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Unformatted text preview: Probbm 56+ 4* L4 " SOM‘HOnS
,———————————————————"‘__________________,________— H10 040°C, Hal, 4 bar) aha (—1o'c,so\, 4 bar) wM‘ViS AS ’98! ‘WﬁS read'ion'? l—Hémimpossibku +0 wwm 3+ 93“an 33001. mach/902,9 no+ naming
jym wt —40°C, SO wL “1.19% buﬂa 0L (gag: o , 5
mo (40 c, Np 4 bar) wi—ﬂ “,0 (4023, So\, 4 m) A54 A53 moan, mi, Mm) Ag I—l,0 62% So), 4 bar)  L 22.3. _.
A94  (Amomsﬂwmwn M5 .. 02.84 J/H ASL, AH—S’mang : — N‘ﬁusmn : — @008 =jag/K
T T 2% Mm‘éﬁ's 53" + 5‘92 * D53 5 To ca\w\OUUL ASSUW, “0:3 J“) 30109 451935 21h?“ 13% : (/1 Mo2)(75.5)(+4o) = 753 a”
Dch : —(,008 J ME = Q1 mo;)(5q,us)(—4o) 7 4%,; 1361353 = A511 + A61z + A613 = “5634.6 If v A NW ——A J‘ €934,57J
[>st — é— ; ,EEL : LE :. 0‘2qu J/K
SUN ’R‘W 9113 )4
A84“, 7 A8835 4’ ASSW: +0.8 .4)
NO’reeS  Aﬂomg 7 O wwariﬂ because 4m \s an ‘Weymibm
wow», MARS Mom; 3.: wvxrswlp 0% wt 0'0 9} To caQookm 5&5va Wad: USL Mabovbjﬁrmm. 44». W an \rercrsibuqvoccssi Uwgmdma (WNme is W M SWTOU/Qinﬂ «XWWQ LS
camer (gym/low he), chM 16 urij 0; 505a Woman P4.17) Use the following data at 25°C to complete this problem:
AH" ~ (kJ mol“‘) reaction 1/: H2(g) + 1/; 02(g) —> OH(g) 38.95
H2(g) + 1/: 02(g) —» H20(g) 3 —241.814
H2(g) _, 2H(g) 435.994
02(8) —+ 20 (g) 498.34
Calculate AH :mﬂon for
a) OH(g) .8 H(g) + O(g)
b) H20(g) _» 2H(g) + 0(g)
9) H20(g) —> H(g) + 0H(8)
Assuming ideal gas behavior, calculate Ammm and AUjmm for all three reactions.
a)
AHjmm (kJ mol1 )
OH(g) —» 1/: H2(g) + 1/2 Ong) —38.95
1/2 1mg) _» H(g) 1/; x435.994
1/: 02(g) —» 0(g) 1/2 x498.34
OH(g) —> H(g) + 0(g) AHjemO" = 428.22 kJ mol—1
AU:eaction = AH:ea:tion ~ An
= 428.22 krmol‘l— 8.314 J mol‘1 K"1 X298.15 K= 425.74 kJ mol—1
b) Amman (k1 mor’)
H20(g) —9 H2(g) + l/202(g)  241.814
H269) —> 2H(g) 435.994
1/2 02(g) —> o (g) 1/: _><498.34
H20(g) —> 2H(g) + 0(g) Amman = 926.98 kJ mol1
AU:eacll'an = AH:eaz:tion _ An
= 926.98 k] morl— 2x 8.314 J mol—1 K‘Ix 298.15 K: 922.02 kJ mol—1
0) AH:eaction Incl1 )
H20(g) —> H2(g) + 1/202(g) 241.814
1/: Hzég) + 1/202(g) —> 0H(g) 38.95
1/; H2(g) —+ H(g) 1/: x435.994
H20(g) > H(g) + OH(g) AHjmﬁm = 498.76 kJ mol—1
AU“ =AH° — AnRT reaction reaction = 498.76 kJ mol"— 8.314 J mol—1 K'1x298.15 K= 498.28 kJ mol" P421) Benzoic acid, 1.35 g, is reacted with oxygen in a constant volume calorimeter tc
form 1120(1) and C02(g). The mass of the water in the inner bath is 1.240 X 103 g. The , temperature of the calorimeter and its contents rises 3.45 K as a result of this reaction.
Calculate the calorimeter constant. AUcombustion : AHcambustinn — 15 1
For the reaction CGHSCOOI'KS) + —2—02(g) —> 7C02 (g) + 311200), = ——2—
AUWW = —3226.87 kJ mor1 ——:x8.3 145 J mor‘K—1 x298.15 K = —3228.11 kJ mol‘1
7" mi! 0
— 5 AU — 1 C MAT
C _ MS R MHZO H10,
calorimeta AT
3
ﬂx3228.11x103lm01‘1 —i£‘}g><1—O%><75.2911m01‘1K'1X3.45°C
122.13 gmol‘1 18.02gmol = 3.45°c
=5.16><1031°C'l P5.4) Using your results from Problems P52 and P53, calculate AS, ASsmaundgngs, and
ASmmI for each step in the cycle and for the total cycle described in Problem P52. 7.62X103J q . 
a—>b AS=AS m... =ﬂ"’"‘=——=873 1 =
m dings T K J K ASrotaI O
b ‘9 C AS = _AS:urrounding: = 0 because qrzversible = 0 AStO‘aI = O
_ _ gm“ 5 —3.68x103J _
C ‘9 d AS — _AS:urraundings — Tb, : “53K— : J K 1 AStatal = 0
d ——> a AS = —ASWWMW = 0 because qmmme = 0. ASMQI = 0 to within the roundoff error. For the cycle, AS = ASme.d.,,gs= ASmml = O to Within the roundoff error. P5.7) One mole of an ideal gas With CV = 3/2R undergoes the transformations described
in the following list from an initial state described by T = 300 K and P = 1.00 bar.
Calculate q, w, AU, AH, and AS for each process. a) The gas is heated to 450 K at a constant external pressure of 1.00 bar. b) The gas is heated to 450 K at a constant volume corresponding to the initial volume. 0) The gas undergoes a reversible isothermal expansion at 300 K until the pressure is half
of its initial value. .a) The gas is heated to 450 K at a constant pressure of 1.00 bar. nRT. _ 1 molx8.3l4 Jmol’lK“ x300 K Vi = I 5 = 2.49x10'2m3
Pi 10 Pa
T .
V] =—f—K = 450 K x2.49><10'2m3 = 3.74X10‘2rn3
r; 300
w: _Pmm,AV = _1ospax(3.74x10‘2m3 —2.49X10‘2m3) = —1.25><103 J 3 X 8.314 J mol‘lK‘l
2 q = AH = AU— w =1.87><103J +1.25x103J = 3.12x103J 450 K 300 K b) The gas is heated to 450 K at a constant volume corresponding to the initial volume. AU=nCMAT=1 molx x150 K= 1.87x103J‘ = 8.43 JK'1 T
AS = nCM 111% =1 molx[%+ 1)X8.314 Jmol‘lK‘len i w = 0 because AV= 0. 3x8.314 Jmol"K‘1
2
5x 8.314 Jmol‘lK“‘ AH=nCAmAT=l molx——2——XISOK= 3.12X103J AU=q=nCMAT=1molx x150K=1.87x103J = 5.06 JK‘1 T
AS=rICym1n—L=1molx 2)x8314 Jrnol'lK‘lxln 450 K
’ 1; 2 300 K c) The gas undergoes a reversible isothermal expansion at 300 K until the pressure is half
of its initial value. AU: AH= 0 because AT: 0.
V
wmm =—q = —nRTln?f= —1 molx8.314 Jmol'lK‘1x300 Kx 1n 2 =—1.73x103J l 3
AS = qreversible = M = JK—l T 300 K P5.8) Calculate ASsummdmg, and ASmm; for each of the processes described in Problem P5.7. Which of the processes is a spontaneous process? The state of the surroundings for
each part is as follows: ' a) 450 K, 1 bar b) 450 K, 1 bar c) 300 K, 0.500 bar a) The gas is heated to 450 K at a constant pressure of 1.00 bar. _ _ 3
q = 3.12x10J:_6.93JK_1
450K = 8.43 JICl — 6.93 JK“ = 1.50 JK‘I
The process is spontaneous. surrounding: _
Tmrraundings Aston = A5 + AS surroundings b) The gas is heated to 450 Kat a constant volume corresponding to the initial volume. —q —1.87><103J _
Assumu in:= = l
"d g Tsunaunding: K
AS =AS+AS =5.06JK‘14.16JK“=0.90 JK“ 10ml surroundings The process is spontaneous. c) The gas undergoes a reversible isothermal expansion at 300 K until the pressure is half
of its initial value. q —1.73x103J _
ASsMrrtmn in =_= 1
" S" T 300 K
ASm, = AS + Asmmg, = 5.76 JK'1 — 5.76 JK“ = 0 There is no natural direction of change in this process because it is reversible. P5.12) The standard entropy of Pb(s) at 298.15 K is 64.80 J K‘1 mol_1. Assume that the 2
£13m:22.13+0.01172£+1.00x10‘5£;.The
Jmol‘lK‘1 ‘ K K melting point is 327.4°C and the heat of fusion under these conditions is 4770 J mol'l. CPM(Pb’12=32.51—0.00301£.
JK‘1 mol‘ K a) Calculate the standard entropy of Pb(l) at 500°C. heat capacity of Pb(s) is given by Assume that the heat capacity of Pb(l) is given by b) Calculate AH for the transformation Pb(s, 25°C) —> Pb(l, 500°C). a) s;(Pb,1, 773 K): s; (Pb,S,298.15 K) 600.55 AH ‘ 773.15 C
j P'“ d[T/K]+ 1m" + I M d[T/K]
298.15 Tﬁuian 600.55 / = 64.80 J mol‘1 K—I
60055 2213+ 0.01172[T/K]+l.00x10’5 [T/K] [WK] d[T/K] +
298.15 + 4770 J mol—1
600.55 K
77“532.51—0.00301[T/K] + ———d[T/K]
600.55 i = 64.80 J 11101—1 K‘1+20.40 J mol‘1 K"l +7.94 J mol'1 K‘1 +7.69 J mol‘1 K'1
=100.8 J 11101"1 K—1 600.55 773. 15 1’) AH: ‘ l Cifidle/Khwmsn J (Wm/K] om! _
298.15 600.55 = 8918 J mol‘1 +4770 J mol‘1 +5254 J mol‘1
=18.94x103Jm01~1 P5.14) One mole of a van der Waals gas at 27°C is expanded isothermally and
reversibly from an initial volume of 0.020 In3 to a ﬁnal volume of 0.060 m3. For the van der Waals gas, = 1 Assume that a = 0.556 Pa m6 11101—2, and that
T {W 2 '
b = 64.0 X 10—6 m3 mol'l. Calculate q, w, AU, AH, and AS for the process. "I for’ a van der Waals gas AU=a ——L— 1
Vm,i Vm,f as shown in Example Problem 3.5 1 1
0.020 m3 0.060 m3 V’ dV “de
+a —2'
V, Vi V —b Vi V =—RT1n(V’—b)—a[—1——i]
Vf 11,. AU=0.556 Pa m6 mol'2[ )x1m01=18.5 J 28.314 J mol‘lK'1X300.15 K 111%
0.019936 1 1 —0.556 Pa m6 mol‘2 ——3————3—
0.060 m 0.020 m j: —2.73X103 J P RT a 2 8.314 J mol“ K“x300.15 K_ 0.556 Pa m6 mol ———————— =1, 5
V,.—b Vf 0.019936m3 (0,020,332 237x10 Pa ____.____ =4.146><10“ Pa
V, —b V} 0.059936 m3 ((1060 m3)2 AH=AU+A(PV)=AU+Pfo—R.K
=18.5 J+(4.146x104 Pax0.060 m3 —1.237><105 PaX0.020 m3)=32.1 J
q=AU—w=2.73x103 J 3
AS = qreversible =w: J 1(1
T 300.13 K There is an W somt. sari ‘m M sows/13 3‘“ 531,4
yaris (bl an? C0 0n\:ﬁ) mold. 5mm (asmr ‘m Warhead) ‘6 Lu” wuss is comm? about W, not wv/«rsiblﬂ M
Stand? in Wimmm' P5 .19) An ideal gas sample containing 2.50 moles for Which C V," = 3/2R undergoes the
following reversible cyclical process from an initial state characterized by T = 450 K and P = 1.00 bar: a) It is expanded reversibly and adiabatically until the volume doubles.
b) It is reversibly heated at constant volume until T increaSes to 450 K.
c) The pressure is increased in an isothermal reversible compression until P = 1.00 bar. Calculate q, w, AU, AH, AS, ASsmoundings, and ASWI for each step in the cycle, and for the
total cycle. The temperature of the surroundings is 300 K. a) The temperature at the end of the adiabatic reversible expansion is calculated.
T f Vf __ = [_]H = (2)1—3 = (2r: = 0.630 T V. l l Tf = 0.630X450 K= 283 K The initial and ﬁnal volume and the ﬁnal pressure are calculated.
nRY]. _ 2.50molx8.314 Jmol‘lK‘1x450K 11,. = 5 =9.25><10*2m3
B 10 Pa
V, =2V,.=0.185 m3
_ T ‘
P, =Pi£——f—=1.00barxl><283 K =0.314 bar
V, 1. 2 450K q = 0 because the expansion is adiabatic.
3x8.314 moi—1K“ 2
5 X 8.314 Jmol"K‘1 2
AS = 0 because qmmibze = 0. ASmmundmgs = 0 because q = 0. ASmmI = 0. AU = W: nCV'mAT = 2.50 molx x(283 K—450 K) = —5.21x103J AHanMAT=250 molx ><(283 K—450 K)=——8.68x103J b) w = 0 because AV: 0. —1 —1
AU = q = nCMAT = 2.50 molxﬂlﬁ—iﬂlixmo K—283 K) = 5.21x103J 1 1
AH =nCP_mAT=2.50 molxslmmg—mql—Ki—xmso K—283 K) = 8.68x1031 The pressure at the end of the process is calculated in order to calculate AS. :r
Pf =13—f= 0.314 barx450K
T. 283K Pf Tf
AS=—ann—+nCPmln—
P. v T. l I = 0.500 bar . . _1 4 0K
= —2.50 molx8.314 Jmol'lK‘1 XhW+ 2.50 molXW—Xln 45 4bar 2 283 K
=—9.63 JK1 +2410 1K1 =14.5 JK1 _ _ 3
ASmrmundin : q = = JK—l
g T 300 K surrounding:
AS =AS+AS =14.5 JK“—17.4JK“=—2.9OJK‘1 total .mrroundings c) AH= AU= 0 because AT= 0. V .
w=—q=—r112T1:17f=—nRT1nﬂ x Pf = —2.50 molx 8.314 Jmol_1K—1X450len: = 6.48X103J P 
AS = —nR 111—] = —2.50 molx8.3l4 Jmol—1K_lxlnM = ._14.5 JKl
i 0.500 bar
3
ASSIIH’Dunding: = —L = 11(—I
‘Tlmrraundings K
“total = Assurroundingx + AS = 7'1 J K—l For the entire cycle,
Wtom1= —5.21x103J + o + 6.48>< 1031 = 1.27>< 1031
11m]: 0 + 5.21 x 1031— 6.48x 1031 = —1.27>< 1031 AUW1= —5.21x1031 + 5.21x103J + 0 = 0 AHWI = —8.68>< 1031 + 8.68x 1031 + 0 = 0 Asmm, = 0 + 14.5 J K"1—14.5 J K”1 = 0 ASmrroungings=AS total = 0 _ J K_I + J K_1 = J I<u1 P5 27) The Chalk Point, Maryland, generating station supplies electrical power to the
Washington DC, area. Units 1 and 2 have a gross generanng capac1ty of 710 MW
(megawatt).’The steam pressure is 25 X 106 Pa, and the superheater outlet temperature
(T h) is 540°C. The condensate temperature (Tc) is 30°C.  u I . I 9
a) What is the efﬁciency of a reversible Carnot engme operating under these conditions. ‘ ' ' 11 efﬁciency of the turbine, which If th efﬁ01enc of the borler is 91.2%, the overa _ ‘ '
iaricludees the CarnZt efﬁciency and its mechamcal efﬁCIency, 1s 46.7%, the efﬁc1enc2;
of the generator is 98.4%, what is the efﬁciency of the total generating unit? (Another 5 0 needs to be subtracted off for other plant losses.)
c) One of the coal burning units produces 355 MW. How many metric tons (1 metric ton = 1 X 10'5 g) of coal per hour are required to operate this unit at its peak
output if the enthalpy of combustion of coal is 29.0 X 10 kJ kg ? b) The efﬁciency of the generating plant is the product of the individual efﬁciencies. E= 0.912 X 0.467 X 0.984 X 0.95 = 0.398 3600 s 12 l
c) The energy output per hour is given by 355 MW X = 1.278 x 10 J hr The heat required to output this energy is 12 —1
q=E""“’"‘=1'278X10 1hr =3.211x10”Jhr" 8 0.398 3.211X1012 J hr"1 _1 =110.7 ton hr“.
29.0x109 J ton The number of tons of coal needed is P529) A reﬁigerator is operated by a 025bp (1 hp = 746 watts) motor. If the interior is
to be maintained at —20°C and the room temperature is 35°C, what is the maximum heat leak (in watts) that can be tolerated? Assume that the coefﬁcient of performance is 75%
of the maximum theoretical value. The coefﬁcient of performance is n. = aux—11$: 3.45
T T ho! cold The maximum heat that can be removed from the cold reservoir is given by input power X coefﬁcient of performance
746 W hp 
1f the heat leak is greater than this value, the temperature in the refrigerator will rise. 2 0.25 hp x x 3.45 = 640 w =640 J s“ ...
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 Fall '06
 Schwartz
 Thermodynamics, constant volume, Isothermal process, reversible isothermal expansion

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