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# HW3

Course: CHE 110a, Fall 2009

School: UCSB

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Department of Chemical Engineering University of California, Santa Barbara ChE 110A Winter 2011 Problem Set No. 3 Due: Wednesday, January 19, 2011 Objective: To understand and perform calculations using the First Law of Thermodynamics, and to assess system energy flows in closed systems. Note: Numerical values for some problems have been changed from those in the book. Problem 7 (thought problem) A friend...

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of Department Chemical Engineering University of California, Santa Barbara ChE 110A Winter 2011 Problem Set No. 3 Due: Wednesday, January 19, 2011 Objective: To understand and perform calculations using the First Law of Thermodynamics, and to assess system energy flows in closed systems. Note: Numerical values for some problems have been changed from those in the book. Problem 7 (thought problem) A friend claims that the work required to compress steam, in general, can be evaluated by looking up enthalpy and other property values before and after the compression in tables (e.g., the steam tables at the back of your book). What is wrong with this assessment? What if the compression is known to be isobaric, specifically? Problem 8 (Smith, van Ness, Abbott, 2.8, page 57) A closed, nonreactive system contains species 1 and 2 in vapor/liquid equilibrium. Species 2 is a very light gas, essentially insoluble in the liquid phase. The vapor phase contains both species 1 and 2. Some additional moles of species 2 are added to the system, which is then restored to its initial T and P. As a result of the process, does the total number of moles of liquid increase, decrease, or remain unchanged? Explain your reasoning. Problem 9 (Smith, van Ness, Abbott, 2.5, page 56) One mole of gas in a closed system undergoes a four-step thermodynamic cycle between equilibrium states 1, 2, 3, and 4. Use the data given in the following table to determine numerical values for the missing quantities, i.e., fill in the blanks. Step 12 23 34 41 12341 U t / J -300 ? ? 3000 ? Q/J ? -2100 0 ? ? W /J -1100 ? 500 ? 800 Problem 10 (Smith, Ness, van Abbott, 2.30, page 60) In the following take 20.8 and 29.1 J mol for nitrogen gas: (a) Five moles of nitrogen at 35 , contained in a rigid vessel, is heated to 240 . How much heat is required if the vessel has a negligible heat capacity? If the vessel weighs , how much heat is required? 75 kg and has a head capacity of 0.5 kJ kg (b) Three moles of nitrogen at 210 is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 30 if the heat capacity of the piston and cylinder is neglected? Problem 11 (Smith, van Ness, Abbott, 2.32, page 60) (a) Find the equation for the work per mole of a reversible, isothermal compression of a gas in a piston/cylinder assembly if the molar volume of the gas is well-described by, where (b) is a positive constant. The gas in part (a) is subjected separately to two isothermal processes that both begin with the system at , . In each, the gas volume is halved by means of application of an external compressive force. In the first process, the force is applied very slowly. In the second, it is applied rapidly and almost instantaneously. Find expressions for the work per mole of these two processes in terms of , , , and . Which is greater? Problem 12 (Smith, van Ness, Abbott, 2.36, page 61) One kilogram of air is heated reversibly at constant pressure from an initial state of 50 and 1 atm until its volume triples. Calculate , , , and for the process, in or / as appropriate. Assume that air is ideal and has 72 . Hint: tables A.1, A.2, and B.1 in the book may be helpful.

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UCSB - CHE - 110a
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UCSB - CHE - 120a
f-l W 3C1 -\-t, e)t' t b wkpvobie Wl1.76 It. Newtonian llqwd fiows slowly into a pia-nar channel, as shown in Fig. P4.76. T he planesthat confine the flow a re parallel, and very wideIn the direction perpendicular t o the page. Hencethe flow is t
UCSB - CHE - 120a
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UCSB - CHE - 120a
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UCSB - CHE - 120a
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,XJA=r ') T -/Q n'A&quot;ttfLMo- v*)=Q*(rrn ? ffi(uut',Mcfw_s=(^/ ^Ar'E n (^f*-v,VA4&quot;3,CYe,t1Jl=U tW * b tnVef,^- t 'np-t'ofi=lf = a eW+TBW- 'oBW;&amp; cfw_ * - L ,)h6\C,MEcfw_A.=l-:cfw_e. 'LOfrel-Dtg(%-W)= g u ,a(W-V-de f4e(A=teW=Cecf
UCSB - CHE - 120c
Hr,'lftZu\$oJ,wrt .4hut(=- NJ 4*.* oIiJcnl2 oJ\lt= of,t'N r= r *rF''A-\ cfw_( ,tn|,rv4=Jn&quot;cfw_r*(Nr *, Vn)itlXd-cD*gY'J/nC (-at)JtN)a = \'rL(a r'rktr^'.7lz-P *lz =I\,r&amp;rcD*s Jtn(- dn)oII- ' &quot; '- l-Xnt\tYcfw_NII= ?QryW
UCSB - CHE - 120c
b uocfw_)/J i+*A 6 4(vu:lr./e=fL=rocu-o(*u*&amp;rcfw_[usin&quot;'tn-@,^,\\e'a\$- cDts#IVA='Tttcfw_rrf'l itf'*r^ n=o +'- l'- ^*L ft='nv/vA *u foNl&quot; I /trF '*T-vhrl'tht&quot;tnY-tJr\ = -iJl - htJ-Dv. r t J X.l: -N* - AcVwnffi^o ,oS/|t&lt;i-J,v
UCSB - CHE - 120c
IVp.Lcl, - Ar,( ^ / 5 t ( t 7 - tz)&quot;ttl'(I l-ltI/-i at'i^nn. CI,Plr q ;' - oL p , p n ( A)tiN ( K-/aGIL4/L5t3/lA7r&quot;11Q, t t\i&gt;'TesII!)eft'Atnl.)z1'7l &lt; )I a :cfw_ t -&quot;-9/'tlo rcl \$Slf7;F ( =tvil/-th r &gt;,y&quot;17i&quot;tt7
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('rnr,/l\bJ^'.*eacukt^IrI,L ltY\-*, - V o urLl ,=*o.fr/aiaf K,z tTrDY :kasvcfw_\u*f ttttn 'dKu1(nNlE)-TF- - P r - ot^fry&quot;f&lt;_ ) vLr-+ c rrNo,=?_.Nr,:&quot;cfw_rtY!.rL'f-+ )1\$,c*@U-Wi/.tf/L-V-= - O1- -f. ' \ n . 'l F: JA + +'/ i
UCSB - CHE - 120c
fiH &quot;u.lb .I,nt/ ^'*'ilr.'^'l'YMr i zv icfw_t\i^\/:cfw_t.t I - .l\tlL\orcfw_. -,'ii , t '|a ! / . . . \ . \f , - \. \-.+ &quot; ,cfw_', ._4 ,I i ,o ' ,-hrl ii 'tp u&quot;l1x'v'lcC J '-i )I.iu'.-tr*\N&quot;v(,-1\n_)r\cfw_-i'; ft'
UCSB - CHE - 120c
(A f - ? fr t -:getf o^&quot;u I t *&quot;tou fr u1.,'r+ UE cfw_,rl,r&amp;*/rnnrtA(r, &amp;[u'*r= J ;: , . % .,f.',|,'l , - t .]r,], .,r0t' ,. , *,n-1 .0,.-!,*,i, l '.1r,.-,/ /ur^trlflru\l l,r.*fii.= t /ri * :lj,t;.-,Nurr,=lXan*L*egiIiu,),1'-Jir+1- -t
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