exam_answers_152A_exam_01_version_a_key

exam_answers_152A_exam_01_version_a_key - CHEMISTRY 152A...

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Unformatted text preview: CHEMISTRY 152A HOUR EXAM l (Version a) Winter 2008 Wednesday, January 30, 2008 Name: TA Section: Student Number: TA Name: Score I 1 . 2 GOOD LUCK! Watch sig figs and units. 3 4 Total (x/100) / . a) l. CONCEPTS AND SHORT ANSWER: Answer all questions (subtotal = 52 pts) p) 1. (a) @F) For an ideal, monatomic gases, Cp = Cv + R /1/ (b) For real, polyatomic gases, why is Cp > (5/2)R always ? . ». w'érqifi'mq/ and ro/é/S'mq/ 77”va ca". 9439,44 €29,099 w/dch/ worm/"7 fir flex 2. (a)Consider an ideal gas at 30 °C in a 1.0 liter bulb connected to an evacuated volume of 4.0 liters by a closed valve. Now open the valve, letting the gas expand adiabatically into the larger volume. For this expansion, give the sign (+ or — or 0) of each of the following: ,5) q w AE AH AS ’ o o o o + (Z) (a) Give a formula for the change in entropy A8 for ,this expansion . - v ' f AS .. 47% [ ‘Z/O) F) 3. indicate which of the following statements is true (T) or'false (F): (a) _Z'__ All real processes are irreversible (b) _/5_ in a reversible cyclical process, the system returns to its original state, but the universe does not. - , (c) A All spontaneous processes increase the disorder of the surroundings (d) L A perfect crystal at absolute zero (0 K) has entropy S=1 (e) L Heating an ideal gas at constant volume increases pressure, but does no work (f) L Thermodynamics concerns changes in state, but not rates of change. A I KEV @ Chemistry 152% Page 2 (lg) 4. For' each of the following forms of matter, give the Standard thermodynamic state(°): (a) pressure of a gas Ira/m (b) temperature 298K ' , . (0) pure solids, or liquids PM”: Ia/I/OV /’9"’/ (d) solution concentration I I11 [5) 5. Consider the following sequence of reactions: AH1 ZB+3/2A ——>C AHz D+3A —>C+3E AH3 F+‘/2A —>E 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: 28 + 3F —> D 25+§¢r9gz A4; ‘3‘tgfl/ngf“ 55%; /Q+/% 'A/L 25*" 2; +3; -—->D AA’xh5A/é +3Aé’A/Z 6. (a) Calculate the boiling temperature T (in K) of liquid bromine Br2| (P=1 atm) ? Br2 (I) —> Br2 (g) For this phase transition, the vaporization enthalpy of Brz is AH° = 28.0 kJ/mole, and Z ( J the entropy change is AS° = 84.0 J/(K mole) 7‘ _— Afl‘ 28.0xng/"V‘ - 156 = 77. o T/zuaé ' . 3 5/3 [25" { g) (b) At this boiling pOint, what is the value of: ASuniver, ASsurr ASS),s o -— 39.th $110 f/(w/t fiw/ 7. Which one of the following species has the highest heat capacity Cv and v_v_h_y? (I) Ne CH4 N2 002 CH4 CzHe ) / 91me a?! <7érru‘, 0' W97! ’ {2' 6,3,3 “what/mi ('Wfl’f’ )n (3) 8. In Boltzman's formula for entropy S: kan, what does!) represent? : *0 a7; _ IV a (Zr/9 mw/g fw/?.W ) may I KZP/gjfaiZ/oy/é’fx' 1k 7% ‘5 S/WZ‘ 5‘74 ‘, Version A 79%) Chemistry 1529 Page 3 {/79 9. For each of the following, give the sign of AS: + evaporating alcohol __:-_ freezing water + dissolving salt in water - compressing an ideal gas at constant temperature 9.0 .69» (z) 10. (a) Name two important greenhouse gases increasing in the atmosphere due to humans. (5‘2 ,€//9_, /V20' C/C For the following, circle the most appropriate answer: (b) The natural greenhouse effect warms the Earth about ( 0.3, 3, (BE 100) °C (c) The warming over the past century has been approximately (0.0, <0.4, >2.0) °C (c) Increasing atmospheric 002 is predicted to cause additional warming in this century of about (0.3, E) 10, 30) °c '(d) The most abundant» ossil fuel is @petroleum, natural gas) ,5 (e) Predicted mea - - o | sea-level rise in this century is in the range of 2 (<0.1m, 5 m, 10m) (f) Current atmosp ~* ~ 2 level is ~ (180, 280 @ 480) parts-per—million ppmv, ' while t natural, ancient C02 levels measured in polar ice cores were about (180, 280 380, 480) ppmv during warm interglacial intervals like the present. 11. State the three laws of thermodynamics: wakfij 9/ fl? th'bf/{t' 1's Gags/éy71 (Z) lst Law: & CowsPIm/{em a/ CM’97 (2.) 2nd Law: £WV) '5/9917’47730145 f/vc’s; /'nrr\e4rey fl: (6749/; c/ 4r hazy“!- (Zl/ 3rd Law: /7,( 5/ a) /?é/)£c/ CV c/ 94;¢/M/- J/é’ 1'; 2.79”. VersionA Y/ a? Chemistry 152@ , Page 4 ll. QUANTITATIVE: Do all four (4) of the following (subtotal = 48 pts) [2) V 1. A balloon filled with 39.1 mol of He has a volume 876 liters at a temperature of 0 OC and a pressure of 1.00 atm. The balloon is heated at constant pressure to a final temperature of 38.0 0C, causing the balloon to expand to a final volume of 998 liters. Calculate q, w, and AE for the He in the balloon. Note: the molar heat capacity of He is 0,, = 20.8 J/( °c mol). 20.8 J - °C mol -q = molar heat capacity X mol >< AT = X 39.1 mol X (38.0 - 0.0) °C = 30,900 J = 30.9 kJ w=—PAV=-1.00 atm x (998L—876 L)=-122Latm>< 151:: = -12,4OOJ=-12.4kJ a . AE‘=q+w=30.9kJ+(—12.4kJ)= 18.5 kJ f; z) 2. A L10. gram sample of copper (specific heat capacity = 0.20 J/( OC 9)) is heated to 82.4 0C and then placed in a container of water at 22.3 0C. The final temperature of the water and copper is 24.9 0C. What is the mass of the water in the container? .Assume that all heat lost by the copper is gained by the water. Specific heat capacity of water is 4.18 J/(g oC) Heat gain by water = heat loss by Cu; Keeping all quantities positive to avoid sign errors: - 4'18] x m x (249°C -22.3°C)= 02‘” g°C g°C x 110. g Cu X.(82.4°C - 249°C) 11m=1300, m=120gH20 Version A 2) Chemistry 152% . Page 5» {/z/ 3. Calculate the entropy change for a process in which 3.00 mol of liquid water at 0 0C is ' mixed with 1.00 mol of liquid water at 100 0C in a perfectly insulated container. Assume that the molar heat capacity of water is Cp = 75.3 J/(mol oC) within this temperature range. . Calculate the final temperature by equating heat loss to heat gain. (3.00 mol)(75.3 J mol" °C-1)(T,f — 0°C) = (1.00 mol)(75.3 J mol'1 °C")(100.°C - Tf) Solving: Tf= 25°C = 298 K Now we can calculate AS for the various changes‘using AS = nCp 1n (Tz/Tl). Heat 3 mol H20: AS1 = (3 .00 mol)(75.3 J mol" K“) In (298 K/273 K) = 19.8 J/K - ' l Cool 1 mol H20: ASZ= (1.00 mol)(75.3 J mol" K") In (298/373) = -16.9 J/K I ‘ AS”, = Asheat + ASm1= 19.8 - 16.9 = 2.9 J/K pa) 4. Calculate AS° for the following reaction: F6203 (S) + 3 H2 (9) 2 Fe (8) + 3 H20 (9) 8° (JIK mol) F6203 90. H2 131 Fe 27 H20 _ 189 AS = 2 mol(27 J K"_ mol") + 3 mol(189 J K-1 mol-1) - [1 mol(90. 1K4 mol") .. . + 3 mol(13l J K“ mol“)] AS = 138 J/K Version A Z) i z i l i g . l i ...
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exam_answers_152A_exam_01_version_a_key - CHEMISTRY 152A...

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