exam_answers_152C_exam_01_version_a_key

exam_answers_152C_exam_01_version_a_key - CHEMISTRY 1 29...

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Unformatted text preview: CHEMISTRY 1 29 HOUR EXAM l (Version a) ‘ Winter 2008 " Wednesday, January 30, 2008 Name: 'TA Section: I Student Number: TA Name: Score 1 . 2 GOOD LUCK! Watch sig figs and units. 3 g 4 5 E Total (x/100) l of” g I. CONCEPTS AND SHORT ANSWER: Answer all questions (subtotal = 52 pts) (1) 1. {(a) ([email protected] For an ideal, polyatomic gases, Cp>CV + R g l l (b), Why is Cv > (3/2)R for r al, polyato ic gases ? l E) VIA/d .0713], 1% 71%/ I’M/’4)” ’ééiz €VV9 é ”/fiw. x’n (wees/9' 5‘” V”9 hr. : l l 2. (a) Consider an ideal gas at 30 °C in a 1.0 liter bulb connected to an evacuated volume of 5.0 liters by avclosed valve. Now open the valve, letting the gas expand adiabatically into the larger volume. For this expansion, give the sign (+ or — or O) of each of the following: (g) q w AE AH AS 0 o o 0 4 l. (b) Give a formula for the change in entropy AS for this expansion E ‘ {@(2/ 'AS :w/évea('%)- l /c) 3. Indicate which of the following statements is true (T) orfalse (F): E " ‘ l M (a) _7; For an ideal gas, E and H can only change if the temperature changes. l (“K (b) _Z: In an irreversible cyclical process, the system returns to its original state, but the universe does not. i (c) j; All spontaneous processes increase the disorder of the system (d) _I_ At constant pressure, the enthalpy change is the heat flow in or out of the system (e) l The maximum work possible in the expansion of an ideal gas corresponds to the reversible expansion. (f) L Thermodynamics concerns changes in state and rates of change. Chemistry 152C Page 2 E 1.5» a (4) 4. For each of the following forms of matter, give the standard thermodynamic state(°): (a) pressure ofa gas has/m (b) temperature 299% . (0) pure solids, or liquids ,owe 9/)? er ”94¢ E (d) solution concentration I M E . E (3) 5. Consider the following sequence of reactions: E AHE2A+5B —>4C+2D AHz E+B ——>C AH3 2F+B —->2D i In terms of the enthalpy changes of the individual reactions (AH1, etc), write an expression for the enthalpy change Aern for the following net reaction: 2E + F —-> A 2£+23 ——> 2c 2416/2 i F +34, *7 D " A/é/Z 2c+p *bflngi/Q “ll/IO/z E 25+ ; a»; 4m”: gala-+49 ~47? 6. (81) (WW (in K) of liquid bromine Brg at P=1 atm ‘ (2/ Brz (I) ‘9 Br2 (g) E For this phase transition, the vaporization enthalpy of Br2 is AH° = 29.5 kJ/mole, and the entropy change is 138° = 89.0 J/(K mole) a 3...— T= 4—4/- : W‘ - Z 3 «2 4/5“ 37.07/w/K ' . 3 3 0 ‘99}‘_I:—/( #39MM/( (3} 7. Calculate the work (w) done when an ideal gas expands from an initial volume of 2.0 liters to a final volume of 5.0 liters against a constant external pressure of 3 atms. ‘0='Bx/AV ly—Vg ; 532:3.o,( w =-(394.)/;.a/) : -? 95/134» [3) 8. In Boltzman’s formula for entropy S= kan, what does 0 represent? mam/fr a/erros'éz; .Or (94,,7/9/eu/ Posy/[4’ ‘il’f’éiyfmg //a;/7{&4 a‘r W139) ,/ c; . 51;“. 544 9/ 9 gfijvggk E VCI'SIOIIA . Chemistry 152C Page 3 33/ (4,9 9. For each of the following, give the sign of AS: : subliming a solid directly to a gas + melting a solid to a liquid - precipitating a salt crystal from a water solution - cooling an ideal gas at constant volume $1.0 .69 10. (a) Name two important natural greenhouse gases: C 01 6447, N2 0 All 0 (2) (4/ dhf ‘ €67; (c) The warming over the past century has been approximately (0.0, 0.3, ® 2.4) °C For the following, circle the most appropriate answer: (b) The natural greenhouse effect warms the Earth about ( 0.3, 3, @, 100) 0C (0) Increasing atmospheric C02 is predicted to cause additional warming in this century of about (0.3, (29 1o, 30) °c , (d) The most abundant fossil fuel is petroleum, natural gas) (e) Complete meltin of the Greenland Ice Cap would raise global sea-level by (1 2 4 10) meters L (f) Currently the global atmospheric COZ concentration is increasing at (0.1, 0.5, 1.0,. 5.0) ppm/yr (parts—per-million) and may be double the natural level by 2050. 11. State the three laws of thermodynamics: (2) lstLaw: Faun/9) c/flc hay/incur 1's Char/(42% oV- 5M4; f5 avg/3%”! (m/éa/dvcl/ o/e;7/05C/. / ('7) 2ndLaW1 EV”) 5W0» 42%va /o»arr’;y in (may; .fle . a’;‘;¢r/c’r ¢/ ,flx é/77/Wrr9, (7) 3rd Law: 7%? @2455 5 WI ca"; jb/ee/ (5;.er g/ 0/( /§4§é‘/lz/4 Pvt/7" /.S gt’r" , Version A Chemistry 152C Page 4 ll. QUANTITATIVE: Do all four (4) of the following (subtotal = 48 pts) ' @— {/7)1. The heat capacity at constant pressure of N20 is Cp (N20) = 38.70 J/K mole. If 88.0 grams of N20 is cooled from 165 0C to 55 °C at a constant pressure of 5.00 atm, calculate q, w, AE and AH for this process. 1 mol N20 ‘ ‘ . , .. ———-,———— = 2.00 mol N20 44.02 g N20 At constant pressure, qP = AH 88.0 gNZO >< AH = nCpAT = (2.00 mol) (38.70 J mol'1 °C'1) (55°C — 165°C) AH= -8510 J = ~8.51kJ= qp w'=;- PAV = -nRAT = -(2.00 mol) (8.3145 J mol'1 K") (-110. K) = 1830 J = 1.83 kJ AE=q+w=-8.51kJ+l.83kJ=-6.68kJ {/Z) 2. A 46.2 gram sample of copper is heated to 95.4 C’C and then placed in a calorimeter containing 75.0 9 water at 19.6 0C. The final temperature of the water and copper is 21.8 0C. Calculate the specific heat capacity of the copper (units of J/(g oC)) assuming that all heat lost by the copper is gained by the water. Specific heat capacity of water is 4.18 J/(g °C) Heat gained by water = Heat lost by copper; Let s = specific capacity of copper. 4. 1 8 _J O g'C X 75.0 g X 22°C = s X 46.2 g X 73.6°C, s = 0.20 J °C'1g'1 1 Version A Chemistry 152C Page 5 215/ {/7/ 3. Consider a rigid, insulated box with two compartments. One compartment contains 0.400 mol of He(g) at 20.0 0C and 1.00 atm. The other compartment contains 0.600 mol of N2(g) at 100.0 0C and 2.00 atm. Heat can flow between the compartments. What is the final temperature in the box at thermal equilibrium? The molar heat capacities are: Cv = 12.5 J/(K mol) for He Cv = 20.7 J/(Kmol) for N2 , Heat gain by He= Heat loss by N2; Since AT 1n °C= AT 111 th 11 th capacities could also be I °C mol'l. K’ e 6 units on the heat ' (0.400 1nol)(12. 5 J °C1‘m01“)(Tf 20. o °C)= (o. 600 mol)(2o. 7 J °C 1 mol 1)(100 0 °C- Tf) 5. 00 T. - 100. = 1240- 12 .4 Tf,Tf=11:40 =77.o °c .4 (I?) 4. Calculate AS" for the reduction of aluminum‘oxide by hydrogen gas: Al203(8) + 3 H2 (9) ----- > 2Alls) + 3H20(9) S° (J/K mol) A|203 51 H2 131 _Al 28 H20 189 féffprav’) " S§%’(“C/) so My 2; (9/) +3: {4%) 572/0 gm) [err/.22} : 2 /s*// + 3/15?) — ;/ -3{/=’/J 7- 5'6 ¢f571—;92-5'/ {1152",}: >4 /777/ K II")!- Version A ...
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