CHE350_2010_midterm 1_soln

# CHE350_2010_midterm 1_soln - ChE 350 F all 2010(100 pts...

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Unformatted text preview: ChE 350 F all 2010 (100 pts possible) Name: U/HOM‘S Exam 1 (show your work on the following pages as indicated) 1. Shown below is X—ray diffraction (XRD) data (using Mo Kc: radiation with X—ray wavelength K=0.0709 nm) from gold at various pressures and temperatures, as reported recently by L. Dubrovinsky, et al. [from “Noblest of All Metals Is Structurally Unstable at High Pressure,” Phys. Rev. Lett. 98 (2007) 045503.] Gold undergoes a phase transition from foe to hcp when heated at high pressure and cooled. Compare for example the patterns (a) and (c), which correspond to gold with fee and hop structure, respectively. Use this data as needed to answer the following questions: (a, 10 pts) Sketch and label the (111) and (110) planes in the unit cell shown below: (b, 15 pts) Calculate the lattice parameter, a, for gold with fee structure at 275 GPa and 295K from the data in trace (a) in the figure: _. a: (I) {chi Wm (c, 10 pts) Sketch and label the (102) and (002) planes in the hop unit cell shown below: . HCP (d, 15 pts) The d—spacing and lattice constants C: and c in an hcp crystal are related as ——= 20000 18000 a ' N il ,8 a 14000 — l E l ll .1 l ' : 12000 ‘1 l\_,_._~.2ﬂL V: D m: 5 235(20) GPa, 295 K .3" c ’8 : N a: E .: 10000 - *- K» g :3 o .— m a. V— I?" D ’— a c s -=, ‘\ s. g a) a “‘3 < N 'E‘ 8000~ .. m n V . b—t 0: l1 5 g 448(20) GPa, 860(10) K a J" ‘ ‘5; ~ 1’ A1; Up! L_,__:~__}L‘\J 4000 - jail I 1.1 § 275(20) GPa, 295 K , u q 2000 — a a “l l l\ ' 18 1'8 26 22 24 26 28 30 32 34 2 8 vFlG. 2. Examples of diffraction pattern collected from gold (a) compressed at ambient tempemture to 275GB] (3P2:> {h} heated Elli 8:304:10} K at 248220) GPa, (c) slowly cooled down to room temperature :11 2361720,}: GPa. and {d} heated again to 1050(15} K at 33320} 6P2}. The inset shows pant of 2D dii’ﬁ‘action image of the sample at 336(3)] GPa and room temperature. The pattern is dominated by diffraction lines of the high—pressure hep-AU phase. 1 W +k2 [2 ~—-—+—. 7 From the diffraction d 2 a” 62 pattern in (c) of Au at 236 GPa and 295K after a heating cycle, determine the latticeparameters a and c: a=©.1\‘5 om ; c= O.%anm ChE 350 Fall 2010 (100 pts possible) Name: Exam 1 (show your work on the following pages as indicated) (e, 10 pts) From the diffraction data, comparing traces (a) and (c) for the foe and hop structures,_ respectively, calculate the % lattice expansion between the close-packed planes (i.e, the d— spacing between the close—packed A, B and C layers in fee and A and B layers in hep): % lattice expansion between close-packed planes: 0 0/0 (f, 10 pts) There was a slight lattice expansion observed when the gold was heated at 237 GPa to 1050K relative to the gold sample at 275 GPa and 295K, (i.e., traces (a) and (d), respectively). The (111) diffraction peak position shifted from 29 = 20.20 (at 295K) to 29 =19.90 (at 1050K). The linear thermal expansion coefﬁcient 05L , is equal to the fractional change in length divided by the temperature change: AL 1 aL =-——— —— L (AT ) L is the initial length (which can be taken as the d-spacing). From the data measured by Dubrovinsky, et al. above, calculate the corresponding 05L that they measured and compare it to the literature value of 05L = 14 x 10'5 / 0C for gold. . — ’5 aL= \.C\‘7>‘\C> 1°C How do the values compare? %\\c\$\‘\\’\u\ \chch \‘305; CADSP. Yb W \WW 'vcdutcﬁ m no" ﬁg, 2. Sketch and note the atom composition in the unit cells below for (a, 15 pts) NaCl (rock salt structure) (b, 15 pts) cubic ZnS (zinc blende structure) CM? 350 Fall 2010 (100 pts possible) Name: wow/u (Nb Exam 1 (Show your work on the following pages as indicated) . Show your work for problem 1: For b : ',,0 3 ‘ ,a (8- m 16 11c) 26? com Z6 :13? (new 16 =32»: \\ 6 \OO 5’ " A F/\ — CQ-(‘>‘%(>CLN’”3 f A Co‘ooocx no“ _’ M. 1': d "1 / 7 . o g .3 t a 23mg ' '2 MW” 2 mum? . _ Loowoqnm3 ' d : Z a @567 o’\ 164‘“ M" d to An N“ d1mc1®znm (kc-,(pxﬁbﬁgftx 0v: (AU—7335.3 OJ: EV‘CLHC’LZ OLLCDfZlDVLghm ‘5 a; :Oqugiﬁv‘o a ., 0191073373 ‘ Ck; C) .056 O “M WWW” W_W,M_W.M..-w_ww—M CL,:O.3LlO\ (\m a 2-—»~W.,...,.w..~ . -..,.n » m» Deg) Z6 a; \‘\° (0075 1‘3’3101? “a d LO .0” OR 0063 ' :O v 107,1‘3 m . ‘> wlﬂ’f—Zb ’ - ZS? > Z 1,. Z ‘ \ O A-Q ~ 7, ‘dzbl'ﬁi om CO'~’L(>’Z,nMﬁ 0» - b w. "L ‘1 \ \’\1>\L X A CD I A Z” «I 2 T" I L ’ - «7’ C 1/ a} (/1 {C1101 7 L0 1\“>3 g ’L ’7— 1 L yo . m "L 00, *7. C - comm” " o, WWW ﬂ» r, W ' a. a _ 0% n m ow ©‘1\5“m :0 ' O W“ gww-M Ww,..,(.ww'~wm- M a“... ChE 350 Fall 2010 (100 pts possible) Name: Exam 1 (Show your work on the following pages as indicated) For (6): MC? (caz'x x; CC gm :9 '3 26.75" 29 :10 15% e c, \(5 316° I 6 :\C3 315 » Loom)an d , 0 102mm a ' 7' ram 003155 I I (E: (5 .lolﬁm I v (3.101 “H—wé « 6 if... c \ o /D o .‘LCbL For (f): Eu“. L\\ W Va; [ND . I \Q 1210\5 \@ Tﬂoob (Lg ’010 L9 ~ \b \° 9 >qﬂ5 a I M iiou’LOSOY'N .’ “L : mega» we“??? . “,5. c I xolL — \ ("‘7 *‘b / C \ ...
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