Polymer A should have the medium T g. It has a larger sidechain than polymer B, and may also have hydrogen bonding, due the -COOH group.

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1 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 Poblem Set 2 Due: Tuesdy, Jnuy 3, :00 AM. Glss Tnsition Tempetue ) The glss tnsition tempetue, T g, is stongly govened by the bility of the polyme chins to move nd otte, which cn be impeded by steic effects o intechin bonding intections. If mobility is difficult, the polyme will tnsition fom fluid to glssy t highe tempetue. Rnk the thee polymes below fom lowest to highest T g. Explin you esoning. (0pts) Polyme B will hve the lowest glss tnsition tempetue. The only intections between chins will be Vn de Wls, nd possibly the steic effect of the F 3 molecule. It lso hs shot, simple me tht cn pck esily. Polyme A should hve the medium T g. It hs lge sidechin thn polyme B, nd my lso hve hydogen bonding, due the -OO goup. Polyme will likely hve the highest T g. Becuse of the N 3 + nd l - ions, the chins cn fom ionic bonds with one nothe. This secondy bond is stonge thn the - bonding of polyme A, with similly bulky sidegoup.

2 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 b) Which is moe likely to cystllize, tctic polystyene o syndiotctic polycylonitile (PAN)? (4pts) PAN will cystllize moe esily. The polystyene hs bulkie sidegoup, nd is lso tctic, both of which ct ginst the fomtion of cystlline polyme. c) You e told the density of specimen of polyme X (PX) is.030 g/cm 3 t 74% cystllinity. Puely cystlline PX is.050 g/cm 3. If you need PX to flot on wte fo you ppliction, wht s the mximum pecent cystllinity llowble? (6pts) The mximum llowble cystllinity is 33%. Fist, find the density of mophous PX, ρ, using the vlues given fo this specimen nd the eqution given in lectue. Tking the density of wte to be ppoximtely g/cm 3, Theefoe, polyme X must be pocessed to hve cystllinity < 33% fo this usge. 2. ystl Stuctue - Peovskite The peovskite cystl stuctue is illustted in Figue 2.6 of lliste. STiO 3 is mteil tht exhibits n idel cubic peovskite cystl stuctue. ) Fo STiO3, which elements e ctions, nd which e nions? (4pts) In the tomic fomul ABO 3, A nd B e both ctions nd O is the nion. Thus, The ctions e S nd Ti, nd the nion is O. b) Wht e the vlence sttes of ech element within this compound? (4pts) The most pobble vlence stte fo S nd O is 2+ nd 2- espectively. hge neutlity dicttes tht Ti must theefo be 4+ nd the vlence sttes e S 2+ Ti 4+ O 2- c) ow mny S, Ti nd O toms e in the unit cell? (4pts) Fom figue 2.6 we see tht thee is single Ti tom t the cente, 8 S toms t the cones contibuting /8 fo totl of S, nd 6 O toms contibuting ½ ech fo totl of 3. Thus, thee e S, Ti, nd 3 O pe unit cell, s the chemicl fomul implies. d) Detemine the coodintion numbe fo S, Ti nd O in this stuctue. (4pts) Fom figue 2.6 nd bit of imgintion, we see tht the S site hs 2 neest neighbos, Ti 6 nd O 6.

3 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 e) lculte the theoeticl density of STiO 3. ompe this with the expeimentl density. (4pts) The lttice constnt of STiO 3 is ngstoms. Using eqution 2. fom lliste nd peiodic tble gives ( * )(g/mol)/ (3.905*0-8 cm) 3 (6.023*0 23 /mol)=5.6g cm Molecul Weight of Polymes Molecul Weight Rnge (g/mol) Men molecul weight M i Fction of pticles in this nge: x i Fction of weights in this weight nge: w i 5,000-0,000 7, ,000-5,000 2, ,000-20,000 7, ,000-25,000 22, ,000-30,000 27, ,000-35,000 32, ,000-40,000 37, ). lculte the numbe-vege molecul weight. (5pts, 3 pts if wong but coect wok) 23,400 g/mol b). lculte the weight-vege molecul weight. (5pts, 3 pts if wong but coect wok) 23,00 g/mol c). This polyme hs epet unit molecul weight of 56 g/mol, compute the degee of polymeiztion (DP). (5pts, 2 pts if wong but coect wok) DP = 23,400 (g/mol)/56(g/mol) = O DP=2300 (g/mol)/56(g/mol) = 42 d) Assuming this polyme is hydocbon consisting of only bon nd ydogen toms, wite down the chemicl fomul of this hydocbon ( mss diffeence of less thn gm between you fomul nd the molecul weight given in pt is OK). Is this hydocbon stuted o unstuted? (A dwing of the hydocbon my help, but is not equied). (3 pts fo fomul nd 2 pts fo unstuted) 4 8 : 2-Butene is the closest mtch. This is n unstuted polyme.

4 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 Molecul Weight Rnge (g/mol) Men molecul weight M i Fction of pticles in this nge: x i Fction of weights in this weight nge M_i*xi M_i* wi 5,000-0,000 7, ,000-5,000 2, ,000-20,000 7, ,000-25,000 22, ,000-30,000 27, ,000-35,000 32, ,000-40,000 37, SUM: 57, Polymophism/Isomoism ) Sketch the cystllogphic plnes fo the following unit cells. Also, indicte how mny toms e locted in tht plne. (Be sue to ccount fo the fct tht some toms my only hve fctionl mounts in tht plne.) (5pts) i.the () plne of n B unit cell ii.the (02) plne of n B unit cell iii.the () plne of n F unit cell iv.the (02) plne of n F unit cell b) At oom tempetue, pue ion hs B cystl stuctue, but undegoes polymophic tnsfomtion t 92 to n F cystl stuctue. These two foms of ion hve dsticlly diffeent popeties due to thei bility to incopote cbon. lculte the pln density long ech of the plnes bove. (5pts) ) i.() plne of B unit cell. Plne with toms is shown below in the unit cell s well s 2D view of the plne. Thee e thee cone toms on this plne, ech of which is shed with six totl simil plnes, thus thee is ½ tom totl on the plne.

5 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 +z +y 3 2 +x 2 ii. (02) plne of B unit cell. Plne with toms is shown below in the unit cell s well s 2D view of the plne. Thee e two cone toms on this plne, ech of which is shed with fou totl simil plnes, thus thee is ½ tom totl on this plne. +z +x ½ +y 5 2 iii. () plne of F unit cell. Plne with toms is shown below in the unit cell, 2D view of the plne is the sme s in pt i). Thee e thee cone toms on this plne, ech of which is shed with six totl simil plnes. Then thee e thee fce centeed toms on this plne ech of which is shed between two plnes. Totl thee e 2 toms on this plne.

6 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 +z +x +y iv. (023) plne of n F unit cell. Plne with toms is shown below in the unit cell, 2D view of the plne is the sme s in pt ii). Thee e two cone toms on this plne, ech of which is she with fou totl simil plnes, thus thee is ½ tom totl on this plne. Thee is lso one fce centeed tom on this plne, which is shed between two plnes. Thus totl thee is tom on this plne. +z ½ +x +y b) Pln densities of ech of the plnes bove.,,

7 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 i. ii. iii. iv. c) Wht is the diffeence between polymophism nd isomoism? (pts) Polymophism is when mteil of given composition hs two o moe possible cystl stuctues. Isomoism is when two o moe polyme molecules o epet units hve the sme composition, but diffeent tomic ngements. d) Diffeent isomes of mteil cn lso hve vey diffeent popeties due to thei tomic ngements. Sketch the following isomes. Descibe some wys tht the diffeent configutions of these polymes ffect thei popeties. (7 pts)

8 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 i. isotctic, syndiotctic, nd tctic polyvinylchloide. Isotctic polyvinylchloide l l l l n Syndiotctic polyvinylchloide l l l l n Atctic polyvinylchloide l l l l n ii. -pentene, cis- nd tns- pentene. -pentene

9 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 cis-pentene tns-pentene The diffeent configutions will ffect inte-chin (between chins) intections. This cn led to bette stcking, which is esponsible fo highe melting points nd cystllinity. e) Dw the two molecules which hve the sme numbe of cbon toms s pentene but e fully stuted nd fully unstuted. (2pts) Stuted hydocbons hve only single bonds, nd thus new toms my be dded only by emoving othes tht wee ledy bonded. Unstuted hydocbons hve double o tiple covlent bonds. Fully unstuted 2 2 Fully stuted Stuctue - Atomic Rtio ) Nme the expected cystl stuctue fo the following mteils using ction to nion dius tios: (int: tbles 2.2 nd 2.3 fom lliste) (5pts) i) MnO ii) F 2

10 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 iii) Nl iv) MgS v) O i) MnO ii) F 2 iii) Nl iv) MgS v) O Mn O l N l Mg S O ood#=6 =>Nl (Rock Slt) stuctue ood#=8 => sl (see fig 2.3 p458) stuctue ood#=6 => Nl (Rock Slt) stuctue ood#=4 => ZnS (Zincblende) stuctue ood#=6 => Nl (Rock Slt) stuctue 0.40 b) Both Fluoite (F 2 ) nd cesium chloide(sl) hve the sme coodintion numbe, but thei stuctues e slightly diffeent. Ignoing the diffeence in tom types nd lttice spcing, how e the two stuctues diffeent nd why? (int: The diffeence my only be seen when comping moe thn one cell) (2pts) Figue 2.3 on pe 458 of lliste shows the sl stuctue. The stuctue fo F 2 will be the sme type except thee will be no 2+ ction in evey othe unit cell (see figue 2.5). Accoding to the chemicl fomul fo Fluoite thee e twice s mny F toms s thee e toms, which is equied in ode fo the totl chge of the stuctue to emin t zeo. c) A smple of MgO hs Si impuities. Fo ech Si ion incopoted in Mg site e ny vcncies fomed in ode to emin chge neutl? (3pts) Yes, fo evey Si 4+ ion in the plce of Mg 2+ nothe Mg 2+ ion must be emoved ceting vcncy in ode to emin chge neutl. hge Neutlity equies the net chge of the mteil to emin t zeo. d) Show tht the minimum ction / nion fo coodintion numbe of 4 is (5pts) oodintion numbe of 4 implies tetgonl ngement of 4 nions ound ction. The ngle fomed between the ction nd two djcent nions is 09.5 Degees. At the minimum c / djcent nions e touching. Using geomety we cn find the eltionship between c nd.

11 MATERIALS 0 INTRODUTION TO STRUTURE AND PROPERTIES WINTER 202 In the illusttion to the left we cn see tht the ngle mked c is 0.5*09.5 Degees = Degees Using the ight tingle constucted we cn see tht unde the conditions of minimum c / : The side opposite c hs length of nd the hypotenuse hs length of + c. sin c c c sin c c sin sin c e) Wht is the tomic pcking fcto (APF) of mteil with coodintion numbe of 8 t the minimum ction to nion dius tio? (5pts) The minimum c / = fo coodintion numbe of 8. At the minimum tion ll tom e in contct with thei neighbos. The stuctue is nged with nions t ll 8 cones of cube suounding the ction t the cente (s seen in t the bottom of Tble 2.2 of lliste). APF= V /V c =(Volume of ll toms in the cell)/(volume of the cell) The cubic unit cell hs sides of length = 2. V cell = 3 3 =8 V =N ction x V ction + N nion x V nion ee: N ction = x =; N nion =/8 x 8= Vction c ; Vnion c 4 4 V APF

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