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Unformatted text preview: _CHAPT" _ER ‘rw'E'g—ve —'sonmo' M's" "71 50' 501417700 ’0" 5”"; i
U=2PV. 12.2 W= PAV= PAAL = (2.00 x 105 Pa)? (16.0 x 102 m)2(0.200 m) = 804] 12.3 We call N the number of molecules, and the average kinetic energy per
molecule is represented by <KE>. The total energy of our system is U= NsKE>=(3Na)%kT=%RT (where we. have used Nails$12). Thus, U=%_(s.31 J/molK) (303 K) = 1.1 x 104 J. Altemately, we could use the result of Problem 1: ‘ U=§~PV=§nRT=§(3.0JRTugiRTasahove. 12.4 The sketch for the cycle is shown at the right. let
us now ﬁnd the net work done during the process:
During the expansion from a to b we have _ Wan a Pa( Vb  Va) = 3(1.013x 105 Parka002 m3)=6101.
Wbc = 0 because AVbc =0 Weand vd Vc) = 20.013 x 105 Pa)(Zx 103 1113), or 3
Wed =  410 J.W(d to a) = 0 volume remains constant V (Literal
Thus, the net work is: Wabcda = 610} 410] = 200 J. (Note: the net work could also be obtained by computing the area
enclosed within one cycle on the PV diagram.) 12.5 (a) W1 AF= W1 A+ WAF (The notation [AF means the Work done from I to
A to F.) The term WM: = 0 because the volume is a constant during
this part of the process. Thus, WiAF= WIA=P1(VA  v1) = (4.05 x 105 Pa)(2.00 x 103 m3) = 810}.
(b) Along path IF, we ﬁnd the work by ﬁnding the area under the PV
curve. This consists of a triangular area plus a rectangular area. Wig =% (0.002 m3)(3.04x 105 Pa) + (1013 x 105 Pa)(0.002 m3) =50? J. (c) W131: = M3 + 1411312. The term W13 equals zero because the volume
remains a constant during this part of the process. So,
wm= W231: = (1.013 x 105 Pa)(2.00 x 10—3 m3) = 203 1.
12.6 The sketches for (a) and (b) are shown below. (c)There is more work done in process (a). We recognize this
from the ﬁgures because there is more area under the PV curve
in (a). Physically, more work is done because of the higher
pressure during the expansion part of the process. 167 12.7 12.8 12.9 12.10 12.11 CHAPTER TWELVE SOLUTIONS P: 1.5 atm= 1.52x105 Pa "
(a) W= PAV, and AV: 4.01113, so W: (1.52 x105 Pa)(4.0 m3) = 6.1 x 105 J.
(b) AV=—3.0 1113, giving W=(1.52x105 Pa)(3.0 m3)=—4.6x105 J. In a constant pressure process, the work done is W= PA V. ForanidealgasPV=nR71andV=£§I,so AV=Vf Vi=§§E(RTi).
This gives, W= PA V= 11 RA T= (0.200 mol)(8.3 1 J/K mol)(280 K) = 465]
(a) We use the ideal gas law as: 21f?— = PfTE/f' After canceling P and solving for the ﬁnal temperature, we have L ﬂ
1%: 13M) = 13(Vi)=4(273.151<)= 1093K. (b) W= PAV=nRAT=nR(T}= Ti)= (1 mol)(3.31 mo] K)(1093K273 K)
or W=6.31k]. (a) W= 0 because the volume remains constant. The system gives off heat. Thus, Q<0. AU: Q— Wand since W: O, we have AU= Q Therefore, AU<0.
(b) Again, W= 0 because the volume remains constant. Q> 0 because the water receives heat. Again, AU: Qso AU>0. The work done by the gas during this process is the area under the
process curve on a PV diagram. This is given by:
W= (area of rectangle) + (area of triangle) =Po(2V0' V0) '1' %(2Vo Vo)(2Po" Po)= lSPOVD Using the result of problem 1, the change in the internal energy of
this monatomic ideal gas is: 3 3
AU= Uf  Ui = 3(2Po)(2Vo) ' §(Po)(Vo) = 45 PoVo
Then, the ﬁrst law of thermodynamics gives: Q=AU+ W= 6P0Vo. 12.12 Using the result of problem 1, the change in internal energy is seen to 12.13 be
3 3 3 3 AU= Uf Ui =31)er — 5PM = 312mm.»  '2'(Po)(2Vo)=0
Then, the first law of thermodynamics gives 0 = Q W, or Q= W.
Since this is a compression process, work is done on the gas by the
surroundings. That is, the gas does a negative amount of work or W<0.
Finally, since Q= Win this case, we conclude that Q< 0, or the gas must
give off heat. ‘
In summary, we find: AU= 0, Q<.0, and W<0. (a) W= PA V= [0.311.013 x105 Pa)](3.00 x 103 m3  3.00:: 103 m3)
=~1521
(b) AU= (2— W We are given that Q=4OOJ. Thus,
AU: 4001(152 J) = 248 J. 168 CHAPTER TWELVE SOLUTIONS 12.14 (a) W= area under the PV curve. MF=%(000200 m3)(3.04 x 105 Pa) + (1.013 x 105 Pa)(0.002 m3)
=50? J.
AUIF= QF ' WiF=418J 507}=89J (b) MAF= MAJ. 0: P1( VA — v1) : (4.05 x 105 Pa)(2.00x10‘3 m3) : 810].
AUis the same as above. Thus,
QAF=AU+ WiAF=89.0J+ 810J= 721 J. 12.15 AUcycle = Qycle  chcle = O (for any complete cycle)
chle = chde = area enclosed in PV diagram = %(4 m3)(6 x 103 Pa) or, dee : 12:;103 J: 12 k].
If the cycle is reversed, then chle = 12 k] 12.16 (a)W= PA V=(1.013x105 Pa)(3.342 x 103 m3) :338 J.
(b) The heat added is: Q=va= (2x103 kg)(2.26 x 106 J/kg)=4520 J.
(c) AU: Q— W=4520J338]=4182]. 12.17(a) AV=AAL=(0.150m2)(~0.20m)=3.0x10‘2m3.
Thus, W= PA v: (6.0x103 Pa)(3.0 x 102 m3)= 180 J. (b) Q=AU+ W= 8.0 J  180] = 188 J (188 J of heat energy are removed
from the gas.) 12.18 W3C = 0 (constant volume), WCA<O(AV<O), WAB>O(AV>O) AU: Q— W gives A U3c= Qgc  W3C < 0 (because QC <0, th=0).
AUcycle =AUAB + AUBC + AUCA=O
Thus, A UAB > 0 since both A UBC and A Uc A are negative. AUCA= Q3A WCA becomes QA=AUCA+ WCA<0 since
both AUCA and WC A are negative. Also, QB=AUAB + WAB > 0 since AUAB> Oand WAB>0~ In summary, Q\3>O, (had), QA<0. WAB>O, WgCrO, WCA<O.
AUAB>O,AUBC <0, andAUCA<0 12.19 (a) W= PAV= (1.013x105 Pa)[(1.09  1.00) x 106 m3] :+9.12x103 J
(b) Q= mLf= (103 kg)(3.33 x 105 J/kg) : 333 J.
AU: Q— W: 333 J9.12x10'3J=—333 J. 12.20 (a) The original volume of the aluminum is:
_£___._ﬂ_l<s__= 3 3
v: p :2] X 103 kg/m3 1.85x10 m.
The change in volume is:
AV= pvo(A:0 =3(24.0x 106 °c 1)(1.85 x 103 m3)(7O.O°C) :9.32 x 106
3
m
The work done is: W= PA v: (1.013 x105 Pa)( 9.32x 106 m3): 0.95 J
(b) Q= chT= (5.0 kg)(900 J/kg “cw/0 °C) : 3.2 x 105 J
(c) AU: (2— w. (1:321:105 J 169 CHAPTER TWELVE SOLUTIONS 12.21 (a) W: PA V = area under PV curve. Wm}: = (1.50 atm)(0.800  0.300) liters, or W} AF: 1.50(1.013x105 Pa)(0.500 x 103 m3) = 76.0 J.
Wig}: = (2.00 atm)(0.800  0.300) liters, or win: 2(1.013 x 105 Pa)(0.500 x 103 m3) = 101 J. WjF= WiAF+%(1.013 x105 Pa)(0.500x 103 m3) = 33.71. (b) We are given that AU: (180J 91 J) = 89.0 J. Thus, Q=AU+ W= 89.0J+ W, giving QAF=89.0J+76.OJ=16SJ, QBF=39JI+ 101 J= 190], and QF=89J+ 88.7]=178]. 293
Effc = = 1 ——£ — 1 3?: 0.488 (or 48.8 96) ——_________
%= 0.300, which gives, Q =0.700Q;.
(a) Therefore, Q = 0.700(800J) = 560] (b) For a Carnot Cycle, eff = 12.23 eff= l 1  %= 0.300, from which Tc =0.700Th = 0.7(500 K) = 350K.
———.____*___ 12.24 We use Effc= lIc as, 0.300=1573 K. Th Th
From which, Th = 819 K = 546 “C. 12.25 The temperatures of the reservoirs are 300 “F = 422 K, and e T 338.7
150°F= 65.6 C = 338.7 K. Effc 1  :17; = 1 Ei'2—= 0.197 (or 19.7 96) 12.26 (a) Eff=g= 2—‘2%£= 0.300. Thus, Q=%%g=667J.
(b) W=Q, Q. Therefore, Q=Q, W=667J200]= 467]. 12.27(a) W=Q1Q=1700J12001=5001. Eff=VK=M— '(b) The work done in each cy h’__590;.L_ 3
(c) P..M—O_3OO 8—1.67x10 w. 1V._Qa'Q_ 9.
12.28(a)EffaQ1_ Q1 1q0.250. With Q = 8000 J, we have Q,=1.07x 104 J.
_ (b) W= Q,— Q =2.7x103 J, and from Pr”: A t , we have
w 2.7 x 103
=F=Tt'00—JK1=053 8
———__________ 12.29 We have W= Q,  Q =200J, and Eff= At H___12 _
Q 500 J..0.4. 170 CHAPTER TWELVE SOLUTIONS If Eff = 0.6Elfc , then Effc =g+2 = 0.667. E _ E 1
But, Effc = 1  Th — 0.667, thus, Th =0.333=3. 12.30 We have, Effc = 1 L; = 1 %= 0.433, and Eff=
Thus, W= Q,(Eff) = 21,000](0.433) = 9.10 x103 J.
3
(a) P=_VtL/=9.1013012 =9.10x103W=9.10kW.
(b) Q=Ql W=21000J9100]=1.19x104j.
The heat expelled in each cycle (which lasts for one second) is
Q=1.19x 104 J. I!
Q. 12.3 1 Work done each second = 1000 M] = 109 J.
_E_ _E_ E11. 9
Eff—Gl—O.33,so car33 33 ~3.0x10‘J
But,also W2QQ,soQ=Q.W=3.Ox109J109J=2.0x1109J. Therefore, 2.0 x 109 J of heat must be absorbed each second by 106 kg 0:
river water.  =_Q= 2.0x1091 = .,
AT mc (106 J)(4186 J/kg°C) 0'48 C‘ 12.32 (a) The change in entropy of the water is: "92}ny 1.00 k  3.33 x105 _ 3
AS—T_ T _ 273 K —1.22x10 J/K. (b) The entropy change of the freezer is +1.22 x 103 J/K. 12.33 Q= mLV= (1.0 kg)(2.26 x 106 J/kg) = 2.26 x 106 J 6
AS=%=2'2637"3 11? =6.1x103 J/K 12.34 The heat generated equals the potential energy given up by the log.
Q=mgh= (70.0 kg)(9.80 m/32)(25.0m)= 1.72x104 J. 4
Thus, AS=%H23—5‘512—1= 57.2 J/K. 12.35 The heat generated is equal to the kinetic energy lost.
Q= (2)(%mv2) = (2000 kg)(20 m/s)2 = 8.00x 105 J. 5
So, AS=Q— MALL: 2.70x 103 J/K. T— 296 K
12.36 (a) Result Possible Combinations Total
all red RRR 1
2R,lG RRG,RGR,GRR 3
1R,2G RGG,GRG,GGR 3
all green GGG 1
(b) . Result Possible Combinations Total
all red RRRRR 1 171 12.42 Entropy change of the hot reservoirﬂASH =  $3 = —~—l =16.O‘I' CHAPTER TWELVE SOLUTIONS ' 500 K
Entropy change of the cold reservoir: ASc= % = iﬁ'ﬁl =26.7JK
and the net entropy change of the system during this irreversible
process is: Astotal = 26.7 J/K  16.0 j/K = +10? J/K> 0. ' K 12.43 The heat discarded is 70% of the input energy for the plant. Thus, the 12.44 12.45 energy going in to the river each second is:
[9: 0.7(25 x108 W) = 17.5 x 108 VS. g_ E _ gt _ 17.5 x 108 1/s _ ,
Also —(t)cAT; or AT‘(m/t)c_ 90 x 106 k —2.8 C.
(—460 s )(4186 —J—kg .C) t
The density of water is p= 1000 kg/m3.
Therefore, 5000 m3 of water corresponds to a. mass of:
. m = (5000 m3)(1000 kg/m3) = 5.0 x 106 kg.
When this mass falls 50 111, its change in potential energy is:
APE= ~ (5.0 x106)(9.80)(50)= 2.45 x 109 J. In other words, gravity does 2.45 x 109 J of work on the water every 
second. Assuming that the internal energy of the water does not
change in the process, we may write (using AU: Q— W): Q= W= 2.45 x 109 J, so I 9 v I
AS=%=———L(§O4i 2715915“ = 8.4x 106 J/K as the increase in entropy each second.
[Prove by contradiction] Assume a quantity of heat, Q ﬂows from the cold object at Tc to the hot
object at Th>Tc (i.e., Q; =  Qand Q; =+ Q) Then: AS = — 7% and Ash: 7% , and the total entropy change
' T  T
of the system is: ASrsAScl 11311 = % + TOE: LEIETh—h)‘ The second law requires AST 2 0. Thus, we must have QTC  Th) > 0. But,
since (Tc;  Th) < 0, it is necessary that Q< 0. Therefore, QC =  Q> 0 while
Q, = Q4. 0, or the heat actually flows from the hot object to the cold  object. Thus, Q; must be positive resulting in a heat flew to the cold 12.46 (a) (101,3:ng  W123=413J 16TJ=+2511 object, contary to the original assumption. (b) Use AUL3 = Q43  W143, with AU1,3=+251].
Thus, 251]= Q43  63.0 J, or Q43 =+314J.
(C) W12341 = W123  W143 = 167]— 63.0J= +104]
(61) W14321 = W143  W123 = 6301 167J=—104J
(e) The internal energy change is zero in both
cases because both are cyclic processes. 173 ...
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 Fall '01
 Shapiro
 Physics

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