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Homework answers / question archive / Question 1: Heat and Thermod namic Processes (55 Points) A e leaned in cla, he ale ceain hemdnamic aniie ame afe a ce deend n he ah aken ding he ce
A e leaned in cla, he ale ceain hemdnamic aniie ame afe a ce deend n he ah aken ding he ce. In hi blem, e ill cnide h mch k i dne, hea i dced, and h he eneg, enhal, and en change ding he flling f diffeen cee.
am, Vini ial=2 L Pfinal=0.75 am, Vfinal=2.67 L.
I. A eeible, adiabaic eanin f 1 ml f an ideal mnamic aing fm Pini ial=1
am, Vini ial=2 L and ending a Pfinal=0.75 am.
a. Pleae da he ee . lme ce f hee f diffeen cee n he ame e f ae. Cleal label hich ce i hich and elain h each ce ake he fm i de. (15 Pin )
PV Diagram ith 4 Cures Labeled: (10 Points)
2.5 Pin f Cec Paablic Ce f Ce 1 (Red Ce )
2.5 Pin f Cec L-Shaed Ce f Ce 2 ( Shld Ge Oienain f Ce Cec )
( Oange Ce )
2.5 Pin f Cec Linea Ce f Ce 3 (Shld Deceae Pel) (Ga Ce)
2.5 Pin f Cec Adiabaic Ce Ling Bel Ihemal Ce f Ce 4 (Ble Ce)
Reason for form of cure 1: (1.25 Points)
F an ideal ga ndeging an ihemal ce, PV=nRT=cnan hld ince he emeae and nmbe f mle ae held cnan. PV=cnan i he fncinal fm f
a aabla, cnnecing he iniial and final in.
F cedi: den m ae ha, becae hi i an ihemal ce, he ce i aablic. Cld h ing mah ih aing aablic elicil.
Reason for form of cure 3: (1.25 Points)
The ein ae ha he ee i
leed in a linea fahin. The iniial and final in hld h be cnneced b a line ih le (P2-P1)/(V2-V1).
F cedi: den m ae ha he endin ae cnneced b a line baed n he m.
Reason for form of cure 2: (1.25 Points)
Thi ce hld be L-haed, hee i fi deceae eicall fm 1 am 0.75 am a a cnan lme f 2 L and hen inceae hinall fm a lme f 2 L a lme f 2.67 L a a cnan ee f 0.75 am.
F cedi: den can ae ha hi i a decibed in he blem.
Reason for form of cure 4: (1.25 Points)
Thi ce eeen a eeible adiabaic eanin, hich i gened b he
fmla Pini ialVini ial = PfinalVfinal , hee =5/3
f a mnamic ideal ga. Gien hi fmla, den can calclae he final lme be 2.3768 L. Gien hi final lme, i i clea ha he adiabaic ce hld lie bel he ihemal ce. Sden can al lgic ha adiabaic eanin hae me ickl deceaing PV ce han ihemal eanin baed n he e.
F cedi: den m ae ha he
|
b. Cme he k efmed b each f he fi hee cee. Pleae h all f k and ae amin. Be e inclde ni. (10 Pin) |
adiabaic ce lie bel he ihemal ce.
Work Performed during Process 1: (3.33
Points)
Thi i a eeible, ihemal ce in hich Pe=Pga a all ime. Theefe, =-nRTln(V2/V1). Ne ha he k i negaie (1 in) ince hi i an eanin.
T can be cmed fm T=Pini alVini ial/nR (Pfinal and Vfinal can al be ed) = 2/.08206
= 24.37 K. Ping hi gehe, ne ge =-1 ml * .08206 L-am/(ml-K) * 24.37 K ln(2.67 L/2 L) = -.578 L-atm. Ane in Jle i al fine.
1.5 Pin f cec k fmla
1 Pin f cec emeae
0.83 Pin f nmeicall cec k.
Work Performed during Process 2: (3.33
Points)
Work Performed during Process 3: (3.33
Points)
T deemine he k nde hi ce, ne m ealie ha he k i he aea nde hee ce. Thi ce fm he bndaie f a aeid ih bae 1=1 am, bae 2 = 0.75 am, and heigh 0.67 L.
Plgging hi in he aeid aea fmla
ield, A= h (b1+b2) = (0.67 L ) (1+0.75
am) = .59 L-am. In he d, he k i
-.59 L-atm.
1.5 Pin f cec aea fmla
k
0.83 Pin f nmeicall cec k
Wk i nl efmed hen he lme f a em change. Th, n k i efmed hen he ee i leed in hi em. Wk i nl efmed ding he ecnd e hen he lme i inceaed. Th,
=-P V=-0.75 am * 0.67 L = -0.5 atm-L.
1.5 Pin f cec k fmla
1 Pin f cec lme change
0.83 Pin f nmeicall cec k
5 Points for correctl stating that Process 3 performs more ork than Process 1. Thi can be cnclded fm he ealain f he k in a b. Ne ha den hld ndeand ha, b dcing me k, I mean ha he em i dcing me k, hich indicae ha he able ale f he abe aniie hld be cmaed.
5 Points for correctl stating that this is because Process 3 has the largest area under
its cure. Thi can al be cnclded baed n he k in a b, , j b cmaing he egin nde he ce in a a.
4 Pin f el aing ha E and H ae ae fncin and h can all be ealaed ing he fmla:
E=nC T= n 3/2 R T = 1 ml * 3/2 * 0.08206 L-am/(ml-K) T (2 Pin)
H=nC T= n 5/2 R T = 1 ml * 5/2 * 0.08206 L-am/(ml-K) T, (2 Pin)
Whee he hea caaciie ae knn f a mnamic ideal ga. All ha diffe amng hee cee ae ha he change in emeae ae. Temeae change can be
deemined baed n he PV=nRT. Wih he change in eneg and k in hand, ne can hen cme he hea baed n he Fi La f Themdnamic.
Process 1: (4 Points)
1 Pin f he 3 Flling Qaniie and
Wk; 1 Addiinal Pin f Cleal Saing
Ihemal
1= -1 = .578 L-atm E1= 0 (Becae Ihemal)
H1= 0 (Becae Ihemal )
Paial cedi f 0.5 inead f 1 in f if k incec abe.
Process 3: (4 Points)
1 Pin f Recgniing Tha, Een Thgh
Pce In Ihemal, T=0, OR, Tha
Since E and H Ae Sae Fncin, The M Be he Same A f Pcee 1,2
1 Pin f Each f he Thee Flling Qaniie
3= -3 = .59 L-atm E3=0 (Becae Same A Pcee 1,2)
H3=0 (Becae Same A Pcee 1,2)
Process 2: (4 Points)
1 Pin f he 3 Flling Qaniie and
Wk; 1 Addiinal Pin f Cleal Saing
Ihemal
2= -2 = .500 L-atm E2= 0 (Becae Ihemal)
H2= 0 (Becae Ihemal)
Paial cedi f 0.5 inead f 1 in f if k incec abe.
Paial cedi f 0.5 inead f 1 in f if k incec abe.
Process 4: (4 Points) 1 Pin f Calclaing he Change in
Temeae be: T = (Pfinal Vfinal)/(nR) (Pini ial Vini ial) /(nR) = (0.75 am*2.38 L)/(1 ml
*.08206) - (1 am * 2L)/(1 ml*. 08206) = -2.60
K
1 Pin f Cec Change in Eneg
1 Pin f Cec Change in Enhal
1 Pin f Cec Hea (Becae Adiabaic)
4=0 (Becae Adiabaic) E4= 1 ml * 3/2 * 0.08206 L-am/(ml-K) *
-2.60 K = -.32 L-am (can be in Jle -32.4 J ) H4=1 ml * 5/2 * 0.08206 L-am/(ml-K) *
-2.60 K = -.53 L-am (can be in Jle -53.7 J )
Man bilgical mlecla ndeg a aniin called denaain, in hich he eeniall nfld fm hei m able, naie, ae. Thi eacin cld be decibed a
Naie (flded) ? Denaed (nflded)
a. Elain h he enhal and en change aciaed ih hi denaain hae he ign ha he d. Bae deciin n h he ein i hicall changing (e.g., ha i haening i bnd, ec.). (10 Pin)
Ding hi ce, he ein i nflding. Ding he nflding ce, i ill beak bnd. Thi naall eie a hea in, meaning ha hee hld be a iie change in enhal. When he ein i nflded, i i me diganied han hen i i flded. A a el, he en hld al ame a iie ign, a i de in he deciin abe.
5 Pin f enhal hld be iie becae ke bnd ae bken
5 Pin f en hld be iie becae he ein i le deed
b. Oe ha emeae ange ill denaain nanel cc? Pleae h all k.
(10 Pin )
Denaain cc nanel hen he fee eneg change i negaie. The en change fa nanei, hile he enhal change difa i. T deemine e ha emeae ange he denaain ill be nane, ne m cme f ha emeae G0. Thi can be led a
HT S
H/ S T
418 kJ/ml/[1.3 kJ/(K-ml)] = 320 K T
In other ords, the protein ill spontaneousl denature for T 320 K.
2.5 Points for stating that denaturation occurs spontaneousl hen the free energ is negatie
2.5 Points for stating the Gibbs free energ equation in terms of the enthalp and entrop
2.5 Points for tring to sole for T using the Gibbs free energ formula based upon the inequalit aboe ( H-T S0)
2.5 Points for giing the roughl (ithin rounding) correct temperature range
c. Se he hea caaci f he naie ae i C (flded) = 2100 J/(K-ml) and ha f he denaed ae i C (nflded) = 2500 J/(K-ml). Cme he al enhal and en change a 2 mle f he ein ae heaed fm 200 K 340 K. Ame ha he hea caaciie, eacin enhal, and eacin en emain ghl cnan a a fncin f emeae in he naie and denaed ae. Pleae h all f k. (15 Pin)
T cme he enhal change, ne ha
H=nC T
if he emeae i changing and ha H= H=418.0 kJ/ml a 320 K, hen he ein nanel denae. Similal, he en change ha cc hen he ein i heaed i gien b
S=nC ln(T2/T1) And S= S=1.3 kJ/(K-ml) a 320 K.
Ping hi gehe, he enhal change ill be
HTot = H200 K 320 K + H + H320 K 340 K = 2 mol * 2100 J/(K-mol) * 120 K + 2 mol * 418 kJ/mol + 2 mol * 2500 J/(K-mol) * 20 K = 1,020 kJ
The en change ill be
STot = S200 K 320 K + S + S320 K 340 K = 2 mol * 2100 J/(K-mol) ln(320 K/200 K) + 2*1.2 kJ/(K-mol) + 2 mol * 2500 J/(K-mol) ln(340 K/320 K) = 3480 J/K
2.5 Points for Correctl Stating Enthalp Formulas
2.5 Points for Correclt Stating Entrop Formulas
5 Points for Correctl Recogniing That Heating Is Done from 200 to 320 K, Then the
Phase Transition Occurs, Then Heating Continues ith a Different Cp from 320 to 340 K
5 Points for the Correct Calculations According to Aboe Math Een If Final Numbers
Are a Bit Off
d. Baed n ei anale, de he ein becme me le able denaain ih inceaing emeae? H d kn? (10 Pin)
Thi can be aneed hgh chemical eaning. A highe emeae, he en change beeen he naed and denaed fm ill becme lage becae he denaed fm ill be able me in me a. The enhal change hld al becme malle (b iie) a highe emeae becae he bnd ill be eakened a highe emeae. Thi mean ha he ein ill becme less stable denaain a highe emeae.
5 Points for Saing Less Stable
5 Points for Eplaining Wh Based on Entrop and Enthalp Changes (Dont Gie These
5 Points If The Dont Mention Entrop and Enthalp )
The mehane (CH4) cmbin eacin (a hn bel) ide a ignifican in f he ld eneg. I i al a ignifican cnib climae change de i dcin f cabn diide.
CH4 (g) + 2 O2 (g) ? CO2 (g) + 2 H2O (l) H = -890 kJ
Negaie, he eacin le gae and cne liid ae.
N, i ill deend n T, ince bh H and S ae negaie ale. (2 points )
The eacin i nane nl if G = H T S < 0, T S > H (3 points )
An inceae in lme cae al ee deceae, eacin hif he eacan ide inceae he ee.
The eacin i ehemic, inceaing he emeae ill cae he eacin hif he eacan ide (endhemic diecin) abb hea.
Adding N2(g), ine ga, ha n effec n eilibim. Adding O2( g) hif he eacin he dc ide.
(i) Bh N2(g) and O2(g) ae added ih changing he al ee emeae f he em. (4 Pin)
Adding gae ih changing he al ee, lme inceae, faing eacan ide.
Adding O2(g) hif he eacin eilibim he dc ide.
Oeall, he diecin cann be ediced.
(K2) a 500K? (14 Pin )
G = -RTlnK = -(8.314 J/K)*(2273K)*ln(1.5*108)
= -357kJ (-357 * 103 J) (4 points )
G = H - T S
S = ( H - G)/T = (-890kJ (-357kJ))/2273K (4 points)
= -234 J/K (6 points )
LnK2 = - H/R *(1/T2 -1/T1) + lnK1
= -890kJ/(8.314J/K) * (1/500K 1/2273K) + 18.826
= 167 + 18.826 = 185.826
K2 = e185.826 = 5.05 * 1080
(Get 2 points if the student knos to use Vant Hoff equation, 4 points for the correct anser)
Sile lfie (Ag2SO3) i a hie calline lid ih l lbili in ae (aingl lble al). Sile lfie dile in ae accding he flling eacin:
Ag2SO3 () ? 2 Ag+ (a) + SO32- ( a )
A 25C, a 100 mL aaed ile lfie ae lin cnain 0.46 mg f diled ile lfie. Pleae ane he flling ein and h k.
Mla eigh f Ag2SO3 = 295.8 g/ml
Mla Slbili ( ) = 0.46mg/100mL * (1000mL/L)*(1g/1000mg)/(295.8 g/ml)
= 1.6 * 10-5 M ( ml/L)
K f Ag2SO3 = [Ag+]2[SO32-] (3 points )
= (2 )2 * = 4 3 = 1.6 * 10-14 (4 points )
Qi f Ag2SO3 = [Ag+]2[SO32-] (3 points )
= (4.810-2 * 50 mL / 150mL)2 * (1.010-5 * 100mL/150mL)
= 1.7 * 10-9 (4 points )
Pin)
Since Qi > K , (G > 0), he lin i eaaed, Ag2SO3 ill eciiae. (3 points )
Peciiae fmain hen he em e-each eilibim:
[Ag+]2[SO32-] = K = 1.6 * 10-14 (4 points )
(3 points for riting the correct equation. 2 points for identifing each conjugate acid-base pairs)
( La Page )
