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CONI'ENTS
Preface •• . . . . . . . . . . . . . . . . . . .
CRA.PI'ER l: Introduction to Statistical Methods.
CRA.PI'ER 2:
CRA.PI'ER 3:
Statistical Description of Systems of Particles ••
Statistical Thermodynamics
CRA.PI'ER 4: Macroscopic Parameters and their Measurement
CRA.PI'ER 5: Simple Applications of Macroscopic Thermodynamics.
CRA.PI'ER 6: :Ba.sic Methods and Results of Statistical Mechanics.
CRA.PI'ER 7: Silllple Applications of Statistical Mechanics •••
CRA.PI'ER 8: Equilibrium between Phases and Chemical Species ••
CHAPrER 9: Quantum Statistics of Ideal Gases ••••••
CHAPrER l0: Systems of Interacting Particles •••••
CRAPI'ER 11: Magnetism and Low Temperatures ••
CHAPI'ER l2: Elementary Theory of Transport Processes.
CRAPI'ER 13: Transport Theory using the Rel.axation-Time Approximation
. . . . .
®.Pl'ER l4: Near-Exact Formulation of Transport Theory •••••
CRAPI'ER 15: Irreversible Processes and Fluctuations •••••••
i
1
ll
16
18
20
32
41
54
64
84
90
93
99
l07
117
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 tillle would
probab]_y approach them. Certain topics which have implications far beyond those called for in
the statement of the problems are not developed further here. The reader can refer to the
numerous treatments of these subjects. Except when new symbols are defined, the notation
conforms to that of the text.
It is a pleasure to thar,k Dr. Reif for the help and encouragement be 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 great]_y appreciate your calling errors or omissions to my attention.
R. F, Knacke
CHAP.PER 1
Introduction to Statistical Methods
1.1
There are 6-6-6 = 216 ways to roll three dice. The throws giving a sum less than or equal to 6
are
Throw 1,1,1 1,1,2 1,1,3 1,1,4 1,2,2 1,2,3
No. of
Permutatiom l 3 3 3 3 6
Since there are a total of 20 permutations, the probability is ~~G = -k
1.2
2,2,2
l
(a) Probability of obtaining one ace= (probability of an ace for one of the dice) x
1 1 5
ity that the other dice do not shov an ace) x (number of permutations)= (b)(l - '6')
5 5
= (5) = .4o2 •
(probabil-
6!
(5!1!)
(b) The probability of obtaining at least one ace is one minus the probability of obtaining
none, or
5 6
1 - (;-) = .667
(c) By the same reasoning as in (a) ve have
l 2 2 4 6!
(b) ('6') 5!2! = .04o
1.3
The probability of a particular sequence of digits such that five are greater than 5 and five are
1 5 1 5
less than 5 is (2) (2). Then multip:cying by the number of permutations gives the probability
irrespective of order.
1.4
(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.
N N.I l N
W (-2) = --------- ( -2)
C!) ! <J) !
(b) The drunk cannot return to the lamp post in an odd number of steps.
1.5
(b) (Probability of being shot on the N
th
trial)= (probability of surviving N-1 trials) x
th 5 N-l l
(probability of shooting onesel.f on the N trial)= (b) (b).
(c) 6
1.6
3 r cl r N m = m = O since these are odd moments. Using n = (p clp) (p+q)
and
4
m =
2 2 2 - ~2 m = (2n - N) = 4 n - 4nN + W--
we find m2 = N and ";f+ = 3lf - 2N.
1.7
The probability of n successes out of N trials is given by the sum
2 2
w(n) = E E
i=l j=l
'W
m
with the restriction that the sum is taken only over terms involving w1 n times.
2
Then W(n) = E w.
. i=l i
(vi-1~
2
) and
2
E v. If ve sum over all i ••• m, each sum contributes
m=l m
W' (n) = (w1 + w2l
N N! n N-n
W' (n) = E .....,...,,.._.__..,.. v w by the binomial theorem •
n=O n!(N-n)! 1 2
Applying the restriction that w
1
must occur n times we have
N! n N-n
W(n) = n!(N-n)! "W'l w2
'We consider the relative motion of the two drunks. With each sillr..i.ltaneous step, they have a
probability of 1/4 of decreasing their separation, 1/4 of increasing it,and 1/ 2 of maintaining
it O'J taking steps in the same direction. Let the number of times each case occurs be n
1
, n
2
,
and n
3
, respectively. steps is
2
The drunks meet if n
1
= n
2
• Then the probability that they meet after N steps irrespective of
the number of steps
vhere we have inserted a parameter x which cancels if ~=n
2
• We then perfonn the unrestricted
sum over n1 n2 n
3
and choose the term in which x cancels. By the binomial expansion,
1 1 1 N 1 2N 1/2 -1/ 2 2N
pr= (4 x + 4X + 2) = (2) (x + x )
Expansion yields
l 2N 2N 2N !
P' = (2) ~ n! 2N-
1/ 2 n -1/ 2 2U-n
! (x ) (x )
Since x must cancel we choose the terms where n = 2N-n or n = N
Thus
1.9
(a)
(b)
(c)
1n (1-p)N-n ~ -p (N-n) ~ -Np n << N ( )N-n -Np thus 1-p ~ e
N!
(N-n)! = N(N-1) ••• (N - n+l) ~ 1f if n << N
() N! n ( )N-n r n -Np "'n -A
W n = n!(N-n)! P 1-p ~ n! p e = n! e
1.10
(a) "'n -t. "' n! = e e = 1
0)
E
n=O
(b)
= n ->..
- " n >.. e ll = L., I
-" n. n=v
(c) n2 = n~ n~ n e ->-. = e ->-. (~) 2 L. ~~ = >-. 2 + >-.
-- -2 2 ...!;) (.6n) = n - n- = >-.
1.11
(a) The mean number of misprints per page is l.
>-.n ->-.
Thus W(n) = -, e
-l
e
n. = n!
and W(O) -1
= e = .37
3
2 -1
{b) P = 1 - 2: ~ = .08
n:::() n.
1.12
(a) Dividing the time interval t into small intervals ~t ve have again the binomial distribution
for n successes in N = t/~t trials
( ) N! n ( )N-n
W n = , (N ) , P 1-p n. -n •
-,.._n "-
In the lilllit ~t ~ o, W(n} ~ n! e as in problem 1.9 vhere )... = ~ , the mean number of disintegra-
tions in the interval of time.
(b) W(n)
1.13
4n -4 =-, e n.
n
W(n)
0 1 2
.019 .(176 .148
3 4 5 6 7 8
.203 .203 .158 .105 .()51 .003
2 2 We divide the plate into areas of size b. 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. Clearzy n << N
so we may use the Poisson distribution
6n -6
W(n) =, e n.
1..1.4
We use the Gaussian approDJDation to the binomial distribution.
W (n) = _ _;l.;;;..__
J;;¥f
(21.5-2COr
e - 2(400/4 = .013
J..1.5
The probability that a line is in use at e:n:y instant is 1/30. We vant N lines such that the
probability that N+l or more lines are occupied is less than .01 vhen there are 2000 trials
during the hour.
N 1
• 01 = l. - r. ___,;;;..__ e
n=O~ cr
vhere - 20CO n = 30 = 66.67,
4
The sum may be approxilllated by an integral
JN¥z° l - 2 2
.99 = -- exp [-(n-n) /2a ] dn
1 ,Ji;' a
J
(n-n+½;& 1 _ 2
12 .99 = - e y dy afier a change of' integration variable.
-1 ~ (-n-2)/cr
The lower limit may be replaced by -eo with negligible error and the integral f'ound f'rcan the
tables of' error functions.
We f'ind
- l N-n~ 3 = 2. 3
a
N = 66.67 + (2.33)(8.02) - .5 ::: 85
1.16
{a) The probability f'or N molecules in Vis given by the binomial distribution
Thus
(b)
(c)
(d)
1.17
If V << V,
0
N ! N-N
W(N) = N!(No-N)!
0
V N V o
(-) (1 - -)
Vo Vo
(N-i)2
Ff-
V
N=NP=N -
0 ov
0
V V ) N - (1 - -oV V
0 0
=
l
:::-
N
~o
i
Since O <<; <<land since N
0
is large, we may use a Gaussian distribution.
0
exp [-(N-N}2/2 d] dN
5
1.18
N N A
R = L. r. = tr_ ~
i -:1. i -:i.
(S. 's are unit vectors)
-:1.
R•R = .f..2 ! S
2 + .f..2 £ £ ~-• ~. = m,2 + .f..2 L. L. cos e.j
i i i I j -:i. -J i I j :i.
The second te:rm is O because the directions are randcm. Hence R2 = Nt2 •
1.19
N
The total volt age i s V = L. v. ; hence the mean square i s
i l.
2 2 where v . = vp and v . = v p.
J. J.
- N - N N
v2=L.v.2 +r. L.V. V.
i l. i f j l. J
Hence P = v2 /R = (ff/R ) i [l+(l-p) /:N'p]
1.20
(a) The antennas add in phase so that the total amplitude is E = NE, and since the intensity
t
is proportional to Et 2, It= ~I-
(b) To find the mean intensity, we must calculate the mean square amplitude. Since amplitudes
may be added vectorially
2 -- --2N2 --2
Et = ~·:!<!. = ~ L. §.i + ~ L. L. §_ . •Sj
-., -., i i / j 1 -
where the S . 's are unit vectors. The phases are random. so the second term is O and
-J.
1.21
Since a = O, the second term on the right is O and
ni
a n.
J
_.!. 1 _ .!.
(A 2)2 = N-2 2)2 .!. -~ (a = N2 lv a n n s
but
_ .!.
(An2)2 = As = Nas
:Equating these expressions we find N = 106•
6
1.22
(a)
(b)
Since
1.23
N
x=I'.s.,
• J..
i
N
but since s. = t, x = L t = Nt
l
i
2 N
2
N N
(x-x) = L, (s.-t) + L, L. (s,-t)(sj-t)
i J.. i J j ~
2 N 2 2
(s.-t) = t - t = O, the second term is O and (x-x) = L. cr = Ncr •
J..
i
N N
(a) The mean step length is t. x = I: s = V.-: Nt
i i i
(b) 2 N 2 N N --.- ----....
( x-x) = L, ( s . - t) + I: I: ( s . - t )( s . - t ) '
i J.. j J i J.. J
( s . -t) = t - t = 0
J..
To find the dispersion (si-t) 2
, we note that the probability that the step length is between
sands+ ds in the range t - b tot+ bis:~.
Hence
1.24
(a)
(b)
1.25
2 1
-fn.b (s. -t )2 ds. b2
( s -t) = J. J.. = -
i t-b 2b 3
2 N b2 Nb2
(x-x) = L, 3 = 3
i
w(e) de = de
2JT
w(e) de= 2JT a
2
si~ e de
4n a
sine de
2
(a) He find the probability that the proton is in e and e + de and thus the probability for the
resulting field.
From problem 1.24 w(e) de= sin/ de
Since b = L
3
(3 cos2e-1) we have !dbl = 6µ cos e sine
a de 83
and
7
(b) If the spin is anti-parallel to the field,
Then if either orientation is possible, we add the probabilities and renormalize
W(b)db =
a3 ,J;' db
12/µ2-µ a%'
( 83~ '
12Jµ2-µ a%
a3Ji db
12 Jµ2-i-µa%'
-2µ/a
+ a3 ,.;-:; ,} db
12 Jµ 2-i-µa %
(b)
-µ/a3
Loo iks b Im eiks
Q(k) = ds e W( s) = = ~ ds -
11 ~ s2+b2
2µ/a3
b
Q may be evaluated by contour integration. For k > O, the integral. is eval.uated on the path
:Fork< O
-kb
residue (ib) = ~ i
kb
-residue (-ib) = ~
8
Q(k) kb
= e
) -lklb Thus Q(k = e
1 L"° ikx N <?(x) = 2rr dk e - Q (k) 1 L0
-ikx Nkb 1 J"° -ikx -Nkb =2rr dke e . +2rr dke e
0
1 1"° -Nkb = - dk e COS k X
'IT 0
This integral may be found from tables.
C?{x) dx = !
7T'
Nb dx
2 _2.. 2
X + Irb
c,>(x) does not become Gaussian because the moments sn diverge.
1.27
From equation (1.10.6) and (1.10.8) we see that
where
We expand Q. (k).,
J.
thus fran (1),
1 L"° ikx (? (x) = 2rr dk e - Q1 (k) •• -~(k)
( ) - 1 2 2
Q. k = exp [i s.k - -
2
(6s. ) k ]
J. J. J.
~ 1L00
N_ lN 2 2 v-(x) = 2rr dk exp(i{L. s.-x)k - 2 L. (.6.s.) k]
i J. i J.
This integral -may be evaluated by the methods of section l.ll. We find
where
1.28
@(x) = &'
N
µ = L. s.
i J.
2 2 exp [-(x-µ) /20]
(1)
The probability that the total displacement lies in the range! to!+ dE after N steps is
@(r) d\: = J ~- J W(~1) W{2.2) ••• W(~) d\1
(d3r)
where the integral is evaluated with the restriction
N
r < L. s. < r + dr
- i=l -i - -
9
We remove the restriction on the limits by introducing a delta f'unction
In three dimensions
or l Leo 3 -i.k•r N (? (r) = --3 d k e - - Q (k)
(2rr) . - -
1.29
For displacements of uniform length but of random direction W(~) = 5(s-~), in spherical
lii7rs
coordinates..W(s) is a properly normalized probability density since
1 27T'J7T'Jc:o s2 sin e ds de dq> 5 (s-t) = 47rf c:o 5(4,ii--t) = 1
0 0 0 !i,,r s
2
O
We find the Fourier transform Q(!).
Q(~ =JJJ a3~ e~·~ w(~ =J2rr1Jc:o s
2 sin e ds de dcp 5(s-t) ei.ks cos e
0 0 0 47r s
2
_ ~17T' sin e de
- t:Cll 4rr
0
i.k-t cos e
e =
From
we have, for N = 3
(?(r) - _l_
- (2rr)3 1 2zr11c:o k2
0 0 0
The angular integrations are trivial and
sine dk de dcp e-i.kr cos e
To evaluate the integral, we use the identities
sin
3k-t Sin kr _ l i kr . k·D cl-COS 2 k-f~
2 - - s n sin :'v ~)
k k2 2
lO
=
1
2 [cos k{r-l)-cos k{r+t][1-cos 2kt]
4k
=
1
2 [3 cos k(r-t)-3 cos k(r+t) + cos k(r+-3-t)- cos k(r-3-t)]
8k
The integral of the first two terms in the last identity is
3 lco cos k(r-t)-cos k(r+t) dk = 3 1= sin2 !(r+t)-sin2 !(r-t) dk
E o k2 4 o k 2
On making the substitutions x = !<r+t) and y = ~(r-t), it follows that
3 lco I r+tj sin
2
x dx _ 3 J 00 I r-tl sin
2
y dy _ 3rr ( I r+tl _ I r-tl)
Ji: 0 2 x2 4 0 2 y2 - Io
The third and fourth tenns may be evaluated in the same way.
; Jco cos k{r+¼)-cos k(r-3-t) dk = -g,.. (lr-3tl-lr+3-tl)
0 0 ~ ~
Ccmbining these results, we have
L
2.2
G>{r) =
1
8rr .t,3
1 3 (3-l-r)
16ir-f;r
0
CHAPrER 2
O<r<t
t<r<3-t
r >3-l
Statistical Description 2! Systems~ Particles
X
p = ~ op =~oE
Hatched areas are accessible to the particle.
11
Since p/ + P/ = 2mE, the radius in
p space is .,;;;;;;- •
(a) The probability oi' displacement x is the probability that cp assumes the required value in
the expression x = A cos (wt+ cp).
-w-(cp)dcp = ~ .and P(x)dx = 2v(cp) dtp = ~
7I'
vhere the i'actor oi' 2 is introduced because tvo values oi' <p give the same x.
Since
ve have
(b) We take the ratio oi' the occupied volume, av, in phase space vb.en the oscillator is in the
p
range x to x + dx and E to E + 6E, to the total accessible
volume, v.
Since
ve have
Hence
For arr:, x, p ~ = 6E
m
.:here ve have used (1).
Thus
2.4
E - P2 + m ,ix2
- 2m 2 (J.)
Ae = 7r ~ r- = ~E , and v = 2:° 6E •
,./mw2
P(x)dx = 6v = 2 6:p 6x
V 2rr 6E
(a)
P(x)dx = ~ dx = ~ dx
7r p 7r / 2 2 2'
"V2mE - mW X
ov/2
From (1) and x = A cos (wt + cp), it i'ollows that E = ½ w2
mA.2 •
P(x)dx = ! dx
7I' '22
VA~-x~
(a) The number of wey-s that N spins can be arranged such that ~ are parallel. and n
2
are anti
parallel. to the fie1d is
and
J.2
Since ve have 1. ( E ) 1. E n.. = - N - - , n = - {N + -)
.l. 2 µH 2 2 2µH
Thus
N! 6E
n(E) = N E
<2 - 2µH)! (~ + 2~)!
2
µH
(b) Using Stirling's approximation, 1n n! = n 1n n-n, ve have
1n n(E) = - .±(N _ ~) 1n .±(1. _ l) _ .±(N + ~) 1n .±(1. + l) + 1n 6E
2 µH 2 NµII 2 µII 2 NµH 2µII
N! -
( c) We expand 1n f(~) = 1n ~ ! (N-~) ! about the maximum, n1
where
From (1.)
ftni_) is eval.uated by noticing that the integral. of f(nJ.) over all n1 must equal. the total. number
of permutations of spin directions,~-
-n1
is found :frcm the maximum condition
d 1n f(n1)
and
1. 1. -- ---..,
N-~-
~ = N/2
Substituting these expressions into (2), we have
1.3
Since
2.5
{a)
If cl. f is an exact differential,
O(E)
Since second deri-va.tives are equal,
and 1 E ) n.. = - (N - - ' J. 2 µH
2 E
2
6E
exp [ - N <2µH) J 2µH
{b) If 71. f is e:x:a.ct, lB -a. f = f(B) - f(A) = 0 if A= B
2.6
(a) not ex.act
o2f'
I o
2r
dX oy eydX
it f'ollws that
J,
2 J,2 2 l (b)
1
ay +
1
{x -2) dx = 3, J,
2 2 J,2 10 J,2 2 J,2 7
1
{x -l)dx +
1
2dx = 3 ,
1
(x -x)dx +
1
x dx = 3
(c) exact
(d) J, 2,2 [ ] (2,2)
71.f= x+Z =l
1,1 X (1,1)
for all :paths
{a) The particle in a state vith energy E does vork, 71.w, vhen the length of the box is changed
to L + dL • This vork is done at the expense of the energy, i.e., "d. W
X X
= -dE.
it follows that Fx = - ~~ •
X 2rr2~2
(b) The energy levels are given by E = -
m
Fx l
:P = L L = - 'E"'"'L
y z y z
2
p = 4ms.2 ( ~)
mV L 2
X
vhere V = L L L
X y Z
Since cl. W = F dL ,
X X
22 2
If Lx = ~ = Lz, by symmetry nx = ny = nz, and using the expression for Ewe have
14
2.8
~ ~ ~
2 l J p dv h pdv + ~ pdv =
C
2.9
From the definition of the Young's modulus, the force is
2.10
2.11
dF-YA dL
- L
F
1
2
L L 2 2
W = - F - dF = - (F - F )
F YA 2AY l 2
l
p = a v-5/ 3, a may be found by evaluating this expressionat a point on the curve. In units of
106 d:ynes/m2 and 103cm3, a= 32. For the adiabatic process,
J,
8
5/3 AE = ~ - EA = -
1
32V- <N = -36 =-36oo joules
Since energy is a state :function,
(a)
(b)
(c)
this is the energy cba.Dge for all paths.
8
W = 321 <N = 22,400 joules
Q = AE + W = -36oo + 22, 4CO = ll,8CO joules
31 P = 32 - 7 (V-1)
W = fi8
[ 32 - 3f (V-1)] = ll,550 joules
Q = ll,550 - 3600 = 7 , 95C joules
8
W = l (1) dV = 700 joules
Q = 700 - 36:::,0 = -29(X) joules
15
CHAPl'.ER 3
Statistical Thermodynamics
(a) In equilibrium., the densities on both sides of the box are the same. Thus, in the larger
volume, the mean number of Ne molecules is 750 and the mean number of' He molecules is 75.
(b) 3 lCXX) 100 l8
P = (4) Cf) = io- 5
3.2
From problem 2.4 (b)
{a) l E ) 1 E l E ) l E ) 1n O(E) = - -(N - - ln -(l - -) - -(N + - ln -( 1+-2 µH 2 NµH 2 µH 2 :;·;µH
where we have neglected the small term ln ~.
Consequently
(b)
~ = ~ ln O(E)
= ..1.... ln ½<1 - ~)
2µH l - ~(l - l)
2 NµR
E = -NµH tanb ~
T < 0 if E > 0
(c) M =µ(n1-n2) = µ(2n1 -N) where~ is the number of' spins alligned parallel to H.
Since
l >i'
~ = 2(N - ~) frcm problem 2.4 (a),
M = Nµ tanb
3.3
(a) for system A, we have f'rom 2.4 (c) ln O(E) =
.I:!!
kT
M = µ(N - ~ - N) µH
where we have neglected the term
In equilibrium, Hence E E1
µ~ = µ'~'
(b) Since energy is conserved, E + E' = bNµH + b'N'µH
Substituting from (a) we find E = µ~(bNµH + b'N'µ 'H)
µ~ + µ'~'
( 2 2
(c) Q = E - bNµH = NN'H b'µ'µ - bµ' µ) from (2)
µ~ + µ'2N'
(1)
ln oE
2µH
(2)
(d)
From (l)
P(E) a: n(E) n' (E -E) where E is the total. energy, E = bNµII + b 'N'µ 'H.
0 0 0
P(E) a: exp [_ f 2 ]
[ 2µ H N
This expression may be rewritten in terms of the most probable energy E, equation (2).
P(E)dE = C exp t(E~!1
2
] dE
where
The constant C is determined by the normalization requirement
co - 2 !-co C exp [- (::~) ] dE = 1
Thus P(E)d.E = - 1- exp [- (E-E)
2
] dE where cr is given above.
J;;'cr 2cr2 '
(e) (3)
(:f) Fram equation (3), N' >> N
E - µ N(bµHN + b'µ'N'H) -b' Nµ~
- µ~ + µ'~' - µ'
N' >> N
hence l\*EI = _µ_.a;;;.~-:F-N-
3.4
The entropy change of the heat reservoir is 68 1 = - ~' and for the entropy change of the entire
system,
Hence
3.5
Q
AS + AS' = AS - T' ~ o., by the second law.
/\es~ Q
'->,;) -v
(a) In section 2.5 it is shown tba.t n(E)a: yN'X(E), where Vis the volume and X (E) is independent
of volume. Then for two non-interacting species with total energy E
0
N +N
n(E) = C nl(E) n2(Eo-E) = CV l 2 Xi(E) x2(E)
17
2 atmospheres.
4.1
(a)
Since
~ =me
CRAPI'ER 4
Macroscopic Parameters and their Measurement
1373
273
dT = 4180 l.n fil = 1310 jou1es;°K
T 273
~s = -~
DB = - ~ = - ~ (373-273) = -1120 jou1es;°K
res T . T
D.stot = ~ + D.sres = 190 jou1es/K
(b) The heat given off by either reservoir is Q = mc(50) = 2.09 X 105 joules. Since the
entropy cha.llge of the water is the same as in (a) ve have
(c) There vill be no entropy cha.llge if' the system is brought to its final temperature by inter
action vith a succession of heat reservoirs differing ini'initesima~ in temperature.
4.2
(a) The final. temperature is found by relating the heats, Q, exchanged in the system. The heat
required for melting the ice is mi.f, vbere mi is the mass of the ice. Since for the other compo
nents of the system, Q = mct.T, where m is mass and c is the specific beat, we have
mi.i + Q(melted ice) + Q(original water) + Q(cal.orimeter) = O
(30)(333) + (30)(4.18)(Tf-273)+(200)(4.18)(Tf-293)+(750)(.418)(Tf-293) = 0
Tf = 283°K
m/
(b) The change in entropy when the ice melts is T• For the other components
T
1 f dT Tf
DB = me T = me l.n T
Ti 1
- m/ g§3_ g§3_ ~ A,
Hence DB - 273 + mwc l.n 293 + mic l.n 273 + mccc l.n 293 = 1 •6 jouJ.es, r-..
18
(c) The vork required is the 8l!IOUD.t of heat it vould. take to raise the water to this temperature.
W = lllr/\v (Tf-Ti) + mccc (Tf-Ti)
= [(750)(.418) + (230)(4.18)](293-283) = 12,750 joules
EQ=cd.T+pcIV by the first lav
RI'
= c dT + V dV Since pV = RI'
l::,S = 1 f cl Q = 1·Tf c dT + r V f RdV
i T T T Jv V
i 1
Tf Vf
= c ln - + R ln - independent of process
Ti Vi
4.4
The spins becane rand~ oriented in the limit of high temperatures. Since each spin bas 2
available states, for N spins ve have
S(T -+ m) = k ln n = k ln ~ = Nk 1n 2 (1)
But since S(T)-s(o) =JT Q!tl. dT
o T
ve have
T r 1
2T dT S(T -+m) = JJ c1(T - 1) T = c1 (l-ln 2)
ffe'1 1
(2)
where we have set s(o) = 0 by the third law. From (1) and (2) Nk 1n 2 = c
1
(1-ln 2)
Nk 1n 2
cl = 1 - 1n 2 = 2 • 27 Nk
4.5
JT£.CT1
The entropy at temperature Tis found by evaluating S(T) - s(o) = T- d.T. Labeling the
0
undiluted and diluted systems by u and d, ve have
T
1 l T dT
su(T1) - s (o) = c (2 - - 1) - = c (1-ln 2)
u ffe'l 1 Tl T 1
T r 2 T dT
sd(Tl) - sd(o) = J1 c2 TT= c2
i'r1 2
sio) = Su(o) by the third lav. Since there are oncy- io as maey magnetic atoms in the diluted
case as in the other, we must have Sd(T1 ) = .7 8tJ(T1 ). Hence
19
CHAPl'ER 5
Simple Applications of Macroscopic Thermodynamics
5.1
(a) pV7 = constant for an adiabatic quasistatic process. substituting pV = v Rr ve have
V l-7
( ...!. ) or Tf = Ti vi
(b) The entropy change of an ideal gas in a process taking it from T1,vi to Tf,Vf vas found in
problem 4.3
t:,S = O for an adiabatic quasistatic process.
5.2
and since 1n 1 = O, it follows that
V l-7
( ...!.. ) Ti' = Ti V 1
(a) W = J p a;, = area enclosed by curve = 314 joules
(b) Since the energy of an ideal gas is only a function of the temperature and since pV = Rr,
3 PcVc PAVA
AE = ~ 6T = 2 R ( ~ - -R-)
= ¾ (6-2) x 1a9 ergs= 6co joules
( c) Q = AE + W vhere AE is calculated in (b) since the energy change is independent of path.
Again the vork is the area under the curve.
Q = 600 + [ 4CO +
1
~ ] = 1157 joules
5.3
(a)
(b) W = J p a!-/ = area under curve = 1300 joules
(c) To find the heat, Q, ve first find the energy change in going from A to c. This, of
course, is independent of the path.
t:.E = c.p = i R(Tc-TA) = i (PcVc - PAVA) = 1500 joules
20
Then since Q = .6E + W
Q = 1500 + 1300 = 2800 joules
(d) From problem 4.3
= ~ R 1n l2 X lO~/R + R 1n 3 X ;a3 = 2.84 R = 23.6 joules/°K
6 X 10 /R 1
5.4
(a} The total system, i.e., water and gases, does not exchange energy with the environment, so
its energy remains the same af'ter the expansion. Since the water cannot do vork, its energy
change is given by the heat it absorbs, 6E = Q = G,_,fT, where ~ is the heat capacity. The
energy of the two ideal gases is, of course, dependent o~ on temperature, .6E = C6T, where C is
the heat capacity.
6E(vater) + 6E(He) + 6E(A) = O
'1l(Tf-Ti) + CHe(Tf-Ti) + CA(Tf-Ti) = O
Thus Tf = Ti, no temperature change.
(b) Let -l be the distance from the left and v the number of moles.
Renee, in equilibrium -l = 3(80-f..), -l = 60 cm.
(c)
vr
Since there is no temperature change, the entropy change is vR 1n V..
J.
R 20 6o p
~ = ~A + ~He = 3 1n 50 + R 1n 30 = 3.24 joules; K.
5.5
(a) The temperature decreases. The gas does work at the expense of its internal energy.
(b) The entropy of the gas increases since the process is irreversible.
(c) The system is isolated, Q = O, and from the first law
6E = -W = - T (V f-V o)
The internal energy of the gas is o~ a i'unction of temperature, .6E = vcv(Tf-T
0
)
Thus vey(Tf-To) = - T (vf-v o)
In equilibrium p = T and since pV f = vRrf , ve have
21
(1)
Substitution in (1) yields
The equation of motion is mx=pA-mg-pA
0
Since the process is approximately adiabatic
and
pV7 =constant= (p
0
+ r) V
0
7
AV 7
mx = (p + !!!:8.) - 0
- - mg - p A
o A (Ax)7 o
(1)
(2)
The displacements from the equilibrium position,:, are small so ve can introduce the coor-
V V
dinate cbange x= Ao + 'I} and expand about : •
7
!_ = -""""'1~- = (~) (1 - ill +
xr V r V
0
V0
(:+'I})
... )
We keep onzy the first and second terms and substitute into (2).
•• 2
m11 = -(p + 5) U 'I} o A V
0
This is the equation for harmonic motion and the frequency may be determined by inspection.
Hence
5.7
½
V = !_ [ (p + !!!:8.) A 2z]
21T o A V
0
m
The volume element of atmosphere shown must be in
equilibrium under the forces (pressure) x (area) and
gravity. Then if n is the number of particles per unit
volume, m the mass per particle, and g the acceleration
of gravity,
p(z+dz)A-p(z)A = -n(Adz)mg
~ dz = - ~ dz, where
dz NA
22
p(z+dz)A
z
p(z)A
Since p = nkT,
(b) pV7 = constant. Substit uting V
Thus
(c) From (1) and (2)
vRT
=-
p '
we obtain
3.E_ _ ...L dT
p - r -1 T
dT _ .Ll:::zl
dz - rR µg
dT
For N2, µ = 28 and letting , = 1.4, dz= -9.4 degrees/kilometer.
(d) Integrating (l) with constant T we have
p = p
0
exp [-µgz/RT]
{e) From (3) T = T
0
- (; ;l) µgz
Thus
5.8
1P~ =
p'
Po
AP
f z __ -__.µc..g...,_d_z_'..---..,....-
0 R(T - Ii:ll ~ )
o r R
I A{P+6P
I
(1)
(2)
(3)
Consider the element shown. 6x is its initial width.
to move a distance s and the right a distance g-ti:i.s.
The wave disturbance causes the left face
Let P and p be tb.e equilibr i um pressure
0
and density, and p = P-P is the departure from the equilibrium pressure .
0
= J. 6. V J. At:.5
tts - v P:P- = - Ex P
0
or
23
vhere we have neglected the pressure difference D.p across the slab. The displacement is found
from the equation of motion.
025
m 2 = PA - (P+l:i.P) A = -b.PA
ot
Since
oP
t::,.p "' dX L:ix and m = p A/::,x
we have cP cp _ P o2g
di= di= ot2
Then
but since the order of differentiation is immaterial,
5.9
i l i ~=-~
ot2 KsP ox2
(a) For an adiabatic process pV7 = const.
oV
Hence dp =
(b)
Substituting
(c)
(d)
5.10
V
,P and 1 oV
Ks = - v ®
1
U = {p KS) ~ = ( 1) 1
vRl' - ~
P = V and P - V , we find
1
u = (zI?-)2
1
u c: T2 , independent of pressure
u = 354 meters/ sec.
Substituting the given values into the equation
we :find
5.ll
vr2a2
cp - cV = -K-
-1 -1 Cy= 24.6 joules mole deg 1 1 = l.14
When the solid, of dimensions x:yz, expands an increment iN, ve have
24
v+av = (x+dx)(y+ay)(z+dz)
To first order
V +av = V + yzdx + xzdy + :xydz
Therefore
For the adiabatic, quasistatic process
dS = (~ ) dT + (~!) dp = 0
p T
( ~ \ -- ~T cc~\ _/ov_\ Using cri) p and the Maxwell relation dl') T = \"2ff) p' ve have
Thus
5.13
o.T
D.T = - D.1' pc ~
p
From the first law cp = T (~)P
l (''v_\ vhere a = V %:r)
p
C
where C = ...E_
p pV
W/2=•a,;/2~~\ =·C~\C:/2
Substituting the Maxwell relation (~)T = {i)p and the definition a= - ¼(t)p we have
~~~)T = -T (~) P rrT = a
2
vr - vr :
5.14
(a) TdS = dE - FdL
(b) From (1) we may read off the Maxwell relation
Since
(c)
(~)T = - (l)L
F = aT2
(L-L)
0
(~)T = -2aT (L-L0 )
=1 T r
bT dT' = b(T-T
0
)
T'
To
25
(1)
(2)
(3)
S(L,T) - S(L0 ,T) = I'r,L (~)T dL = I'r,L -2aT(L'-L
0
)dL' = -aT(L-L
0
)2
0 0
Hence S (L,T) = S (L ,T ) + b(T-T ) - aT(L-L )2
0 0 0 0
(d) In the quasistatic adiabatic process DB = 0
S(T ,L) + b(Tf-T) - aTi(L~-L )2 = S(T ,L) + b(Ti-T) -aT (L -L )2
0 0 0 ~ 0 0 0 O f f 0
Hence
b-a(L -L )2
T = T i o
f i b-a(L -L )2
f 0
Thus if L? Live bave Tf > Ti, i.e. the temperature increases.
(e) Fram the first law
Substituting (2) and (3) ve find
~~~T = -2aT (L-L0 )
Hence CL(L,T) = C(L
0
,T) + I'r,L ~~~)T d[, = bT - aT(L-L
0
)
0
(f) S(L,T )-S(L ,T ) = fL (~) dL' = fL --2a To{L'-Lo) dL' = -aTo(L-Lo)2
o o o JL T JL
0 0 2
1T C dT 1 1T bT' - aT 1 (L-L )
S(L,T) - S(L,T ) = ~ = o dT'
o T T T T1
0 0
= b(T-T) -a(L-L )2 (T-T)
0 0 0
'!'bus S(L,T) = S(T ,L) + b{T-T) - eT(L-L )2
0 0 0 0
5.15
(a)
(b) From (1)
Hence
cf. Q = TdS = dE - 2cr t dx
dE 2ot
dS = T - T dx
(o~\ dx + cos\ d.T = ! co~\ dx + ! co~\ <fr - 2ot ax
di)T ~/x: T ax)T T af)x T
Equating coefficients, we have
26
(1)
Fran (1), ve ca.n read off the Maxwell relation
Since cr = cr -ctr,
0
If' the :film is stretched at constant temperature
(c)
By the first law
E(x) - E{o) = 2-to- X
0
W (0 --t x) = - J F dx = - Jx 2ot dx' = -2otx
0
dE zfY dv
dS = T + T
Hence (i)T dv + (l)v dT = ¥ (i)T dv + ~ ( ~),, dT + zf;{' dv
Equating c oefficients, we have
From (1), we mey- read off the Maxwell relation
Thus
If' one mole is produced
5.17
By equation (5.8.12)
substituting p = n kT (1 + B2 (T) n] we find
~ - p + n~ dl32 - p = n~ dl32 > 0
\oV)T - dT dT
27
(1)
5 .18
(a)
Since
ve have
(b) From the first law, dE = Td.S - pd.V. Hence
:f'rc:m (5.8.12)
0 = T cos~ -p and (~\ = ~
. av)E oV)E T
( c) For a van der Waals gas
2 vRI' v a
P=~- v2
Thus
5.19
(a) From (p + 't)<v-b) = RT
we have (
rlT>~' - -RTc ~ + 2a3 = O
CV T (v -b)2 V
C C
2 D 2RI' 6a J c --,;:=O
2 = (v -b) 3 V
T C C
9 VC
Solving, we find a = 8" RI'cvc, b = 3 ·
(b)
(c)
Rr
Substituting a and bin (1) yie1ds pc= 3. _£ ff VC •
(p' i :,2) (v' -½) = ~ T'
To find the inversion curve, ve must have
(l)H = ~p (~ <I) p - 1) = 0
28
(1)
or
From the van der Waal.s equation
dp = RdT + (- R:r _ 28: '\ dv
v-b (v-b)2 ,.3)
thus
v-b V
T
: (v;b)2 = b
On i:>Jiminatillg v and puttillg the equation 1n tems of the dimensionless variables of problem
5.19, it follows that
5.21
(a)
We have
T'
6-3/4
1 (ovD d ..s d -a a - - ""°" - - a' - , - V o-8 dT - dT
p
µ = (~)H : = µ' :
Substituting these expressions into ( l) ve find
29
(9, 3)
p'
(1)
(2)
(b) Fran (2)
5.22
1T~
T T
0
= r ,s a'diS'
J •' µ'C r 13
0 l + ~
V
T = TO exp J ,s µ, C ,
[
r,s a'd,S J
0 l + :____,J2_
V
The system may be represented by the diagram
w
(a) By the second law
41 42 .6S=T - T~o
For max:IJmJm heat, the equality holds and since '½ = q1 ..Jfl, it follows that
(b)
5.23
We have the system
(a)
(b) "3'J the second law,
Thus
It follows that
l
41 Ti
w=Ti-To
ql
V = 11.9
(c) The maximum amount of vork vill be obtained vben Tf = JT
1
T
2
' •
Fram (1)
30
(1)
To freeze an additional mass m of water at T0 , heat mL must be removed from the ice-water mixture
resulting in an entropy change 001 = - ~. The heat rejected to the body of heat capacity C
-o { Tf dT Tf
increases its temperature to Tf With an entropy change .6.82 =C IT T = C 1n T. By the
mL Tf V O 0
second law 001 + t::.S2 = - T + C 1n T ~ O. For min:ilnum temperature increase and thus minimum
0 0
heat rejection the equality holds and it follm.s that
The heat rejected is Q = C(Tf-T
0
)
5.25
The probability that the system occupies a state With energy Eis proportional to the number of
states at that energy, n(E). By the general definition of entropy, S = k ln n(E), we have for the
probability
P ex: n(E) = e
8/k (1)
To raise the weight a height x, an amount of energy, mgx, in the form of heat, Q, must be given
off by the water. The temperature change is found from
Q = C'AT = -mgx
t:il: = -mgx/c
vhere C is the heat capacity of 'Water. For a mole of liquid and x = l cm, t:iJ: is on the order of
10-4 °K. Hence we may approximate the water as a heat reservoir and find for the entropy vhen
the weight is at height x
S=S _2:.._=S
o T o
0
vhere S is the entropy at height x = O. Thus (1) becomes
0
-mgxjkl'
P(x) = Ae 0
sjk
Here we have absorbed the te:rni e into the constant A wbich is found to be kTjmg by the
usual nomalization condition. Then the probability that the veight is raised to a height Lor
more is
1
co
1
co kT -mgxjkr -mgLjkr
(? = P(x)dx = -2 e O dx = e 0
L L mg
Substituting L = l cm and the given values yields
C? = e-1018
31
In the procesS:S a -+ b and c -+ d no heat is absorbed, so by the f'irst law 'W = -6E, and since
~ = mT, ·whereC is the heat capacity, we have
wa"'b = -(¾-Ea) = -vc;, (Tb-Ta)
We~= -(Ed-Ee)= -vc;, (Td-Tc)
Hovever, in an adiabatic expansion TV?'-l = canst.
T V 7-l
wa,b = -vc;, Tb (1~ = -vc;, Tb (1- <v:) ) Hence
7-l
wc,d = -vc;, Tc G: -1) = vc;, Tc (1- (:~) )
The volume is constant in :process ~c so no vo~ is :performed.
Thus
CB'.API'ER 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-E/kr. Hence the ratio of the :probability oi' being 1n the i'i.rst excited state to the
probability oi' being in the ground state is
e -3~/oo
--Hw/~
e
=
(b) By the dei'ini tion of mean value
6.2
The mean energy per particle 1s
e ==
Hence i'o~ N particles,
µ He -µR/kr - µHeµ.H/kl'
e-µifi'ki:r +eµH)\!r
E = -NµH tanh I
32
1 + 3e -iiwjkr
1 + e -15.wjkr
6.3
(a) The :probability that a spin is parallel to the field is
eµRjkr l
p = e -µHflci' + eµR/kT = 1 + e - 2µR/k.T
vhich yields T = k J.n[fyl-P] (1)
For P = .75 and. H = 30,CXX) gauss, T = 3.66°K
(b) Fran (1), w"'ith µ = 1.41 x lo-23 ergs/gauss, T = 5.57 x 10-3 °K
6.4
The power absorbed is :proportional to the difference in the number of nuclei 1n the two levels.
This is,
·w'here N is the total number of nuclei. Since µR << kr, ve may expand the exponentials and keep
onzy the first two tenns.
(1 + ~ - 1 + ~)
Hence, n - n .., N ___ kT"'-__ __,;;_
+ - 1+~+1-.@
kT kT
=
Thus the absorbed power 1s proportional. to T-1 •
:p(z+dz)A
____ ..._ ___ z + dz
The volume element of at2!1osphere shown must be 1n
equilibrium under the forces o:f the pressure and
gra:n ty. Then i:f m is the mass per :particl.e, A
the area, and g the eccel.etation of granty, we
...,_--+-------~ z mn(z)Adzg p(z)A
have
p(z+dz) A - p(z)A =~dz A= - mn(z) A dzg
Substituting the equetion of state, p =~leads to
dn(z) - - 5 n(z)
dz - ~
Thus n(z) = n(O)e-mgzjkr
33
6.6
(a)
E
As T approaches O the system tends to the low energy state, while in the limit of high tempera
tures all states became equal.ly probable. The energy goes :from the low to the high temperature
oE) (b) The specific heat is, C.., = (~ , i.e., the slope of E.
V
T
(c) _ N \.l + _ ~~: ~:::1 tl+e - J
and
To find the temperature at which E changes limit, we
evaluate T vhere [ J
- N e:l+e:2
E-Ne: +- ---e: - 1 2 2 l
which yields
34
The heat capacity is
and f1v -+ 0 as T --. o, T -+ cc •
(a) The mean energy per mole is
E = N
A
-OJ,,.m -e;...., Oe /A.I. + 2e:e ,~
l + 2e -e:jkT
2 NA e:e -e:/kT
l + 2e-e:jkT
(l)
where NA is Avogadro's number.
(b) We find the molar entropy from the partition function z.
-e: .j.kT
t = ~ e 1
, the sum being over states of a single nucleus.
i
Thus
and s = k (1n z + ~E)
NA
Clearly, Z = ~ where
(2)
(c) As T -+ O, the nucleii go to the ground stat~, and by the general definition of entropy
S=Nklnn=Nklnl=O
A A
as T -+ =, all states become equally probable, n -+ 3 and
It is easily verified that expression (2) approaches these limits.
(d) From (1) we have
6.8
The polarization, ~ is defined a.s N ~ where
N is the number of negative ions per unit volume,
e is the electronic charge, and i is the vector
from the negative to the positive ion. Then the
35
CV
0
0
T
£, 0
> e
0
a
aean x component of Eis
p = !Pl cose = Ne It! lt_xl = Ne X
X - -
vhere x .nd e are shown in the diagram. The positive ion can have energy + e;a or - e~a ,
and since there are two lattice sites for each energy ve have
ela ~ _ ~ ~
(~) e 2 + (- ~) e 2
F =Ne---------------
x et'.9. ~ _ ~ ~
e 2 + e 2
6.9
(a) The electric field is found from Gauss's Theorem
Ia~ • £ da = lm-q
A
_ ~ tanh e!a
- 2 2ld'
vhere q is the enclosed charge and£ is the vector normal to the surface, a.
By symmetry,
The electric potential is given by
Ea= E 2rrr L = lm-q
-
E(r' =_gs,~
!.I rL -
U(r) = ir ~--~ = - f!l ln ~o
Since U(r) =
Thus
-Vat r = R, ve can evaluate q.
0
0
V R A
E(r) = r ln r ! ,
0
U(r) = V ln !,_
- ln R ro
ro
(b) The energy of an electron at position! is qU(!) = -eU(r) vhere e is the electronic
charge. Since the probability of being at! is proportional to the Boltzmen factor, ve have
for the density
( 0
-ev ~ ln(R/r )
= p(O) !... o
r
6.10
2 2
()
2 mw r a The centrifugal force mw r yields the potential energy - -
2
-. Since the probability that
a molecule is at r is proportional to the Boltzmann factor, the density is
p(r) = p(O) exp [mw2r 2/2kT) (1)
(b) Substitutingµ= NAm, the molecular veight, into (1) and evaluating this expression at r
1
and r 2 ve have
6.ll
The mean separation, x, is found from the probability that the separation is between x and x+dx,
_ U(x)
P(x)dx cc e kT dx
Where u(x) = u [<!)
12
- 2(!)
6
]
0 X X
It is easily verified that the minimum of the potential is at x = a. Since departures from this
position are small, ve may expand U(x) about a to obtain
36 U 2 252 U
U(x) ""' -U + --0 (x-a) - 0 (x-a)3
0 2 3 a a
vhere ve have kept only the first three terms. It follows that
[
36 U (x-a)
2
252 U (x-a)3]
P(x)dx = C exp - a~o + a~
0
d.x
We have absorbed the irrelevent constant term, e- Uc/k'I', into the normalization constant C.
This constant is eval.uated from the usual. requirement
1
00 100
[ 36 U0 (x-a}2] . [ 252 U0 (x-e.)3]
P(x)dx = C exp - 2 exp
3
dx = l
co O a kT a.k.T
The predominent factor in the integrand is exp t36 U
0
(x-a)2/a~] so the second factor may be
expanded in a Taylor's series as in Appendix A.6. Furthermore, the region of integration can be
extended to -co since the exponential is negligably small except near a.
Thus (1)
vhere ve neglect all but the first two terms. The first integral is evaluated in Appendix A.4
vhil.e the second is O since the integrand is an odd function.
Then 6 cu;1 C=- _?,._
a iT kT
By definition, x = J~ x P(x) dx, and to the same approxillla.tion as in (1) we have
[
36 U0 2]( 252 u0 (x-a)3)
exp - - 2-- (x-a) l. + ------- dx
a kT a3k.T
The first term in the integrand is a Gaussian times x and the integral. is just a. The second
may be eval.uated by making the substitution~= x-a
37
- 0 Ku~½ X =a+ -7rk1'
Noting that the second integral is O since the integrand is an odd function, and evaluating the
first by Appendix A.4, we find
Thus
1 dx
a= - dT =
X
6.12
.15 ak .15k
au + (.15)ald' ~ ~
0 0
since U
0
(a) Let the dimensions of tl:e box be x, y, z. Then the volume and area are
V = xyz
EJirninating x,. we have
A - 2yz
V = yz y + 2z
A = xy + 2yz + 2xz
av av . From the conditions for an extremum, ~=oz = O, it follO'ws that
(A-2yz) (y+2z) - 2yz (y+2z) - y (A-2yz) = 0
» kl'
(1)
(2)
(A-2yz) (y+2z) - 2yz (y+2z) -2z (A-2yz) = 0
l l
A2 A2
Subtracting these equations yields y = 2z. Then substitution in (2) gives y - - z - -
1 - E ' - 2J3'
A2
and from (1) we find x = ./3 .
(b) From equation (1), we have
dV = yz dx + xz dy + xy dz
The equation of constraint is
g(x,y,z) = xy + 2yz + 2xz - A= 0
and Adg = A(y+2z) dx + A(x+2z) dy + A(2x + 2y) dz= 0
Adding (3) and (4) and noting that the differentials are nO'W independent, we have
yz + A(y+2z) = 0
xz + A(x+2z) = 0
xy + ;\.(2x+2y) = 0
We multiply the first equation by x, the second by y, and the third by z, and add.
3xyz + A(2xy + 4xz + 4yz) = 0
Substituting (4) yields
A=-~
2A
(3)
(4)
(5)
(6)
and substituting ;I.. into equations (6) gives
l - ~ (y + 2z) = 0
l - i ( X + 2z) = 0
1 - : (2x+ 2y) = 0
The first two equations show that x = y and the third yields x = ~- Then substitution in the
Alfa Alfa Alfa
first and second equations gives x = .,,/3 , Y =
13
, z =
2
.fj
6.)3
S + S = -k L. P (l)ln P (l) - k L. P (2)1n p (2)
l 2 r r s s r s
But since L, P (l) = L. P (2 ) = l, we may write
r r s s
s + s = -k L. L. p(l)p (2)ln p (1) _ k L. L. p (2)p (l) 1n p (2)
1 2 rs r s r s r s s r
= -k L. L. p (l)p (2) 1i.n_ p (1)+ 1n p (2~
rs r s l:- r s J
= -k L. L. p (l)p (2)ln p (l)p (2)
r s r s r s
r s
= -k L, L, p 1n p
rs rs
since P (l) p (2) = p
r . s rs
6.14
From the definition, we have
S + S = -k L.P (l)ln P (l)_k L. p (l)ln p (l)
1 2 r r s s r s
(1)
SubstitutL".lg P (l) =LP
r rs
s
and P (2 ) = L P in (1) we have
s rs r
s1 + s2 = -k r, r, P 1n P (1)_k r, r, P 1n? (2)
rs rs r rs rs s
= -k L, L, p 1n p (1)? (2) (2)
rs r s r s
For the total system, the entropy is
s = -k L, L, p 1n p
rs rs r s
Then from (2) and (3)
s-(s
1
+s
2
) = -k r, r, p 1n p + k r, r, P 1n P (1)p (2)
rs rs rs rs rs r s
= •k LL P ln rs r s p (l)p (2)
r s
39
(3)
(4)
(b) Since -1.n x ~ -x + 1 or l.n x ~ x-1, we have
Fran (4),
But since
6.15
l.n p (l)p (2)
r s
- 1
p (l)p (2) (p (l)p (2)
s-(s1+s2) = k Z: I: p 1n r P 5 ~ k r, r, p r s
rs rs p rs rs rs rs
L. L. P (l)p (2 ) = 1, it follows that
·r s
r s
s-(sl+s2) 5 k r, Lo Pr(l)Ps (2)_1 = 0
r s
(a) Since S = -k L. P l.n P and S = -k Lo P (O)l.n P (O) we have
r r o r r ' r r
S-S =kL. \_p lnP +P lnP (O)_p lnP (O)+P (O)lnP (OJ (l)
o rLr r r r r r r rJ
vhere we have added and subtracted the term k Lo P 1n P (o).
r r We show that the sum of the last
two tems in (1) is zero.
Since
Then using Lo P = 1, ve have
r r
r
)
-,3E
P (0 = e r;z
r
1n P (O) = -f3E - 1n Z
r r
-!: p ln p (O) = f3 L P E + L. P ln Z = ~E + 1n Z
r r r r r
r r r
Similarzy Lop (O)ln p (O)= -13 Lo P (0)3 - Lo P (C)ln z = -p3 - 1n Z
r r r r r r r r
Addi:cg, we obtain
_ z p 1n F (0) + :z ? (0)1.n p (0) = 0
r = r r r r
Thus (l) becom:::s
p (0)
s-s = k L, r r r r r P
0 -
P (0) P (0) •
[
-P 1n p + p 1n p (O)] = k Lo p 1n _r __
r r
(b) Since! 1n -f-- ~ T - J.,eq_uation (2) yields
r r ~O)
consequentzy S ~ s.
0
s-s ~kl'. P r o r P
r r
40
= 0
(2)
CHAPI'ER 7
Simple Applications of Statistical Mechanics
7.1
(a)
th
We label the positions and momenta such that r .. and p.j refer to the j molecule of type
-iJ -1
i. There are N. molecules of species i. Then the classical partition function for the mixture
J.
of ideal gases is
Z I 1 exp[- ~l (:2.1/+. • ,+E_lNl 2) • • •
The integrations over~ yield the volume, V, while the E_ integrals are identical. Since there
The term in brackets is independent of volume, consequentzy
ln Z' = (N1+ ••• +Nk) 1n V + 1n (constant)
and - lo 1
p = §" dV 1n Z' = (Nl+ ••• +~) fN
pV = (N1+ ••• +Nk)kT = (v1+ ••• +vk) RT
(b) For the i
th
gas, piV = viRT, hence by (1)
p = I: p.
i l.
7.2
(1)
(a) By the equipartition theorem the average value of the kinetic energy of a particle is
- 3
€ = 2 kT.
(b) The average potential energy is
u
Then
iL mgz e -i3mgz dz
f Le -pmgz dz
0
On carrying out the differentiation, we find
u=kT+ mgr.
l - emgLjkr
41
7.3
{a} Before the partition is removed, we have on the left, pV = vRT, After removal the pressure
is
_2vRT_~
Pr -(1+1i)v - 1+b
(b) The initial and final entropies of the system for different gases are
Si a vR r N:v + i ln T+o~ + vR r ~:v + i ln T+aJ
Sf a vR r (l;:lV + ~ 1n T+o~ + vR r (l;~lV + ~ 1n T+o~
Here one adds the entropies of the gases in the left and right compartments for Si while sf is
the entropy of
Then
two different gases in volume (l+b)V.
!::,S = S -s . = vR 21n __.__.._ - 1n - - 1n - = vR 1n t V(l+b) V bV J
f l. NAV NAV NAV
(l+b)2
b
(c) In the case of identical gases, S. is again the sum of the entropies of the left and right l.
compartments. Sf is the entropy of 2v moles in a volume (l+b)V.
sl.. = vR 11n ..,;!,__ + .11n T+cr 7 + vR !in bNV + .11n T + crl L NAv 2 ~ L Av 2 ~
Thus
2
= vR 1n (14~)
7.4
(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
or
The total volume is found from the equation of state
V l RI' l V 2RI'2
V=--+--
pl P2
Thus the final pressure is =
42
(c)
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 ~ extended, the total energy would be -Nila). Since the
segments are non-interacting,
Since the total energy is additive,
where U is t he energy of interaction, the eq_uipartition theorem still applies and
7.7
I:f the gas is ideal, its mean energy per particle is
2 2
PX Pv e = 2iii + 2m = :k!r
and the mean energy per mole becc:mes E = NA:k!r
Thus
43
For classical. harmonic motion the mean energy is
- 2
1 2 2 E-L+-mw x - 2m 2
By equipartition, each of these terms contributes½ kT.
Hence
7.9
(a)
2
P"'X o:T
The mean elongation is found by equating the gravitational and restoring forces
cix. = Mg, thus
(b) By the equipartition theorem
½ a (x-x)2
or (x-x)2
(c)
or
7.10
= ! kT
2
kT
a
M -Jkr a' - 2
g
(a) Let the restoring force be -ax. Then the mean energy of N particles is
- 1 72 1 2
E = N <2 mx + 2 ax )
By equipartition - 1 1 ) E = N (- kr + - kT = N kT 2 2
Thus oE
C =?!r=Nk
(b) If the restoring force is -ax3, the mean energy per particle is
J_ 72 1 4
E:= 2 mx + 4 ax
The kinetic energy term contributes ½ kT by equipartition. The mean of the potential energy is
found from
=
J~ exp [- ~] ~ dX
J~ exp [- ~] dX
44
where we have made the substitution, y = 131/
4
x.
Thus d O Jco 4
/4 1 1
4 = - ¥ (- ¼ in 13 + 1n ...., e -czy dy) = 4p = 4 kT
Hence
1721"""'1i'. l 1 3
€= 2 mx + 4 ax = 2 kT+ 4 kT= 4 kT
Then the total energy is E = NE and
7.11
Since the :probability of excitation is proportional to e - kT where n is an integer, it is
clear that the two parallel components are negligibly excited at 300° where i:iw
11
>> 300 k.
The perpendicular component is excited since hw
1
<< 300 k; and since 300° is in the classical
region, the equipartition theorem holds and we have for the mean energy per mole
1 ~ 1 22
E = NA (2 mx + 2 mw l x ) = NA kT
Thus
(a) Consider a cube of side a. The force necessary to decrease the length of a side by .6a is
K !::,;a.
0
and therefore the pressure is t::,.p = K 6a/a2•
0
The change in volume is !:::,.V = -a2.6a.
Hence K = (
2 ) 1 a 6a a
= - a3 - K .6a/a2 = Ko
0
(1)
(b) The Einstein temperature is eE = hwjk. Since w = ~Kd1m
I
where m is mass, we have from (1)
(2)
The length a is found by considering that the volume per atom is µ/ pi-IA whereµ is the atomic
weight, p the density, and NA is Avogadro's number.
1/3 1/3
a=(_H._) =( 63•5 ) =2.3Xl0-8cm
pNA (8.9)(6X1023)
Thus
Substitution of m = µ/NA and the given values yields
GE"'"' l50°K
45
7.13
'We have Mz = N
0
g µ
0
J B/71). By equation 7.8.14, ~ (71) for J =½becomes
Thus
B_21(71) = 2 (coth 71 - 12 coth ~2) = 2 cash n - cash t~f~1
sinh 71 sinh
= 2 cosh
2({/2) + sinh
2(//2)
2 sinh 71/2) cosh (71 2)
sinh
2 ( /2)
sinh (71J2) cosh (71/2)
-
cosh
sinh
The energy of the magnetic moment in the field His
E = -1:_•1! = -µH cos e
where e is the angle between the magnetic moment and the direction of the field or z-axis. Then
the probability that the magnetic moment lies in the range e toe+ de is proportional to the
Boltzmann factor and the solid a.ri..gle 2rr sine de.
Thus P(e)de « epµHcose sine de
and
w J 'IT SµH cos e sin e de(µ cos e) ·'o e
1'1z = N µ
0
=
0 Z J n eSµH cos e sin e de
0
or No o 1n r -;r ;3µH case . e de
H ~ e sin = e ,., o
i:,µ:! -;3µH
e - e
= No ~ 1n 2 sinh ~µH _ No [ • . " ]
~A ~ H - . . -... "µH µHpcosn ,.., µH - sinh pµH H ~,-, pµ F.B SJ.Il.H ...,
We have (1)
46
1ere by (7.8.14)
If Tl << 1 and J >> 1
-where we have used
ll l l l l ~ B (Tt) = - (J + -) coth (J + -)Tt - - coth - Tl J J 2 2 2 2
in such a way that JTt >> 1, B
3(Tt) becomes
Bi Tt) = J f coth JTt - ½ (~~ = coth JTt
X -X
1
JT)
coth x = e + e = (l + X + •••~ + ?l - X + •••~
(1 + X + ••• - 1 - X + •••
1
::::: - for small x
X -X X e - e
becomes
whereµ= gµ
0
J is by (7.8.2) the classical magnetic moment.
:i-,1z = N0 µ ~oth pµH - P~
,
(a) The energy of a magnetic moment in a field is E = -1:. • ~- For spin 2 atoms
(b)
n(z2)/n(z
1
) > 1 since n2 > :a:1 .
(c) Ii' µH << kl' we have
and
7.17
+ (l - µH R + ! ~ 2,::2) 2..... 2 µ ? ...,
1 2 2 2
+ (l - µHl~ + 2 µ Hl (3)
Then (1)
To find the fraction,;, of molecules with x component of velocity between_-:;; and v, we must
integrate the distribution between these J.illlits, i.e.,
47
J
; P,} .l
s = .± g(v )dv =f (-~-/ -(mv 2/2kT)
e x dv
n _-:;, X X -Jti!!. 2rrkT
m
X
Making the change of variable, y = ~ vx, we have,
1 Jfi' (2/) 2 2 s = - e - y 2 dy = L 1 e -(y / 2 ) = 2 erf 42
Ji;; --12 ,/~ 0
Then from tables s = .84-3.
In problem 5.9 we found that the velocity of sound is
1
u = czm)2
µ
where , = C/Cv and µ is the atomic weight. Since µ = N~,we have
1 l
u = czm) 2 = (Z:kT) 2
NAm m
.l
The most probable speed is ~ = (2kT/m) 2 • Thus
For helium, 1 = 1.66 so that u = .91 ~ and the fraction of molecules with speeds less than u is
1 -::i 2
,., - ,,,!,. mv
1 1°91v 1 ·91(2kT/m)
2
2 2 - 2kT
s = n O F(v)dv = 47!" O (;kT) v e dv
.l
Making the change of variable y = (m/ld') 2 v, we have
4trr 1.91./2 2 - 2/2 s =~
0
y ey dy
(2rr)2
l a _ 2/2 This integral :rn.ay be evaluated by noticing that integration by parts of
O
e y dy yields
1 a 2;~ 2 /21 a 1· a 2 2/2 e-y ~ dy = ye-y + y e-y dy
0 0 0
Since the integral on the right is the required one, we have,
The integraJ. on the right may be evaluated from tables of error functions. Thus we finds = .r,.
48
7.19
(a) V = 0 by symmetry
X
(b)
2 kl' by eq_uipartition V =-
X m
(c)
2 3 2- 2-) = 0 (v V) = (v + V V + V V
X X y X Z X
7.20
(a) TT-7vY = ! Jco r{tl dv =J"" 4n
n Q V 0
l
(d)
(e)
(f)
3/2
C21rm"ld')
(v 3 v) = v 3 v = 0
X y X y
)2 2 - - 2 2
( V + bv = V + 2b V V + b V
X y X X y y
= (l+b2) kT
2 m
V 2 V 2 = (k'I')
X y m
2 mv
ve 2kT dv
2m 2
By appendix A.4 we find n:Tv; = (7rkl') , a.".ld since v =
(l/v) = ~ = 1.27
(1/v) 7T
1 2 2 ~ {b) Since the energy is € = 2 mv , we have v = 2€/ m and dv = d€/ ...; 2m€. On substituting these
relations into the speed distribution F(v) dv, it follows that
3 €
- - l - -
F(€)de = 27m (nkT) 2 €2 e kl' de
7.21
rhe most probable energy is given by the condition ~(e) = 0 or from problem 7.20b
E l €
1 1 - kl' €2 e - kl' = 0
~ e -kl'
Thus
.... 1
E =2kl'
.... 1..
The most probable speed is V = (2kl'/m) 2
Hence
7.22
(a)
1 -2 2mv =kl'
V
V = V (1 + ~ ) = V
O C 0
since v = 0
X
(b) Therms freq_uency shift is given by
(.6v) = rms
But 2 2
'V = V
0
~
- l
2 -21 2
= v -vJ
~+ ~x+~
2 Since v = 0 and v = kT/ m by equipartition, we have
X X
Thus (~v)
rms ~
l
2 kT 2 2 2
= v +-v -v~ O 2 0 O me
Vo
C
(c) The intensity distribution is proportional to the distribution of the x component of
velocity.
I(v )dv = I exp
0
where I
0
is the intensity at v=v
0
•
7.23
2 2
[
- me (v-v 0 )
2kT
2
Vo
(a) The number of molecules which leave the source slit per second is
18
= 1.1 x 10 molecules/sec
where A is the area of the slit.
(b) Approx:il!lating the slit as a point source, we have by (7.u.7), the number of molecules
with speed in the range between v and v + dv which emerge into solid angle an is
A~ (v) d3z = A[f(v)v cos e][v2 dv an]
cos e ~ 1 for molecules arriving at the detector slit; hence the total number which reach the
detector is 13Jnv2
e- - 2 -dv = (1)
where we have used the results of appendix A.4 and ps = ~. Letting L = 1m. be the disiance
between source and detector, the solid angle nd is approx:il!lately A/L2• Substitution of the
given values into (1) then gives 1.7 x 1011 molecules/sec arriving at the detector slit.
(c) In the steady state, the n1.nnber of molecules entering the detection chamber per second is
equal to the number being emitted. From (1) we find
or - A 4 -8 Pd = Ps 2 = 2. x 10 mm of Hg.
TIL
The rate of change of the number of particles in the container is given by
dN
dt = - cpA
l -= - 4 nvA
50
Since the pressure is proportional to the number of particles, we have
dp
-= dt
vA
- 4Y p
or p(t) = p(O) exp t * ~
where p(O) is the original pressure.
p(t)/p(O) = 1/e is
Hence the time required for the pressure to decrease to
7.25
As in problem 7.24, we have
or
4v
t -
VA
where p(O) is the original pressure. The mean speed, v, is for nitrogen at 300°K,
v = J~ k:T' = .!n5 x 105 cm/sec
7f m
Thus _ 4(4n/3)(1c?) 10-6
t = (.475 X 105)(1) 1n 10-l =
4007 sec
1
(a) Since the rate of effusion is~= p/(27r m kT)2,and the concentration is proportional to the
pressure, it follows that
(b) For the uranium separation, we have
1
I I (238 + 6 19 )2
c'235 C'238 = c235 c238 235 + o 19 =
The rate of change of the number of particles inside the container is
dN l - NA j8 k:T' ;>,.:r,r
dt = - 4 nvA = - 4V -rr m = - "-Jm
thus defining i... Since pressure is proportional to the number of particles, we find after
integrating
For Helium gas p/p
0
= 1/2 at t = 1 hour. Su:t,stituting we have
;>,.=~ln2
51
(1)
(2)
where ~e is the mass of the helium molecule. From (l) and (2) it follows that the ratio of the
Ne and He concentrations in terms of the initial concentrations is
Since
~e/~e = (~/~e)
0
exp [-, .. ~/.;~e' - 11,/m;;J ~
= (~/~e) 0 exp [-1n 2 (J~./,e' -~ ~
(~/~e)
0
= 1, we have at t = 1 hour,
I (1-J~/,e,) (1- j~/~e' )
7fe-' 11He = 2 = 2
(a) The rate of change of the number of particles in the left side of the box is
dNl(t)
dt = (no. entering/sec)-(no.escaping/sec)
Since the rate of effusion is f nvA, we have
dNl (t) Av r ;J - I: '1
dt = Ii.il,72° LN2<t)-N1<tj =: r-2N1<t2J
where N is the total number of particles.
Thus
Integrating,ve find
(b) Tbe initial entropy is
Si= Nl(O)k (1n 2N:(O)
In final equilibrium, N1 = N2 = ~
From the equation of state,
P1 (0) + P2 (o) V
and N = ----- -2 kT
Sf = Nk (ln ~ + ~ 1n T + v
0
)
52
(1)
_ Av t
e V
7.29
(a)
{b)
Inside the container v =Oby symmetry.
z
The velocity distribution, <1>(:!), of the molecules which have effused into the vacuum is
<I>(v) d3v = f(v) V d3v - - - z
where f(v) is the Maxwell distribution. 2 .... mv
J~dvxf~ dv f 00
dv
z
<I> (v) V f 00
dvz
- 2kT 2 e V
y O z z z
Thus 0 V = = z f
00
dvxf"" dv f 00
dvz <l>
2
(v) mv
z
-co -co y 0 f 00
dvz
- 2kT
e V z
0
The second equality follows because the integrations over v and v are identical in the numera-x y
tor and denominator and therefore cancel. The integrals over vz are tabulated in appendix A.4.
3
7.30
(a)
~ (21;)2
VZ a ½(~) aJ~ 1;;'
In t:i.!lle dt, molecules with velocity ccmponent v which are in a cylinder of height v dt z z
·with base at the aperture can escape from the container. Molecules of higher speed can escape
from a larger cylinder than can molecules of lower speed, and a proportionalJ.y larger share of
the fast molecules effuse. Thus the mean kinetic energy of particles in the beam is greater
than in the container. J d3v ! mv
2
<l> (2::)
- 2
(b)
l 2
EO = 2mv = J d3
! 9 (~)
Since <I>(_!) = :r(;:) e and d3v 2d . e de dqi , have V COS = V V S1!1 we
mv2
i oolrr/2 ro1T(v2 1 2 - 2kT J, dv sine de d~)(2 mv) e v cos e
€ =--------------------------0 f 00f-rr/J,-rr (v2 dv sin e de dcp)
O o O
2 mv
l J"" ~ -2kT - m -.? e dv
2 0
= ---------- = 2k.T
mv2
e - 2kT dv
where the integrals over v are evaluated by appendix A.4. Since the mean kinetic energy in the
. - 3 4-the enclosure is e;i = 2 kT, it follows that e:
0
= 3 e;i.
7.31
When a molecule is scattered, it commutes momentum l:.p= 2mv to the disk. The force is z z
l:.p / at. To find the total force, we must multiply l:.p /at by the number of molecules with z z
velocity in the range (v, dv} which strike the disk in time dt and then integrate over all
,v
velocities. Clearly only molecules in the cone of half-angle e = sin-l R/L will strike the
0
disk.
F = J d3~ [A~ (!) at] ~!z
= J:°"J:ei2rr(v2dv sine de dcp)(A f(v) v cos e d_::)(2m ;/os 8)
I 1 le J°" 4 = (2mA.)(2rr)(- 3 cos3e)
0 v f{v)dv
0 0
The integral can be expressed in terms of the mean square speed inside the container, i.e.,
! °"4 2
v f(v)dv = ~
0
1 2 1 3 )
8
0 1 2 [ 3 (s-i.,.., -1 _RL)] Thus F = 3 nm v A ( - 3 cos e
O
= 3 nm v A 1 - cos .,_....
But by eq_ui:i;,artition we have 1 2 -
3 nmv =nkT=p
Hence F = pA [1 - cos3 (sin-l f)]
CHAPI'ER 8
Equilibrium between Phases and Chemical Species
8.1
By equation (8.3.8), we have the proportionality
(?{v,T) dV dT o: exp [ -G
0
(V,T)jk.r
0
] dV dT
where V and Tare the volume and temperature of the portion of substance of mass M, and T is
0
the temperature of the surrounding medium.
minimum at V = V 1 and T = T yieldf
Expanding the Gibbs Free Energy G (V,T) about the
0
(1)
54
where the derivatives are evaluated at T = T and v = v. Since we are expanding about a mini.mum
. ,
the first derivatives in (1) are O. By (8.4.7)
Since this is Oat T
Thus
and at T = T = T,
0
cG
(dT 0)
V
( c2GD
"T2 C:
V
From (2) we also find the mixed derivative
cG T cC
(~) = (1 - ~)(~) = 0
ar v T av T
By the equation preceding (8.4.13)
cG "s
( "-v o) = (T-T ) (¾-)
0 T O C T
at T
= (T-T )(els~ - (ffe) =
o ?,v2 Jr oV T
----(~p) ..... , at T = T , V = V
oV T o
(2)
By definition, K = 1
V K
Substitution of these results into (1) yields
CV )2 1 ..... 2
G (V,T) = G (V,T ) + - (T-T
0
+-:::-- (V-V)
0 0 0 2T0 2VK
and since V = r~/p
0
and CV = McV where ~ is the specific heat per gram, it follows that
(?(v,T)dV dT • exp t :;/ (T-Tf - 2M:0
k'.1'
0
(V - Mjp0 )j dV dT
where we have absorbed the constant term G (v, T) into the normalization factor. Performing tl::e
0 0
usual normalization, we find
(?(V,T) dV dT =~~;,,:;/exp tMc
_V_ (T-T )2
2.kT 2 o
0
8.2
(a) For the solid, l.n p = 23.03 - 3754/T and for the liquid l.n p = 19.49 - 3(£3/T. At the
triple point, the pressures of the solid and liquid are equal.
55
hence
(b) Since 1n p = - t + constant vhere t is the latent heat, we have
t ~lim t· = 3754 R = 31,220 joules/mole suu a ion
tvaporization= 3c;63 R = 25,48() joules/mole
(c) tfusion = tsubl:ima.tion - tvaporization
= 31,220 - 25,48'.) = 5740 joules/mole.
The latent heat of sublimation is equal to the sum of the latent heats of melting and vapori
zation, i.e., t = t + t. Since t = T6V (dp/dT), the Clausius-Clapeyron equation,where vis
S m V
the volume per mole, it follovs that
(:)s a To(vg~s- vsol) ~o (vliq- vsol)(:)m + To(vgas- vliq)(:)J
or
8.4
Since I = ~ and l::,.S -+ 0 as T -+ 0 by the third law, it follovs that : -+ 0 as T -+ O.
8.5
(a)
(b)
Q/1 moles/sec evaporate and are svept
Q pm"/
E = Rr
r
out by the pump.
or
QRI'r
pm= LY
Then by the equation of state
We find the temperature at the pressure p from the Clausius-Clapeyron e~uation
m
d"D L L
dT = T(vgas - vliquid) :::::s T vgas
where we have neglected vliquid since the volume per mole of the gas is much greater than that
of the liquid.
Hence
Then using v = Rr/p we find gas
T = T {
m O\
1 c!... - !...)
R T T o m
o m
-1
TR p )
- T 1n ;;-
-o
56
8.6
The intensity, I, is given by
Differentiating with respect to T and dividing by I,we have
Since pis the vapor pressure, the Clausius-Clapeyron equation yields dp/d.T.
dp L L
dT = T(vgas - vliquid) ""T vgas
Substituting into (1) and using v = Rr/p gives gas
1 dI L 1
I dT = RI'2 - 2T
Differentiating the molar latent heat yields
(1)
(1)
(s2-s1) is the molar entropy difference between the phases.
Thus
~A [os2 osl 7
(~) T = T tdP) T - (dP )~
But (os/c)p)T = -(ov,loT)P, a Maxwell relation, and since the coefficient of expansion is
a = -1/v (cv/oT) ,it follows that p
Differentiating at constant p gives
(~P • (s2-s1 ) + T ~~\ - (~)J
T(os/'oT) is the specific heat at constant pressure.
p
Thus
(~ = ¥ + ( C - c"O )
p P2 -1
Substituting (2) and (3) into (1) yields
Since
d.Ta:-f.. = -T(a2v2 - a1.v1) fT +} + (c - c )
P2 pl
57
(2)
(3)
8.8
The bar moves because the higher pressure underneath it lowers the melting point. The vater is
then forced up along the sides to the top of the bar and refreezes under ~he lo.rer pressure.
To melt a mass dm of ice, heat -\'..am must be conduc,ced through the bar.. The heat passing through
it per unit time is
J:! = -K: (be) m
a
vhere mis the temperature drop across the bar and x: is the thermal conductivity. Letting x be
the position in the vertical direction, we have dm = p.(bc dx), and equating the rate at which
l.
heat is conducted to the rate it is absorbed in the melting process ve have
-x: (be)~= t pi (be~)
mis given by the Clausius-Clapeyron equation
d"D ll.p t
dt ""m = -T-,(-v1,-1a_t_e_r ___ v_i_c_e,...)
vhere v is the volume per gram or 1/p. Then since ll.p = 2 mg/be,
Substitution in (1) yields
dx
dt
From the Clausius-Clapeyron relation, dp/d.T = L/TAv we find
Ja: = JD.; dp
(1)
We know L(cp) and ti.V(cp), and we can find dp/dcp from the curve of vapor pressure vs. emf.
Thus JT d.T' Jcp ~vf co 'j d:p -, = . I (-) dcp'
273.16 T o L 9 d~
where 273.16°K is the temperature of the reference junction of the thermocouple.
Then
8.10
We divide the substance into small volume elements. Tle total entropy is just the sum of the
entropies of each of these, or
S = Z: S. (E.,N.)
il. l.l.
where Ei and Ni denote the energy and number of particles in the i th volume. If' we consider
58
an interchange of energy between the i th and j th volumes,we have since the entropy is stationary
but since dE. =
l.
-dEj, we have
clS
cs.
l.
2;E.
l.
Similarly, if we interchange particles
cs.
m.plies
].
dNi
cs.
~ oN.
J
dS = 0
= 0 =
;sj
= cE.
J
els. 2s .
l. dE. + 0i dE. ~
o~i l. J
J
or T. = T.
l. J
or
8.ll
(a) The chemical potential can be found frcr:n the free energy byµ=~:. He consider a small
0 1{
volume element at height z over which the potential energy may be considered co?J.stant. 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. fu..'"Plicitly, the partition f'unction will be the free
particle f'unction times the factor e-~-rnez.
Thus 8 "~ 1n ~free e -;,mg~ + ~ (Ei'ree + mg~ " k [1n 2free + "freJ
Hence the free energy is, letting N be the number of particles in £::.V
F = E - TS
= ¾ 1llcr + Nmgz - ~ ~ ~; + ¾ Lri T + c J
and _ cF _ .2. '"'i'
µ - 6N - 2 -"'..c + mgz - kT
where we have put t::.V /N = ld'/p .
(b) Si.riceµ is independent of z,
~ kT d-o
dz =mg+pd; =O
Integration yields
8.]2
a:o
or . ---= -
p
:p(z) = p(O) e
The relative concentration remains the same. The dissociation is described by
nco ~ o
--2- = K (T)
nee~ n
2 2
Generally, when the number of particles is conserved, equilibrium is inde:pendent of v,
59
8.13
The intensity of the I atoms in the beam is proportional to their concentration 7: in the oven.
Since dissociation proceeds according to
We have
which gives
by the e~uation of state.
into (1)
or
nI-2
n_ = K (T)
l.2 . n
K (T)
= ~ = ~(T) (1)
The total pressure is p =Pr+ Pr which yields on substitution
2
Thus, if the total pressure is doubled, the ratio of final to initial intensities is
-1 +Jl + 8Kpp
-1 +Vl + 4K p
p
Sir..ce KP is small we may expand the square roots to obtain
~f -1 + 1 + 49
~-,,--"C""""-~- = 2
c[J_ -1 + 1 + 2K n
J. p-
8.14
For chemical equilibrium we have
wheren is the number of molecules per unit volume. Since ni = :pi/ kT it follows that
bl+b2+ ••• +b
8.15
7'h tn "" · .,. ~ 1-, -',, = -./ •• e .:.i_ s.., c.,r a~ dE +p dV.
••• "O
b
m
-m. = (kI') m K (T) = K (T) n :,:i
Letting d(:pV) = V dp + p dV we have
-d. G, = dE + d(pv)-v dp
= d(E + pV) - Vdp
= d.H - V dp
where the last follows from th2 definition of enthalpy, H = E + pV. Then if pis constant
-a. Q = d.H
i.e., in a constant pressure process, the heat absorbed is equal to the change in enthalpy. In
a chemical reaction,Q = 6E = L b.h. since enthalpy is an extensive quantity.
i J. J.
60
8.16
B'J the e~uation preceding (8.10.25)
Lb. 1n ~- I = Lb. 1n .ni
i 1 1 i 1
Substitution of ni = pijki' into (1) yields
(1)
L b. 1n P. = L (b. 1n ~. ,
i 1 1 i , 1 1
+ b. ln kT)
1
or
where
Thus
By (8.10.30) ~ 1n ~it
piVi
by 1 = ILkT we have
1
b
p m = K (T)
m p
1n ~(T) = L (b. 1n ~-'+bi 1n kT)
_ i 1 1
dTd 1n K (T) = L (b. dTd 1n ~-· + bTi)
p i J. . 1
€i
= kT where e:i is the mean energy per molecule.
b.
Then multiplying Ti
d b. p.V. b.h. 1
dT 1n K (T) = L J.2 Ci. + ; 1
) = L :t.
1
2
p i kT 1 ~·i i kT
where h.' is the enthalpy per molecule. Multiplying and dividing by Avogadro's number gives
1
d b.h . .6H
dT 1n Kp(T) = L ..1:...1:. = -
i RI'2 Rl'2
where hi is the enthalpy per mole.
8.17
We let~ be the fraction of H
2
0 molecules which are dissociated.
portional to the number of molecules, we have
PH 2p0
2 2 =----
The total pressure is the sum of t~e partial pressures.
Substituting (1) and (2) into
we find
~(T)
"Dt3 =------
(1-g )2 (2+;)
61
Since the pressure is pro-
(1)
(2)
(3)
8.18
In problem 8.16 it was shmm that
Lili = Rr2 9:_ 1n K (T) dT p
Using ~(T) from problem 8.17 would give M for the dissociation of 2 moles of ~O since the
reaction is 2H20 -)2H2 + o2• Hence for the dissoc~ation of l mole
2
M = ~ ~ 1n 1),(T)
where ~(T) is given by e~uation (3) in 8.17. Since s << 1, we have
p<3 - P_53 K (T) = ___ s __ _
P (1-s/ (2+s)- 2
Hence
2 3
Lill= RT 9:_ 1n Ps =} R1'2 9:_ 1n s
2 dT 2 2 dT
is found by plotting 1n s vs. temperature using the given values and measuring the slope
of this curve at 1700°. We find ~T 1n s = .oo64. Thus at 1700°K
M = ¾ (1.99)(1700)2(.cc.64) = 56,000 cal/mole
8.19
N
(a) The partition function is Z = ~ wheres is the partition function for a single molecule.
For a li~uid we assume a constant interaction potential-~ and that the molecule is free to move
in a volume N •
V
0
Thus
where the integration is
thus defining s ' .
Hence
s = .l:...}' d\:i d3r e -(f3i /2m)+j3~
h3 - -
0
crver N,e,v
O
and all .E•
S = N,e,v
0
/TJ h\ J d3_E e-f3i/2m = N,e_ ef3~ v
0
!;'
0
(b) For an ideal gas, the partition function is
The Helmholz free energy is
Fg, -kT J.n Zg a -kT ~ tg(~Ng_ J.n Ng)
= -kT ~g 1n V g s I - Ng 1n Ng + N~
62
(1)
where we have used Sterling's approximation ln N ! ""'N ln N - N
g g g g
clF
Then µ = ~ = -kT (ln V ~' - 1n Ng) g oNg g
(c) For the liquid, we have by (l)
Ft = -kT (1n [N-f., el3TJ v
0
~ J N,e_ - 1n N,e_, 0
= -kT (N,t 1n v
0
~
1 + N,t t,T} + N,e_)
clF,t
µp = ~ = -kT (ln V ~ 1 + t,T} + l)
1.; oN,t o Therefore
(d) When the liquid is in equilibrium with the vapor, the chemical potentials are equal or
Thus
but since pjkr = Ngl'vg we have
V
__£ = V ef:>TJ+ l
N o g
E.... = e-t,T}
kT ev
0
(2)
(e) We calculate the molar entropy difference from the Clausius-Clapeyron relation.
where in the second equality we have assumed that the molar volume of the liquid is much less
than that of the gas. Since pv = RI' we have g
Then using (2) in this relation we find
The molar heat of evaporation is
(f) From (3) ir = l + ~ •
&, = R (l + ~)
L = T6s = RI' (l + fu')
""~=NAT} if TJ>>kT
Taking the logaritbm of (2), we find
kT V
1.+.!J_=ln-=ln~
kT pV V
0 0
where v is the volume per molecule of the vapor.
g
63
(3)
Hence (4)
when v and v are evaluated at Tb·. g 0
(b) In (4) -we can take v to be the volume per mole instead of the volume per molecule since we
have the ratio vg/v
0
• To est:iJnate the order of magnitude of L/RI'b we
4 3 4 -8 3 23 3 and v ::::: - m- N ::::: - 'TT (. 5 x 10 ) ( 6 x 10 ) ::::: 10 - liters /mole.
o 3 o A 3
Thus L 20 -:::::ln--:::::10
~ 10-3
(h) w e find L/r~ from eq. (4).
1/T0
let v ::::: 22.4 liters/mole g
L/ T0
V V calculated experimental 0 g
water .018 liters/mole 31 liters/mole .83 cal/ gm°K 1.45 cal/ grn°K
nitrogen .035 6.4 .37 .64
benzene .C87 29 .16 . 27
These results are remarkabzy good in view of the very simple theory considered here.
Quantum Statistics of Ideal Gases
9.1
configuration no. of states (a)
0 € 3€ M] BE FD
xx 1 1 -- (b)
xx 1 1 --
xx 1 1 -- (c) -€13 -3€/, -4:i,
~ = e + e + e
X X 2 1 1
X X 2
, 1 .J..
X X 2 1 1
9.2
(a) For an FD gas, we have
-a~E
1n z = aN + Z:: 1n (1 + e r) and nr =
r
Hence
[ -a-pE ;1
s = k [ ln z + pE] = k r + ~ 1n (1 + e r) + P~
64
1
but since N = I: ii and E = I: n €, we :find r r
r r r
1
Fram n = -----
r 0:+/3€r
e +l
S = k [r. n (ct+ /3€ )+ Lo 1n (1 + e -ct-/3~r~
r r r r J
it follows that
a+ /3€ = 1n (1-n) - 1n n r r r
-a-/3€ n
e r r
= 1-l'!
r
Substituting (2) and (3) in (1) we find
s = k G nr c ln(l-llr) - 1n :n~ + r. 1n (1 + :n.:. ~ r 1-n r r
= -k z:: In 1n :n + c1-n ) 1n (1-~ ~
r L> r r. . r:J
(b) For the BE case
-a-/3€
1n Z = aN - Lo 1n (1-e r) and
r
(1)
(2)
(3)
(4)
The calculation of S Froceeds in exactly the same way as in (a). We quote the result.
s = -k r. :n 1n ii - (1+n) 1n (1+n)
r r r r r
(c) Fram (4) and (5) we see that in the classical limit where iir << 1
s ~ -k z:: :n 1n :n
r r r
for both FD and :BE statistics.
9.3
The partition function in sN/NI where
s = Lo
it it it
X y Z
[
~11
2
2 2 2] exp -
2
(it + it +it )
m X y Z
(5)
(1)
. The sum is over all FOSitive it it it when we use the standing wave solution, equation (9.9.22). xyz
The density of state is now greater than in the traveling wave case, and by (9.9.25)
If' we assume that successive terms ins differ little from each other, we can aFFrox:il!la.te (1)
with integrals.
Thus
/31'12 2 /31'12 ·2
L cc
- =--K io:, --K
z: 2m X 2m X (~ dK) e "" e
K ::()
7f X
0
X
L 1
X (271m)2 by (A.3.6) = 2mi 13
3
Hence V 2 s = 3 (2rr m kT)
h
vhich is identical to (9.10.7).
(a) The chemical potential can be found from the He.1Jnholtz free energy, F = -kT 1n z.
N
F = -kT 1n fi :::e -kT [N 1n s - N 1n N + N]
3
vhere ve have used Sterling's approximation, 1n NI :::e N 1n N-N ands V 2
= 3 (2rr m kT) •
h
Then
-:sF •
µ = ~ = -kT 1n .s..
oN N
(1)
(b) The partition function for a single molecule of a 2 dimensional gas vith binding energy -e:
0
~' = L exp [- i3i'i\K 2+tt 2 )~e: ] :::e e
13
e:0 }Jcc exp
~ 2m X y 0
K K . ~
X y 2 2
is
i3e: -
vhere A is the area. Thus s' = e o s3 vhere ve put A = -:S .
2
Hence
i3e: -
~ o. 3
µ' = -kT 1n N' = -kT 1n e fi'
(c) ~ eq_uilibrium µ = µ' and by (1) and (2)
or
2
i3E:
l - e o 1;3
N - N'
N'
(2rr m kT)
. 2 n
Letting n' = N'/A and since pjkT = N/v, ve find
9.5
(a)
1
By equation (9.9.14) if L = L = L = L = v3 we have
X y Z
66
- 1 13h
2
2 2 A I --( K +K ) - dK dK L 2m X y 2rr)2 X y
(2)
(b) p = -r
€ (v) r
cc 2 21-~ (_!:.) = 2rr 1S. ~ 3 (n 2 + n 2 + n 2) oV m 3 X y Z
(1)
(2)
(c) The mean pressure is -P = I: n P where ii is the mean number of parti cles in state r.
r r -r r
Thus P=g_ L.n € =_g_3 VE
3V r r r
(d) The expression for pis valid for a system in which the energy of a c;_uantum state is given
by (1). For photons,
Hence
€ = r
and
(e) The mean force on the wall is found by multipzying the change o::' momentum per collision,
1 -2mv, by the number of particles which collide with the wall per unit time, 0 nvA. Hence the
pressureis
-2 2 Letting v ~ v we have
1 ?
since N(~ mv-) = E.
9.6
- 1(-)1- 1 - 2 P = A 2mv (ti nvA) = 3 nmv
(3)
Equation (3) is the exact result as shmm in section 7 .14.
( l\·cP) 2 -- .,.,_ 2 _ :;_2 . (a) The dispersion is ~ µ µ
- ? 4 s2 2
By .,.,ro'ol0 •m 9 5 .,.,- - - -- and,_-~ ~-?~ ~i·~ • . s to ~_,_._n.d .,.,_ • !,'=•,!,'- a~ , --=- u
/ ~
Hence
(b)
4 ,;-o- 2 ) = 2 t.., n € by (2 in 9.5
gv r r r
--2 4 ?-? 4~
(c.p) = 2 (E- - E:-) = _.2 (c.E)
9V ';3V
' --2 ~2
Since E = - ~ca. 1n Z and (c.E) = 0
2 1n z, it follows that
,.., ot3
67
(6p)2
4 c2 4 dE
=--lnZ =
- 9V2 cf3 9V
2
orl
but by (2) in - 3 -9.5, E = 2 V P•
Thus (6p)2 2 cp 2k.T2 bP
=
'3"l cl~ = oT 3V
(c) Substituting p - ~ kT - V , W? find
--? 2 2kT2 N -2,
(IT kT) 2
(6p)- /p = - (- kT) ~ = 3N '3"l V T V
*9.7
The probability that a state of energy€ is occupied is proportional to the Boltzmann factor
-€/kT
e 0
• Since the vibrational levels have a spacing large compared to kT, we see that
-€/kl' 0
e O is small and we may neglect this contribution to the specific heat. On the other hand,
the spacing between rotational levels is small so that these states are appreciably excited.
For a dumbbell molecule the rotational energy should be
where A is the moment of inertia and w the a.ngul.a.r frequency. We have neglected rotation about
the axis of the molecule because the moment of inertia is small. At room temperature, we
may use the equipartion theorem yielding €r = kT
0
• Adding the energy of the translational
motion gives for the total energy
€=et+ Er= i kTO + kTO = i kTO
Thus the molar specific heat is
The reaction proceeds according to
and the Jaw of mass action becomes
21ID =
= K = n (1)
From the results of section 9.12, we can immediately write dmm the quantities ;. In the
notation of that section,
68
l t3€' ~ kr
- ! t,nw
~2
V [2rr(2m) kT]2
R2 2 H2
- h3
e
2fi
2
e
3 t3e~ ~kr
1 .
- - f,·i\ w_o
= y_ [2rr(2M) kTJ2
2 --
~D
2 2 e 2 e
2t? 2 h3
l f,€' ~kT
1
s V [ 2rr ( m-tM )s:r ] 2 e HD - 2 t3n"1m
HD - h3 112
e
where we have assumed that molecules are predaminantzy in the lO'W'est vibrational state and have
treated the rotational degrees of freedom classicaJJ.y.
The electronic ground state energies€' are approximatezy equal for the three species since
this quantity is altered onzy slightzy by the addition of a neutral particle to the neucleus
(to form D). Thus
(2)
Similarzy, the moments of inertia are of the form
A = !_ µ * R 2 = !_ ~m2 R 2
2 o 2 m
1
-lill2 o
where R
0
, the equilibrium distance between nuclei, depends primarizy on electromagnetic forces.
It is then approximatezy the same for the three species, and we have
-2 m M mM -
2
= ~ (mmM+M)
2
~
2
~
2
~ = <2) (2) (m+M) 4 (3)
FinaJJ.y, the frequency for two masses coupled by a spring of constant K is
Again, K must be the same for the three molecules, a.~d
w =
HD
(4)
Substituting the ~'s into (l) and using the results (2), (3), and (4) yields
Since 7I = 7) , we hz.ve
2 2
or
9.9
~ l(m·+M)2l
L mM j
%D -=
N
exp
In a thermalJ.s" isolated ~uasistatic process, there is no entropy change. Hence by the first lav,
TdS=O=dE+pdV
Substituting E = Vu and p =,½ u where u(T) is the energy density, we have
- - l-o = u dV + V du + 3 u oY
Thus -l 8v.!t rN'
3 V' V J du 1Tf dT'
= = 4 '.fr - 4 since u ex: T.
li. Tf
Integrating, we find - .,;. 1n 8 = 4 1n -
3 Ti u T.
or
9.10
:L
Ti
T =f 2
(a) By the fundamental thermodynal!lic relation
TdS=dE+pdV
Then substituting E = Vu and ii 1-= 3 u, we have
- - 1- 4 - -TdS=uoY+V~+ 3 urN=ruoY+V~
du But since u is onzy a function of T, du = dT dT, a.."ld. it follows that
dS 4 ~ dV .L y d.u dm = -3 . .,.
T T dT -
We also have ds = (~s) dV + (c8) dT
oV T 2;T V
Equati.Ilg coefficients in (1) and (2) yields
4u
(~s) = -3 -T and
oV T
(b) Since (c2sfev 6T) = (d2S/'cT QV), it follows from (3) that
70
(1)
(2)
(3)
or
Thus 4 - - 4 1n T = 1n u + C where C is the constant of integration. Finalzy u ~ T.
The energy of the black body radiation in the dielectric is
where c' is the velocity of light in the material. c' is related to the velocity of light in
vacuum by c 1 = c/n, n being the index of refraction.
0 0
Thus
2
- 7r E =V-15
3 4
(kr)4 n 3 = 4cr n0 VT
(ct1)3 0 C
Here CJ is the Stefan Boltzmann constant, cr = n2k
4/6o c2
i'i3•
16 CJ n 3 vr3
0
Hence =------
C
(1)
Substituting v, the volume per mole, for Vin (1) gives the heat capacity per mole,~'· The
classical lattice heat capacity per mole is~= 3R. The ratio of these two quantities is
c' 16crn 3 vr3
V o -=
CV 3Rc
For an order of magnitude calculation, we can let v = 10 cm3 /mole and n = 1.5. At 300°K, we
0
9.12
(a) The total rate of electromagnetic radiation or :pover is
2 2 h.
P = @(4n r ) = 4rrr oT ·
l.L
= (4n)(l02)(5.7',<10-5)(106) · = 7.2xio22 ergs/sec
7.2 x 1a15watts
(b) Since the total energy/sec crossing the area of a sphere of radius r' = llm must be the
same as that crossing the sphere of radius r = 10 cm, we have
or
71
ll 2 4 / 2 = 5.7 X 10 ergs/sec c:m = 5.7 X 10 watts c:m
(c) The power spectrum bas a maximum at
tiwmax he
- ----=3 kr -krA.
max
Thus
(a) The power given off by the sun is
4 2
P = cr T (4n R)
0
Half the earth's surface intercepts radiation fram the sun at a:ny time. Renee the ~ower that
the earth absorbs is
The power that the earth emits is
In equilibrium, the power incident upon the earth is equal to the power it radiates, or P.= P
1 e
Thus
(b) From Eq. (1)
9.14
T = T /R
o "-J 2L
T = (5500)
(1)
We consider the liquid in equilibrium with its vapor (at vapor pressure p). By (7.12.13), the
flux of moJ.ecules incident on the liquid is p/ ./27r m kr' , and this must also be the flux j(
of particles escaping from the liquid. Since J{ cannot depend on the conditions outside the
liquid, this is the escaping flux at all pressures.
Thus J{ = p
.J27r m kr'
The mass of a water molecule ism= 18/6.02•1023 = 2.99-10-23. 0 Thus for water at 25 C where
4 2
P = 23.8 mm Hg = 3-17•10 dynes/cm we have
)( 3.17•104
= (27rX2.99 X 10-23x1.38>0..0-l~ 298)½
= 1 _14 X 1022 molec~es
sec•cm
At the vapor pressure p(T), the flux of molecules of the vapor incident on the 'Wire is by
(7.ll.13), cp. = p/ ,J2rr m kr'. When the metal is in equilibrium 'With its vapor, the emitted flux
).
must equal the incident flux, cp = cp .•
e i
Then when the temperature is T, the number of particles
emitted from the 'Wire per unit length in time t is
N = p(T) 2rrr t
-J2rr m kr
1
But N = 6M/m where liM is the mass lost per unit length and m is the mass per molecule.
Thus 6M = p(T) 2rrr t
m ,./2:rr m kr •
or p(T) =~ ~
where ve have usedµ= mNA and R = NAk.
v = o by symmetry.
X
By definition,
J d3y f(!) v}
= J d3y f(:~:)
(1)
At absolute zero, the particles fill all the lowest states so that the distribution f(y) is just
a constant a.nd cancels in (1).
2 12
= µ. By sylllllletry, VX = 3 V
and (1) becomes
9.17
(a)
At T = O all states are filled up to the energy µ. Hence
the mean number of pa..-ticles per state is just 1 and we have
E = f µ e:p (e:) d€
0
1 1
By (9.9.19) V (2m)2 2
p(€)d€ = ~ 3 € de;
4?T ii
where the factor 2 is introduced since electrons have two spin states.
3 3 3 5
E = v2 (2m)2 Jµ €2 d€ = v(2m.)2 µ2
2n' ii3 0 5rr2 ii3
73
(b)
Since
"tl.2
µ = 2m
2
3
(3rr2 i) by (9.15.10), it follow-s that
E = i Nµ
(c) Substituting the expression forµ in (1) we have
2 2
3 2 3
E = io (3rr2) ! (i) N
(1)
(2)
As particles are added, each new particle can go only into an unfilled level of bigher energy
than the one before. Hence the energy is not proportional to N as it vould be if particles were
added to the same level.
(a)
(b)
Differentiating E in 9.17 (c) gives
- 2
2
/ 3 ~2 5/ 3
(~E) -~ cu N
avT - 5 m (;,) p = -
Again from 9.17 (c) we find
2 2/3
"O = g ! = (3rr)
- 3 V 5
"tJ..2 N 5/ 3 - (-)
m V
(1)
22 . 3
(c) For Cu metal, N/v = 8.4 X 10 electrons/cm • Substituting into (1) where m is the mass of
the electron, we find p:::::: 1r? atom. 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 o°K.
9.19
From the general relation TdS = cl. Q = CdT, we have that if the heat capacity is proportional to
a power of the temperature,C a: Ta
6S = J C; a: J Ta-1 dT a: Ta
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.
Renee the ratio of the final and initial entropies is
sf 400
-=-=2
Si 2CQ
(b) The energy of the radiation field inside an enclosure is proportional to T4, and the heat
capacity has a T3 dependence.
Thus
9.20
(a) At T = O, the total number of states within a sphere of radius tt must equal the total
F
number of particles N.
Thus
or
2 _V_ (}± 7r tt 3) = N
(2rr)3 3 F
2
><2 2 3
µ = ~ (3rr i) (1)
The number of particles per cm3, N/v.,is NAp/µ -where NA is Avogadro's number, p = .95 gm/cm3,
andµ= 23. From (1) ve find the Fermi temperature TF = µ/k = 36,0C0°K.
(b) To cool 100 cm
3 or 4.13 moles of Na from 1°K to .3°K, an amount of heat Q must be removed.
-where~ is given by (9.16.23),
Thus Q = (4.14)(; R)(~
2
~') 1i·3
T dT = -.0022 joules
Hence .0022/.8 = .0027 cm3 of He3 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 x.._ = Nµ 2/kT -whereµ is
MB m m
the magnetic moment. In the actual case, most electrons cannot change their spin orientation
-when an external field is applied because the parallel states are already occupied. Only
electrons near the top of the distribution can turn over and thus contribute to x. We expect
then, that the correct expression for Xis \re with N replaced by the number of electrons in
this region, Neff.
-where TF is the Fermi temperature. 2
µm
Thus X = Neff kI' = (1)
(b) Since TF is on the order of 104 or 105, (1) yields a X per mole on the order of 10-6. If
electrons obeyed MB statistics, X -would be changed by the factor
75
\m
-- =
X
This is on the order of lCO at room temperatures.
9.22
When the field His turned on, the energy levels
of the parallel spins shift by -!lmH vhile the
levels of the antiparallel spins shift by -tµmH·
The electrons fill up these states to the Fenni
energy µ as shovn in the figure. The magnetic
moment is detennined by the number of electrons
vhich spill over from antiparallel states to
KINETIC AND MAGNETIC
EN:::RGY
µ
te
Density of
States
parallel states to minimize the energy. This is (µmH)p(µ
0
) where p(µ
0
) is the density of states
at the Fenni energyµ. Consequently the magnetization is
0
9.23
!:!H
V
N
where n = V
If we choose the zero of energy to be that of the electrons in the metal, the partition function
of an electron in the gas outside the metal is by (9.10.7)
s = 2v (27r m k.Ti e-vjkT
h3
where the factor 2 occurs because electrons have two spin states. The chemical potential µ
0
is
:!Ound from the Helmholtz free energy.
F = -kl' 1n Z = -k.T [N 1n ~ - N 1n N + N]
= ~ = -k.T 1n 1 = -kl' 1n 2V3 (2nm kT)~ e-vjkT
~ ~ N ~
The chemical potential inside the metal is justµ= V - ~, and in equilibrium the chemical
0
potentials inside and outside are equal, i.e.,µ=µ.
0
Thus
Hence the mean number of electrons per unit volume outside the metal is
l
n = TI = g_ (271" m 'kr)2 e _q,/kl'
V h3 (1)
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
q,e = (1-r) q,i (1)
The electron gas outside the metal may be treated as a classical ideal gas from which it follows
that the incident flux is
~
- ]l 3 -
- 2 1 - 1 2 2 _q, 8kT
•i - 4 nv - 4 ~3 (2mn l<T) e /l<T l""'
where we haved used (1) of problem 9.23. The emitted flux is then given by (1). 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
where e is the electronic charge.
9.25
= 4"rr m e~l-r) (kT)2 e-q,jkT
h
There are (27r)-3 d3_!t states per unit volume in the range (1t:d1t) and with a particular direction
of spin orientation. Since the mean number of electrons with one value of K is ~/3(€-µ) + ~
If we take the z direction to be the direction of emission, the number of electrons with wave
vector in the range (~:d~) and of either spin direction which strike unit area of the su....-f'ace
of the metal per unit time is
(1)
Then the total current density emitted is given by summing (1) over all values of K which will
allow the electrons to escape. That is, the part of the kinetic energy resulting from motion in
the z direction must be greater than V, or ~2
K
2/2m > V 1 where we use the notation of problem
0 Z 0
77
9.23. In addition, we assume that a fraction r of -electrons of all energies are reflected.
Thus the electron current density is
J = e~ = 2e(l-r) 1= dit 1= dit 1=
e ( )3 X Y 2rr ...., ...., ",./2mv )11
(m /m) dit z z
[
~11
2 2 2 2 1 + exp -;:;::-(K +K +K )
where e is the
Hence
=i X y Z
electronic charge. The integral over K is of the form z
2 2
= J xe ;x -+a = _ ½ 1n [l+e -x -+a]
e-x -+a+l
Since V
0
>µand, in general, I~ (µ-V
0
)1 >> 1, we can use 1n (l+x)::,,, x to expand the logarithm
and obtain
J = 2e (1-r)
(2rr)3
F.a.ch of these integrals is
(a) Since the integrand in (9.17.9) is even, one can write
= 2
where we have integrated by parts. Since the first term on the right vanishes
I = 4r = X dx
2 'J , X o e +1
(1)
(b) This expression can be evaluated by series expansion. Since e-x <lone can write
-x
~ = ~ = e-xx [l - e -x+e -2x-e -3x •. •]
ex+l l+e-x
= ~ (-l)n e-(n+l)x x dx
n=O
Hence (1) becomes
I = 4 ~ (-itf co e -(n+J.)x x dx = 4 ~ ( -J.): Jco e ""'Y y ey
2
n=O o n=O {n+l) o
Ti.le integral is, by appendix A.4, eg_ua.J. to unity. Thus
(c)
C' C ,,,·· -.......
, \
Fig. l Fig. 2 Fig. 3
One can use the contour C of Fig. (1) to write
J 2 l
C z sin 71" z
co '-J.'n co (-it
dz= 27T:i. I: ~ = 2i I:
n=l n
2
71" n=l n2 (3)
J 2 dz = -2i S
C z sin 71" z
or (4)
Here the integrand ~ 0 for z ~ co. The sum in (2) is unchanged if n ~ -n. Hence one could eg_ually
veil write the symmetric expression
J dz = -2i S
2 .
C' z sin 71" z
along the contour C' of Fig. (J.). Adding (3) and (4) one gets then
J 2
dz
= -4i s
C+c' z sin nz
(5)
(6)
But here the contour can be completed along the infinite semi-circles as shown in Fig.(~ since
the integrand vanishes as lzl~co. Except for z = O, there is no other singularity in the
enclosed area so that the contour can be shrunk down around an infinitesimal circle C surroun
o
ding z = 0 as shown in Fig. (3). For z ~ O,
Hence
l
z2sin 71" z
l l l 2 2 -l
= 2 l ~ 3 = 3 [l - -r;rr z ••• ]
z [m - 0 z ••• ] m
l 1 22 l 71"
= 3 [l + b 71' z • • • J = 7 + 5z +
7fZ 7fZ
f 2 ~z = < -2n:i. Hl)
C z sin m
0
79
Thus it follows by (5) that
or
9.27
(a) We expand eikx in the integral
J(k)
=!~
i-rr2 - T = -4iS
_ ~ ( -l)n _ 7r2
S = n~l n2 - 12
Interchanging the sum and integral yields
(1)
(2)
(b) Putting x=z in (1) to denote a complex variable, we evaluate the integral on the closed
contour sho'Wll in the figure. Here C' is the path where z is increased by 27ri and C" is a small
circle around the singularity at i-rr. Then the integral on the complete path is zero since there
are no singularities in the region bounded by the contour.
C'I
C"
C
But the integral is Oat x =±cc and cancels along the imaginary axis. Hence
=J
ikz e dz
(3)
where both integrals are evaluated from left to right. Along C' vhere z = x + 27ri, we have
since ex+27ri = ex
eikx dx
80
The last integral is just J(k). Thus f'rom (3) we have
ikz ikz
(l-e -2rrk) J e dz = J e dz
C (ez+l)(e-z+l) C" (ez+l)(e-z+l)
(4)
To evaluate the integral around the small circle c", we put z = 7Ti + t; and expand the integrand
in tem.s oft;
J
ikz d e z ikl; -7Tkf e I' = e-1Tkf [l+iks+- •• ) a" i'
= e (-t;)( t;) d~ -1;2 ~
C" C"
= e -7Tk (9+c2m) (-ik~ = 2-rrke -7Tk (5)
where we have used ex= .l+x+ ••• and the residue theorem. Then combining (4) and (5) we find
7rk
= sinh 7r k
(6)
(c) Expanding (6), we find
2 4
( ) 7rk (1rk)"' 1(-gk)
J k = sinh 7r k = 1 - ---r + 3 0 - • • •
Then com!)aring coefficients with the series in (2) we can read off the Im's. We find r2 = 1r2/3.
(a) According to equation (6.9.4), the probability of stater of the system is
-pE -aN
pr c:: e r r
Then the mean. number of particles in the states is
L. n s exp [-p(nl€1 + n2€2 + ••• )- a(n1+n2+ ••• )]
r n = s L. exp [-~(nl€1 + n2e2 + ••• )- a(n1+n
2
+ ••• ))
r
where the sum is over all n1, n2, ••• withcut restriction.
Letting
we have
l = L. exp [-i,(n
1
€
1
+ n
2
€
2
+ ••• )- a(n
1
+n2 + ••• ))
r
n=
1 'a~ 1nZ
s 13 "'s
(1)
where the partial denotes taking the derivative where €s occurs explicitly. Evaluating the sum
(1) in the usual way, we obtain
81
Thus
and
{ -{~.l~ (- -<n"il•2i\ ..
1-e )} 1-e })
-a-Be
1n l = - L. 1n (1-e · r)
r
(2)
(b) Using the same argument vhich leads to (9 . 2 .10 ) v ith the function £. ve obtain
---=-2 1 c,n
(lms) = - 'i3' c/
s
where again the differentiation is performed where es occurs explicitly. Thus
a:-lj3e
2 e s ~ 1
("- ) -- ----- - ,.., ( ) w.u = ns- 1 + -_
s a+i3e
(e s -1)2
= ii (1+n ) s s
n
s
The correction term in (9.6.18) is absent sho,n.ng that if on1~ the mean number of particles is
fixed, the dispersion is greater than if t he total number is known exactly.
(c) For FD stati stics, the sums in (2) become
l = ~ + e -(a:+i3e1J
bl'
-a-;3e
= L. b. (l + e r)
r
~hus 1 " l 1
:J. = - - ~ 1-'1 = a:+5e s p oes . s
+ l e
and
1 ens
a-lj3e
(lm)2 =
s
- 2cl V e --~ = a-lj3e = ns n -l p oe s (e s+1)2 s
- (1-:n) = n s s
82
We can remove the restrictions on the sum over states by inserting the Kronecker o :f'unction
where the sum is over n = 0,1, ••• for each r. Letting o: = o:+io:' and interchanging the sum and
r -
integral yields
where
This sum is easily evaluated as in section 9.6 and
-a-f,€
lnt (!!) =~ 1n (1-e - r)
r
(l)
(2)
To approximate (1) we eA-pand the logarithm of the integrand about o:' = 0 where it contributes
appreciably to the integral.
1n e<PT l (g) = ~ + 1n l (g)
= (a+ia')N + 1nZ(o:) + Bl(ia') + ½ B2(ia 1
)
2
+
where
i(N+J\)o:'
Hence e
As in section 6.8, we optimize the a:pprox:i.mation by setting
or
and (3) becomes
N + o 1n ! (o:) = 0
l:g ,2
o:N aN "":2 20:
e-- [ (g) = e l, (o:) e
Substitution into (1) yields ~ 1 ]3 12
o:ri ,z ( ) f 7f - - 0: z = e ~ a e 2 2 aa'
4T
or
-a-f3€.,..
1n z "" aN + bl (a) = o:N - I 1n (1-e -)
r
where we have used (2). The vo.lv.e o: is determined by (1.:.)
83
1 2
- - B o:' + 2 2
(4)
(5)
which gives
CHAPI'ER 10
Systems of Interacting Particles
10.1
2 2
Since cr(w) dw cc K dtt and for spin waves ro = AK , it follows that
Then by (10.1.20)
Substituting the dimensionless variable x = ~1'.m we obtain
The integral is just a constant not involving~. Hence
10.2
(a) By (10.1.18)
where 3V 2
2iT
2 3 ill
1 C
for ro < ~
for ill > ~
But since , we find
In tenns of the dimensionless variables x = ~:,m and y = ~~, this gives
1n z = YNn - _2li Jy 1n (1-e -x) x2dx
~ y3 0
This expression can be put in terms of the :function D(y) by integration by parts.
84
Thus
where keD = ~wD.
(b) The mean energy is
Thus E =
= 'Y ~ - 3N 1n (1-e-y) + ND(y)
-eJT
= ~ - 3N 1n (1-e ) + ND ( e Irr) i1wD D' ~ (1)
= -N71 + ~:r D(y) = -N71 + 3N kT D(eJT) (2)
(c) The entropy may be found from the relation S = k[ln Z + l3E]. Then from (1) and (2), we
have
s = Nk
= Nk
[-3 1n (1-e-y) + 4D(y)]
-eJT
[-3 1n (1-e ) + 4D(eJT)]
For y >> 1, the upper l.illlit in the function D(71) may be replaced by co.
D(y) = 1_ !co ~dx
y3 o ex-1
h
The integral is shown to be 7T '/15 in tables.
Hence
4
D(y) = 7T 3
5Y
y >> l
For y << 1, we can expand ex::= l+x in the denominator
D(y) =; ~7) x2a.x = 1 y << l
From the results of' problem J.0.2, we find for y >> 1, or T << eD when we use y = ~~ = eJT
N .. ...._4T3
1n z = @¾ + -"'"--
.<>.:.J. 5e 3
D
'srr4 NkT4
E = -N~ + - 5- ~
D
4,n-4 c!-/
S = - 5 Nk eD
For y « 1 (T >> eD) we find, using ey :::::: 1 + y
10.4
1n Z = ~ - 3N 1n er111 + N
E = -IiTj + 3N 1d'
s = Nk [-3 1n er/1' + 4]
- 10 The pressure is given in terms of the partition function by the relation p = ~ oV 1n Z where by
equation (1) of problem 10.2
-eJT
1n Z = ~ - 3N 1n (1-e ) + ND (eJT)
Thus
-eJT
- _ N £!1 + 3N kT e
P - ov -eJT
where oD(GJT)
dD(eJT)
1-e
In terms of,= -(v/eD;(deJdT), this becomes
p = N .s.!l + 3,NkT D(e frr)
oV V D" ~
, 4 4
- 2!1. 37T ?' N kT
p = N 2V + -5- V e 3
D
For T » eD, D( eJT) ~ l
p _ N ,Q:! + 3?' N kT
- oV V
Comparing the expression for the energy E, equation (2) in problem 10.2, vith equation (1) in
problem 10.4, ve obtain
- on {E+Nn)
p - N = + ?' -------~-- oV V
86
Thus a=
10.6
(a) Fran electrostatics,
(b) The induced dipole moment in terms of the polarizability a is
(c) The energy of a dipole 1
2
in a field Eis
u = -g
2
·~ a: p
2
E a: R-6
By equation (10.3.8) and (10.3.16), the log of the partition function is
l.n Z l.n ~! ~~ ~ + N l.n V + ½ NnI (~)
where n = Njv. The first two terms are the ideal gas partition functions while the last is the
contribution of the interaction. Since the interaction is represented by an additive term in
1n z, it must also be represented by an additive term in the entropy, S = k [1n Z - ~ ~~ 1n z]
Thus the entropy per mole is
where
and
S(T,n) = S (T,n) + S'(T,n)
0
S'(T,n) = k [½ n I(~) - t3 ~(½ n I(B_V]
To show that A(T) is negative, we must show that the derivative c/c~ [I(~)/S] is positive.
We have
but
(~ ;:,. Jo:,
= c~ o
-~ue -~u -e -~u+l = 1 _ (~ > 0
~u e
since (1-+f3u) is just the first two terms in the expansion of e~u. Thus the integral is positive
and A(T) is negative.
Also
10.8
From the arguments of section 10.6 ve can write down the partition function for a two dimensional
gas
z = 4 ('arm/ z
N. h2/3 U
vhere 1~
1n zu = N 1n A + 2 A I(/3)
I(/3) = J'\e -/3u_1)27rR dR
0
Denoting the film pressure by Pr, ve have
1 cl
Pr = i cA ln Z
Pr N 1 if ------I kT - A 2 A2
or
l
= - - I. 2
10.9
(a) One can write dovn the mean energy from (10.7.3) and (10.7.5)
E = -Ngµ H. sj = -NS lcDc B (x)
0 1 Z S
Here N is the total number of particles. The consistency condition is
We solve this equation in the given temperature limits using Fig.(10 .7.1)
T << T
C
(1)
(2)
Here the slope, kT/ 2n JS, of the straight line is small and the intersection with B (x)
s
is at large x vhere B (x) == 1. Thus (2) becomes
s
and
T == T
C
or 2n JS X=-w
E = -NS kT (2n JS)(l) = -2n NS2 J (3)
In this region kT/2n JS is large and the intersection is at small x. We cannot, however,
approx:il!late Bs(x) by the linear term ½(s+l)x {eq.10.7.12) since this vould not cross the
88
line (kT/2n JS)x. Hence ve must evaluate the second term in the expansion of B (x). For
s
small x,
1 X x3
coth x ~ x + 3 - 45 X << 1
Substitution in Bs (x) = ½ 8s+½)coth (s+½)x - ½ coth ½ j yields
B6 (x) ~ ½ {<s<½) [cs4)x + }<s<½)x - fycs<½)
31-½ E + ~ · ~~ }
Bs(x) ""½ (S+l)x - (s2+s+½) (sit;) x3Then (2) becomes 2~JS x = ½ (s+l)x - (s
2
+s+½) (sit~) x3
or
•
2
= 52 ~<½ [½ - 2n JS~+lJ
In the energy (2) we substitute B (x) to first order and use (4)
s
E = -NS kT x Bs{x) = - ts(s+l) kT x2
= _ 5NS(S+l)kT [ 3kT ]
s2+s+½ L -2n JS(S+l~
T >> Tc
T ""T
C
(4)
(5)
From (4) we see that x=O at kT = kTc =} nJS(S+l), i.e., the energy becomes Oat this point,
and larger T gives imaginary x in this approx:i.:mation.
Hence E=O T >> T
C
(6)
(b) From equations (3), (5), and (6) we have
~o T << T
C
C = ~ = 15N k
2
T ~- ! ~ T ""T
2fr nJ(S2+s+½) 3 kT
C
0 T>T
C
( c) C
89
CHAPI'ER 11
Magnetism and Low Temperatures
11.1
The mean magnetic moment is
2'.(-oE_/oH)
- r lolnZ
M = ----"""t,E=----- = ~ oH
-t,E
where Z =Le r.
r
11.2
2'.e r
r
The change in entropy per unit volume is given by
H
f;;B =f O (~) dH
o oH T
But from the fundamental thermodynamic relation dE = TdS - MdH, we have the Max1,ell relation
oFr
(~) = (--2)
oH T dT H
where Sand M are the entropy and magnetic moment per unit volume. Since M = XH and
0 0
X = A/(T-e) we obtain
oM AH Cy/) =
-(T-e)2 C H
Thus
H H 2 0
AH A
f;;S dH = -
0
= J (T-e)
2 ? 2
0 (T-a)-
11.3
The change in the molar volume when the field is increased from H=O to H=H is
0
f H "
6v = o o (g)
cH T ,P
But from dE = TdS - p dV - MdH, we see that
where v and Mare the molar volume and magnetic moment.
we find
AB a H
0 =------
[T-e (l-faP) )2
0
90
dH
Since M = XH and x = A/[T-e (l;ap) J
0
and
H
£:;.v = { o _A_e_0_a_Ha._H __ = _ __ A_e"""0_a __ ..,,.
o [T-0
0
(1-+a p)J2 [T-0
0
(1~)]2
u.4
H2
0
2
(a) From the relation d.E' = TdS - pdV + HdM where E' = E - v'Ff-/&r we obtain
cos) = c~M)
~ V,T aT V,H
In the normal state, M = VXH"" 0 since X"" O. Thus (oS/oH)V,T ""o. In the superconducting state,
4-rrM
B=H+-=0
V
or
Thus Mis independent of temperature and (cS/oR)V,T = 0
(b) Since the heat capacity is given by C = T(osjoT) we bave fram (11.4.18)
V dH S-S =r.-::H-
s n Ll'1f d.T
where His the critical field.
( c) When T = T , the critical field is o. Thus
C
VT dH 2
C -c = ~ (-) s n Ll'1f d.T
11.5
From (1) in problem 11.4 and H=H [1-(T/T )
2
) we obtain
C C
But since C ==aT + bT3, we have n
VH
2
3VH
2
C -C = - _c_ T + c T3
s n 27rT 2 27rT Ii.
C C
(1)
(1)
To satisfy the condition C /T ~ 0 as T ~ 0 we must set the coefficient of T in (1) equal to 0 1 . s
i.e., a= VHc
2/2rr Tc 2.
91
Thus
ll.6
(a) The entropy is related to the heat capacity, dS = Cd.T/T. Then since at T=Tc and T::0 the
entropy of the normal and superconducting states is the same, we have
Fram which it follows that
f
Tc C d.T f Tc C dT
s(Tc)-s(o) =
0
T =
0
+
(1)
(b) Since dE = Cd.T, the energy at T=T and T::0 for the normal and superconducting states is
C
Tc 7T 2
En(Tc) - En(0) = f 7T dT = +
0
T ,_,n 4
f C 3 u..i..
Es(Tc) - Es(0) = ctr dT = +
0
But E (T) = E (T) and by (1) a= 3,/1' 2• n C S C C
Thus
ll-7
7T 2
E (0) - E (0) = -~ s n '+
From the relation dS = CdT/1' we have for the entropy in the normal and superconducting states
T
Sn (Tc) - Sn (0) = f c ,dT = )'Tc
0
Tc ctr 3
S
6
(Tc) - S (0) =J efdT = _c
s O 3
2 Since the entropy is the same for both states at T=0 and T='rc' it follows that a=3r/l'c •
Hence at T=T
C
92
CHAPl'ER 12
Elementary Theory of Transport Processes
12.1
(a) 6
12.2
(a) ,E,
(b) 6
(b) ,E,
(c) 6
( C) {,
(a) The equation of motion for an ion in a field C: is
Hence
By ( 12. l. 10) ,
and
and
mx = e!
e e t 2
X=m 2
t2 = 1= t2e -t/r d.; = 2T2
0
e! 2
X =-T
m
(b) The time to travel a distance xis
t=~; =£ T
Thus the fraction of cases in which the ion travels a distance less than xis
f ,/2' T -t/T dt -V2 s = e - = 1-e = .757
0 T
12.4
The differential cross section is given by
cr(n) = -2rrs ds
d!2
where d!2 is the ele::nent of solid angle ands is
the "inpact parameter" as shown.
From the sketch, s = (a
1
+ a2) sin q>
Since q> = 90-e/2 and dn = 2rr sine' de', it follows that
ds
an =
-(a
1
+a2 ) sin 81/2
47!" sin e'
93
Hence a(e)
12.5
ds
= -2rrs an=
(a +a )2 sin e 1/2 cos e 1/2 1 2
2 sine'
By arguem.nts similar to those of section 12,1, one has that the probability that a molecule
travels a distance x without collision is
p = e -x/~
where tis the mean free path. Letting P = .9 and t = (ff ncr)-l yields
-ln .9 = x ./2 no = x ./2' pajkI'
Letting x =20 cm and T
-4
= 300°K gives p = 1.14 X lJ llllll Hg.
12.6
The coefficient of viscosity is
(1)
where we have used t = ( ~ n cr ) -l. To determine the cross section, we not e that the volume
0
per molecule is A/NAp where A= 39.9 gms, the atomic weight, and p = 1.65 gm cm-3, the density
in the solid state. T::us the radius of' the molecule is
and
From (1) we find 11
the observed value,
12.7
r =
3 A 1/ 3
(47r N P)
A
= 4rrr2
= 5.6 x 10-15 cm2
00
-- 1.1 x ,,0-4 gm cm-l sec-1 • Th ·h t· f th 1 ult d 1 ~ t ..,_ · en .., e ra io o e ca c a e va ue 0.1. 11 o
-4 -1 -1. 11 = 2.27 X 10 gm cm sec ,is
l 1
The coefficient of viscosity is proportional to T2 since 11 = 'IIN/ 3 ..f2 a and v = (8kr/71Ill)2 •
0
Thus if' the temperature increas s, the terminal velocity decreases. The velocity is independent
of pressure.
12.8
The force on the inner cylinder is the pressure p times the area A= 2rr RL,
'"'U .9- '
= pA = 2rrRL 11 ~
We make the approxillla.tian that the air at the outer cylinder is moving with it, i.e., at
velocity ro(R+o) ~ cuR, and the air at the inner cylinder is stationary.
Thus
and the torque is G = ..7R
(b) The viscosity is given by Tl = mv/3 ,.[2 a •
0
Since air is ma.inzy composed of nitrogen, we
let m = 28/6.02 X 1023 , the mass of a nitrogen molecule. -15 2 0 Putting a ~ 10 cm and T = 300 K,
0
we find
and
12.9
(a) We let
T) = 2 x 10-4 gm cm-l sec-l
G = 8 dyne cm
wherex,y, and z are to be determined. Letting L = length, T = time, and M = mass, the
dimensional equation becomes
But since C = FRs, it has dimensions MT-2Ls+l
Thus
which gives x+(s+l)z = 21 x+2z = 0 1 y + z = 0,
On solving this system of equations, we find x = 4/1-s; hence the cross section a is proportional
0
to v4/(1-s).
(b) The viscosity is given by T) = riv/3../2 a
0
and here a
0
~ v4
/l-s. Since velocity depends on
the square root of the temperature, we have
-2
1
) 4;tr.-s) C..+71V2(s-1) T) ~ T T = 'I!° h
12.10
(a) Consider a disk of radius rand width t::;x.
at the point x. There is a pressure difference
2 .C:.p across the disk giving force -rrr t:.p and a
. ou(x,r) viscous retarding force 27rT t:.xT) or • Hence
the condition that the disk moves without
acceleration is
95
'
\ I
\. __ ,.,
~x-1~1-
But since t::.p ~ - cpfx) t::.:x., where the minus sign is introduced because p(x) is a decreasing
ox
function of x, we have
2 op(x) _ 2JT 6U(r,x)
71T ox - r7J or
Assuming that the fluid in contact with the walls is at rest, it follows ihat
For a liquid, ~~
f
u(r)
du'
0
1 ::,_~/...,.\ r, r ~J r'dr' ox a
The mass flowing per second across a ring of radius r' and width
dr 1 is pu(r')2rrr'dr 1 where pis the density. Then the total mass flow is
f a Ja 2 2 M = pu(r')2rrr'dr' = li £..... (p -p) (a -r' )r'dr'
2 TJL 1 2
0 0
4
= ~ ~ ~ (pl-p2)
(b) For an ideal gas, the density is p(x) = (µ/RT)p(x) whereµ is the molecular weight. Thus
the mass flow is
.
Since Mis independent of x, we obtain
L 4 p2 • f 1f µa 1 M ax = - -s m p dp
0 TJ p
1
w},..ich yields
4
• 7r ~ a ( 2 2)
M = 16 TJRT L P1 - P2
12.ll
Let T(z,t) be the temperature at time t and position z. We consider a slab of substance of
thickness dz and area A. By conservation of energy, (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 unittime through the surface at z+dz). Since the heat absorbed per unit time is
me oT/ot = pA dzc oTjclt, where c is the specific heat per unit mass and p is the density, we have
pA dz c ~ = A [Qz -Qz+dJ
• A [-• ~ - ~• ~ - • ~(~)0]
and
u.12
The wire gives off heat fR per unit time and length. In order that the temperature be indepen
dent of time, a cylindrical shell between rand r -kir must conduct all the incident fR heat.
Thus
or
12.13
T
-1 o dT = r2R f b dr
T 2rrK a r
2
6.T = T--'l' = I R 1n ~
o 27rit a
(a) The heat in.flux per cm height of the dewar across the space between the walls is
Q = -2rrr
where r is the radiu.s, r ~ 10 cm.
al'
Kdr
We let oT t:1r _ 298 - 273 _ 42oK/
°ar ~ t:.r - 10.6 - 10 - cm
by (12.4.9)
The specific heat per molecule is c = (3/2)k while cr
0
2 -8 2 2
= 47rR = h7r(lO ) en.
(1)
The mean velocity
- 7~ 0 5 is v = (8kT/ 7l!ll)-I • Choosing an intermediate te~perature T = 286 K, we find v = l.2 x lO cm/ sec.
Substitution of these values into (l) yields
Q = l.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 roug~ proportional to the number of molecules per unit volume n. Thus
the heat influx is reduced by a factor of l.O when
n = 1 ~ 10
14
molecules/cm3
l0./2 L cr
0
ham p = nl!:T, we find p ~ 3 x 10-3 mm Hg.
97
(d) cr - ~ 1 Jµ kT gives cr l l x io-l5
-
3
.frr 'I} NJ' l = •
2
cm ,
2. 8 -8 -8
(e) cr = 7T<l gives ~ = l. X 10 cm, d2 = 3.0 x lO cm.
l2.l5
2
cm •
The gases are uniformly mixed when the mean square of the displacement is on the order of the
square of the dimensions of the container. We have, after N collisions,
2
z = r.
i
since the second term is zero by statistical independence.
Then
2 22 2 22
Z = 3 v -r N= 3 v rt
where N = t/-:.
2
Since -r = ,f,fo = 1/ ,/2 n cr v, one has on neglecting the distinction between v2
and v,
z2 = -~-J v kT)t = ~ kT J~ kT t 3412\ crp 5 CTp 7r m
2 h 2 -l~ 2
Substitution cf z = 10 · en, cr"" lC "cm, and the given temperature and pressure yields
sec.
12.16
A molecule undergoes a momentum change of 2µv z in a collision ,.;i th the cube where µ = rnM/m+.'-1,
the reduced mass. Then the force is given by the momentum change per collision t:imes the
number of molecules which strike the cube per unit time, summed over all velocities.
,z ~ 1/2 2100 2
......:Ttot = 2µn (27r) L vz
0
e
We may approximate the exponent since v >> V. Thus
r3m )2 - -(v -V
2 z dv z
- ~ (v -v)2
2 z
e (1-1-f)mv v)
z
and
m8 1/2 2f°" 2 - f3m V 2
-.5tot=2µn(2rr) L vz e 2 z (1+/3mV"zV)dvz
0
The first integral is just the force on a stationary irall as in section 7.14. The resistive
force is then
- 2 =µnvVL
f3m 2 --v
2 z
e
by Appendix A.4 and 7.u.13. Since we expect M >> m, µ. ""'m, and the equation of motion becomes
M ~ = -mn v VL2
dt
V = V
O
exp t ~ n v l~
The velocity is reduced to half its original. value in time
13.1
By (13.4.7)
M 1n 2
C = - 2
mn v L
CHAPl'ER 13
Transport Theory using the Relaxation-Time Approximation
't" V z
2
The value of the integral :is unchanged if v 2 or v 2 is substituted for v 2 • Hence since
X y Z
v 2 = v
2 + v
2 + v
2
, we obtain
X y Z
e
2 J a3 ~ 2 ael = - - v -rv 3 - e
The quantities-rand ogfc,e are functions o~ of v. Thus the substitution d3v = v2dv sine de d~
yields
99
The f'u.nction g becomes the Maxwell velocity distribution
3/2 c~t '£!)
2 mv
- 2kI' l
e (- kr)
In terms of ; = (2kl'/m)1/2,. this is
~ __ -3/2 ne-(v;v)
2
0€ - 2rr .... 5
V
Substitution into (1) of problem 13 .1 with s = v ;v yields
2
cr = ~ {-r)
m cr
where
co 2
8 f -s 4 "' {-r)cr = yJ;
0
ds e s -r(vs)
]3.3
Since € = ½ mrf = ½n (u}+ u/ + Uz
2
), we have
Thus
og _$ o€ _$mu
~ - 0€ dUX - 0€ X
T) = -mf d3u $._ u 2 u -r = -m2f d3u $ u 2 u 2
-r - oU __ z X - 0€ Z X X .
rn· spherical coordinates, the components of velocity became
u = u sine cos cp u = u cos e
X Z
and d3E = rrdu sin e de dcp
Therefore 2 r 2rr 2 f 1T 3 2 f"" o 6 T) = -m cos cp dcp sin e cos eae dU ~ u T
JC') 0 0 O€
2 J1r 2 4 [(X) c 6 = -m 7T
O
[cos e sine - cos e sin e]de dU f u -r
= -m2 4
1r f"" dU $ u6 -r
15 0 0€
But since f 7rdcp f 2rrsin e de = Im-, it follows that
0 0
TJ = - ~
2
J1rdcpf ~in e def dU $ u6-. = 5 0 0 0 O€
100
(l)
Substitu+.ing the expression - J i = -2rr 2
ne -(u;v)2
from problem 13.2 where;= (2kI'/m)1/2 into equation (1) of 13.3 yields
-where
13.5
Ptttting '
( ' )
T] 15..f;
T] = nkT (,)
!
co 2
-s ds e
0
T]
s = u;v
~ -)-1 ( "le. n cr v into TJ = nkI' ,
0
and using v = (8kT/7!lll) 1/ 2 gives
This is larger by a factor of 1.18 than the mean free path calculation (12.3~18) and smaller
than th-2 result of the rigorous calculation (14.8.33) by a factor of l.20.
13.6
We assume the local equilibrium distribution
/3 3/2 1 2
g = n(~) exp [- 2 pmv] (1)
where the tenperature parameter /3 and the local density n are functions of z. Then in the
notation of section 13.3,
df(o) df(o)
-~ on ~ ~ ~= - dt
0
= on ~ - /3
0 0
cg 0 dz(t) og ~ dz(to) n o
= - dn 6Z dt - ~ z dt
0 0
where the last follows since dz ( t
0
) /dt
O
= v z. This expression is independent of t'. Thus
(13.3.5) becomes
df(o) J"" -t'/-r
M =-w- e dt'
0
or (2)
Using (l) this expression becomes
181
ll on 3 ot3 mv
2 ~
f = g-g v z... Ln: oz + 2p oz - 2 c1~
It remains to determine on/clz for which we use the condition
or
v = l: J d3v f v = 0 z n - z
J 3 J 3 2 11 on + 1.. 06 - mv
2
c)~
d 2- g v z - d ::. v z -rg Ln az 2p az 2 ~ = 0
The first integral is zero since the integrand is an odd Iunction of v.
z
J d3y V / g [¾ ~ +(¾ ½ - m;
2
)~] = Q
(3)
where we have assumed T independent of v. By symmetr'J this equation also holds if v 2 in the
z
2 integrand is replaced by vx
or
or v y
2 Then since v2
= v 2 + v 2
+ vz2
, we have
X y
on substituting (l) arid d3y = v2dv sine da d9, and canceling co.Cl!Ilon factors. The integrals
are tabulated in appendix A.4
or l on l 05
n dZ = ~ oz
Substitution of the equation of state n = p,!'kI' = pp yields
or
The pressure is independent of z.
13.7
C:o -:=- = 0 c:.z
Then ( 3) c...'1d
c1s
f = g-gv z -r oz
(r) give
[~P -m;~
g
+ gv z -r dT l,2. _ l:_ mv-27
T dz l? 2 :riJ
(4)
Since there are no external forces and f is independent of time, the Boltzmann equation becomes
of of f-f(o)
V • ~ = VZ a'z = - -r
J.D2
where f(o) = g is
where f(l)<< g we
Thus
the local Maxwell distribution of the previous problem.
have
V ~ = V ~ on + V ~ ~ = - f(l)
z clz z on dZ z Op dZ ..
f = g-v ,. r $ ~ + clg ~ 7
z lEn oz ~ oiJ
Letting f = g+f(l)
This is identical to equation (2) of 13.6. The ·rest of this calculation proceeds as in that
problem.
13.8
The heat flux in the z-direction is given by
where
The integral over g
= ! mf d3v f v v2
2 - z
f=g+-z- - .2._!~ gv-r clT ~ j
T oz 2 2 kT
is O since the integrand is an odd function of v. Consequentzy
1 2 z
m,- 1 clT 1 ~ 3;~ f 3 2 2 -~ r.. 21
Qz = 4 T dZ 1.:1 (27r) J d Y v v z e ~-t3mVJ
We can exploit the synmietry of the integral by replacing v 2 by v2 /3. Then the onzy angular
z
de-oendence occurs in d3v = v2dv sine de dcp whic.h yields 47TV
2dv when the integration over angles - -
is performed. Thus we obtain
/ 1 2 1 J
'Tr !llT clT [ ~ 3 ~ I. Jco 6 -~ Jco 8 -~
Qz = 3 T ~ L(2rr) J L5 o dvv e -;3m o dvv e
The int2grc.ls are tabulated in appendix A.4 •
. ~ = IT m,- oT C (m3)
3
/ ~ ~-~
"',z 3 T oz E 2rr J t lo 7r
2
7
/
2
-~m 105 ( pm) ..., 32 1r
nk2r,, -:,, _2_. 1 oT
2 ~ ,- oz
13.9
Th~ mean free path argument has
l - 4 nk
2
T
K = -3 ncv t = - -- -r
7r m
since t = v,- and v = (8kT/7TII1) 1/ 2 • Thus the constant relaxation time approximationyields a
103
result greater than the mean free path argument by ~ 2
= 1.96.
7T
13.10
From problems 13.4 and 13.8 we find
~ = (5/2)(nksr/m)T = 2 ~ = 5 C
T) n kl' T 2 m 3 iii
where c = (3/ 2)k, the specific heat per molecule. The simple mean free path argument gives
C
T) m
Experiment shows that the ratio (K/ TJ)(c/ m)-l lies between 1.3 and 2.5 (see discussion after
(12.4.13)) so that the constant relaxation t:L..-ne appro::d . .mation yields better agreemer.:t with
experiment than does the mean free path argument.
13.11
In the absence of a temperatur.e gradient, the eqv.ilibrium distribution of a Fermi-Dirac gas is
(1)
In the given situation, f is independent of time but ,rill depend on z because of the temperature
gradient oTfe z along the z direction. ·rhis 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 t he mean drift velocity v to zero.
z
Thus the Boltzmann equation becomes
V
z
of + e C ( z) cf =
dz m ovz
f-f
0
T
We ,rill assume that the temperature gre.dient is sufficie..'ltzy small so that f ca.n be written
f = g+f(l), /l) << g whe!'e /l) !'epresents a s:::iall departure from local equi.lioriu."D. •
Thus
.,.(1)
J.
T
It is convenient to write this in terms of the dinensionless parameter
Thus
) 1 2 )
X = ~(€-µ = ~(2 IllV -µ
~= ov
z
~ (€-µ) ~ = i ~ i
clg ox
dX ~=
z
104
5mv og
. z dX
(2)
(2)
Then from ( 2) we find
It remains to find C from the condition that v = !}'d3v f·v = 0 where f is given by (4).
z n - z
The integral over g is O since
the Fe!"!li level, we have
the integrand is odd. Letting T = TF = const. for electrons near
or (5)
The second integral can be expressed in terms of n, the total number of electrons per unit
~ 1 cg
volume. Since ox= ~v cv
z z
Joo 2 C l }'
00
d l
00
1 J '00
·...,, dvz v ~ = - dv v £5.._ = - [ v g] - - g dv
Z dX pm Z Z dV /3m Z /3m Z
-co z -co -co
where we have integrated by parts. Since g=O at ±co, it follows that
J 3 2 clg 1 J 3 n dvv -.;;-::-:=-- dvg=--
- z ox pm - pm (6)
Proceeding to the evaluation of the first integral in (5), we note that x and ogfc,x are functions
2 2 2 only of v so that we can replace v by v /3.
z
The substitution d3v = v dv sine de d~ yields
4n v2 dv when the integration over angles is performed.
Thus J d3
!_ v2
2
x ~ = ~7T l"° dv v
2
x fx
Using x = 13(½ mv2-µ) this becomes
~ c1-//2foo dx x(x+/3µ)3/2 ~
3 pm -co dX (7)
From the properties of the Fermi distribution, we know that og/ox ~ 0 except where€~µ or
x ~ O. Ti e integrand contributes appreciabzy to the integral only L'l this region so we can
expand the factor x(x+/3µ) 3/ 2 about x = 0 to approximate the integral. Also, the lower limit -13µ
may be replaced by -co with.negligible error.
x(x+pµ)3/2
and,
2
~ (t3µ)3 / 2 (x+ l ~)
2 /3µ
X
e
105
from (1)
Hence (7) becomes
=.!!.(-) dxx(x+pµ) =---µ xe dx+- x e 0-,,- 2 5' / 2 !co l / 2 rx 23/2 3/2ml/2 [!co X 3 la, 2 X dxj
3 ~:m -a, X 5Tf2 t3fi3 -co (eX+l)2 2/3µ -c::o(eX+l)2
(8)
These integrals are the r1 and r2 defined by e~uation (9.16.9). It is shown there that r
1
= O
2
and r
2
= 'IT / 3. Combining the results of (5), (6), and (8), we obtain
and (4) becomes
f = g-v T ~
z F dX
Having found the distribution f, ve can nov calculate the heat flux Q.
z
where
Qz = n <vz(½ mv2
)) =½nm (vzv
2
)
(v v
2
) = -n
1 J d3v v v
2
f
z - z
(9)
The integral over g is zero since the integrand is odd,
Hence Q = - m-rF 1 9.§. ~ d3v v '2..2 x~ - 1 (2µm
3
) l /
2f d3v v 2,l~ 7 (10)
Z 2 t3 dZ ~ - Z dX 3 n fi3t3 - Z d~
The calculation of these integrals is similar to the steps leading to (8), Again v 2 can be
z
2 .
replaced by v /3 and the integration over angles performed to yield 41r, The first integral in
(10) becomes
4nJ= 6 d 2rr 2
7
/
2 1= 5/2 ~ - dv V X ~ = - (-) X (x+pµ) dx
3 OX 3/3m X
0 ..0:,
The second integral in (10) is
J d3V V 2V2 ~ = 47Tf a, dv V6 og
,.. Z dX 3 dX
0
1c6
Here we have neglected the squared term in the expansion since it is much less than 1. The
integral is I
0
+ (5/2 ~µ) r
1
= 1.
Thus -rr2 nk2.r
K =---T
3 m F
13.12
Combining the results of the previous problem and (13.4.12), we find
2 k 2
K/cr = 1L_ (-) T el 3 e
(12)
we find
o I -u -1 -1 At 273 K this becomes K crel = 7.42 X 10 erg cm deg or in more standard units
I 6 6 -6 -1 0
K crel = • 8 x 10 watt obm deg • Experimental values at 273 Kare
element Ag Au Ql. Pb Pt Sn w
6
Zn
K/crel X 10 6.31 6.42 6.09 6.74 6.85 6.88 8.30 6.31
watt ohm deg -1
CHAPrER 14
Near-Exact Formulation of Transport Theory
l4.l
The mean rate of momentum loss per unit volume due to collisions is, by (14.4.22)
(6E) = J J J d3! d3! 1 an' ff1 vcr (v,e')(6E)
! !1 n
(1)
where we assume central forces. We first perform the integration over angles by changing to
center of mass coordinates.
Thus (6p) = m (v'-v) - - -
107
becomes, since v = c + ~ V and v' = c + ~ V'
m - m _
(A;e_) = µ(V' -V)
mml
whereµ= -- , m being the mass of the ions and~, the mass of the heavy molecules. Since
m-tml
the mean values vx and vy are zero, it is clear that (Apx) = (Apy) = O. Hence we are left with
(Ap ) = µ(V ' - V ) and z z z
dn' cr(V,e')(Ap ) z
(2)
V' is given in terms of V and the scattering angles. Using the coordinate system illustrated
z z
in Fig. 14.8.1, we see that
V z' = V'cos(y' ,!) = V cos (y' ,!)
since V' = V. Inspection of Fig. 14.8.1 yields
V cos (Y',~) = v[cos(y,z) cos e'+sin e'cos ~' sin(y,z)]
= V cos e'+ V sin 0 1 cos cp' sin (y,E._)
z
and µ(V '-V ) = µV (cos e'-l)+µV sin e'cos cp' sin (Y,!) z z z
Substituting into (2) we obtain, since J~os cp'dcp' = O,
0
For hard spheres, a is constant and
27rµVzcr J'\cos e'-l)sin e'de' = -4n µVzcr = -µVzcrt
0
where a. is the total cross section. Equation (1) then s:L~plifies t o
t.,
where
-(Ap z) =µat ff V zv ffl d3! d3~1
! !1
(3)
Here u is the mean z component of velocity of the ions. To avoid carrying repititious constants
we let
II$
a= 2 and (4)
l'J8
f can be simplified by making the approximation that v >> u, then
z
We again express! and !i in relative coordinates
Thus
m
V=C+-
1-V
m+ml -
v =c-~V
-1 - m+m -
1
On completing the s~uare this yields
where
Also
[
(oml -o;lm)
(a-+al) .£ + (a+a ) (m+m )
1 1
(a-+al) i2 001 v2 +--
a+al
K = C + -
(~ - a1m)
V
(a+al) (m+ml)
(5)
(6)
a
= K + - 1
V (7) z a+a1 z
We wish to put the integral (5) in terms of y and the variable~ defined by (6). It has been
shown (14. 2 .6) that d3y1 d3y = d3£ d3v. In terms of K we have
d3£ d\ = Jd3~ d3y
It is easy to check from (6) that J=l. Thus (~ becoraes
3/ 2 i2 _1_ v2
- (0.pz) "µat nnl c~:) ff e-(~) e al->a ~+2a u Kz+2u ;:;l vJ
The first two terms in the brackets give integrations over odd functions and hence do not con
tribute to the integral. Changing to spherical coordinates gives
( )
- ~ (001)5/2 ~ Jex, -(a-+<-v )x2 J [J°' -ool v2 f n
- 6pz - n3 at nn1 (a+ai) L7T
O
e ... ~,
0
e a-+a1 v5dv
0
27T cos
2e sin
.:he integrals are found in appendix A. 4 and we obtain
109
mml
whereµ - -- The momentum balance equation is then - m-+ml.
which gives
Since j = neu = o el C: , we have
14.2
ne e + (6.p ) = 0
z
(a) We will assume the distribution function
f = f(o)(l+qi)
where
It is shmm in problem 13.6 that the condition v = 0 yields the relation z
and that the pressure is independent of z.
(b) The Boltznann eq_uatior: (ll".7.12) becomes
~Co) of CO
)
DI = ~ + ~
cf(o) .p
~=""-0
(1)
(2)
(3)
:Because the situation is tir.ie indepencier:t and there are no external fo:::-ces, we are left with
or (~f3 ~;) f(o) v
2
(1 - ~;v-j = ff d3yl. dn 1f(o)fl. (o)Vo 6~ (4)
on using (2),(3), and (J.4.7.10).It remains to determine the :f'unction ~- The left side of (4) leads us to expect that a good
approx:i..n:.a.tion will result if we choose
110
where A and a are constants. ¢ must satisfy the restrictions (14.7.14) to be a good approximation.
These become J d3:::: f( O) V z (l-av
2
) = 0
r d3v lf (o) V V (1-av2) = 0
J - - z
The first and third integrals are zero because the integrands are odd . S::..rnilarly only the v
z
component of v in the second relation will give a non-zero integral. He use this condition to
deten:iine a. That is,
on USi..."lg V
z
and
= V COS
ki'
-=-=a
m
mB
a= 5
0 = A V (1- m,5 v2 )
z 5
2
3/2 / - ~~v
(:) v
0
e dv
(5)
We will use the procedure of section 14.8 to evaluate the constant A. Multiplying (4) o;:; i and
'I'he integral on the left is ee.si.l~y ev~luated. 1·le have
r d3 """ ( ::) [ 2 2Bn _ 2 2 (Br.:)
2
V .;. - - ~ - V V + -::- V
vv - z :;) z ; z
n 26m 1
2
pm
2 J 2rr 17T 2 = - - (-·-)5n(Qm) + (-;:-) d~ cos 9
f3m 5 ,., ) 0 0
sin
(6)
(7)
rco mB 3/2 8
9d9 n (2rr) v e
'-' 0
dv
The int2gral o:f the second term in (7) was evaluated previously. The third term is easily shown
to be 7n/ 5pm. ~hus the integral becomes
n 7) 2 n (1-2 + - - - -pm 5 - 5 f3m (8)
We split the right side of (6) into an integration over the angle n' and over~ and ~
1
• We
lll
define
J = fu,an' V cr 6[vz (1- r v2J
Then using (8), equation (6) becomes
(9)
f"" 2-
To evaluate J, we express 61v (1-av 1)1
1...Z j
in center of mass coordinates (we have reintroduced
a= pm/5). By (14.7.11),
Using v = c + ½ y_ \v1 = 5:.. - ½ y_, we have
2 '2 L ? o:2 b 2 V
2
v
2
(1-o;v ) + v12(1-av1 )=(cz + 2'z)(l-ac--o:5:..•y - 4 V )+(c2- 2,)(1-ac -+a.:::"Y.. - a 4 )
2c - 2o; c c2 - ~ c v2 - o;V C•V
z z 2 z z- -
Then c = c' and ly_l = l-y_ 1 I gives
6jv (1-av
2
~ = -a (V ' c•V' - V c•V)
L. z .:J z z- -
(lC)
CaIT'Jing out the product in ( 10) •,;e find
2rr 7r
J = -af dco' f sin e'dB'aV 1c (V 'V '-V V )+c (V 'V '-V V )+c (V ' 2-v 2~
0
· J
0
[x z x z x y z y z y z z z J
where the coordinate system is shown in Fig. 14.8.1. We notice that, except for constants, the
first integral is (14.8.7). The second integral involves or..ly a change of the index x toy but
is otherwise the same as the first. Consequently, by (14.8.20), these terms yield
2 o; cr VV (c V + c V)
2 TJ z xx yy
(ll)
where cr is given by (14.8.21).
T)
It remains to find
f 2-rr J-rr 2 2 -a dcp' sin e'de'a Ve (v ' -v ) z z z
0 0
(1 ?)
fy (14.8.18)
V ,2_v 2 2 2 (A A) 2 2 2 A A)2 2 z z = Vz cos e'+2V
2
V sin e'cos e'cos~' .S.'~ + V sin e'cos ~'(t·~ -v
2
112
and f 2rrdcp'(V 12-v 2
) = 2rr V 2(cos201-1) + -rrv2 sin2 01(;·~) 2
z z z - - (13)
0
To eliminate s, we write
or
and (13) becomes
The last term in J is then
(14)
and (ll) and (14) give
3 aav (c V V . + c V V
? t) J = 2 + C V - - C
TJ xxz yyz z z z 3 (15)
- J a a V (V (c•V) - C t)
- 2 TJ z - - z 3
We now evaluate
The evaluation of thi s integral is straightforward though tedious. In the center of mass system
2 2 ac (c•V) vz(l-av ) = C -ac C -z z z --
This expressicr.. is multiplied by J in
3
2(mS) =n -27r
a ? l - 4 C v- + V
z 2
the i~tegrc.nd .
(16)
a 2 Ct a V2..v - 2 V C - 2 V c •V - E z z z z
.,,(o)_~ (o) is clee.:r-ly C.!l e-~ .. e:; flL~Cti8:: -'-1
c e.nd V; hence we need keep only even terms in v (l-crv2)J. This yields explicitly
z
I=(f'a.3c d3v f(o)fl(o)(J2 a a V) ~ 2y 2-ac2c 2..v 2_ ~ 2v2y 2_.g (c 2y 2y 2+c 2y 2y 2+c 2y 4) JJ' - - TJ tz Z Z Z i+ Z Z c. X X Z y y Z Z z
2 v2 a 2 2_2 a 2 4 a 2 2 ?~
-c - + ~
3
C V + l2 C v + 7 C v --Y-z 3 z z O z z (17)
i::
Since f(o)f
1
(o) has the si.!llple fozr.: (16) all these integr~ls are the standard ones encou..>1.tered
:i..'1 this theory . We quote the result of integrating over c. In (18) the terms appear in the
ll3
order that they appear in (17)
I = [ i a] [ n2
(~/] [ ~~2 l 5/2]
(-) X
~
~v2 2
J d3v -~ [vv 2 _ 5a vv 2 _ a v3v 2 _ £ v 2v3- y_ + 5a v3 + £.... v5 + a v3v 2]
- e O TJ z 2~ z 4 z 2 z 3 6~ 12 b z
The integral becomes on subs~ituting a= ~/5,
~r vv 2 3
J d3v e -~ [ _z - ~ V 2v3 - v + m8 v5]
- TJ 2 bO z b 60
!
00
5 - ~t f 7!" (cos2e ~ _ _2 2 1 II$v2) =
0
V crTJe dV 2rr
O
- 2- - bO vcos e - b + bC5""" sin e de
~v2
8 !CD -~
= ~ (- 3) II$ v7crri e ay
0
We let s = ½ -/mi V and
= ~ (-8 m/l)(.J~y ½f ds .-s
2
"a ~) 57
-
102471" _l_
= - -n- (II$)3
where cr is given by (14.8.26). Substitution into (18) yields
TJ
2
I __ 32 n cr
- 25 7!"1/2(II$)3/2 TJ
1/2
A = _ 25 (!!!:IT) o~
32 n cr ~ dZ
TJ
and by (9)
t _ 25 (!!!:IT) l/\ (l- ~ v2) o~
~ ? - ~ z ) oz
j_ ncrTJ
To find the coefficient of thermal conductivity, we eV2luate the heat flux
Q = n (v (1
2
mv2 )) z z
= ½n J d3y :f(o) (1+1>)vzv2= ½-n, J d3y f(o)~ vzv2
= k Af d3v f(o) v 2v2(1 - II$ v2)
2 - z 5
= ½m A G~)2 -~((~~~
ll4
(18)
Q = _ nksr2
A = 25
z m 32 cr
TJ
Thus
l/2
(..1L) ~
!!$5 dZ
= _ 25 ~ ~7r kl' dT
32 - ln dZ cr
TJ
Therefore the coefficient of thermal conductivity is
(c) :By (14.8.32)
Renee
l4.3
(a)
(b) :By (12.4.2), the simple mean free path argument gives
K C R
- = - = l.50 -
TJ m µ
Hence _the Boltzmann equation gives a value for K/TJ greater than the mean free path calculation
by a factor of 2.5
(c) ~ X lQ-7erggm1aii"] Ne A Xe
TJ
Calculated l.55 .780 .237
Experimental J.. 55 .785 .250
14.4
Let t i ng f = n(:)
3
/
2
exp [ - ½ Bmv
2
], we have
Performing the integration over angles yields
3/2 = 2 - ~
2
[ 3/2 2 ]
= 47!Il ci) l V dv e ln n (~ - ~
and by appendix A. 4 this is
ll5
--- ln-+-lnT--ln-+-r N[ V 3 3 m 3]
-v N 2 2 2rrk LJ-
Comparison with (9.10.10) shows that except for a constant, this is -S/k where Sis the entropy
per unit volume of an ideal gas.
(a) Differentiating the function
H = J d3
_!. f 1n f
yields dH J 3 elf ) dt = d :!.. dt (ln f +l (1)
elfjdt can be found from the :Boltzmann equation. 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
elf =ff d3v an' Vcr (f 'f' - f f) at ..:.J. 1 1
(2)
Substitution of (2) into (1) yields
dH = JJJ a3v d3v an 'Vcr(f 'f' -f f) (ln f+ 1) dt - -1 1 1 (3)
This integral is invarient under the interchange of:!.. and ~l because a is invariant under this
interchange. When the resulting expression is added to (3) we have
(4)
Similarly, this integral is invarient if the initial and final velocities are interchanged because
cr is invarient under the inverse collision, i.e.,
Thus ~ = !fff d3v 1 d3v ' an' V'cr (f f-f 'f') (ln f 'f'+2) dt 2 - -1 1 1 1
( 4 8) 3 I 3 I 3 3 But V=V' and by 1 .2. , d :!._ d _::1 = d :!._ d 3
1
dH = !fff d3v d3v an' Vo (f f-f 'f' )(ln f 'f' + 2) (5) dt 2 - --1 1 1 1
Adding (4) and (5) gives
~~ = - ½J J J d3v d~ an' Vo (r1
1r 1 -f1f)(ln r1 'f'-ln f 1f) (6)
116
(b) Letting f 'f' = y
1 and f 1f = x, we see that since (ln y - 1n x)(y-x);::: O, the integrand
in (6) is greater than zero wiless f 1
1f 1 = f
1
f. Hence dH/dt < 0 where the equality holds when
the integrand identica.lzy vanishes.
14.6
The equilibrium condition is
1 2
1n f = Af B mv + B mv + B mv + c-2 mv
X X y y Z Z
This can be written in the form
2
1n f = -A'(v-v) + 1n C'
- --0
(1)
where the constants A', C' and the constant vector v
-A' (v-v 1
contain the constants in (1).
Hence f = C' e - --o
v is clearly the mean velocity which can be set equal to zero without loss of generality. To
--0
determine A', we find the mean
€
Letting € = (3/2):Id', we find A'
energy of a molecule.
J d3! ½ rrrlf 4n f ""v2dv ½ rrN
2 e-A 'v
2
0
J d3
! f
= m/2kT.
3 m
= 4 A'
The constant CI is easily found from n = J d3! · f which
3/2
gives C' = n(~kT) showing that f is the Maxwell distribution.
CHAPTER 15
Irreversible Processes and Fluctuations15.1
Let the friction force depend on a, v, and T) in the manner
Then the dimensional equation where L = length, T = time, and M = mass becomes
Thus x+y-z = 1, z=l, -y-z = -2. Solution of this system of equations yieldx x=y=z=l, and
f cc T)aV
117
15.2
Since NAk = R vhere NA is Avogadro's number and R is the gas constant, we have from (15.6.11),
t
RI'
NA= 2
37111a(x)
or t
7 -1 0 -1 23 Substituting R = 8.32 X 10 ergs mole K and the given values yields NA= 7.0 X 10 molecules
-1 mole •
15.3
(a) The energy of an ion in the field at point z is - e C z.
Hence ~)
nfzT
f3e! (z+a.z) e
f3ei' z
e
::::: 1 + f3e f. dz
(b) The flux is given by
J =-Don~ D[n(z+dz) - n(z)]
D oz - dz
Then from (1) J = -nD f3e ! D
(c) J = nv = nµ c
µ
( d) J+J = D µ
-nDi, e.l+nµ! = 0
Thus I:!:. e
D kT
dv 1
d.t = -rv + iii F' (t)
Substitution of v(t) = u(t)e-rt into (1) gives
Hence
- l t au - r .,..,, ("'") -=-e t " dt !D.
1ft ... ,
u = u(o) + iii e1 " F'(t')dt'
0
Since u(o) = v(o) it follows that
15.5
v = voe -rt + ~ e -rt J t e rt' F' ( t' ) dt'
0
(1)
(1)
(2)
By dividing the time interval t into N successive intervals such that t = NT, we can replace the
integral in (2) of 15.4 by the sum of integrals over the intervals of length T betveen ~ and
118
Then t' = kT + s where k is an integer and dt' = ds.
Hence -rt -,NT N;l l J,. ,y{k,-+s) F 1 v-v e =· e L, - e' (k,-+s) ds
o k=O m o
N-l ( ) JT
= L, e-,,- N-k ¾ e's F'(kT+s) ds
k=O 0
In the integral, s <Tor e's< e'T~ l since,- is chosen so that T << ,-1 . It follows that
-')'T v-v e =
lf,. where C\_ = m F'(k,-+s) ds.
0
15.7
We integrate the Langevin equation
0
N-l
= L, y = y
k=O k
dv l
dt = -')'V + m F' (t)
-1 over a time,- which is small compared to, but large compared to the correlation time,-* of
the random force
l J,. v(,-) - v(o) = -')'V(o),- + iii
O
F'(t')dt' (1)
The time interval t is divided into N smaller intervals such that t = NT. Then in the k
th
subdivision t' = kT +sand dt' = ds. Hence (1) becomes
vk = (l-n) vk-l + ¾f ,.F'[(k-l)T+s] ds
0
We can find the velocity at t = NT by an iterative procedure. Defining Gk as in problem 15.6 we
have
vl = (1-,,-) V +G
0 0
v2 = (l-')'T) vl + Gl
(l-,,-) [(l-')'T)v + G]
0 0 + Gl
= (1-')'T) 2 V + (l-,T) G + Gl
0 0
Continuing this way we see that
)N )N-1 )N-2 VN = (1-rr Vo+ (l-')'T Go+ (1-,,- Gl + ••• + GN-1
Si.nee,,-
N -~,- N N
<< l, e-,,- = (e' ) ~ (1-rr) it follows that
ll9
or
since rr << 1. This result agrees with 15.5.
15.7
Hence
and
Then as t ~o.,, v ~ O.
15.8
From 15. 5 we have
ds = 0 Gk = G = J (F' (k+s) )
0
__ N~l _ _ N;l -rr(N-k)
y - t... Yk - t... e G=O
k=O k=O
v-ve-rt =Y=O
0
v = v e-rt
0
- N-1 - N-1 N-1
i2=I: yk2+r r y y k j
k=O k / j
It 'WaS
Thus
shown in 15.7 that yk = O.
- N-1 ( ) - - N-1 y2 = L e-27T N-k
k=O
G 2 _ G2 -2rt ~ 21Tk
k - e ~ e
k=O
Where we have used NT= t. The sum is just the geometric series and may be
y2 = a2 e -2rt ~-e
21
~""' G2tl-e -
2
rtJ
1 2n 2rr -e
easily evaluated.
where e27T-l""' 27T by series expansion since )'T << 1. To find G2 ve note that
-v+ 2 2 2 1-e - 2rt
{v-v e-'v) = Y = G
o 2rr
Letting t ~o., yields, by equipartition,
2 kT G
2
V =-=-m 2)'T
120
From problem 15.5 we ~ve
and (l)
where we have changed the integration variable from s tot' to conform to the notation of the
text. G2 is independent of k allowing the choice of' k==O in (1).
G
2
= \JTdt'fT{F'(t') F'(t"))
0
dt"
m o o
The integrand is the correlation :f'unction of F 'which depends only on the time difference s = t "-t' •
Making this change of' variable, we bave
- J'r -r-t'
G
2
= \ dt'f (F'(t')F'(t'+s))
m o -t' 0 J
T T-t'
ds = \ dtj K(s)ds
m o -t'
vhere the last equality follows because (F'(t')F'(t'+s))
0
= (F'(o)F'(s))
0
= K(s). Since K depends
only on s, the integration overt' can be performed easily by interchanging the order of inte
gration and reading off the limits from the figure.
G
2 ~ ~2 [1_:ds L:K(s) dt' + 1:~, f-;(,) dt]
Since T >> T* and K( s) ~ 0 for s >> -r*, s can be neglected compared
to T and the limits replaced by ± co.
Thus
121
T
t'
2
By the results of 15.8, G = 2kI' n/m. It follows that
7 = 2zDiT Jet) K(s) ds
-<C
15.ll
(a) We integrate the Iangevin equation
a..v 1
dt = --yv + m F ' ( t)
over the interval~-
Jv(-r) JT
dv = -7 V dt'
V
0
O
+ ~JTF'(t')dt'
0
Since Tis small, ve can replace v by the initial velocity v in the first integral on the right
0
and obtain
l!T l:::.v = v-v = --yv T + - F'(t ')dt I
0 o m
0
= -rv -r + G (l)
0
and l:::.v = - 1 V
0
T since G = o.
(1:::.v)2- = (7 v -r) 2
- 21 v T G + a2
0 0
""' G
2
= 2k.Trr /m (2)
where ve have used the result of 15.8 and kept only terms to first order in -r.
(b) Fram (1), keeping terms to lowest order in T and noting that odd moments of Gare o, ve find
(!:::.v) 3 = -3, V
0
T G
2
(!:::.v)4 = ~
2 ~ 2 Since G « -r, G should be proportional to T. Hence (!:::.v)3 and (!:::.v)4 are proportional to -r2 •
(c)
15.12
From (2), (.6.v) 3 = - 6kr r2-r2v /m.
0
From eq. (2) of 15.4
= (v2(o)) e--,,t +~iol .-rti\rt• F'(t')~ (v(o)v(t))
2 rt -rt ft rt'
= (v (o)) e- + ~ (v(o)) (e F'(t')) dt'
. 0
But since F'(t ') is random, the second term on the right is O. By equipartition
122
Thus
~ -rt {v(o)v(t)) = - e m
and since (v(o)v(-t)) = (v(o)v(t)), we have for all t
(v(o)v(t)) = ~ e-,ltl
m
15.13
Squaring equation (2) of 15.4 yields
2 2 2.,,... -2rt It .,,+ I -2,t J t 1 t (t '+t •)
v (t) = v
0
e- iv+ 2v
0
7 e 1 v F'(t')dt' + ~ dt' ey F'(t')F(t")dt'1
· o · m o o
(1)
since (F'(t')) = O. The correlation function in the integral is only a function of the time
differences= t"-t'. Ma.king this change of variable yields
f
tdt'f\,(t'+t") (F'(t')F'(t"))dt" =ftdtft-t~s e1 (2t'+s)K(s)
0 0 0 -t I
where K(s) = (F'(t')F '(t'+s)) = (F 1{o)F 1 (s)). The evaluation of this integral is similar to
that of problem 15.10. Thus interchanging the order of integration and reading the limits off
the diagram of problem 15.10 with t substituted for T yields
J\tf t-t~s e'(2t '+s)K(s) =J O
ds It dt 'ey(2t '+s\(s) +J \s r t-:t 'e1 ( 2t '+s\(s)
Q -t I -t -S O J 0
Jo . ( 2rt -2ys) ft ( 2,(t-s) )
= ds eysK(s)e
2
-e I ds e)'5K(s)e
2
-l
-t i' 0 )'
Since K(s) is only appreciable when ys << 1, we can put e1s ""l in the above and find
J o ( 2,,-t 1 ) f t 2rt ds K(s)e
2
_ -- + ds K(s) (e -l) =
-t I o 21 f t ( 2)'t 1) f a,
-t ds K(s ) ~e
21
- ...,, ds K(s)
Substitution in (1) yields
2 (1-e -2,t) la, v (t) = v/ e -
2rt + -"---
2
-,._- ds K(s)
2)'m -0::,
-2--
as t -'>a,, v (a,) = kl' /m by equipartition and
-2-- ~ l !co
v (co) =-=-
2
dsK(s)
m 2rm ...,,
Hence , = ~ J~ ds K(s) as in (15.8.8)
123
15.14
(a) We will denote the Fourier coefficient of F'(t) by F'(m) and that of v (t) by v (m). That
is
J
oo • t
F'(t) = eJ..lO F'(m) am
~ f co ·wt
and v(t) = _..,e
1
v(w) dw (1)
Substitution in the Langevin equation gives
iwv(m) = -tV(m) + ! F'(m)
m
or () 1 F'()
V ID m(7+iID) ID
Denoting the spectral densities of F' and v by JF 1 (m) and Jv(m), we have from (15.15.u)
2 2 2 7T' I
1
2
= m (r -fo)) e v{m)
or JF,(co) = m2(,l-fo)2) Jv(m) (2)
(b) By (15.10.9) K(s) = (V(o)V(s)) = kT e-rls l
m
(3)
Thus JvCco) = ~ J~ K( s) e -iIDs ds
• !, [r ~ e (-w-+, )s ds + f e (-W>-7 )sds]
3v(a,) •:, [c/:;,,,j - (-7=:io,~ " :, [ x.,2J (4)
(c) Substitution of (4) into (2) gives
l
JF(w) = TI mkT y
or JF,C+)(co) = 2JF(m) = ~ mkT r
This result depends on co<< (T·;(- )-1, where T;: is the correlation ti.me of the fluctuation force,
since in deriving (3), one used a time scale such that -r >> ,.*. Consequentl;y', IDT ::::: 1 >> w-r*.
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