Midterm Solutions - Purple

Midterm Solutions - Purple - Chem 112A Midterm Oct. 29,...

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Unformatted text preview: Chem 112A Midterm Oct. 29, 2009 NAME: K01 ‘ PERM: Instructions: The last page of the exam contains some useful information. Please verify that your exam contains 6 pages (5 of exam and 1 of information). Show all work and box your final answers to receive full credit. Question I, 10 points (2 points each): True or False: Provide a brief justification or no credit. a) AG for a reaction at constant temperature is zero if AS is zero. e; M A Is; - _ b 7' zC/o [Tut '7 Nb)"THE—Wfiflfflfifié“(Tfi'fifijfitéfiféfifi éXCEédTliEiil‘C‘fEfi'Si—Z"fifth?‘ififéffizflfiéncrgy ‘ TE ‘ wag—er Mi qLO 2> w>I$E c) For all isothermal changes in state for an ideal gas, q=0. [ L J/ LL4 4’ Pk pro/h CW/lo’i 0’70 ‘aswi C1710 jgw Pgolhbnmcl (EU. QWQMJ/COM/éfl. .d) AG: 0 for all cyclical changes in state. ‘9‘ C7 is or Simii \Cmgilovx \fom/~ e) The following three quantities are all extensive: temperature, entropy, pressure. ' "r P" (“m lemma. " A 9 mixer/u l we, Question II, 15 poins: You throw a 1 kilogram hot (400K) block of solid lead (Pb) into a swimming pool. The water in the pool starts out at 300 K. Assume the swimming pool is so large that the lead is unable to change the temperature of the water and assume that the swimming pool (including both the lead and the water) is thermally isolated from the rest of the universe and is held at a constant pressure of 1 atm. Calculate ASPb, ASwm, ASunivem. What is the final temperature of the lead and how do you know that the change to this final state will actually happen? The specific heat capacity of lead is 0.150 J g"1 K'1 The specific heat capacity of water is 4.184 J g'1 K'1 we ‘Ct fl“ @w/ioohgws Z Q'ODOJW” S“ WW“ a7) 3 “6/3 T//(; ‘TW 3, M \ -: 3 DD K M; Ego) 17an it; u/ H20 5 Ctmmfl It Sfm 0L6“ €00: éAlswpD Quest10n III, 10 points: The standard enthalpy of fusion and entropy of fusion for pure mercury (Hg) are 2.292 kJ mol‘1 and 9.78 J K“1 mol'l, respectively. The densites of solid and liquid mercury are J approximately 15.6 g mL'1 and 13.5 g mL'l, respectively. What temperature does. mercury melt at when it is under a pressure of 1 atm? How much work does mercury do on its surroundings (on a per gram basis) when it melts under 1 atm of pressure (answer in Joules please)? Question IV, 40 points: Consider the following thermodynamic cycle consisting of three reversible steps. You start with 0.5 mol of an ideal gas at a pressure of 1 atm and a volume of 12.5 L — refer to this as state “A”. Step 1 involves doubling the volume via an isothermal expansion to arrive at state “B”. The second step is a constant volume heating back to the initial pressure to arrive at state “C”. The third step is a constant pressure cooling that reduces the volume back to the initial conditions and returns you to state “A”. Remember that “C— = 2R (molar heat capacity) for an ideal gas. V 2 a) Draw the thermodynamic cycle on a P-V diagram. Label all the states and steps clearly on the diagram and make it clear what values of P & V the states correspond to. b) Calculate q, w, AU and AH for each step and for the overall cycle in ,loules. Fill in the table with , w. .youranswers,-andshowallworkcaherejsmorer_o_om.onwtllenextpage),-7.-fl..7V V, , 7. ,V ICADMMHM“ Eflfamfi‘ldw 34:34. écs 4 Mgr/3 ‘ BE: 13H: 0 w? 'v\ WW JQ‘*Qj2/\/I\u ’ (05 W3 (9' g’L/WW’KBLEOS KM”; 3 I???“ "3% I970? qt’w CAEZDB Orz. Question V, 15 points: From the following data taken at T = 298.15K and p = 1 atm, determine AHf°(FeO(s)) and AHf0 (Fe203(s)) ( 0 means “at standard conditions”: T = 298.15K and p = 1 atm). ® Fe203(s) + 3C(graphite) —> 2Fe(s) + 3C0(g) Aern = +4926 kJ/mol Fe0(s) + C(graphite) —> Fe(s) + C0(g) AHm=+155.8 kJ/mol C(graphite) + 02(g) —> C02(s) AHm= -393.51 kJ/mol C0(g) + 1/202(g) -> C02(g) AHm=-282.98 kJ/mol Q/om AHf/AAZ Kj/lyol : BALHOCCDB’AHFOCFCZOSB 7> All; (11205} :2 “‘33”! til/revel 9% New 2 KS? lav/p01: Mime ' All; @103 Question VI, 10 points: Consider isothermal changes in state for an ideal gas. Derive a general expression for AG — AA (the difference between the change in Gibbs free energy and the change in CONSTANTS: CONVERSION FACTORS: 1 atm = 760 torr = 101.325 kilopascals 1L atm = 101.3 J R = 8.3145 J mol'lK‘1 = 0.08206 L atm K'1 mol'1 kB=1.38x10‘23JK~1 NA = 6.022 x1023 atoms/mo] me = 9.109 x 10'31 kg mp: 1.673 x 10'27 kg EQUATIONS : dE = dw + dq NOTE: dW and dq are inexact differentials and should have a slash through them H = E + PV G = E — TS + PV 'Ai-l E —"TS dq = deT dq = CVdT dw = —Pde dS = dqrev T 503,13) —S(T1,P) = f 9:1ng S = kB ln(§2) ' ideél g‘é‘sl'? PV = nRT = inRT 2 rev.,isorherm. = —nRT1n(V2 /V1) qrev.,isoflzerlu. = +nRT1n(V2 / S(T,P2) — sang) = ann(Pl /P2) W ...
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Midterm Solutions - Purple - Chem 112A Midterm Oct. 29,...

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