Homework 7A (due Wednesday 9PM 2/22/12) Q1) Suppose 18 grams of water (one mole) at 298K falls on a large slab of ron (our heat reservor) that s mantaned at 423 K. Calculate the entropy change for the water, the ron and the unverse (whch s just the sum of the system and ts surroundngs, assumng they are adabatcally solated from the rest of creaton.) All of ths happens at 1 Atm. Assume the heat capactes are constant over the applcable ntervals. Cm, p ( ce) = 37 J Cm, p ( water ) = 75.3 J Cm, p ( steam) = 33 K mol K mol J K mol And for water H m, melt ( H 2 O) = 6.kJ Hm, vap ( H 2O) = 4.1kJ mol mol a) Outlne the transformaton that must take place to the water. b) Wrte the entropy term for each step. You may assume that the heat capacty of water and steam (although dfferent) are ndependent of temperature. And determne the total entropy change for the water to steam transformaton. c) Descrbe the physcal transformaton of the ron slab. d) Determne the entropy change for the ron slab. (All heat that went nto the water had to come from the ron slab). e) Determne the total entropy change and verfy that the change s consstent wth he Clausus Inequalty. Q2) Consder a system that s a mxture of 4 moles of N 2, n moles of H 2 and (8- n) moles of O 2. he gasses behave as deal gasses and do not react. a) Wrte a general expresson for the entropy of mxng as a functon of n. n S = n ln n ot b) Fnd the value of n whch maxmzes the mxng entropy c) Fnd the entropy at the maxmum value of n. Q3). he entropy has been defned as tme s arrow. Explan ths defnton usng thermodynamc reasonng. Utlze n your argument the statstcal nterpretaton of entropy. 1/9
Q4) Explan the Carnot engne s relevance to the effcency of heat engnes. Explan also the relatonshp between entropy and the effcency of a heat engne. Use a drawng of the engne f you thnk t helps to explan ts functon. Q5) he Helmholtz (A) energy and Gbbs energy (G) are sometmes called free energes. In the old lterature the Gbbs energy s called the free enthalpy. Why are the Gbbs and Helmholtz energes referred to as free? In your explanaton, you may want to recall the defntons of A and G n terms of U, H, and S. Explan also why the Gbbs energy G s so often dscussed n chemstry and bochemstry texts. Q6) Ammona s a common metabolc by-product. It s also very toxc so terrestral anmals convert ammona to urea: NH 2 CONH 2. Consder the synthess of urea from ts consttuent elements: 1 + + + C gr N g O g H g NH CONH s 2 2 2 2 2 2 2 he table below shows entropy values at for =298K and heat capacty values for each reacton component. Substance ( 1 S JK mol 1 1 1 ) CP ( JK mol ) C(gr) 5.69 8.64 N 2 (g) 191.49 29.12 O 2 (g) 25.3 29.36 H 2 (g) 13.59 28.84 Urea(s) 14.6 93.13 Calculate the entropy change for ths reacton at =31K. Assume all heat capactes are constant between 298K and 31K. 2/9
Q7) he shells of marne organsms contan calcum carbonate CaCO 3, largely n a crystallne form known as calcte. here s a second crystallne form of calcum carbonate known as aragonte. Physcal and thermodynamc propertes of calcte and aragonte are gven below. [hs compares wth the graphte to damond reacton n the text example 6.13] Propertes (=298K, P=1atm) Calcte Aragonte H f (kj/mole) -126.87-127.4 G f (kj/mole) -1128.76-1127.71 S f (J/mole K) 92.88 88.7 C P (J/mole K) 81.88 81.25 Densty (gm/ml) 2.71 2.93 a) Based on the thermodynamc data gven, would you expect an solated sample of calcte at =298K and P=1 atm to convert to aragonte, gven suffcent tme. Explan. b) Suppose the pressure appled to an solated sample of calcte s ncreased. Can the pressure be ncreased to the pont that solated calcte wll be converted to aragonte? Explan. 3/9
c) What pressure must be acheved to nduce the converson of calcte to aragonte at =298K. Assume both calcte and aragonte are ncompressble at =298K. d) Can solated calcte be converted to aragonte at P=1 atm f the temperature s ncreased? Explan. [Assume the reacton enthalpy and entropy wll reman constant wth temperature.] Q8) One vessel contans 1 mole of gas A and 1 mole of gas B at pressure Pot = 2Bar, at ntal volume V A. A second vessel contans 1 mole of gas A only, at volume V, also at pressure P = 2Bar. A valve separatng the two vessels s B ot opened, mxng wll occur. Both A and B are deal gases, and the ntal pressure n both vessels s 2 bar. Use the pressure dependence of the chemcal potental of each gas. here s a pston on top that can allow the volumes of the two contaners to vary to mantan pressure at 2 Bar, and the temperature s held constant at 298K. a) Determne the composton of the gasses n the two vessels after the valve connectng them s open. b) What thermodynamc crteron dd you use to determne equlbrum composton? c) What s the pressure of the gasses n the two vessels? d) How has the total volume, V + V, changed? A B e) Determne the change n the free energy of each gas f) Determne the change n the total Free Energy. g) Determne the change n the total entropy. h) Determne the change n the total enthalpy. ) How much heat was transferred to the system (and n what drecton)? Q9) he pressure dependence of G s qute dfferent for gases and condensed phases. Calculate for the process (C, sold, graphte, 1 bar, 298.15 K) to (C, Gm sold, graphte, 25 bar, 298.15 K) and Gm for the process (He, g, 1 bar, 298.15 K) to (He, g, 25 bars, 298.15 K). By what factor s the change n m G greater for He than for graphte? he densty of graphte s about 2.2 g/cc. [For 4/9
smplcty assume He s an deal gas. Assume graphte s ncompressble over ths pressure range.] 2) One mole of an deal monatomc gas undergoes the followng transformaton from an ntal state P=1 bar and =3K. Calculate q, w, U, H, and S for each process. a) he gas s heated to =45K at a constant pressure of P=1. bar. b) he gas s heated to =45K at constant volume. c) he gas undergoes a reversble sothermal expanson at =3K untl ts pressure s P=.5 bar. 3) Calculate the entropy change f 2 moles of lqud ammona at P= 1atm and =233.2K are heated to = 473K. he normal bolng pont of ammona s 239.7K. he heat of vaporzaton of ammona s Hm, Vap = 23.2 kj he heat mol capacty of lqud ammona sc 74.8 J P = and s constant from 233K to mol-k 239.7K. he heat capacty of ammona gas between 239.7k and 473K s gven n able 2.4 of your text. You may use that heat capacty over the range of the gas (.e. from 233K up.) Q1) Fve moles of an deal monatomc gas contract adabatcally and rreversbly when subjected to a constant external pressure of 1 atm from an ntal volume V1=2L and ntal pressure P1=.5 atm. Assume the contracton ceases when the system reaches equlbrum. Calculate S. sys Q11) A sample consstng of 2.5 moles of an deal gas at 298K s expanded from an ntal volume of 1.L to a fnal volume of 5.L. Calculate G and A for: a) an sothermal reversble path; b) an sothermal expanson aganst a constant external pressure of.75 bar. c) Explan why G and A do not dffer. 5/9
Q12) hs s an openended queston, and wll requre you to make several assumptons about how much reacton s actually takng place, and how the energy s transferred. Background nformaton: A hot pack can be made by a soluton of sodum acetate (NaAc) n water. he solublty of NaAc at 1C (~bolng) s 18 g NaAc per 1 grams of water. At 2C the solublty s 2 grams NaAc per 1 grams of water. You can look up the solublty on the web at other temperatures. he reacton yelds -17.4 kj/mole of NaAc. Let us assume: A hot pack weghs about 12 grams (a few ounces) and conssts of about 8 grams of NaAc and 4 grams of water. Assume that the hot pack has a heat capacty of that of water ( 4J/gram). he human hand s about 31C (core temperature s 37.C for humans). he hotpack s generated by placng t n bolng water where all the salt dssolves and remans supersaturated even at room temperature or below, untl t s dsturbed by bendng a metal dsk n the soluton at whch pont the NaAc ppt. out and soldfes releasng the heat. So the queston s: How much heat s released and how warm wll the hot pack get f used around C? Would ths devce be able to warm your hands? Be clear about the assumptons you need to make to determne the amount of heat avalable that wll flow nto you hands. Q13) A box as shown n fgure 6.2 of your text contans a volume VL on the left and the same volume VR on the rght. 1 mole of He s on the left and one mole of Ar s on the rght. A sempermeable membrane separates the two parts. Only the He may pass through the membrane. he temperature s kept constant (@3K). What s the concentraton of He on the left after the system comes to equlbrum? What crteron dd you use to determne ths? How does the presence of the Ar on the rght mpact ths result? Explan. What s the pressure nsde the two parts (left and rght) of the contaner before and after equlbrum? In the prevous steps the volumes were fxed. Now suppose, pror to any He movng out of the left compartment, that the rght compartment has a pston and 6/9
a mass on t that gves the pressure needed to keep the volume the same as the orgnal volume, VR. And now the He s allowed to move across the membrane. What wll the new volume of the rght be relatve to the old volume? he volume of the left compartment, VL, does not change. he temperature s kept constant. [If you can t work t exactly, you can argue approxmately based on the constant volume results and get pretty close.] Be sure to show what ntal assumptons (and equatons) you have to solve ths problem. 7/9
Chemstry 456 (Bagley 154, 1:3 AM) Informaton that wll be avalable to you on the second exam. Law I U = q + w U = U (, V ) U U du = d + dv V V Law II: ds = q rev S = S(, V ) S S ds = d + dv V V Comb I,II: du = ds PdV hermodynamc Equaton of State H P S CP S V = ; = P P P V = V S P = V Dalton's Law V P = χ P; P = P ; G = n µ A A A A P Gmx = R n ln Po Calculus Identtes: P x f Z Z = dx x x y dx dz dy = 1 dy dx y dzx z 8/9
Chemstry 456 (Bagley 154, 1:3 AM) d( yz) d( y) d( z) = z + y dx dx dx dx dy dx x y x = ; = dz dz dy z z y a a a hermal expanson and compresson coeffcent 1 V 1 V β = κ = V V P P vdw Gas EoS: P R a = V b V 2 Reacton Info o o rxn f o o rxn f Int P, rxn P, m o o H = H rxndx H = ν H S = ν S dn = n n = ν dx m C = ν C Gas Constant: R = 8.3 J / mol K R =.82 L atm / mol K R 298K = 2.48 kj / mol 11J = 1 L atm 1 1 1 5 Atm bar = Pa ( K ) = ( C) + 273 m 9/9