54.
(
)
W = 1 " #S = k ln W = 1.38 $ 10!23 J K ln1 = 0
For four heads:
For 3 heads, 1 tail:
(
)
W = 4 " #S = k ln W = 1.38 $ 10!23 J K ln 4 = 1.91 $ 10!23 J K
(
)
W = 6 " #S = k ln W = 1.38 $ 10!23 J K ln 6 = 2.47 $ 10!23 J K
For 2 heads, 2 tails:
For 1 head, 3 tails:
(
)
W = 4 " #S = k ln W = 1.38 $ 10!23 J K ln 4 = 1.91 $ 10!23 J K
(
55.
)
W = 1 " #S = k ln W = 1.38 $ 10!23 J K ln1 = 0
For four tails:
From the table below, we see that there are a total of 26 = 64 microstates.
Macrostate
6 heads, 0 tails
5 heads, 1 tails
H
H
H
4 heads, 2 tails H
H
H
H
3 heads, 3 tails
T
T
T
2 heads, 4 tails T
T
1 heads, 5 tails T
0 heads, 6 tails T
Possible Microstates (H = heads, T = tails)
H
H
H
H
T
H
T
T
H
T
T
H
T
T
H
H
H
H
H
H
H
T
T
T
T
T
T
T
H
H
H
T
T
T
T
H
H
T
H
H
T
T
H
H
T
T
H
T
H
H
T
H
H
T
T
T
H
T
T
H
H
T
T
H
H
H
T
T
H
T
H
H
H
T
H
T
T
H
T
T
H
T
H
H
H
H
H
H
T
T
T
T
T
T
H
H
T
H
T
H
H
T
T
H
T
T
H
T
H
T
H
T
T
H
H
T
H
T
T
H
T
H
T
H
H
T
T
H
T
H
H
T
H
H
T
T
H
H
H
T
T
T
H
H
H
H
H
H
T
T
T
T
T
T
H
H
T
T
T
T
H
H
T
H
H
T
H
T
H
T
H
T
T
H
H
T
H
T
T
H
H
H
H
H
T
T
T
T
T
H
H
H
T
H
T
H
H
T
T
H
T
T
H
T
H
H
T
T
H
H
H
T
T
T
H
H
T
T
T
T
H
H
T
H
H
T
H
T
H
H
H
H
T
T
H
T
T
T
T
H
H
T
H
T
T
H
T
T
H
H
H
H
H
H
H
H
T
T
T
T
T
T
H
H
T
H
T
H
H
T
T
H
T
T
H
T
H
H
T
T
H
H
H
T
T
T
H
T
H
T
H
T
T
H
H
T
H
T
T
H
H
T
H
T
T
H
T
T
H
H
H
H
T
H
T
H
H
T
T
H
T
T
H
H
T
H
T
H
H
T
T
H
T
T
H
H
H
H
H
H
T
T
T
T
T
T
H T H H H H H
T
H
H
T
T
H
H
H
T
T
T H T T T T T
(a) The probability of obtaining three heads and three tails is 20 64 or 5 16 .
(b) The probability of obtaining six heads is 1 64 .
56.
When throwing two dice, there are 36 possible microstates.
(a) The possible microstates that give a total of 7 are: (1)(6) , (2)(5) , (3)(4), (4)(3), (5)(2),
and
(6)(1). Thus the probability of getting a 7 is 6 36 = 1 6 .
(b) The possible microstates that give a total of 11 are: (5)(6) and (6)(5). Thus the
probability of
getting an 11 is 2 36 = 1 18 .
(c) The possible microstates that give a total of 4 are: (1)(3) , (2)(2) , and (3)(1). Thus the
probability of getting a 5 is 3 36 = 1 12 .
57.
(a)
There is only one microstate for 4 tails: TTTT. There are 6 microstates with 2
heads and 2
tails: HHTT, HTHT, HTTH, THHT, THTH, and TTHH. Use Eq. 20-14 to calculate
the entropy change.
Number of
Microstates
1
6
15
20
15
6
1
"S = k ln W2 ! k ln W1 = k ln
W2
W1
(
)
= 1.38 # 10!23 J K ln 6 = 2.47 # 10!23 J K
(b)
Apply Eq. 20-14 again. There is only 1 final microstate, and about
1.0 ! 10 initial microstates.
29
"S = k ln W2 ! k ln W1 = k ln
W2
(
)
1
$
% = !9.2 # 10!22 J K
29 '
( 1.0 # 10 )
= 1.38 # 10!23 J K ln &
W1
These changes are much smaller than those for ordinary thermodynamic entropy
(c)
changes.
For ordinary processes, there are many orders of magnitude more particles than we have
considered in this problem. That leads to many more microstates and larger entropy
values.
58. The number of microstates for macrostate A is WA =
10!
= 1. The number of microstates
10!0!
for
macrostate B is WB =
10!
= 252.
5!5!
(a) "S = k ln WB ! k ln WA = k ln
WB
(
)
= 1.38 # 10!23 J K ln 252 = 7.63 # 10!23 J K
WA
Since !S > 0, this can occur naturally.
W
(b) "S = k ln WA ! k ln WB = ! k ln B = ! 1.38 # 10!23 J K ln 252 = !7.63 # 10!23 J K
WA
(
)
Since !S < 0, this cannot occur naturally.
"
60. The required area is $ 22
&
103 W h # " 1day # " 1 m 2 #
= 61m 2 ! 60 m 2 . A small house with
%
$
%
$
%
day ' & 9 h Sun ' & 40 W '
1000 ft2 of floor space, and a roof tilted at 30o, would have a roof area of
2
1m "
2
)! cos130o "$ !# 3.28
$ = 110 m , which is about twice the area needed, and so the
ft &
%
&%
(
1000 ft 2 #
cells would fit on the house . But not all parts of the roof would have 9 hours of sunlight,
so more than the minimum number of cells would be needed.
77. We need to find the efficiency in terms of the given parameters, TH , TL , Va , and Vb .
So we
must find the net work done and the heat input to the system. The work done during an
isothermal process is given by Eq. 19-8. The work done during an isovolumentric process is
0. We also use the first law of thermodynamics.
V
!Eint = 0 = Qab " Wab # Qab = Wab = nRTH ln b > 0
ab (isothermal):
Va
ab
bc (isovolumetric):
!Eint = Qbc " 0 # Qbc = !Eint = nCV (TL " TH ) = 23 nR (TL " TH ) < 0
bc
bc
cd (isothermal):
!Eint = 0 = Qcd " Wcd # Qcd = Wcd = nRTL ln
cd
Va
Vb
= " nRTL ln
Vb
Va
<0
da (isovolumetric):
!Eint = Qda " 0 # Qda = !Eint = nCV (TH " TL ) = 23 nR (TH " TL ) > 0
da
da
W = Wab + Wcd = nRTH ln
Vb
Va
Qin = Qab + Qda = nRTH ln
Vb
Va
! nRTL ln
Va
= nR (TH ! TL ) ln
Vb
Va
+ 23 nR (TH ! TL )
"
#
V
ln b
$
%
&T !T '
W
Va
Va
%
=
=
= ( H L )$
Qin T ln Vb + 3 T ! T
TH + $ Vb 3 & TH ! TL ' %
*
2( H
H
L)
$ ln V + 2 ( T ) %
Va
* H +, a
(TH ! TL ) ln
eSterling
Vb
Vb
"
#
V
ln b
$
%
Va
$
%
= eCarnot
$ Vb 3 & TH ! TL ' %
$ ln V + 2 ( T ) %
* H +, a
Since the factor in [ ] above is less than 1, we see that eSterling < eCarnot .