
9-15
At the vapor pressure p(T) , the flux of molecules of the vapor incident on the wire is hy
(7.11.13),
<p
= p/ *J2tt m hr'. When the metal is in equiUhrium with its vapor, the emitted flux
must equal the incident flux,
®
Then when the temperature is T, the number of particles
cp^.
e =
emitted from the wire per unit length in time t is
N =
P(T) gnr t
VarmS
But N = £M/m where dM is the mass lost per unit length and m is the mass per molecule.
*
^m
Thus
V27T m hT
p(T); =
or
/§L
^
rt N 2nu
where we have used u = mW^ and R =
9.16
v
= 0 "by symmetry.
3
Jd v f
By definition.
(v)
(
J d^v f
1)
(v)
At absolute zero, the particles fill all the lowest states so that the distribution f(v) is just
™
a constant and cancels in (l) .
and
(l)
1
2
The maximum speed is at g mv~
%
becomes
v
2
X
k
v
a
dv
3
_ 1 v 2 _ £
" 5 F
5
=
2
v
krr
£
m
,
dv
9.17
(a)
Hence
At T = 0 all states are filled up to the energy |i.
the mean number of particles per state is just 1 and we have
PM.
E =
ep (e) de
/
Jo
By (9.9.19)
,
= oJL
p(e)de
2 p
2
1
klT
(2m)
I i
e
2
de
3
x
fi
where the factor 2 is introduced since electrons have two spin states.
3
v
\/
E
>m v2
xaai.
(
r
2
%2
Jo
3m~
73
2
1
x
3
By symmetry, v
= M*
£
2
V
2
oo
Since
2
u = |L
f
2
(3tt
by (9-15.10), it follows that
|)
E =
(c)
(1)
tyi
|
Substituting the expression for p. in (l) we have
2
2
5 -
2 3
Jj
3
£
(37T )
N
(|)
<*)
As particles are added, each new particle can go only into an unfilled level of higher energy
than the one "before.
Hence the energy is not proportional to IT as it would "be if particles were
added to the same level.
9-18
(a)
Differentiating E in 9-17 (c) gives
2/3
5 - - <i>
(b)
Again from 9-17
(c)
5
£m V
a 5/3
(
1)
we find
^
2
- _ £I
(3tt )
3 V
5
s '
m
V
"5
(c)
22
electrons/cm 0 .
For Cu metal, N/V = 8.1 x 10
the electron, we find p ~ 105 atom.
Substituting into (l) where m is the mass of
This enormous pressure shows that to confine a degenerate
electron gas, a strong containing volume such as the lattice of a solid is required even at 0°K.
9-1
From the general relation TdS = ~i Q = CdT, we have that if the heat capacity is proportional to
a power of the temperature, C « T
s>
T
a_1
dT
<x
Thus the entropy is proportional to the same power of the temperature as is the heat capacity,
(a)
The heat capacity and entropy of conduction electrons in a metal is proportional to T.
Hence the ratio of the final and initial entropies is
ff _ iioo _
S.
200
(b)
The energy of the radiation field inside an enclosure is proportional to T^, and the heat
capacity has a T
Thus
dependence.
ff _ ,2000
S
~ '•1000
i
7
9.20
(a)
At T = 0, the total number of states within a sphere of radius
k
F
must equal the total
number of particles N.
Thus
2
(2tt)
3
3
7r k
(|
F )
3
N
=
1
2 N 3
',
or
“F
2 2
Substituting p = fi k
p
= (3tt
y)
/2m, we find
2
fi
^ = ST
,_2 N>, 3
(37r
V}
(
The number of particles per cm 3 , N/V, is N p/p where
is Avogadro’s number, p = .95 gm/cm 3 ,
A
and p = 23 .
(b)
1)
From (l) we find the Fermi temperature T_ = p/k = 36,000°K.
F
To cool ICO cm 3 or 4.13 moles of Na from 1°K to
Q = vj
c
3°K, an amount of heat Q must be removed.
.
dT
y
2
where
is given by
(
= | R
9 . 16 . 23 ),
2
_
Thus
Q = (4.14)
Hence .0022/. 8 = .0027 cm
3
of He
R)(4t^
u
3
J
^-)
(y
—^ Jp
,
.
.
,
3
T dT = -.0022 joules
/
)
2.
are evaporated.
Note that a large amount of substance is
cooled by a very small amount of helium.
9.21
(a)
If electrons obeyed MB statistics, the susceptibility would be
the magnetic moment.
A
where p is
x^ = Npm2/kT
m
In the actual case, most electrons cannot change their spin orientation
when an external field is applied because the parallel states are already occupied.
electrons near the top of the distribution can turn over and thus contribute to X.
Only
We expect
X^ with N replaced by the number of electrons in
then, that the correct expression for X is
this region,
eff
=
(?)N
= (£-) N
'
p
where T_ is the Fermi temperature.
F
Thus
(b)
F
Np
X = N
Since Tp is on the order of 10
'T.
m
kT
eff kT
or 1C?
(l)
1)
yields a X per mole on the order of 10
electrons obeyed MB statistics, X would be changed by the factor
75
(
If
This is on the order of 100 at room temperatures.
KINETIC AND MAGNETIC
ENERGY
9.22
When the field H is turned on, the energy levels
of the oarallel spins shift ty
while the
mH
-p.
levels of the antiparallel spins shift "by +u H.
m
Density of
States
The electrons fill up these states to the Femi
The magnetic
energy p as shown in the figure.
moment is determined "by the number of electrons
which spill over from antiparallel states to
This is (p H)p(p ) where p(u ) is the density of states
m
Q
Q
parallel states to minimize the energy.
Consequently the magnetization is
at the Fermi energy p .
Q
M = 2
But since p(p ) =
Q
^
,
jj-
^ p(pj
H/V
we have
M _ 3
y _
B -2
and
^m_
N
where n = ,
„0
9.23
If we choose the zero of energy to he that of the electrons in the metal, the partition function
of an electron in the gas outside the metal is ty
(9 .
10. 7)
VoAT
£
-
% (ar m Xlf e'
ir
where the factor 2 occurs because electrons have two spin states.
ribund
The chemical potential p
from the Helmholtz free energy.
-hTlnZ = -kT [N In £ - NlmN + N]
F =
^o
=
1
=
111
N
=
M26
^ ^3
The chemical potential inside the metal is just p = V
Q
potentials inside and outside are equal, i.e., p = p .
Q
76
VqAT
)
- $,
and in equilibrium the chemical
is
Q
Thus
flh-
Hence the mean number of electrons per unit volume outside the metal is
3
n = ^ = i- (2?r m 1st) 2 e
-5
'
^
(1 )
9.24
Consider a situation in which the metal is in equilibrium with an electron gas.
Then the emitted
flux is related to the incident flux by
*
(1-r)
e =
«>.
(1)
The electron gas outside the metal may be treated as a classical ideal gas from which it follows
that the incident flux is
.
1-1
= 5 nv =
where we haved used (1) of problem 9.23.
2
(2rm IT)
5
e“
$/<kT
8 IT
7im
The emitted flux is then given by (l).
But since the
emitted flux cannot depend on the conditions outside the metal, this is also the flux where
equilibrium does not prevail.
Thus the current density is
J = e®
v q ft~r (M) 2
e
=
-
-
)
where e is the electronic charge.
9.25
There are
(27t)
3
d\ states per unit volume in the range (k dx) and with a particular direction
of spin orientation.
Since the mean number of electrons with one value of
k
is
+ j]
o 2
where e = h k /2m, we obtain
f( K )
A
d3 K
i
A-
If we take the z direction to be the direction of emission, the number of electrons with wave
vector in the range
(jc:ds)
and of either spin direction which strike unit area of the surface
of the metal per unit time is
4>(k)
d 3jc = 2f (jt) v
z
d3£
Then the total current density emitted is given by summing (l) over all values of
allow the electrons to escape.
(l)
k
which will
That is, the part of the kinetic energy resulting from motion in
the z direction must be greater than V
,
or h
2
2
/2m > V , where we use the notation of problem
Q
77
9.23.
In addition, ve assume that a fraction r of electrons of all energies are reflected.
Thus the electron current density is
J=e* 6
flkjfo) **
r oo
dK
(&r)
3
1 + exp
§~( K X2+Ky2+K z
)
“ft*
is of the form
The integral over
where e is the electronic charge.
2
2
xdx
2
1+e
““
h
j
Since V
Q
»
x
-x -kz
x *a
= - i In [l+e“
]
’
xe
2
-cz
e
-X -KZ.,
+1
> p. and, in general,
*
1
t; fi
**
(h-V
|g
q
)
I
exp
» 1, we can use In (l+x)= x to expand the logarithm
and obtain
„«2 2
-pn k /2m
P(n-Vo )
2e (l-r)
T _
fr
e
3
dK
*
m 2 k 2/2m
n»
-P»
e
/
dK.
)
1/2
Each of these integrals is (2nm/pfi
hy appendix A. b, and
)
% me (l-r) ^^2
j =
e
-4>/bT
9-26
(a)
Since the integrand in (9.17*9) is even, one can write
x 2
p« x 2
ex.
x
„
“"
p~
-r
/
2 ”^
e
/
X
-» (e +l)
= 2
X
e
where we have integrated "by parts.
Jo (e X+l) 2
2
+ 4
+l-
o
1
0
x dx
X
e +l
Since the first term on the right vanishes
—m
h(h)
x dx
X
e +l
k
l
o
(
This expression can he evaluated hy series expansion.
x
X
e +l
e
-x
l+e"
=
Z
x
-x
= e x [1
X
(-l)
n
e-^
- e
n+1 ) X
n=0
Hence (l) becomes
78
-x,
+e
x dx
-2x
<1 one can write
Since e
-e
-3x
1)
!
... J
Ip = *
S
(-l)
n
f e “( n+1 ) x x dx = k 5 Jtil
u o
n=0
The integral is,
"by
z
4
2
n=0 (n+l)
.
2
4
e_y y ay
J
n=0 (n+l)
appendix A. 4, eq.ua! to unity.
I
0“
n
Thus
1
±+
.
2
C
2
O
1_ * * #
.
4
3
= 4s
p
2
(2)
(°)
e
Fig. 1
Fig. 2
One can use the contour C of Fig.
(l) to write
dz = 2nl1
E
dz
or
/
Here the integrand -* 0 for z -»».
z
2
a 5
-
n=l n
Sin 77 z
Z
Fig. 3
(3)
n
n=l
7T
= -2i S
00
.
sin 77 z
The sum in (2) is unchanged if n -» -n.
Hence one could equally
well write the symmetric expression
r
J
g,
z
dz
T
S
(4)
one gets then
(5)
-4i S
(6)
sin 77Z
Z
,
' 2l
sm 77 z
Adding (3) and
along the contour C’ of Fig. (l).
CHC
dz
2
But here the contour can he completed along the infinite semi -circles as shown in Fig. Q since
the integrand vanishes as
Except for z = 0, there is no other singularity in the
|z|-»«.
enclosed area so that the contour can he shrunk down around an infinitesimal circle C
ding
z =
0 as shown in Fig.
For z « 0,
(3) •
1
z
2
1
sm tt z
.
z
2
[7TZ
1
y
r TT^Z^ . . .]
-
2 2
I
[1 +
77 Z .
g
r-,
1
=
.
2
z
sm 77z
.
1
77Z
.
3-Z
“
=
]
TVZT
Hence
/ ^
\ /7T\
= (-2771) (g) = t
79
12 2
Ll - g77 z
TTTT*
1
'
r.
2
177
+ 77 +
51
.
-1
,
. . . j
o
surroun-
Thus it follows by
(5 )
that
JUS
3jj? =
00
,\n
/
(- 1 )
_ _ v
or
n=l
and by (2) , I
:
_ TL
1
12
r\
/
n
2
= 7r /3
9.27
(a)
We expand e
in the integral
,00
e
j(k) =
-0=
1
ikx
X
.
dx
e
x
(e +l)(e" +l)
®
x
e
eo
,
dx
X
(e +l)
2
\m
/
x
ikx
,
dx
(
2
1)
m
X
mI
m=0
e
(e +l)
-=
(ik)
y
x
Interchanging the sum and integral yields
l
;<*> -
m=0
(b)
\m
eg
m
e
(e
X
X
+l)
m
2
x
00
,
=
dx
v
L
m=0
(2)
Putting x=z in (l) to denote a complex variable, we evaluate the integral on the closed
contour shown in the figure.
Here C* is the path where
circle around the singularity at i7T«
z
is increased by 27Ti and C" is a small
Then the integral on the complete path is zero since there
are no singularities in the region bounded by the contour.
But the integral is 0 at x = + ® and cancels along the imaginary axis.
e
ikz
.
e
dz
*
2
(e +l)( e- +1 )
(
ikz
Hence
.
dz
(3 )
e+l) (e+i)
C’+C"
where both integrals are evaluated frcm left to right.
since e
x+?ni
^ikz
e
1
( e +l) (e
= e
Along C' where
X
* g ik(x+2ni)
j
dz
-f
u
+l)
-00
(e
+l) (e
X
ikx
dx =
e
-27Tk
dx
(e+l) (e
+l)
80
+l)
z =
x + 2iri, we have
The last integral is just J (k) .
(l _ e
-2nk)
e
f
C
Thus from (3) we have
(e
z
1^ 2
dz
e
z
+l)(e" +l)
(e
ikz
z
,
dz
(M
z
+l)(e" +l)
C'
To evaluate the integral around the small circle C", we put
in terms of
+
(;
and expand the integrand
£
r
J
e
ikz
z
-V
dz
r
r
_-7rh
e
z
J
(e +l)(e" +l)
C"
b
e
ram
_-nk
e
dE
c"
= e
where we have used e
e
^ -®
—
[l+ikC+...]
f
J
ikx
X
di
c"
-rck
7rk
= 27rke"
|cH-(27ri) (-ik)|
= l+x+... and the residue theorem.
^00
(c)
z = 7ri
_
.
dx
x
,
27rk e
l-e
(e +l) (e~ +l)
-rrk
(5)
Then combining
(L)
and (5) we find
77k
sinh -n k
_27Tk
(6)
Expanding (6), we find
W
_
- 11 _
~ sinh 7T k
Then comparing coefficients with the series in
2
+
350
we can read off the I^'s.
(2)
We find I
= 77/3.
9.28
(a)
According to equation (6. 9 .It-), the probability of state r of the system is
-3E
P
« e
r
r
-cSS
r
Then the mean number of particles in the state s is
Z n g exp
n
n e
2 2
+ •••)- a(n 1+n2 +. . .)
-3 (n 1 e 1 + n e
+ ...)- a(n +n +. ..)
?
1 2
[-p(n e
1 1 +
=
Z exp
[
where the sum is over all n.,n , ... without restriction.
Letting
Z = Z exp [-3(11^ + n2 e 2 +. .) - a(n 1+n2 +. .)
.
n
we have
S
1
= - —
3
]
(l)
— In 2
3
8€
.
^
s
where the partial denotes taking the derivative where
(l) in the
.
usual way, we obtain
81
occurs explicitly.
Evaluating the sum
2 = ^E
exp [-(cft^e^n^^E
[-(a+pe^n^
exp
(2)
-a-Be
In £ =
Thus
-
E In (l-e
r
—
"
and
(h)
1
'^
=
s
8
£
^
1
5:
.
s
a+j3e
s
-
1
Using the same argument which leads to (9*2.10) with the function
we obtain
8n
,
(An
v
51
1
s
= - U ss'r
Se
e
where again the differentiation is performed where
e
occurs explicitly.
g
Thus
a+Be
S
(e
= n
(l+n
s
-l )
2
= E
(i + i-)
2
s)
The correction term in (9*6.18) is absent showing that if only the mean number of particles is
fixed, the dispersion is greater than if the total number is known exactly.
(c)
For FD statistics, the sums in
(2)
become
r
Z
-(cf+Be-JI
[1
=
E In (l + e
~
In
-(ostpeJ]
I
J L1 + e
+ e
=
-a-Qe
Thus
J
• * *
r
)
^
3 0€_
i-
=
a+Se.
+ 1
and
(An)
2
,
=
1
on
a+3e
e
s
P ci"
C£+3e
r
(e
= n
.
5
(1-n
s
82
-
n 2f 1_
s
0
n \2
+1)
s')
s
—
\\n
s
v
>
y
j
9.29
We can remove the restrictions on the sum over states by inserting the Kronecker 6 function
Z = Z exp [-p(n e +np e +. ..)!
1 1
2
= Z exp [-?(n e +n e +.. .) ]
1 1 2 p
B2
N,Z n
exp
f
[
(K-Z n^) (a+ia' ) ] da’
Letting a = a+ia' and interchanging the sum and
where the sum is over n^ = 0,1,... for each r.
integral yields
Z =
£(a) da*
fef
2(a) = Z exp [-(g+ge )n
where
1
(atf-£e
1 -
2
)n
2
. . . ]
This sum is easily evaluated as in section 9.6 and
r
In 2 (a) = -Z In (l-e
r
To approximate
(l)
(2)
)
we expand the logarithm of the integrand about a' = 0 where it contribute
appreciably to the integral.
In e^
2 (a) = aN + In 2 (a)
B (ia')
= (a+ia')N + In 2 (a) + B^ia’) +
^ 2
= ON + In£( a) + iCw+B^a’
8^ In 2
where
2(S) =e
a2I
£(a)
4B2a' 2
i(ir+B- )a l
L
As in section 6.8, we optimize the approximation by setting
=0
N + Bj
h +
= 0
43„q: t2
2
e^1 2 (a) = e 0111 2 (a) e
and (3) becomes
Substitution into (l) yields
—
e°®£(a)
i Ba
r _12
~
e
'
2
2
,
,
° -7T
-a-ge
In z ~ aW + In 2 (a) = cfi* - Z In (1-
where we hs.ve used
(2)
The value a is determined by (h)
83
-
|
B
g
a'
2
2
+ ...
+
...
IT
w
'
a+P€
r
e
)
'
Sa
r
E
which, gives
r
= IT - Z g— In (l-e
v
+ v- In t a
oa
r
= 0
n
1
-
CHAPTER 10
fstems of Interacting Particles
Since
cr(u)
and for spin waves m = Ak
du «
,
it follows that
to « o2 dm
cr(m)
Then hy (10.1.20)
P"
JF™
v2
w (pbm)
“
^ //
J
2
1
m2 dm
(e^-1)
°
Substituting the dimensionless variable x = phm we obtain
CO
e
_
x
x
J72 Jo (ex -1)
Hence
The integral is just a constant not involving g.
p“3/2 k rp3/2
cc
p OO
(a)
Ey
(
10 . 1 . 18 )
In Z = PNt]
In (l-e ^
lXD
(^(m) dm
)
-J
3V
0-2 3
where
“
2
for m <
2tt c
a
D (°)
=
for m >
But since
J
c
2
V = 6rr N
(
—
)
,
vs find
In Z = PNt} -
In (l-e
/
S 5m
‘
)
m
2
dm
0
“D
In terms of the dimensionless variables x = p-nm and y = pbrn^, this gives
In Z =
^ ^
-
3
f In (l-e"
x
2
)
x dx
o
y
This expression can be put in terms of the function L(y) by integration by parts.
84
^2=y
ry
"
-
^- e x )
111
pr
r
"
f
X
o
e -l
= y £“~ " 3H In (l-e y ) + HD(y)
/t
*"9
^
= gj- - 3N La (l-e
+ ND (Sj/r)
)
where k©_ = ttcu .
D
D
(b)
The mean energy is
S.-^laZ.-fc^lnZ
3”* <°p e
E =
-Ni)
l-e
=
(c)
-Ntj
r
+
y
D(y) =
+
y*”
-Nr)
y
3j**y
X
'J o
+ 3H
e -l
y^
have
S = Nk [-3 In (l-e
-y
)
-0
= Nk [-3 In (l-e
» 1, the upper limit in the function
\ey -:L
W D^yT)
(2)
The entropy may be found from the relation S = k[ln Z + pE],
For y
(_
Then from (l) and (2),
+ lt-D(y)]
T
n/
u
)
+ hD^^/T)]
may be replaced by «.
D(t])
Dw=%r#
y^ J o
e -1
k
The integral is shown to be 7T /15 in tables.
.
k
D(y) =
Hence
—y
5y
For y
« 1, we can expand
e
J
y
»1
~ 1+x in the denominator
D(y) =
5
y
= 1
f
y
«i
'-'o
From the results of problem 10.2, we find for y
l»z.g +
» 1, or T « 0^ when we use y =
3
58
d
85
= 0^/T
t—
k
=
,T
S = -Nr|
mr k
37r
.
+
3
s
For y
=_^m(i-)
« 1 (T » 0^) ve find, using ey « l + y
§
In Z =
3 =
-Ifr]
S =
m
^
+ N
3N In 9
-
+ 3H KT
+ 1+3
[-3 In
10 . +
1
The pressure is given in terms of the partition function hy the relation p =
^
In Z where In-
equation (l) of problem 10.2
-0 /t
In Z =
Thus
p , „
|
. 3S
r/
)
(e^T)
+ ND
aDtej/r)" "dfSj/r)'
(- + ^2) + »KP
S(eJ/r)_
U
V
—
dV
—
T
SDOSj/T)
iD ^ s
°
3H In (l-e
g i ft—
l-e
where
g
-
T
i^dx
(9^a7"
^
3
+
e
-1
¥ *v« i
d0^
5 -
-
In terms of y = -(V/0._) (d0.VdT) , this becomes
k
For T
« D D^T) -»y (|-)
S
,
p
,
»0
,
»
+
^
and
p
^ = 1N ia
cV
CV
For T
-
3
Sl; 7 N KT
5
?
V 0^
.
"3
DCSjy/T) -*
10.5
Comparing the expression for the energy E, equation (2) in problem 10.2, with equation (l) in
problem 10. 1+, we obtain
5 . „ §a * ,
^
86
“
Thus
--7 S> * K? TV
k
/Sp\
k(
/SE\
(
w> v
t
(a)
From electrostatics.
(h)
The induced dipole moment in terms of the polarizability a is
«
E
R
P 2 = OE « E
(c)
The energy of a dipole
in a field E is
u = -P *E = p E « R
2
2
By eauation (10.3.8) and (10-3-16), the log of the partition function is
+ N In V +
In z = In
where n = W/V.
|
Ital
(p)
The first two terms are the ideal gas partition functions while the last is the
Since the interaction is represented by an additive term in
contribution of the interaction.
In Z, it must also be represented by an additive term in the entropy, S = k
In Z
- p
le
^z
.
Thus the entropy per mole is
S(T,n) = S (T,n) + S'(T,n)
0
where
S'(T,n) = k
[§
n l(p) - P
(§
n 1(B))]
S'(T,n) = n A(t) - a [§ 3
and
= n
2
|
l(p) - p
^I
(p)]
^ (^)
To show that A(T) is negative, we must show that the derivative 0/0B [l(p)/s]
is positive.
We have
®
but
'
-Bue"
pu
-e'
«
S’
hi"
Pu+ =
1
l
e
pU
> 0
1
p
since (l+Bu) is just the first two terms in the expansion of e
.
and A(T) is negative.
A(T) =
-
I
J
jl -e
_Pu 1+pu]
87
(
2
hmR
dR
Thus the integral is positive
10.8
From the arguments of section 10.6 we can write down the partition function for a two dimensional
gas
N
i— (i!B)
2 - NJ < 2 >
h e
z
%
o
InZ^
where
= N In A +
1(3) =
Denoting the film pressure by p
,
(t)
PU
-l)27TR dR
2
N
A
W
or
2
(e'
^ 1(3)
we have
hT
where n = N/A and B
r
|
1 N
2 2
A
’ “ + B (T >
2
= I.
^
10.9
(a)
One can write down the mean energy from (10 • 7 • 3 ) and (10.7*5)
S = -Ng|i H.
^o i
= -NS kDc B (x)
S“
jz
s'
'
(1)
The consistency condition is
Here N is the total number of particles.
BsW
=
2^S
X
(2)
We solve this equation in the given temperature limits using Fig. (10. 7.1)
T
«T
Here the slope, kT/2n JS, of the straight line is small and the intersection with B (x)
g
is at large x where B (x)
s
= 1.
Thus
W^ x
,
1 =
and
E = -NS 3jT
(2) becomes
or
2n JS
x =
2n JS
kT
JE
= -2n NS2 J
(^i#)
hT (1)
a
(3)
T = T
In this region kT/2n JS is large and the intersection is at small x.
approximate B (x) by the linear term ^(s+l)x (eq.10.7.12
g
88
)
We cannot, however,
since this would not cross the
line (l£r/2n JS)x.
small x,
Hence ve must evaluate the second term in the expansion of B (x) .
,3
x
x
1
coth X-> — + -- T-T
.
.
X
3
89
^5
s
X
«1
.
.
,
For
CHAPTER II
Magnetism and Low Temperatures
11.1
The mean magnetic moment is
-PE
ZC-SEyBH)
r
M =
-£E
Z e
-PE
vhere Z = Z e
e
r
1
S In Z
P
PH
r
11.2
The change in entropy per -unit volume is given hy
°
“ -/
®
<tl>
J
T
o
But from the fundamental thermodynamic relation dE = TdS
-
MdH, we have the Maxwell relation
an
<t> T
vhere S and M
q
c
-
(?)
PT
H
are the entropy and magnetic moment per unit volume.
Since M
q
= XH and
X = A/(T-s) ve obtain
SM
,
^ST
AS = -f °
Thus
^o
Ov
~ ~
'
H
^
(T-e)
AH
\2
(T-0)
,
2
dH =
-
—
j
(T-s)~
-§2
11-3
The change in the molar volume when the field is increased from H=0 to H=E
H
Av = r ° (f|)
J o
But from dE = TdS
-
p dV
- McLH>
is
dH
T,p
ve see that
© _=-©
T,p
where v and M are the molar volume and magnetic moment.
ve find
S7
^ T,H
AS
o
a H
[T-S (l-K*p) ]‘
o
90
T,H
Since M = XH and X = a/[T-0 (1-H2?) ]
o
H
Av
and
Ada
o
o
--L'
o
[T-S
AS a
Hflff
E
o
(Ha p)]
2
[T-e
(l-iarp)]
2
2
o
11.4
(a)
From the relation dE' = TdS
pdV + HdM where E' = E
-
'
(ST
^V,T
In the normal state, M = VXH ~ 0 since X = 0.
^
or
— = V
M
- JE
Thus M is independent of temperature and
(SS/^iH)
we obtain
V,H
Thus (gS/oH)^.
B = E +
(b)
2
- vfl /8rr
^
« 0.
In the superconducting state,
= 0
= 0
v,T
Since the heat capacity is given by C = T(oS/cT) we have from (11.4.18)
s _ S
s
r
C
C -r
s
- m
n
8T
fa.
<S
_c
S
s
V
ffl
n
dT
.
1
n'
VTH /d^x
5tt 'dT'
47T
',
(i)
2'
cur
where H is the critical field.
(c)
When T = T , the critical field is 0.
Thus
c
7Tc
c _c
s
n
=
'dT'
4tt
H.5
From
(l)
in problem H.4 and E=E
c
[1 -(t/T c )~]
we obtain
VE
c -c
s
n
= -
3VE
T +
2nT.
c
jr-
T
3
27iT
3
But since C =aT + bT , we have
To satisfy the condition C /T -> 0 as T ->0we must set the coefficient of T in (l) equal to 0,
s
i.e., a = VE
2
c
/2tr t/.
91
(a)
Then since at T=T
The entropy is related to the heat capacity, dS = CdT/T.
c
and T=0 the
entropy of the normal and superconducting states is the same, we have
m
m
a c C ffl
a cCdT
s(T )-s ( „,
=
o
o
^r-/ -V
_/
7 dT =
a = 3 yh r
From which it follows that
(h)
Since dE = CdT, the energy at T=T
n
c)
= E (T
£
c)
T
p c
m
7T
E (T
n c)
‘
E
n^ = v o ^
dT =
E (T
-
E
3 dT =
s
But E (T
and T=0 for the normal and superconducting states is
c
c)
and by (l) a = 37A
s
=
(0)
f
Off
2
-f-
•
C
Eg (0) - E (0) =
n
-
From the relation dS = CdT/T we have for the entropy in the normal and superconducting states
T
c
S (T
n
s
s (
T
c)
c)
" S
- S
n
(0)
s(
=
0) =
7dT =
Jfo
T
p C
J
card's =
Since the entropy is the same for both states at T=0 and T=T
Hence at T=T
%
c
s
"
or
3
C
92
i
‘ 3
^c
ca*3
—
o
,
it follows that 0=37/T
.
CHATTER 12
Elementary Theory of Transport Processes
12.1
(a)
6
(1)
6
(c)
6
t
(*)
l
(c)
£
12.2
(a)
12.3
(a)
£ is
The equation of motion for an ion in a field
mx = e £
x =
Hence
—T
e£
By (12.1.10)
t
t
d
(h)
z
e t
x =
tV‘A 2E , 2 ,2
2
,J
-
X =
and
—T
—
and,
e£ T 2
m
The time to travel a distance x is
t =
-/§T
**
=
^
T
Thus the fraction of cases in which the ion travels a distance less than x is
pvr T
i
=
e
—
-t/r
dt
'
.
= 1-e
-y/T
12 .
The differential cross section is given "by
a(n) = -2*.
§
where ca is the element of solid angle and s is
the "inpact parameter" as shown.
From the sketch,
s
as
d£
= (a^ + a^) sin <p
(a
l
+ a
\
2)
cos
cp
d®
—
Since ® = 90-9/2 and da = 27T sin S' d9', it follows that
ds _
da
"(a^g)
sin g */2
% sin 9'
93
=
.
757
(a +a
I
Hence
sin 6 '/2 cos 9 '/
)
---
-
-
sva ?
:
i (v*2 >
By arguemnts similar to those of section 12.1, one has that the probability that a molecule
travels a distance x without collision is
x/C= pe
P _
where £ is the mean free path.
Letting P = .9 and -t =
(
V IT na)
yields
"
-In .9 = x 1/ 2 ' na = x ^2' pa/kT
mm Hg.
Letting x =20 cm ana T = 300°K gives p = l.lt X 10
The coefficient of viscosity is
l
s
n
3
where we have used -t =
(
n a
m
8 kT
v„t.^5-J2K
“
/
.
(!)
°o
To determine the cross section, we note that the volume
)
per molecule is A/N^p where A = 39-9 gms, the atomic weight, and p = 1.65 gm cm
_A_n
,3
^WT N, p
„
and
a
(
Q
1 ) we find rj = 1.1 X 19"^ gm cm
the observed value,
the density
Thus the radius of the molecule is
in the solid state.
From
,
71
= 2.27
= 5*6 x 10
=
1
^ cm2
Then the ratio of the calculated value of
sec
tj
to
X 10 ^ gm cm ^ sec \is
71
,/tj
'cal'
= .U8
‘obs
The coefficient of viscosity is proportional to T 2 since
t]
= mv/3
i/p'
Thus if the temperature increas s, the terminal velocity decreases.
a
Q
ana v = (SkT/W) 2 .
The velocity is independent
of pressure.
The force on the inner cylinder is the pressure p times the area A = 2m RL.
= pA = 2ttRL 7]
^
We make the approximation that the air at the outer cylinder is moving with it, i.e., at
velocity
m(R+S) = cnR, and the air at the inner cylinder is stationary.
Su
Thus
cdR
~
cir
and the torque is
G =
^R =
The viscosity is given "by rj = mv/3 VI" a
(h)
5
IP
.
o
Since air is mainly composed of nitrogen, we
.
2
let m = 28/6.02 X 10 ^, the mass of a nitrogen molecule.
we find
= 2 x 10
7]
and
Putting
ct
q
= lO
-1 '’
cm
2
and T = 300°K,
-4-1-1
gm cm
sec
G = 8 dyne cm
12.9
(a)
We let
a
where x,y, and z are to he determined.
«
o
Letting L = length, T = time, and M = mass, the
dimensional equation becomes
V
2
= L
L
S
But since C = FR , it has dimensions MT
t
2
Thus
L
which gives
x+(s+l)z = 2,
vVc Z
X
I^C
2
‘2 S
*^
s+1
L
x+(s+l) z -x-2Zj^t+z
= L
t
x+2z = 0, y + z = 0.
On solving this system of equations, we find x = 4/l-s; hence the cross section a
is proportional
-
tovVf 1 5 ).
{"0
)
The viscosity is given hy t) = nrv/3c/2" o
and here a
®
7^
the square root of the temperature, ve have
^ c
=
12.10
(a)
Consider a disk of radius r and width Ax
at the point x.
There is a pressure difference
2
Ap across the disk giving force 7rr Ap and a
viscous retarding force
27rr
Axq
^
•
Hence
the condition that the disk moves without
acceleration is
95
^+7)/2(s-l)
5
.
Since velocity depends on
2
But since Ap =
-2jtt Ax
Ap =
7ir
t]
Ax, where the minus sign is introduced because p(x) is a decreasing
-
function of x, we have
7TT
2 5p(x)
§-* u
—
=
2ir TT\
OX
Assuming that the fluid in contact with the walls is at rest, it follows
"that
§p(x
1
—
ox
du’ =
2t|
u <, r ' x ^)
P2 P 1
dp
,
„
,
For a liauid,3 w2 =
ox
L
.
.
-
—
dr' is pu(r , )27rr'dr
'
where
The mass flowing per second across a ring of radius r* and width
Then the total mass flow is
the density.
p is
M =
p^r^BHr'dr'
/
=
(h)
1 dp(x)1 / 2 "
r a )
5^ -it <
=
(P^g) I
r
5 5
(p i _p
(
a2 “ r
P
'
)
r,dr
2)
For an ideal gas, the density is p(x) = (p/RT)p(x) where p is the molecular weight.
Thus
the mass flow is
(r'
2
-a
2
)
r'dr'
-Sfe p W
=
Since M is independent of x, we obtain
.
M
rL
/
dx =
_ 71
S
which yields
7T
pa
k
5 t)BT
a
a
tiRT
L
'
/_
^
P1
"
cL\
P2
'
12.11
Let T(z,t) be the temperature at time t and position z.
thickness dz and area A.
By conservation of energy,
We consider a slab of substance of
(the heat absorbed per unit time by the slab
is equal to (the heat entering the slab per unit time through the surface at z) minus (the heat
leaving per unit time through the surface at z+dz) .
me dT/dt = pA dze dT/dt, where
c is
Since the heat absorbed per unit time is
the specific heat per unit mass and p is the density, we have
96
by (12.4.9)
=
k
——
3^
ff
The specific heat per molecule is c = (3/2)h while a
/2
is v = (83ST/7im)~
.
v
o
= 4ttR
Q
2
= 4-7r(lO
_s 2
)
cm
2
.
The mean velocity
Choosing an intermediate temperature T = 286°K, we find v = 1.2 X l£p cm/sec.
Substitution of these values into (l) yields
Q = 1.2 watts/cm.
(b)
The thermal conductivity remains approximately constant until the pressure is such that
the mean free path approaches the dimensions of the container, L = .6 cm.
At lower pressures,
the conductivity is roughly proportional to the number of molecules per unit volume n.
the heat influx is reduced by a factor of 10 when
= lO1^ molecules/cm^
n =
10a/2
From p = niff, we find
L c
q
p«3xl0mmHg.
97
Thus
12.14
(a.)
a /a
0-0
Kg/Vti —
(c)
D /D
(d)
a = —^—
1 =
2
2
1 =
1 2
(^/rjg) (ug/i^) /
2u1
/ti
t]
3^
— r
71
gives a
^
V
2
a = 7id“ gives
(e)
1h2
^ cm2
8-8
= 1.1 X 10
1
-
= 1.8 X 20
cm, d
a
,
= 2.8 X 10
2
3*0 X 10
2 =
^ cm2
.
cm.
12.15
Tlie
gases are uniformly mixed when the mean square of the displacement is on the order of the
We have, after N collisions,
square of the dimensions of the container.
1
1 F J
since the second term is zero hy statistical independence.
,2
Then
= V
C
*
z
2
z
2 2
2
t=-VT
2
3
2 2
2 2 2
= - v T n =
V rt
2
'
2
where IT = t/c.
.
-2
Since t = -t/v = l/i/2 n a v, one has on neglecting the distinction between v‘
2
and v ,
2
z
o
Substitution cf
z
=10
k
2
=
V = V| M
J^U
05
sM
cm , a = 10
-1=;
'
J
j
foxi
cp vir m
2
cm , and the given temperature and pressure yields
1
t as 10 sec.
12.16
A molecule undergoes a momentum change of 2(iV^ in a collision with the cube where
the reduced mass.
jj.
= raM/mtM,
Then the force is given by the momentum change per collision times the
number of molecules which strike the cube per unit time, summed over all velocities.
^
V2
(H
^2tt'
We may approximate the exponent since v
)
* rv /
» V. Thus
98
f<vv):
d
? v .-V) 2
-
(
l/2
/mBx 7
2
t
= 0
L
2tm
^
4.
and
Aot
?V
p®
P
w
Jo
:
V
-ffmv-
(l+gnrv V)
_
§E v 2
2
2
\ e
z
(l+gmv^V) dv^
The first integral is just the force on a stationary wall as in section 7-l^*
force is then
l/2
= 2pn
_
2
L
(g)
27T'
-
p®
BmV v, 3 _
/
gm v 2
2
z
lA
. I*n (g)
v
27r
The resistive
I- L2
gm
= pm v V L
hy Appendix A. 4 and 7-H-13*
» m, p » m, and the equation of motion "becomes
Since we expect M
V = V
—
dV
^ = -m v VL
M
exp
o
t i
2
n v l2 *1
The velocity is reduced to half its original value in time
o
M In 2
- t2
=
mnvl
CHAPTER 13
Transport Theory using the Relaxation-Time Approximation
13.1
By (13^.7)
°el
=
^J
Sg
1
v
ce
The value of the integral is unchanged if v
x
v
2
= v
x
2
+ v
y
2
+ v
2
2
2
T V
or v
2
z
is substituted for v
y
2
Hence since
.
z
,we obtain
’
z
e"
f
2 TV
—
d^V 3g
- ae
J
,3
el
/
-r
2
-
3
o
The quantities t and cg/Se are functions only of v.
2
Thus the substitution d J v = v dv sin
yields
2
a
,
el
=
— JqP
=—
3
/
p7T
2ir
dcp
*J
sin
in SdS
/
j'
x v^ dv =
-
e
2
J"
Q
99
^ r v^ dv
(1)
9
d9 dcp
13.2
The function g becomes the Maxwell velocity distribution
2
3/2 - 5I_
23£T /
- n
e
5e
1
-v
^2rr
2
In terms of v = (PkT/m)^ ,. this is
S
$£ = -stt" 3/2 S£
"(V*)
*5
v
Substitution into (l) of problem 13*1 with s = v/v yields
ne
a =
where
=
(t)
—
m
\
(t)
'
'a
,
ds e“
/
S
t(vs)
s'
^o
3v7r
13-3
2
2
= in (U
Since e = \ mU
^
x
c.
+ U
y
2
+ TJ
2
we have
z ),
)U
cte
x
TllUS
T1
1
= -m
—
T d3 U
u
U
u
z
a
J
x
I
2
2
= —-m
t —
u
U
X
u.
*
xu.
U
oU^
f u.d3TJu
J
s u
I
v— U
/
x
cte
Se
2
2
u
U
t
X
z
In spherical coordinates, the components of velocity became
U
x
U
cp
z
= U cos 0
dij = I^dU sin 0 d0 dcp
and
Therefore
=
tj
^7^
2
-in"
/
Jo
2
= -m tT
2
= -m
pir
But since
= U sin 0 cos
/
Jo
dcp
p 27T
/
r
cos“ ^ do
o
P
2
5
sin' © cos^od©
I
5
U
/
Jo
^o
[cos 0 sin 0 - cos 0
o1
f”
sm 0jd0
W
r
6
dg
% P dU^J
U
jQ
.
.
fl
/
TT
Sg
t
6
tt
U
t
.
sin 0 d0 =
t
(
it follows that
o
2
r
1
= -
P 7T
o 2?r
sin 0 d0
dcp
/
O
Jj o
100
6
m_
f dU || U x = - 15
1)
13A
Substituting the expression
V
|£ =
oe
from problem 13.2 where v = (2hr/m)
1//2
ti
ne
2
" (U/
^
-
into equation (l) of 13.3 yields
= nJdT <t)
r\
16
where
s
ds e
s^ t(vs).
U/v
15 ^TT
13-
Putting
t —
(
*72" n
o v)
q
^ into
2 2
and using v = (ShT/nm) / gives
= hkT t
rj
'fn
n =
T
“7
m KT
=
.
1*3
VnikT
o
0
This is larger by a factor of 1.18 than the mean free path calculation (12.3.18) and smaller
than the result of the rigorous calculation (14.8.33)
"by
a factor of 1.20.
13.6
We assume the local eouilibrium distribution
/gnu
3/2
exp
g = n(g^)
r
1 Dpmv2-,
]
(
where the temperature parameter 3 and the local density n are functions of z.
1)
Then in the
notation of section 13 .3 >
df(°)
dt
df^°)
dt
_
1
where the last follows since dz(t
'
( 13 - 3 * 5 )
) /dt
o"
o
.
z
Sn
St
Sg
-S
Sn
^z
Sg
Sn
Sn
5z
_
dz
Sg
^3
^o)
S3
St
Sg SB
dt
}z
Sg Ss
z
v
dz
^c)
dt
z
This excression is independent of f.
becomes
Af =
or
= v
Sg
Sn
_
f = g -
v
df
(o)
df
dt'
+
|s
S3
Using (l) this expression becomes
101
8z
(2)
Thus
1
n
z
Sn
Sz
3_ S§
mv
2B Sz
2
2
SB
Sz
It remains to determine Sn/Sz for which we use the condition
1
or
d v v
Z S v z -Jf„3
z
p d,3
J
n
z
2
1 Sn +
^ 3
Tg
/d3 -v f v z
= -
v
2
/
,
mv
S3
=0
SB
^
The first integral is zero since the integrand is an odd function of v
'3
d v v
2
g
z
_n
fl Sn
[n Sz
3_
[l Sn
3_
Sz
2B
[n
on
2
or v
x
2
d v v c
2g
SS
\2 B
m SB
w
W—
2 Sz
-
>
|SS
2
= v
+ v
x
+ v^.
2
,
in the
we have
t
r
,3
d v v s =
/
m SB
"
6
2
3
2
2
J
2
1,
-2^v
4
2
mv
2 )% z_
2
Then since v
.
'3
fz 1 _
By symmetry this equation also holds if v
where we have assumed t independent of v.
integrand is replaced hy v
Sn
.
I
^
dv = 0
substituting (l) and dJ v = v dv sin 9 d 9 do, and canceling common factors.
The integrals
are tabulated in appendix A.k
fl Sn
[n Sz
3_ Ml
2B SzJ
1^
r
8
2_v
5 /2
m S3
15
/- ,2_>
2 Sz
IS
Bur
Sn
1
=
Sz
3
S3
Sz
3m'
1
n
or
1
Substitution of the equation of state n = p/kT = pB yields
1
/'-n
(p
S3
+ ?
SF
dP
—
or
_ 1 S3
Sp>,
„
SI
S Sz
^
0
=
C.Z
The pressure is independent of z.
Then (3) and (r) give
2*
.
r
d3
.
f = s-sV
5
iv
" 2
SI |fe
1
T
1 mv
ffl
5
dz
[2 " 2
2
hT_
Since there are no external forces and f is independent of time, the Boltzmann equation becomes
Sf
I
*
Sv
= V
Sf =
z Sz
102
"
f-f^
0)
where f
where
^
= g is the local Maxwell distribution of the previous problem.
f^'« g we have
^
v-
Thus
8g Sn
,
.
of 13 . 6 .
( 2)
Sg
.(1)
56
^
[
dp
z
^
S^.
L& ^ ¥ s£J
f = g-v T
This is identical to equation
8g Sn +
v
= v
z ciz
Letting f = g+f
The rest of this calculation proceeds as in that
problem.
13.8
The heat flux in the
is given by
z -direction
Q„ = n
2
(§ mv ))
<v
= i m
= \ nm (v v
2.1 d^v— f v zv
f = g +
where
2
)
2
/
gv T
jz_
T
M
Sz
1 mv
5.
2 ~ 2 kT
The integral over g is 0 since the integrand is an odd function of v
„
Qz =
T ^
2
dependence occurs in d Jv = v dv sin 9 dS dcp which yields
is performed.
Thus we obtain
3/2|
o
_ ZT 51
^z " 3 T =z
r
r-
5 /
_ J o
[fW
'
{5-Bmv^j
z
We can exploit the symmetry of the integral by replacing v^
*3
22j2mv
r 3’
„20v
„ 20 e
d „
v v
—
^
1 5t
n
T
L
hit
^
1,
6
„~ " 2
dw e
Consequently
.
2
2
.
by v / 3 -
Then the only angular
wv2 dv when the integration over angles
^
1«2
2
-3m
8 • 2
/
aw e
Jo
rhe integrals are tabulated in appendix A.V.
3
/iiO\ 3/2|
T
ST
Sz
5
hk T
TT 13T
r 15 _ ,2 .T/2
5
-E»
[ -§ " %)
9/2
'
TT
(fj)
j
9
ST
Sz
m
5
Hence
k = o
2
2
nk T—
IT
m
T
13.9
The mean free path argument has
k =
— 0
1
v ncv
-c,
3
since £
VT and v = (8kT/7im)
1//2
.
2
k
nk T
7T
m
= -
T
Thus the constant relaxation time approximation yields
103
= I.96.
result greater than the mean free path argument by
13.10
Prom problems 13-4 and 13-8 we find
k
T]
where
c = (3/2)k,
2
(nk T/m)T = 5 k = 5
= (3/2)
n Iff t
2 m
3
The simple mean free path argument gives
the specific heat per molecule.
il
£
m
—
r|
Experiment shows that the ratio
(k/t)) (c/m)
£
h
lies between 1.3 and 2.5 (see discussion after
(12.4.13)) so that the constant relaxation time approximation yields better agreement with
experiment than does the mean free path argument.
I3
.H
In the absence of a temperature gradient, the equilibrium distribution of a Fermi -Dirac gas is
^
= e(c)
(27r)
1
(
P( £ “!1 ) + 1
e
3 3
fi
1)
In the given situation, f is independent of time but will depend on z because of the temperature
gradient oT/ciz along the z direction.
This temperature gradient alone would cause the electrons
to drift, but since no current can flow, there is a redistribution so as to set up a field <£(z)
to counter the effect of the temperature gradient and reduce the mean drift velocity v^ to zero.
Thus the Boltzmann eauation becomes
f
z
8f
e £ (z)
Si
m
f-f
Sf
5v
We will assume that the temperature gradient is sufficiently small so that f can be written
("0
-
(ll
-/
v
‘
t-f
= g+f'
', f'
Thus
'
« zg where f
(2 )
represents a small departure from local eauilibrium.
v
3g
z 3z
,
,(l)
e£ og
m ov
(2)
It is convenient to write this in terms of the dimensionless parameter
2
x = p(e-ji) = P(w mv -u)
Thus
8g _ 5g 5x _ 8g /
5x5z
5z
5x' e
8g
_ 5g
\
BSx
5x
srZ = ^ srZ =
'
5 b _ x 5g
^'Sz
T
104
z
If
(2 )
83
Sz
Then from
(2)
It remains to find
dB
dz
r r- X + H3„ e fce
|1
f = g - v t !§
z
dx
we find
'P
£ from the condition that v^ = i
J
The integral over g is 0 since the integrand is odd.
(4)
d 3 v f -v^ = 0 where f is given hy (L).
Letting t = t„ = const, for electrons near
r
the Fermi level, we have
= 0
or
r a 3v v 2 x
+ H
e
z
dx
J
d 3—
v v
—
P dz
2
(5)
dx
z
The second integral can he expressed in terms of n, the total number of electrons per unit
_
,
volume.
.
Since
dg
dg = 1
y—
^
pmv dv
dx
2
z
,
dv
—
pm
3 = 1
v 2 ttdg
'
z
dx
z
where we have integrated hy parts.
.
dv V
i
'
1
— Vz°g]r —
Bn,
dg
2— = i
pm
dv
nr
z z
-00
z
-
r
g dv
-oo
Since g=0 at ± °=, it follows that
2 dg
1
n
r 3
- —
dvv
/dvg
— z ^=-—
— & = pm
8m,
dx
'3
(6)
Proceeding to the evaluation of the first integral in (5), we note that x and dg/dx are functions
only of v
2
so that we can replace v
2
*5
2
hy v /3.
The substitution d°v = v dv sin 9 d 9 dtp yields
% v dv when the integration over angles is performed.
'3
2
2 dg
dg
%
d~V v
Thus
x v2 =
av v x ^
-
1
-=-
-
dx
z
3
.
2
Using x = p(— mv -p.) this becomes
iH t±)
3
5/2
2
dx x (x4i3n) 3 /
'pm'
(T)
From the properties of the Fermi distribution, we know that dg/dx = 0 except where
x = 0.
= 4 or
The integrand contributes appreciably to the integral only in this region so we can
expand the factor x(x+p p) 3^
may be replaced by
-co
2
about x = 0 to approximate the integral.
with negligible error.
.
2
x(x+p^i) 3 /
and.
e
« (Pii ) 3 /
2
2m3
(2rr)
3 3
ft
X
(
+l)
2
105
(x+
|
from (l)
Also, the lower limit -p p.
Hence (7) "becomes
ze { i.)
/2
3
r
j
dx x(x+p3
and I
= 7T /3 .
2
2 3/
u
^
and. 1
P»
^
3
X+
(e
£
-«(e +l)
l)
defined "by equation (9.16.9).
^
2 x
x e
X
2
3/2nl/2
2
37:
These integrals are the
2
«
i/2
|_i)
It is shown there that
1=0
Combining the results of (5), (6), and (8), we obtain
(2pm)
1/2
1
Be
Sz
3
ne £
_
m
and (^) becomes
f = g-v
x
t_
F dx
z
n h3 p
Having found the distribution f, we can now calculate the heat flux 0
^
Q
<v
= n
z
v
(v
'
where
2
r
I.
)
z
2
mv )^ =
2111
|
'
z (|
= —
n
^
V V^
2
2
d Jv v v
—
/
/
.
<
f
z
The integral over g is zero since the integrand is odd.
Hence
Qz = -
~f
±
§f
v^2 x|j -
d 3v
J
/d 3v v/v2
±
^
(10)
The calculation of these integrals is similar to the steps leading to (8).
p
replaced by v /3 and the integration over angles performed to yield
',3
-
The integral is I
+
d v v
Hence
The second integral in
(
/
J
‘
3H
2
2 2
v
B»
x
Sg =
S
3
2 2 Sg
v -r2 =
d v v
~
-
Z
Sx
9
% f dvv6 rSg2
—
8x
J
,
/
3
(5
3//2
jr
10 ) is
'3
/
<X
x2 ^
c
S
^
2
D
106
e
1//2
3//2
4
-
g
2
ft
3
x
x
(e +l)
/29 p)l 2 = 5
m'
— — -———
5*2
- -
5/2
x(x+gn)
r®
-1/2
5/2 5/2
0
2 '
n ' m '
”
where we have expanded as before.
f «b>
-
p
can be
The first integral in
(10) becomes
w r- -
Again v^
=:
-
n"‘
2
/6 sn.
u
vyva
(
5x
1+
2
3ti-
3tt
^(e X+l) 2
dx
Here we have neglected the squared term in the expansion since it is much less than 1.
integral is I
q +
(
5/2 Pu)
The
= 1.
I.
v
d,3.
v ,
Thus
v r-^= .
dz
z
m Obdi
O
Combining the results of (10), (ll), and (12) and using q =
q = _
^z
_Z i M
5
p dz
3
^
2
2 nT
^
3
dp
^3 dz
m
7T
nk.
3
m
O —3/2
/2m)(3rr n)
, we find
.
"3
2
Thus
(fi
_n_
2JL_ + ^2
71
2 2
2_2
m P
m p J
t£ nk^T_
m
F 1
7T
_
12 )
(
ph-
3?r
dT
T
F dz
2
^
13-12
Combining the results of the previous problem and (13.4.12), we find
"/‘.l =
At 2T3°K this becomes
k/o
^
= 6.68
X 10
^
K /°
el
1
watt ohm deg
X 1C)6
.
erg cm
(
2
deg ~ or in more standard units
Experimental values at 273°K are
Ag
Au
Cu
Pb
Pt
Sn
W
Zn
6.31
6.42
6.09
6.74
6.85
6.88
8.30
6.31
element
K / CT
X 10
7-42
e j_ =
f i )2 T
watt ohm deg
CHAPTER 14
Wear-Exact Formulation of Transport Theory
14.1
The mean rate of momentum loss per unit volume due to collisions is, by (14.4.22)
(Ap) =
/
/
v v
where we assume central forces.
d 3v d 3 v
/
dfl*
ff. Va (V,S’)( A?)
1)
We first perform the integration over angles by changing to
center of mass coordinates.
Thus
(
n
(Ap) = m (v'-v)
107
"becomes 7• since
v = —
c +
—
m—V and —v* = —c + —m V*
*=•
(ae) = n(v’-v)
where p =
j
m “being the mass of the ions and m^, the mass of the heavy molecules.
the mean values v
(Ap
z)
p(V
=
>
z
-
x
and v
are zero,7 it is clear that (An
'
y
’
= (Ap
y
= 0.
)
Hence we are left with
V ) and
z
a(V,9 ')(Ap)
dn'
V
x}
Since
is given in terns of V“
z
(2)
Using the coordinate system illustrated
and the scattering angles.
in Fig. 14.8.1, we see that
V
since V’ = V.
= V'cos(V',z) = V cos (V',z)
'
Inspection of Fig. 14.8.1 yields
V cos
= V cos(V, z ) cos S'+sin 9'cos
(V* ,z)
cos 0'+ V sin S' cos
= V
cp'
d(V 2 '-V z ) = pV z (cos 0'-l)+pV sin 9'cos
and
&
cp'
sin(V,z)
sin (V,z)
cp'
sin (V,z)
o 2m
Substituting into (2) we obtain, since
i
cos
/
= 0,
cp’dcp’
pm
>2n nir
p(V '-V }o(V,0')sin 9'd9'd<j>'=2mpV
/
z
u o
o
/
z'
(cos 9' -l)a(v,0')sin 0'd9'
For hard spheres, a is constant and
p7T
2mpV
where
a,,
z
c
/
(cos 9'-l)sin 9'd9' =
z
Z
>
=p a
t
Jj
V V ff
z
d°v d-v
(3)
x
—
3m
-
mS.3/2
where
.
3/2
f = n(g) '
e
"3 S
2
2
2
(
2
v +v +(v -u)
X
2
y
Here u is the mean z component of velocity of the ions.
= vf
2
)
To avoid carrying repititious constants
imjg
mg
cc
o
3
1
Hi
we let
z
Equation (l) then simplifies to
is the total cross section.
<AP
pV a = -pV a,.
-47T
J0
cs^
108
=
—
(M
» u, then
f can be simplified by making the approximation that v
„
n0
3/2
a^a n 3/2
nr
'
/Q!\
f =
2auv
2
-CCV
e^
3/2
x
z
*
v
_
e
n(^)
'
e
7T
2
^ (l+2a u vj
-av
Equation (3) becomes
-
(Ap
)
= nCTj.nn
—
(
2
-(ov -kz,v
2
)
d^
(1+22 u v )V V d^v
e
z
z
(5 )
77
We again express v and v^ in relative coordinates
1
—
——
v
— = —c + m+m^ V
—
av
rhus
2
2
+
x
= (a-ta,)c
2
—
v.
—l
2 (am -a m)
+
— c-v +
m+m.
= —
c -
—— V—
(am^-tcnm
2
)
v
On completing the square this yields
(am -a,m)
(aw^)
=
+
-
(af+a
1'
-
(m+m^)
(af+ct^) (m+m
2
2
+ a-jn
1
)
2
(m+m
1 )
a+a^
(am
m
= c
z
z
1)
(m+m^)
= K
z
z
(6)
-
- a m)
m
+
va
4
x V
(a-fa^) (rn-nn^)
z
m+m^
—
(cm,
V
m+m^
+
a^m)
-
1
- “ - + (ct+a
v
(am
2
+
iv
where
Also
2
aa-,
?
= (a+a, )
1
)
(am -a m)
1 1
- 7
—— V
—
z
a-,
=
K
+
z
a-K2
1
(7)
z
We wish to put the integral (5) in terns of V and the variable K defined by (6).
shown (14.2.6) that d^v^ d^v = d^£ d^V.
In terms of K we
a^a
-(Ap
z
)
nn
=
^2
2
pp
1
Thus
-(a-ta )K
6
1
(^ becomes
~a,+a
V
aa,
|l+2a u K +2u
JJ
7T
ha^v
eve
d\ = Jd 3K d^V
d3 c
It is easy to check from (6) that J=l.
It has been
z
—- V
a+a
V V d 3—
K d^V
—
.
1
zj
z
The first two terns in the brackets give integrations over odd functions and hence do not con-
tribute to the integral.
1.
-(Ap
z
)
=
/
cr
t
nn
1
Changing
to spherical coordinates gives
C
aa,
2
1
5/2
r
(on^)
-(a** ^k
-~Z
Q7+a
~~
j(a-KLj
%
4?7
"
iTdK
n
'1
„
V
2
^77
c
„5
V^dV
/
w o
rhe integrals are found in appendix A. 4- and we obtain
109
27T
cos 3 sin 0 d9
_
/»
x
2uu
pz
^
8
a
ua
3
t
5/2
(° p j)
ml
t
r
%^
(a-tc^)
m1
’
(a-ta^)
“ 3//
^
/^L\ %
LW /
3_
lm (i + i_)'
N
m-Hn,
m.
'm
7r
= - - ua x nn.
8
1
men
where p.
m+m.
The momentum balance ecuation is then
1
ne£ + (Ap^) = 0
which gives
u =
£
|
Ca.ii
Since
j
= neu = Cg^ £
,
V 2 u kT/m'
we have
£
ne
n CT
i t V2|i it'
3 -fn
T
el
14.2
(a)
We will assume the distribution function
f = f(°)(l-HJ>)
f (°) .
where
nW (SgM)
(
3/2
_
exn
^(z)
v2
1)
(2)
It is shown in problem 13-6 that the condition v^ = 0 yields the relation
1 Sn _ 1 83
n 5z
p
(3)
and that the pressure is independent of z.
(b)
The Boltzmann ecuation (14.7.12) becomes
„(o)
( °)
_ 8f
+ „
v
8f
dr
.
,
—
at
(o ^
m
8f
8v
(o)
=
Because the situation is time independent and there are no external forces, we are left with
3f
dz
or
/ */°) 8n
(°)
= V
z V
(|g If)
n
§z
f
(o)
f^S3 _ f (°)
+ 3
2 3
8z
2
3z
an-fWf^oJvo a»
\ (l -
(k)
on using (2), (3), and (14.7.10).
It remains to determine the function <J>.
The left side of (4) leads us to expect that a good
approximation will result if we choose
* = Av
z
(1-ay
no
)
0 mist satisfy the restrictions (it. 7-1*0 to he a good approximation.
where A and a are constants.
These become
r
J
_
\
,
d^v
~ f
/
v (l-av~) = 0
'
z
fdW°) - v z 1-csv^) = o
(
'
d^v
J
f^ v2v z (l-orv 2
= C
)
The first and third integrals are zero because the integrands are odd.
Similarly only the v^
component of v in the second relation will give a non-zero integral.
We use this condition to
determine a.
That is.
0 = A v (l-
and
^v
z
We will use the procedure of section it
2
(5)
)
to evaluate the constant A.
.
Multiplying (t) by 4> and
integrating over v yields
k lt£
d3X f<0)
V& f
-
V
2
2
)
. A
6?Zl
fj"
W
fV
2
)
*
(o)
r d^v
f' '
/
2
2Sm
z
2
= £r
pm
-
'Sm
The integral of the second term in
to he
7n/5f3ni.
+
(
V
z
?
/
+
—
/S2 \
(
pir
dcp/
i
V
)
V
z
5)
•
(7)
Z
p
n27r
#)
fv ]
We hav
2
- -
V
|V
(1-
7
(
The integral on the left is easily evaluated.
2
a[v
2
"cos a sin 9d9
7 ) vas evaluated previously.
/
n
m3'
2rr
3mv
3/2
v° e
dv
The third term is easily shown
Thus the integral becomes
(1-2 + J) = | SIpm
5 pm
5'
'
We split the right side of ( 6 ) into an integration over the angle
111
(8)'
and over v and v
.
We
define
a\ a\
SE_) j
X Xx
J =
Then using (8), equation
V a a|j
ffl'
J
(1-
z
^V
2
)
becomes
(6)
^ M)n
_ AI
1
"2 M
5z
(o)
(i 3-)- _EL_
(-l.
Sz
5
171
To evaluate J, we express A jv^l-av
2
in center of mass coordinates (we have reintroduced
)!
a = 3m/5).
A
Jv z
By (14.7.11),
(l-QV
2
)
=v z ’(l-av' 2
Using v = c + i V > v^ =
(l-ov
2
)
+ v. (l-av
_LZ
'j
-
c
C
1
+ V
)
'
lz
(l-OV 1
)=(c
+ wV
X z)
Z
=
1
{
|
V
-
-V (l-QV‘ )
lz
z
2
(l-ac^-ac'V
-
2a c
z
|v|
)
(l-QV 1
2
)
i V, we have
= 2c
Then c = c* and
£
,2
2
c
-
V )+(c
r
x
-
c V
^
z
2
z
2
-
hr (l-ac“+ac-V
zxz
-
)
-
a r~)
aV c»v
z
gives
2
A jv (l-av )l
= -a (V
Z
i
'
z
c-V'
V c«V)
-
z
Carrying out the product in (10) we find
J = -a
pcXT
nTT
sin S'de'cV
do'
Jo
•
(“
c
J
’
(V ’V ’-V V )+c (V
’V '-V V )+c (V
v
v
z
x
z
z
z x'
z y'
y z y
xv
where the coordinate system is shown in Fig. 14.8.1.
?
-V
z
)
'
We notice that, except for constants, the
The second integral involves only a change of the index x to y hut
first integral is (14.8.7).
is otherwise the same as the first.
Consequently, by (14.8.20), these terms yield
a a
|
2
where
P
W
Tj
z
(c
v
V
x x
+
cV)
yy'
is given by (14.8.21).
It remains to find
r^
r
-a /
On
dcp’
j
o
sin S'dS’a Vc (V
Z
Jn
1
Z
p
-V
Z
)
By (14.8.18)
V,/ -V ~ = V
z
cos 3'+2V V sin 9'cos 0'coscp'
z
112
(|_-z)+ V sia 0'cos q'(f -z)
T
dcp
t
'
(Y
2
z
-v
2
= 2n V
)
2
2
2
(cos 0'-l) + ttV
sinV(t-z) 2
To eliminate g, we write
= 1 = (Syzj
z_
*s2
+
‘
+ (£*z)
(ij*z)
,
/O a.x2
= l-(l-z)
(l*z)
and (13) becomes
dc?’(V
’
2
2
-V
)
= 7TV
2
2 ,
2
2
2
(2 cos 0 -sin 9'-2)+rrV sin S’
2
= (-37T V
+ 7TV
2
2
sin 0'
)
The last term in J is then
~
2
a o V cfv 1
2
\Z 3
T\
and (ll) ana (it) give
p
2
|cmV(cVV
+ c V
2
xxz + cVV
zz
y y z
-
v
c
r|
V
= I “
Cv Cs-v) - c
a
rT)
z 3 '
3
z
We now evaluate
=
a3 v d 3
r r
Z
f^) v^ci-ov2
^
j
)
It
The evaluation of this integral is straightforward though tedious.
d 3 v d3 v^ = d 3 c
d^
3
,(0)^ (o); = 2,m6\
n (— ) exp
= n
2 ,mS\ J
(—
-
exp
)
In the center of mass system
by (14.2.6)
pm
2\
2,
,
(v +v
„
(c
-6m
,
2
V„
c
± )
V \
+ -j7*j
<c7i
v„(l-av
2
)
= c^-csc^c
2
-
ac„(c-v)
-
“ cj2 +
|
V,
This expression is multiplied by J in the integrand
c
-
|
„(o)7
•
i
1_
and V; hence we need keep only even terms in v (l-av^)j.
l.jpc ad
2
"z
^
in this theory.
has the simple
(o)7
2
| y_ c-V - g V
-
*/
„
.
.
is clearly an even iimcn ion in
This yields explicitly
^zSWSVl
r^f^d a cv)
-c_
Since f ^°^f 1
2
four.
2
I- + ?o.Z c
3
3
V
+ 55 e s V*+ I
:
-
Zr 2 2
' -
V
l
(17)
(16) all these integrals are the standard ones encountered
We Guote the result of integrating over c.
113
In (l3) the terms appear in the
order that they appear in (17)
I =
2
a
»
31
2
(f)
'Hi
X
2
''pm'
(18)
2
SmV
W
d^V
2
-
VV
vv
2pm
z
2
-
z
2
— + |2L
y3 + 2_ y5 + « y3y
12
£ vv
5jS
2
- - V
v
r V^V
¥
z
2 z
v
z
3
The integral hecomes on substituting a = pm/5,
2
PmV
,„ T 2
Ta
d^V e
PmV
r
W -^vV.vr
+
g v5
2
•>7T
V5 a e
dV 2tt
2
1
mPV
+
cos 0 ^
60
7m6 ^2_ _2.
cos 9
sin 0 d0
.
I
,
2
PmV
~
-
<-
1 *> sy % 1
We let s =
^
=
\Vgmj
vhere a
dV
V and
!(-« -e)/-=\%
“ ' 102%
15
T
r
2s \
4s e_s2 a
Hrr
W,
11
7
n
(%
is given by (12.8.26) .
Substitution into (l8) yields
7^F
32
T -
n~
Z
25
and by
s
I
(9)
A = -
5TT ^
25
"
1//2
8P
/HOT\
sj
n
-
4°)
i . _JS_ (y)
_
l/2
v,(l-
¥ v2 II
8z
)
5
32 no
To find the coefficient of thermal conductivity, ve evaluate the heat flux
Qz = n
= hn
v (| mv
'
<
2
z
))
F d^v f^°^(l+<S>)v v2 =
1
P , 3 -(o)
'
;
= ^mA j d v f
.
1
= g111 A
.
_5
(mp)'
112
2 2,,
v^v(l
nS /
35
T d^v f^°^$ v v 2
—mP— v
2X
)
Q2 -
-
1/2
*!
M
« -=5- (JL)
m A
S
2
V
32 a
7
25 k_ hr kg
32 — Aj m
5g
n
Therefore the coefficient of thermal conductivity is
K
25 k
I
32 -
a/
tt
kg
m
7
(c)
By (it. 8. 32)
2
= r- a
a
3
1
Hence
o
75 k_ tt kg
St a y m
K
)
it.
-1
(a)
15
i/nmkg
l7Tkg
5
m
2t
y
L
a
TJ
=
J
k
TE
R
3-75
(h)
By (12. t. 2), the simple mean free path argument gives
- = - = 1.50 m
d
B
Hence the Boltzmann equation gives a value for
k/t\
greater than the mean free path calculation
hy a factor of 2.5
— X 10”^erggm' aqg 1
5
He
A
Calculated
1.55
.780
• OJ
Experimental
1.55
.785
.250
Xe
T)
t
It.t
Letting
2-i
pmv
2
Performing the integration over angles yields
2^2 r In n
1=
and by appendix A.t this is
115
Comparison with (9-10.10) shows that except for a constant, this is -S/k where S is the entropy
per unit volume of an ideal gas.
14-5
(a)
Differentiating the function
H =
3
J d v f In f
^ = J"d3 v || (in f +l)
yields
Sf/5t can he found from the Boltzmann equation.
(l)
We shall assume that there are no external
forces so that the distribution function should depend only on the velocity and time.
Then the
Boltzmann equation becomes
=//d3 v1
Substitution of
(2)
V0
dfi*
(f^f - fxf)
(2)
into (l) yields
ff =
fff
d3
® Va(f 1 'f -f^ (In f+1)
,
Ii
‘
(3)
)
This integral is invarient under the interchange of _v and v^ because a is invariant under this
interchange.
When the resulting expression is added to
f| = |
fff d3I d3l!
(3)
we have
Va(f 'f’-f f)(ln f^f+2)
1
1
clfl'
(4)
Similarly, this integral is invarient if the initial and final velocities are interchanged because
a is invarient under the inverse collision, i.e.,
a(v v -» v^'v’) = ff(v^'v’ -»v v)
x
Thus
|| =|
fff
1
V'a
^Zp'
But V=V» and by (14.2.8), d3 v‘ 3 v^ = d3 v d3 v
ft =
if// d3z d3ii
(f^-f^f ')(ln ^'f+2)
1
Va
(vfi
,f,)(ljl f ,f *
i
+ 2)
(5)
Adding (4) and (5) gives
ft
=
3
’
ifff d3 v d vi
dil*
Va
n6
(f^f-f^Hln f^f'-ln f^)
(6)
(b)
Letting
= y
and
= x, we see that since (in y - In x) (y-x)
in (6) is greater than zero unless f^'f * = f f.
> 0, the integrand
Hence dH/dt < 0 where the equality holds when
the integrand identically vanishes.
Ik.
The equilibrium condition is
C—
xx + Bymvy + B mv + d
2
In f = Af B iv
z
2
mv'
(1)
z
This can be written in the form
In f = -A'(v-v
+ In C*
o')
where the constants A’, C’ and the constant vector v
-A (v-v
)
'
Hence
contain the constants in (1).
f = C' e
v^ is clearly the mean velocity which can be set equal to zero without loss of generality.
determine A', we find the mean energy of a molecule.
.3
—
e
1
2_
d v r iv f
- 2
,,
47r
3
k
47T
d v f
r5//2
V7T A
(| m)
|
=
_
e
“2
/
/
1
v dVTir
rv
3
2a
dv e
2
e
-A'v
2
"A ' v2
m_
5 A’
R fZ/2
Letting e = (3/2)kT, we find A’ = m/23£T. The constant C* is easily found from n =
3/2
showing that f is the Maxwell distribution.
gives C’ = n( ^a-)
/
d3 v f which
27ThT
CHAPTSE 15
Irreversible Processes and Fluctuations
15-1
Let the friction force depend on a, v, and q in the manner
f -
aVV
Then the dimensional equation where L = length, T = time, and M = mass becomes
MLT'
2
-
1
(LftHD- )
Thus x+y-z = 1, z=l, -y-z = -2.
V-V
1 )*
-
Solution of this system of equations yieldx x=y=z=l, and
f = qav
U7
To
15.2
Since N k = R where
is Avogadro's number and R is the gas constant, we have from (15.6.11),
-2- t
2
(x > . JS- t -
37niaR
3TTTia
A
RT
N
A
or
Substituting R = 8.32 X 10
mole
ergs mole
^
2
37rna(x
)
°K ^ and the given values yields
2
= 7*0 X 10 ^ molecules
.
15.3
(a)
The energy of an ion in the field at point
^
(b)
Pe ^ C 2
nfz+dz'l
Hence
-
e
~
z is
^2
-
-
peT z
-
eZ z.
)
1 +
•
'
^*
2
(1)
The flux is given by
T
J
Then from
D "
D[n(z+dz) - n(z)]
„ 8n *
"
dz
8z
= -nD pe£
(l)
j
(c)
= nv = nu. £
d
(a)
J + J = -nDp e / + mi£
D
d
Thus
H
= £_
D
M?
=0
15 .k
S=
Substitution of v(t) = u(t)e
7
F*(t)
1)
into (l) gives
dt
= ± e rt F*(t)
'
'
m
u = u(o) + i
Hence
(
/
e7 °
F'Ct^dt’
Since u(o) = v(o) it follows that
= v eQ
Jrt
f
e
7t
F*(t-)dt«
(2 )
15-5
Ey dividing the time interval t into N successive intervals such that t = Nt, we can replace the
integral in
(2)
of 15.I by the sum of integrals over the intervals of length t between t^ and
118
Then t' = kT + s where k is an integer and dt' = ds.
-v e'^ = e"
Hence
7NT
T
^
e
-7T(H-k) 1
< t or e y
S
r
T
)
7S
e
m
k=0
In the integral, s
^ kT+S
m f
k=O
^o
°
=
^
N-l
(kT+s) ds
F*
pI(kT+s) ds
TT
< e' = 1 since t is chosen so that t
v-v e"
7T
=
°
« 7 “1
.
It follows that
G,
k
k=0
N-l
=
Z yk = y
k=0
where
Gj^.
=
F'(kT+s) ds.
15.7
We integrate the Langevin equation
= -yy + i F’(t)
^
dt
m
'
over a time t which is sma ll compared to 7
x
-1
'
large compared to the correlation time t
"but
the random force
of
^
v(t) - v(o) = -tv(o)t + i
T F'CtOdt’
(1)
^o
Then in the k
The time interval t is divided into N smaller intervals such that t = Nt.
subdivision t* = kT + s and dt' = ds.
Hence (l) becomes
v = (1-7t) v _ + ijT F*[(k-l)T+s] ds
k
k x
We can find the velocity at t = Nt by an iterative procedure.
Defining G
k
as in problem 15 .6 we
have
+ G
v = fr-rO
X
v
v
(1-7t)
2 =
G
1
[( 1-7 t)v
= (1 -7t)
= (1-7t)
1 +
q
2
v
q +
(
G ] + G
o + q
1
1-7t) G
q
+ G1
(
1-7t)
Continuing this way we see that
1
Since 7T
« 1, e
= (e
v = (1 -tt) * v + ( 1-7t)
q
N
N
N
= (1~7t)
.
N_1
G
o +
it follows that
119
N_2
G
1
+...+
G^
‘7T
\
or
<S
V N
''
'
+ e rr(l" 1> 2 *
-O
o
V'
'
**
*
¥ e-
71
%+•••+ =„-!
£
%
/T ^ N - k
G
)
.
k
k=0
since /t
« 1. This result agrees with 15.
5.
15-7
G =G =
k
N-l
Hence
Y -
J F’(k+s)) ds = 0
(
N-l
\
,
S yt . E
k=0
5 .
k=0
v - v e _rt = Y = 0
and
o
yt
v = v e
o
Then as t -»o, v -» 0.
15.8
From 15.5 ve have
-w
2
N-l
Y
=
Z
—w2
y
.
N-l N-l
+
k=0
It was shown in 15-7 that y = 0.
k
?
Thus
¥
.
e-
2
-1 = 27T by series expansion since
- t,
(v-v e 7
o
ri-e-1 ~
27t
7T
«
2
,.2
. y
v2 =
and
Y
=
^m
=
^m (l-e”
2
'
¥
e
27Tk
k=0
2 7T
„2
2rt
l-e"
27T
M = G2
m
2rfc
).
1
120
2 l-e-
To find G
. 0
7T
G
27T
2rt
Lr
1.
Letting t -»® yields, by equipartition.
Thus G
o
The sum is just the geometric series and may he easily evaluated.
/
2
y
*
l- e
2tt
k
"^) I5 • 7
e‘
where e
y.
* / j
k=0
Where we have used Nt = t.
S
Z
2rr
2
we note that
Since (AT)
2
= Y
2
-
2
2
= Y
Y
,
it follows immediately from the central limit theorem that
2
e'^)
(v-v
l/2
,11/2
r
f
]
T l1
exp p (v > t v ) =
0
2
—~
l
Jbtl ]
,
e
m(v-v
exp
-7t%2
J
o
-
2t&T( l-e"^)
2kT(l-e“
2rfc
)
15* ID
From problem 15.5 we have
=
G
k
1
,2
at*
2
m
F'Ckr+t’) at*
m/
J o
^o
(f* (k*r+t *)f* (kT+t") ) at"
/
^o
where we have changea the integration variable from s to t* to conform to the notation of the
text.
G
2
is inhepenhent of k allowing the choice of k=0 in (l)
g
2
= —
2
m
at*
/
<F'(t*) f* t") )
/
at"
^o
^ o
The integrana is the correlation function of F'vhich aepenhs only on the time aifference
s
= t"-t*.
Making this change of variable, we have
g
2
= —o
r
at*
(F*(t*)F’(t*+s))
f
Jo
where the last equality follows because (F'(t’)F'(t *+s))
0
as =
^ f at wF-t K(s)as
*
m'-'o
= (F'(o)F'(s))
*
o
= K(s).
Since K aepenhs
only on s, the integration over t' can be performea easily by interchanging the orher of inte-
gration anh reaaing off the limits from the figure.
*
2
g
=
O
^ ^ -T
I
m
as
I
pT
k(s) at
/
w -S
1
+
1
ds K(s)(t+s) +
Since t
» t* ana K(s)
-» 0
/
/
for s
T
as
P T-S
k(s) at
/
ds K(s)(t-s)
» t*,
s
can be neglectea camparea
to t ana the limits replacea by ± «.
G
2
=
^
/
K(s) as
121
By the results of 15 . 8 , G
2
= 213 rr/m.
It follows that
?
r
7 =
afe
K(s) ds
15.11
We integrate the Langevin equation
(a)
f « -TV-ifft)
over the interval t.
r
(
T)
F v dt'
dv = -7
F'(t')dt
/
Since t is
s mall ,
we can replace v hy the initial velocity v
in the first integral on the right
q
and obtain
Av = v-v^ = -7V t + i f F’(t')dt
q
= -/V T + G
(
o
and
Av
(Av)
2
= -7 v t
= (7 v t )
q
'
since
q
2
-
1)
G = C.
2
27 v q t G + G
» G2 = 2l3?7T/m
(2 )
where we have used the result of 15.8 and kept only terns to first order in t.
From (l), keeping terras to lowest order in t and noting that odd moments of G are 0 , we find
(h)
(Av) 3 = -37 v t
= G
(Av)
Since G
2
2
lj.
a t, G
'
should he proportional to t
= -
From (2), (Av)
.
Hence (Av)
3
and (Av)
4.
613?
7 t v /m.
15.12
From eq.
(2)
of 15 .h2
(v(o)v(t)) = (v (o)) e
=
^
F'(t’)dt^
e
2
_rt
(v (o)) e
+
^— (v(o)) f
'-I
^ F'(t')) dt'
7
o
But since F'(t') is random, the second term on the right is 0 .
122
(e
2
are proportional to r
OO
O
(c)
2
G
By equipartition
.
(v (o) } = 3£T/m
<v(o)v(t)> =
Thus
f e-^
and since (v(o)v(-t)) = (v(o)v(t))., we have for all t
S e vW
(v(o)v(t)) =
15-13
Squaring equation (2) of 15 * ^ yields
v2/vft]' = v
2 -2/t
\
and
o
2
+ 2v
e
v (t) = v
2 ~ 2yt
e
+
m
o
6
-
a
2
difference
*F (t ')dt
'
'
t +t
+
7 ^ t +t
'
e
~~
f dt '[
2
m w o
-
e
^F (t )F(t")dt"
,
,
'
")(F'(t )F (t n)> dt'
,
,
(1)
o
The correlation function in the integral is only a function of the time
s
= t"-t'.
f
dt'
Jo
^,
.
e7
dt 'f e 7 ^
f
w
in
since (F'(t')) = 0 .
nt
-2rt
ea
e
jfo
Making this change of variable yields
7(t’+t")
(f
'
t ')F (t
'
'
') >dt"
=
F
dt
that of problem 15 . 10 .
'
ds e 7 ^
^ -t
O
where K(s) = (F'(t ')f '(t '+s) ) = (F '(o)F
/^
'
2t,+S
)K(s)
’
The evaluation of this integral is similar to
(s) ) .
Thus interchanging the order of integration and reading the limits off
the diagram of problem 15.IO with t substituted for t yields
\t—t'
2t +S
ds e 7 ^
^K(s) =
J -t
_
dt'/
I
o
nO
ot
ds
f
2 t ,+s
+
)j^( s )
I"
d -s
-t
>
(
dt .-7
'e
/
o
ds e
vs
/S
»
re
K(s) 1
27t -
e
" 27S
'
1
M-
ds K(s)
(°
gn -l)
27
«
j
2rt
w 1 in the above and find
-i)
f «.*.)-*•iilil
/
!
(t)
= v
2
e'
2rt
27
+
r
t
J.
f
27m
ds K(s)
as t-s-co^ v («) = kr/m by eauipartition and
v (-) =
^ = -A-
/
ds
K (s)
27m
Hence
7 = PmVP
ds K(s)
123
^K(s)
t
2 Ht-s)_iS
ys
re
<ds e 7 K(s) 1
27
Substitution in (l) yields
v
2 ^ ,+S
7S
1, we can put e
r\as K(s) hfgni =
+
r
/
dt 'e 7 ^
Jo
O
-t
Since K(s) is only appreciable when 7s
ds T
as in (15.8.8)
,27t
27
K(s)
15.1^
(a)
We will denote the Fourier coefficient of F'(t)
F'(t) =
Substitution
f e^
0
F'(m) dm
and
"by F'(cjd)
v(t) =
f e^
and that of v (t) by v (m).
That
^ v(w) du
1
in the Langevin equation gives
iwv(m) = -yv(as) + — F'(m)
v <®) =
or
m( 7 iiu)
Denoting the spectral densities of F' and v hy J
and J
(as)
x
J
f
,(cd)
= | |F’(m)|
2
= m
J .(®) = m
p
(b)
2
2
-to
v (as)
)
^
)
Jv (co)
2
2
(-/
(os),
=| |m(r+4fl)) v(cd)|
2
-to
(7
2
V
|
2
|
^ e" 7,si
f K(s) e' ^ ds
1
v
(co)
=
|p
(-ixD+7 )s
;
T
Substitution of
2
K(s) = (v(o)v(s) ) =
Ey (15.10.9)
J
(c)
we have from (i5 .i5 .ll)
(b)
/
\
v'
'
1
1
(7-iu)
(-7-im)
_ KT
r
a
e
(-ios- 7 )s.
27
2 2
L 7 --to
.
into (2) gives
«*=.(“)
J
or
This result depends on co
KT
2nm
,
27im
ds +
« (t*)
f/
+
^(cd)
= J mKT 7
mfcT 7
= 2J (m) =
F
|
where t* is the correlation time of the fluctuation force,
since in deriving (3), one used a time scale such that
t
» T*
.
Consequently,
cdt
=1
» un*.