Exam1Solutons.nb 1 G03 Md-Term Exam #1 Solutons 1-Oct-0, 1:10 p.m to :5 p.m n 1 Pupn Ths exam s open-book, open-notes. You may also use prnt-outs of the homework solutons and a calculator. 1 (30 ponts, Mean=6,StdDev=) An deal gas wth f degrees of freedom moves through a cycle n the P-V plane gven by the below dagram: P 0 P 1 P o 0 V o V o a) For each leg of the cycle, descrbe the physcal nature of the process, and calculate the heat flow n or out, and the work beng done. Express all of your results n terms of P 0 and V o. 0Æ1: W H0 Æ 1L = 0, snce DV = 0. DUH0 Æ 1L = UH1L - UHL = = f 8 P - P < = f P = QH0 Æ 1L f 8 P 1 V 1 - P < V Physcally, the system's pressure s ncreasng whle at constant volume, whch requres a ncrease n temperature, whch s suppled by the heat flow we just calculated. 1Æ: For ths segment, t's easest to frst fnd W(1Æ) geometrcally, snce t's the area under the lne segment (well, the negatve of the area, snce we're ntegratng -P dv ).
Exam1Solutons.nb WH1 Æ L = -9P + = = - (If you found the equaton of the lne P(V) and then ntegrated, that's also fne.) Agan, we compute the dfference n nternal energes and use that to fnd Q: DU H1 Æ L = UHL - UH1L = f 8 P V - P 1 V 1 < = f 8P 0 V 0 - P < = 0 so QH1 Æ L = DU - W = 0 -- = Physcally, the gas s expandng n a way that has that energy at equal to the energy at, and so the temperatures are equal at 1 and. (Ths s not the same as sothermal expanson. Why?) Work s done by the gas n the expanson (hence the negatve W), and Q s just the amount of heat requred to balance that work. Æ0: WH Æ 0L = -P 0 DV = -P 0 HV 0 - V 0 L = P DUH Æ 0L = f 8P - P 0 V 0 < = - QH Æ 0L = DU - W = - -- = - f +1 P Here the gas s beng compressed, so work s done on t, but t s also coolng, so heat must flow out. b) Fnd the net work and the net heat flow n the cycle, and comment on the sgns of each. Dong the fnal accountng, we fnd W NET = WH0 Æ 1L + WH1 Æ L + WH Æ 0L = 0 + - + P = - and Q NET = QH0 Æ 1L + QH1 Æ L + QH Æ 0L = + +- f +1 P = + whch s good, snce we must have W NET + Q NET = 0, snce we returned to our ntal pont (wth the same nternal energy) n performng ths cycle. Ths cycle represents an
Exam1Solutons.nb 3 "engne" extractng Q NET per cycle from the envronment, and usng t to perform the equvalent amount of mechancal work. (0 ponts, Mean=18,StdDev=5) Consder the combuston of one mole of hydrogen wth 1/ mole of oxygen under standard condtons: H + 1 ÅÅÅÅ O Ø H O How much of the heat energy produced comes from a decrease n the nternal energy of the system, and how much comes from work done by the collapsng atmosphere? (Treat the volume of the lqud water as neglgble.) The text (p. 35) tells you the "enthalpy of formaton" for one mole of water s -86 kj. (The negatve sgn of course means that energy s released.) To fnd how much comes from work done by the atmosphere n "collapsng" nto the volume formerly occuped by the hydrogen and oxygen, note that t's P V where V s the volume occuped by the ntal 1.5 moles of gas. We could fnd ths volume, put n atmospherc pressure and calculate (f you dd so fne), but t's even easer to wrte PV = nrt f W = 3 R H98 KL = 3.71 kj Usng ths result n the defnton of enthalpy gves the change n the nternal energy (mostly keepng track of sgns): DH = H f - H = U f - U + PHV f - V L = DU - PV I so - 86 kj = DU - 3.7 kj f DU = -8.3 kj 3 (0 ponts, Mean=16,StdDev= The multplcty of a N-spn two-state paramagnet wth N UP spns s gven by WHN, N UP L = j N y z. kn UP { Use Strlng's approxmaton (v1.0) to fnd an approxmate formula for the multplcty when N UP ` N.
Exam1Solutons.nb It's easest (but not necessary) to work wth the log of N! then re-exponentate, so log WHN, N UP L = log N!-log N UP!-logHN - N UP L! ª N log N - N - N UP log N UP + N UP - HN - N UP L loghn - N UP L + HN - N UP L = N log N - N UP log N UP - HN - N UP L loghn - N UP L = N @log N - loghn - N UP LD - N UP @ log N UP - loghn - N UP LD = N Alog N - logni1 - N UP ME - N N UP log N UP ª N Alog N - logn - logi1 - N UP ME - N N UP log N UP ª N N UP - N N UP log N UP f WHN, N UP L ª I en M N UP N UP ª N UP - N UP log N UP N = N UP log en N UP (Yes, ths s just about the same as Problem.19) (30 ponts, Mean=,StdDev=9) An nsulated lter contaner s ntally parttoned nto 3 lter and 1 lter sub-volumes. The 3 lter sub-volume contans 1 mole of He gas wth nternal energy 1 kj; the 1 lter sub-volume contans 3 moles of He gas wth nternal energy kj. Calculate the change n the entropy after the partton s removed and the system reaches thermal equlbrum. Descrbe n physcal terms the source(s) of ths entropy change. You may assume that the He s always gaseous, and behaves as an deal gas wth 3 degrees of freedom. HINT: Snce you are asked only for the change n the entropy, use the propertes of logarthms to elmnate as many common factors n the Sackur-Tetrode equaton as possble. To solve ths, use the Sackur-Tetrode equaton for the entropy of an deal gas wth nternal energy U: SHU, V, NL = Nk9 loga V N 1 j k h p mu y 3 N z { E + 5 = But you are only asked for the change n the entropy, that s, (takng explct note of the fact that t's He on both sdes of the partton, that s, no mxng of dstngushable atoms): DS = SHU = 1 +, V =, N = L - SHU = 1, V = 3, N = 1L - SHU =, V = 1, N = 3L Notng that many of the terms n the expanded logarthms wll cancel n computng the dfference, and
Exam1Solutons.nb 5 that N k = nrl, ths dfference can be wrtten as DS = R loga = 91 R loga j 16 y k { j 16 y k { A I 16 E - 1 R loga 3 j 1 y 1 k 1 { E - 1 R loga 3 j 1 y 1 k 1 { M E M E = 1 R log 9 = + 3 R log 9 = A 1 = 1 R loga 1 3 A 3 I 1 1 1 M E j 1 y k 3 { A I 16 I 3 3 M E E - 3 R loga 1 j y 3 k 3 { E= + 93 R loga j 16 y k { E + 3 R loga33 E = R loga33 E = 5 R log 3 Numercally, the value of ths entropy change s D S = 5 R log 3 = H5 molesl 8.3 J mole J log 3 = 5.6 K K E E - 3 R loga 1 j y 3 k 3 { Physcally, there are two sources of entropy ncrease: the volume ncrease for each sub-volume, and the heat exchange between the two sub-volumes. Note that the ntal temperatures (energy per partcle) dffer by a factor of 9(!), and n fact correspond to ntal temperatures of about 100 K and 1000 K. E=