" 1 = # \$H vap. Chapter 3 Problems

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1 Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius Clapeyrn equatin n the vapr/liquid bundary: " ln " 1 = \$H vap R % \$H = R & ln ( ) ln( ) vap 1 1 where is atmspheric pressure, is the triple pint pressure, is the vaprizatin temperature at 1 atm. and is the temperature at the triple pint, which we dn t knw. We als assume that ΔH vap is independent f temperature. On the /L line hwever, d d = "H m where "V l s is the vlume difference between "V l s the slid and liquid phase. his vlume difference is much smaller than the vlume difference between the liquid and the vapr. herefre, the /L line is nearly vertical and as a cnsequence, ~ m. ln( 11325) ln( \$ 8.4 %18) As a result "H vap = R \$ & 285kJ /ml

2 rblem hase α f species A transfrms int phase β at 55 K and 1 atm. he heat capacities f A,! 5 3,! 4 3 in the structures α and β are, respectively, C = 8.8" 1 and C = 2.38" 1. Using the hird Law and assuming that a and b are pure crystalline elements, calculate the entrpy and enthalpy f transfrmatin at 55 K and K at 1 atm pressure. Frm the third law, we knw that = "! = K; ( if they are pure crystalline elements). " \$ = \$ C C = ", \$, d d = 2.93*1 = 7.95*1! 5! 5 3 = 55K, " \$ % % \$ =! = 8.37J / ml * K \$ G "! ( 55K) = = \$ H"! % \$ "! " % \$! H (55K) = J / ml! " (K) = \$ "H \$% (K) = "H \$% (55K) + & 55 (C,% ' C, )d =117.2J /ml 2

3 rblem in can exist in either the metallic β-n phase r in the semicnducting α-n phase. Cnsider the phase transitin frm n in the β-phase t the α-phase. he enthalpy change "H \$% assciated with the transitin at 1 atm. is -2.1 kj/ml. β-n is in equilibrium with α-n at ne atm. pressure and =238.8 K. At 1 atm. the equilibrium temperature is =233.8 K. Estimate the vlume difference between the tw phases. Which ne has the higher density? Explain yur assumptins. Apply the Clausius equatin: d d = " \$% "V \$% We will assume that Δ and ΔV are independent f temperature and pressure. " \$% = "H \$% = &8.8 J.K &1.ml &1. Frm the data pints, we have d/d=-2x1 6 a/k. hus "V \$% = " \$% d /d = 4.4 cm 3 /ml. he β-phase is denser than the α-phase. 3

4 rblem Yu wish t evaprate Au cntacts nt a simple integrated circuit in an evacuated thermal evapratin system. Au has a biling pint f 313K and an entrpy f vaprizatin at 1 atm. f 16.8 J.ml -1.K -1. If the desired vapr pressure f Au in the vacuum chamber fr the depsitin prcess is 1-7 atm, estimate the temperature t which the Au surce must be heated in the thermal evapratr. tate yur assumptins. Using the Clausius Clapeyrn equatin, d = "H vap d = " vap vap d where we assume R 2 R 2 that the vaprizatin enthalpy (= vap Δ vap ) des nt depend strngly n temperature and pressure ( vap =313K). Integrating the Clausium Clapeyrn equatin, we have: ln( 2 ) " ln( 1 ) = \$ vap vap 1 " 1 ' & ). By plugging in this equatin 2 =1-7 atm, 1 =1 atm, R % 1 2 ( 1 =313K, we btain 2 =1389K 4

5 rblem 1-hw that at the triple pint the sublimatin enthalpy f a material is equal t the sum f its melting enthalpy and its vaprizatin enthalpy. 2-Belw the triple pint (-56 C) the vapr pressure f slid CO 2 is given as: ln( (atm) ) = " with in K. he mlar heat f melting f CO 2 is 8.33 kj. Making sme simple assumptins, calculate the vapr pressure exerted by liquid CO 2 at 25 C and explain why slid CO 2 is referred t as dry ice. ketch the phase diagram f CO 2 and place the =1 atm line and the =25 C line with respect t the triple pint. 1-If H is a state functin, we can g frm s t v by tw different rutes with the same ΔH. he tw different rutes are direct sublimatin r melting and then vaprizatin. At the triple pint all the phases can exist in equilibrium and these three prcesses can ccur. herefre "H sv = "H sl + "H l v. 2-Using the expressin, we can calculate "H sv = \$R % & ln = \$R % We btain &1/ "H sv =25.9kJ/ml. Using the infrmatin abut "H sl, we btain "H l v = "H sv \$ "H sl =17.6 kj/ml. We nw use the Clausius Clapeyrn equatin t estimate the vapr pressure f the liquid at 25 C using the triple pint as the starting pint: ln & 25C % ( = " )H l *v 1 \$ "56C ' R 3K " 1 & % (. \$ 217K ' he pressure at the triple pint is btained frm the expressin fr ln() given in the questin: -56 C =5.21 atm. We btain 25C ~75 atm>>1 atm. CO 2 is called dry ice because at rm temperature it sublimates directly frm slid t gas withut prducing a wet liquid phase. he phase diagram is as fllws: 5

6 rblem he ice f an utdr skating rink is at the ambient temperature f -2.5 C. Calculate in a the minimum pressure (applied fr example by the blade f a skate) required t melt the ice. At C, the specific vlume f water is 1 cm 3 /g and that f ice is 1.9 cm 3 /g. he heat f fusin is J/g Ice is less dense than water therefre increasing pressure lwers the melting temperature. he prblem asks t find the pressure where ice will melt at -2.5 C=27.5 K. We knw that at 1 atm and 273 K water and ice are in equilibrium. We use the Clausius equatin t calculate the pressure variatin f the melting pint: d d = "H sl \$ " = "H 27.5 sl d % "V sl "V sl J /g \$ 27.5' 1 = + ln& ) = a * 337atm. "9 1 "8 m 3 /g % 273 ( 6

7 rblem ure ice, initially at -2 C, is cmpressed adiabatically and reversibly until it melts. 1) ketch n a / phase diagram hw yu wuld find the slutin (i.e. the path f the cmpressin and shw n the diagram hw yu wuld identify the melting pressure and temperature graphically). (1 pints) 2) Calculate the temperature and pressure at which melting starts? (2 pints) slve this prblem yu will need a few apprximatins: - he temperature ranges are small enugh that the mlar vlume f water is apprximately cnstant (be careful, this is nt the same as stating that α=~) - Remember that ln(1-x)~-x if x<<1 C =2 kj.kg -1.K -1 ; ριχε=92 kg.m -3 ; αιχε=15 ppm.k -1, L m =334 kj.kg -1, ρ water =1 kg.m -3, β~ " & Hint: t slve questin 2, yu may find useful t calculate % ( \$ " ' 1) he adiabatic and reversible cmpressin heats the slid. In a - phase diagram, the cmpressin path will intersect the /L equilibrium line. 2) he simplest way f thinking abut this prblem is t think abut finding the intercept between tw curves: the /L equilibrium curve and the () functin during cmpressin. he /L equilibrium curve is given by the Clapeyrn equatin. Yu already knw frm class and prblem sets that it s very steep and clse t being a line. Yu can use the simplest apprximatin t derive the () functin fr the /L equilibrium: d d = "h L and assume that yu are interested in this functin ver a small range f ( v L \$ v ) temperatures therefre n the right side f the equatin is apprximately cnstant and equal t m =273 K. Yu can als integrate this equatin rigrusly and yu can cnvince yurselves that if yu are integrating ver a small interval, assuming ln(1-x)~-x takes yu t the same result. Hence, the equatin fr the /L equilibrium line is: 7

8 / L L ( ) = + m m " 1 1 L " ( ) m ρ water >ρ ice, the slpe is negative, as it shuld be. ( ) where is the pressure at = m : =1 atm. ince Nw we need t find AC (), which is the trajectry in the / plane during the adiabatic cmpressin. ince the cmpressin is adiabatic and reversible, we can write: d = " & % ( d. \$ " ' Integrating this equatin will give us AC (). " & Using a Maxwell relatin, we have % ( = " & % ( \$ " ' \$ "V ' " & We use the recipe discussed in class t find % ( : \$ "V ' d = " & % ( dv + " & % ( d = " & % ( [)Vd * +Vd] + " & % ( d \$ "V ' \$ " ' \$ "V ' \$ " ' V d = " & % ( )Vd +... \$ "V ' I am nt interested in the term in d as what I wrte is sufficient fr me t identify!! \$ &. "!V % " & We have % ( )Vd = C \$ "V ' hence " & % ( = " & % ( = C \$ "V ' \$ " ' )V " & One f yu nticed a much easier way t find % (. We have seen in class that \$ " ' d = C d "Vd. herefre, =cnst d=, frm which we quickly btain " & % ( = C \$ " ' )V In any case, integrating this equatin using the apprximatins allwed in the prblem and again assuming that the interval ver which we integrate is small, we get AC ( ) = + C "V i he prblem is nw reduced t finding the intersectin between tw lines: L AC ()= /L () m m " 1 1 ( ) = C ( L " i \$V i). ( i ) where is the pressure when = i = 271 K: =1 atm. ( ) m lving fr, I get =271.5 K and plugging int any f the expressins fr, I get ~225 atm. V 8

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