key_exam_2_06 - GRADE ll 1(10 pts}3elow is the phase...

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Unformatted text preview: GRADE / % ll 1) (10 pts)}3elow is the phase diagram for sulfur showing two kinds of solid sulfur — rhombic sulfur (Sg (Rh) and monoclinic sulfur (Sg(mc)) , liquid and gaseous sulfur Sg (Rh) has a greater density than 330110) and is more stable at high pressure For each of the lettered locations identify the material (rhombic sulfur, monoclinic sulfur ,etc) and phases present. For example, point A is solid rhombic and liquid sulfur in equilibrium. a) b) B 5540 w. w/5a5L/0 114.5. 4:62 C Soup 729 M“? D smug) MC. 0W7 55 E ens Aim 60W W 7’37". F GAS/:40? 6Q (lg-P.) ( 5 pts) REEZE DRYING is the process of vaporizing a solid without going through the liquid state b lowering the pressure. Can monoclinic sulfur Sg(rnc) be freeze dried at 100C"? Explain “d” BETH Maw Cm €057“ A?” Wang; c. AT 79< Mm»: ( 5 pts) At one atmosphere heating rhombic sulfiu‘ at any temperature rather quickly will cause it to melt and not go through the solid/solid transition to monoclinic sulfur but rather melt to form liquid sulfur [ you must keep heating or the liquid will return to monoclinic until the normal melting point is reached]. If this happens at many pressures use a dotted line to roughly add this “metastable melting point” to the phase diagram below. (Even if you can’t draw the line) Explain why quickly heating may cause unusual features on a phase diagram ( what is a fimdamental requirement of all phase diagrams?) (+q)N6,Q rE'Qw/ugM-Aw dam/37‘! Tia N. Baffin/’7’ 44.4514 €Nc’bW 77% F‘s—,2: 53(KL)$aoio ”3758(02- Qsaezo 2) A 50 g block of gold ( Cp= 0.13 J/gK) at 100 C is placed into contact with a 150g block of almnjnum (Cp= 0.90 J/gK ) at 200C. They are in an evacuated, perfectly insulated container. a)(5 pts) What is the final temperature at thermal equilibrium? f. m) a»... ”*Qma‘r' Tum” “°°°“3(’1 ’ > “4,0951” 5 thCQaLAT Lars/7;: 204.8 " ’503Cv‘7‘M11—«ZOQ = So (~13)Cfi—-Icn) T7, = (75‘1‘6/ 2‘7 #7:, b) (10 pts) Show the energy transfer is spontaneous using AS(total). If you cannot do part A use m as the final temperature which is not correct but use it anyway. b51504 w A an A“ C? (in) cf 45 a: 5°66 1332) ii. 19.8. as = (gal-fling . a . (95C 373 Choose number 3 or number 4 or number 5 —— cross off the ones you do not want graded. If none is marked we will grade number 5. 3) One of the relationships we found fiom the fundamental equation dA = -SclT—PdV is that e = . 6V T Use this to develop an equation that shows how A changes with V a)(5 pts) for a very slightly compressible solid Ge A : '-* WV 5/4 : uPAl/é‘ (“Naif/iv 0L 0%” b)(5 pts) foranideal gas l9 at *1 RT U $5bf0 _. 7 fat/9 ° "777175; F bAzwmw’fln .3 m, “arr/£4 65,) : “7/2712”? (3) 4) (10 pts) Given 6—3 = fl and the vanderWaals equation can be written P = fl _ _a2 6 6T (V-b) V 1’ \/ Use these two rela ionships to develop an equation for AS with V changing for a vanderWaals gas. $3.2: —a’0:_&-. (an V*b/lSé We a5 3,8,. ,. I: ll / 3LT U’b (55 g aflflefi’) a. lad-o we EFFé’cT. U (01"?) 5475: 12-4... the) 5) (10 pts) Given the fimdamental equation dH = TdS +VdP Show for an ideal gas - : ' ’- "e . (Hint: divide by dP at constant T) (how does H change with P when T is constant for an IDEAL GAS?) ’ u 6'1 .5 Tell/Lg”) )L =0 Fomwgfl" “60/057“ 7‘ = M30 6) (Spts) In the three piston problem a) reversible expansion 1)) expansion against a constant pressure and c)expansion against a vacuum) the q of the surrounds was said to be always reversible but same q for system may be irreversible. Explain why the surroundings are reversible. Tfi sqeeatwbl‘ww 14“ 5" MM’Z dmpfl‘gfi’f) 73 7‘15 57$filh Tear ANfi cmw’e’ W ? ’5 ”W’WTE‘MM .96 ms» fie’ofiag/guf. A defies: CAP fl" LAKE duke/my 7) (lOpts) Ethanol has a vapor pressure of 115 mmHg at 34.9C and a AHvap= +40.5kJ/mol. What is the normal boiling point of this liquid? ( assume 1 atmos — 760.0 mmHg, R=8.31J/molK). a lag.) 2.54 2:1 1?! a 7*;-r2’ filfig = #05603 £2 -(3q.7+2,73) "5 8-3‘ f/hplic l-5’7 “ 4874(T1“303 .5033 L $8ZTL= 41377 7;, ——- L56>¢{0 8) Lime—CaO-- is made by calcining or roasting limestone CaCO3 to form CaO and C02 CaCO3 —+ CaO +002(g) a) (5pts)Use AG to show this reaction is NOT spontaneous at 298K. (limestone is stable) (+5) 55-: .. ,9. gm») .. <3»... 6:0!) sieéay) {— 0374)] .. 23-1130] (+9 b) (Spts) Roughly estimate the temperature when this reaction will turn spontaneous.(what assumgtions a?» “121% be .. M! - Tag 12.6 = M“/bs <*-' 3°77” ““570 l) c)(5pts) Without doing the calculation list the equations and information you would need to do a very exact dete n ' a tion of AG at a temperature far above 298K say —-2000K ' cmpd AHfO AGfO Sfo C02 (g) -3 93.5kJ/mole —394.4kJ/mole 213.4J/moleK CO (g) -110 -137 198 CH30H(1) «239 "166 127 H20(g) —241.8 -228.6 188.8 N0(g) 90.25 86.55 210 N02 (g) 33.18 51.31 240 N20(g) 82.05 104.20 219 CON2H4 (s) [urea] -345 -445 102 N113 46.1 ~16.45 192 H2 0 0 130 CaCO3 -1 200 —l 130 83-7 C30 -635 -604 40.0 (55 9) (1 Opts) In your own words ( using an example would be helpful) BRIEFLY define the 2nd and 3ml laws of thermodynamics. What are Chapters 4 and 5 all about? EndlawyN/DTMML 19/145565 MK!) VIMGT/DN @ WW CM 56 mpgsséo 73 19a LoTZk @ 77a? Dike’éf/‘éw f3 ALWA75 ha 729: Diagcnop (442/ Ufa—512’. (0 1166 D/pe’of/au (LAN 36 ?¢£D/¢7t29 Fm D5T>O bé—(‘QQ><6 AAC'GU <0 mus) my fidfi Nw'nfimg {Quar/éadf Ta ¥f~> 7%43 6;) 5e. [LE-«1L wllflté 1L6 # of INDEfifin/WN—r Mmmflé a? 47% 46W»: aux/nesrwflzw M = @9st P @117 50:55—17 (3-3! :1” 6%23V (m = “Gm flag 'fi 6M =5: SaQT~ few WWW ...
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