CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013
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1 CHEM-443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal units (where apprpriate), lgically presented (include nn-mathematical language if necessary t cnvey yur intent clearly and transparently), and crrect/accurate. Yu are free t use yur Equatins Handbk, Calculatr, blunt/sharp writing instrument, and brain (yur chice). If yu have any dubt(s) abut the meaning/intent f a questin(s), please ask immediately s yu d nt wander ff n an incrrect path! Please leave all respnses t be graded n the exam sheet. Wrk n scratch paper will nt be cnsidered. Prblem Pints 1 /10 2 /30 3 /20 4 /20 5 /20 Extra Credit /5 Ttal /100
2 1 (10 Pints) Matching / Fill-In. Prvide the prper respnse frm the right clumn in the blanks prvided in the left clumn. Keep in mind that sme blanks in the left clumn may require multiple selectins frm the chices in the right clumn. 1. A Legendre Transfrmatin is a reversible transfrmatin and leads t infrmatin lss. True r False? H, False A P A =P B (P = pressure) 2. At equilibrium cnditins fr a tw-phase (phase A and phase B), tw chemical cmpnent (i and j) system at a particular temperature and pressure, what 3 relatinships hld? A,Q,J B!F % #!P U =!V % #!G H 3. At equilibrium, the Gibbs Free Energy is maximized under cnditins f cnstant temperature and entrpy. True r False? H, False 4. Entrpy is a rigrus definitin f chas in a system, and maps in a unique, ne-t-ne fashin with the chas functin. True r False? _H, False 5. The chemical ptential f a species A in an ideal gas mixture is K that f the pure species at the same temperature and ttal pressure. C state functin D maximum E isthermal expansin 6. The Third Law f Thermdynamics guarantees that we can reach 0 Kelvin (abslute temperature) and realize a perpetual mtin machine; the nly trivial limitatin is that we have nt fund the right material. True r False? H, False F!T % #!P S =!V % #!S P 7. Ttal entrpy change fr a reversible adiabatic prcess must be zer. True r False? G, true 8. Internal energy f a system is minimized at cnstant temperature and pressure. True r False? H, False 9. The thermdynamic ptential that reaches an extremum under equilibrium cndtins f cnstant vlume and temperature is R. 10. A Maxwell Relatin assciated with a pure fluid, P-V wrk nly system, dh(s,p ) = TdS + VdP is F. G true H false I adiabatic cmpressin J. µ i A (T, P) = µ i B (T, P) K less than L intensive M isthermal cmpressin N enthalpy O mininum P zer Q T A =T B (T=temperature) R Helmhltz Free Energy S. T.!T % #!V!A % #!P S U =!P % #!S P =!V % #!G H
3 2. (30 Pints) Fr the fllwing reactin, FeO(slid)+ CO(gas)! Fe(slid)+ CO 2 (gas) the equilibrium cnstant (expressed in species partial pressures) at tw temperatures has been determined t be: Temperature 700 Celsius 1200 Celsius K P A. Using this data and necessary assumptins, calculate at 700 Celsius the fllwing:!g R 0,!H R 0,!S R 0. Slutin: At 700 Celsius (973.15K) we can find the value f the Gibbs free energy as: ln(k P ) = ΔG rxn RT ΔG ln(0.688) = rxn J (8.314 ml K )(973.15K) ΔG rxn = 3026J /ml The temperature dependence f equilibrium cnstant can be used t btain the enthalpy f reactin, assuming that this value des nt change ver the range f temperatures cnsidered here. ln(k P (T 2 )) ln(k P (T 1 )) = ΔH 1 1 R T 1 T 2 ΔH = (8.314J /ml K) K K ΔH = (8.314J /ml K) K K ΔH (T = 700Celsius) = 19004J /ml The Entrpy is thus: ΔS rxn = = ΔH rxn J ml K ΔG rxn T = 19004J /ml 3026Jml K B. Calculate the mle fractin f carbn dixide gas in the gas phase at 700 Celsius. Fr this prblem, we treat the gases as ideal. As discussed in class and the textbk, we can write the equilibrium cnstant in terms f partial pressures f the ideal gases:
4 K P = p CO 2 / p CO2 = x CO 2 p CO / p CO x CO x CO2 1 x CO2 =.688 x CO2 = = P ttal P ttal =.688
5 3A. (20 Pints) One mle f an ideal gas at 300K is isthermally cmpressed in a cnstant pressure prcess. The cnstant external pressure is P external =2.494 x 10 5 Pa. The initial vlume is 25.0 L and the final vlume is 10.0 L. The surrundings are at temperature 300K. Calculate the fllwing three thermdynamic prperties fr this prcess:!s Ttal,!S system,!s surrundings. Slutin: Fr the system, we need t cnstruct a reversible path. Since we have an ideal gas, isthermal prcess, this is as fllws: du = dq rev + dw rev = 0 dq rev = dw rev = pdv = RTd(lnV ) Thus, ds = dq rev T = nrtd(lnv ) = nrd(lnv ) T ΔS sys = nrln V 2 = (1ml)(8.314J /ml K)ln 25 = 7.62J / K 10 V 1 Fr the surrundings, we need the actual irreversible heat generated frm the system s perspective, which is taken as reversible frm the large surrundings perspective. The surrundings are taken t be at 300K. The actual prcess is irreversible. The external pressure is initially higher than the equilibrium pressure that the system wuld have n its wn, s the cmpressin takes place irreversibly. dq = dw dq = p ext dv = (2.494x10 5 Pa)dV q = (2.494x10 5 Pa)(V 2 V 1 ) = ( 15L)(2.494x10 5 Pa) = ( 15L)(2.494x10 5 Pa)( atm) = -37 L - atm The surrundings entrpy change is: ΔS = q sys 37L atm J = T 300K L atm =12.5J /K Ttal Entrpy Change: ΔS Ttal = = 4.88J /K
6 4A (10 Pints) On a graph with x-axis being entrpy and y-axis being enthalpy, plt the steps f a Carnt Cycle. Recall that the Carnt Cycle cnsists f (nt necessarily given in crrect rder) adiabatic expansin, adiabatic cmpressin, isthermal cmpressin, and isthermal expansin. Make sure the stages are in the crrect rder and label yur graph axes and indicate each f the 4 steps f the Carnt Cycle n yur plt; these will be required fr full pints. 4B (10 Pints) On a graph with x-axis being entrpy and y-axis being internal energy, U, plt the steps f a Carnt Cycle. Recall that the Carnt Cycle cnsists f isthermal expansin, adiabatic expansin, isthermal cmpressin, and adiabatic cmpressin. Make sure the stages are in the crrect rder and label yur graph axes and indicate each f the 4 steps f the Carnt Cycle n yur plt; these will be required fr full pints.
7 5 (20 Pints) Thermdynamics f stretching a rubber band (plymer). This is a simple argument fr cnsidering the restring frce assciated with the retractin f a rubber band t its riginal length after it is stretched. Starting frm sme initial equilibrium length, ne can cnsider the reversible (quasi-static) wrk f stretching the rubber band t be: dw rev,sretching = f dl where f is the restring frce (frce fr retractin) exerted by the rubber band upn being stretched by a length dl. Including this term in the expressin fr the First Law gives: du = TdS! pdv + f dl A. Based n this expressin, what are the inherent variables fr U, the internal energy? U = U(S,V,L) B. Use an apprpriate Legendre Transfrm t generate a thermdynamic ptential that is dependent in part n mre practical, experimentally cntrllable variables such as Temperature (T) and Pressure (P). Keep in mind that yu will still have mre than tw inherent variables that this ptential will depend n. What is the name f the ptential that yu transfrmed t? new functin = U(S,V,L) TS (-PV) d(new functin) = du(s,v,l) SdT TdS + PdV + VdP = TdS pdv + fdl SdT TdS + PdV + VdP d(new functin) = VdP SdT + f dl This is Gibbs Free Energy, G(T,P, L) C. Using yur knwledge f the definitin f the Gibbs Free Energy, determine the expressin fr:!f % The expressin shuld be in terms f f, temperature (T), and #!T $ #!H!L % ' Slutin Frm the expressin fr the ttal differential f the Gibbs Free Energy, and the definitin f the Gibbs Free Energy, we have: =? p,l. f = G L (H TS) = L = H L T S L We can nw use a Maxwell Relatin fr the partial differential f entrpy: S L = f T p,l Thus, we btain the requested partial derivative as: H L = f T f T p,l
8 Extra Credit (5 Pints) It can be shwn that the partitin functin f an ideal gas f N diatmic mlecules in an external electric field, ε, is: Q = [q]n k with q = C B T sinh µε N! µε k B T Here, T is temperature, k B is Bltzmann s cnstant, µ is the diple mment f a single mlecule, and C is a cnstant independent f ε. The partitin functin, Q, relates t the Helmhltz Free Energy thrugh the fllwing equatin: A = k B T ln Q = k B T ln [q]n N! Using this infrmatin alng with the Fundamental Thermdynamic Relatin fr the ttal derivative f the Helmhltz Free energy: da = SdT PdV (Nµ ) dε where µ is the average diple mment f a mlecule in the directin f the external field, ε, shw that at cnstant temperature and vlume: Slutin: µ = 1 A N ε T,V = 1 N ε k [q]n BT ln N! µ = µ cth µε k BT k B T µε = k BT N ( Nln(q) - ln(n!) ) T,V ε = k BT N ε Nln(q) ε ln(q) = k BT q = k BT Ck BT 1 sinh µε µ + csh µε q ε T,V q µ ε 2 k B T k B Tε k B T = k BT Ck BT 1 sinh µε µ + csh µε µ q ε 2 k B T k B Tε k B T T,V ( ) = k T ( ) T,V B T,V 1 sinh µε µ + csh µε ε 2 k B T k B Tε k B T = k B Tε sinh µε = k B T 1 + µcth µε ε k B T k B T = µ k BT + cth µε = µ cth µε k BT µε k B T k B T µε
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