Chem 453/544 Fall 3 /8/3 Exam # Solutions. ( points) Use the genealized compessibility diagam povided on the last page to estimate ove what ange of pessues A at oom tempeatue confoms to the ideal gas law to within ±%. Assume that oom tempeatue is 98 K. The citical tempeatue of A is T c 5.95 K so that the educed tempeatue being consideed is T 98 /5.97 ~. The isothem coesponding to this educed tempeatue is shown in ed on the coelation diagam to the ight. Ideal volumetic behavio in this epesentation coesponds to a compessibility facto of Z. Geate than % deviation is not obseved fo the T isothem until a pessue of ~ in educed units. Fo A, P c 49.3 ba so that this educed pessue coesponds to P P P 54 ba. c Thus, at oom tempeatue A can be descibed to within % accuacy using the ideal gas e.o.s. ove the pessue ange 54 ba. Aside: That the ideal gas law should be easonably accuate up to this exteme pessue might seem stange. Howeve it should be emembeed that at a tempeatue of ~T c such a high pessue does not mean high densities. In the pesent case, a pessue of 54 ba coesponds to a educed density of.4 high but still fa fom liuid like.. (5 points) The nd viial coefficient of K can be epesented by the euation: 3-4 6 ( T ) / cm mol 3.7.98 (K / T ) 3.37 (K / T ) Use this expession and coesponding states ideas to estimate the nd viial coefficient of O at 98 K. Thee ae two ways that one could make this estimate. In eithe case the essential idea is that unde coesponding conditions, which, in the case of vial data (the P limit), means the same educed tempeatues, the educed viial coefficients of the two gases should be eual. Fo educing vaiables, one can eithe use the Lennad-Jones paametes listed in Table.7, which was my oiginal intention, o one could eually well use the citical paametes listed in the st table of the data sheet. The elevant values ae tabulated below. Species ε / k / K b / cm 3 mol - T c / K c / cm 3 mol - O 8 57.9 54.6 73.4 K 64 7.9 9.4 9. ~ P Calculations using both methods ae outlined below:
Chem 453/544 Fall 3 /8/3 i) Using LJ paametes: Hee the coespondence is: * O ( T ) K ( T ) ( T ) whee Tx kt / ε x and The viial coefficient of O at 98 K can theefoe be estimated fom * * ( T ) bo ( T kt / ε ) O O * x x / b x. whee * ( T * ) is obtained fom the paameteization of expeimental K data via: * * [ T T ( k / ε O)] [ TK T ( ε K / k )]/ b K Using the tabulated infomation, a tempeatue of 98 K coesponds to a educed tempeatue of * T 98 K /8 K.5 fo O. Fo K, this educed tempeatue coesponds to a eal tempeatue of T 64 K.5 43.3 K. Using the paameteization of (T) povided fo K, at this tempeatue: K ( 43.3K) 9.7 cm 3 mol - K and * ( T * ) 9.7 / 7.9. 47 Finally, the viial coefficient of O desied is: 3-3 - O( 98 K).47 57.9 cm mol 4.7 cm mol ii) Using citical paametes: Hee the coespondence is: O( T ) K ( T ) ( T ) whee T x T / Tcx and x x / cx. The viial coefficient of O at 98 K can theefoe be estimated fom O( T ) co ( T T / TcO) whee * ( T * ) is obtained fom the paameteization of expeimental K data via: [ T T / TcO] [ TK T TcK ]/ ck A tempeatue of 98 K coesponds to a educed tempeatue of T 98 K /54.6 K. 93 fo O. Fo K, this educed tempeatue coesponds to a eal tempeatue of T K 9.4 K.93 43.6 K. Using the paameteization of (T) povided fo K, at this tempeatue: ( 43.6K).89 cm 3 mol - K and ( T ).89 / 9.. 9 Hee the estimate fo the viial coefficient of O is: 3-3 - O( 98 K).9 73.4 cm mol 6.8 cm mol The expeimental value of (98.5) fo O is -6.5 cm 3 mol -. In this case at least the estimate obtained using the citical paametes is much close to the expeimental value.
Chem 453/544 Fall 3 /8/3 3. ( points) The acentic factos of methane, ethane, CO, and HF ae.8,.98,.5, and.37, espectively. Rationalize these values in tems of the natue of the intemolecula inteactions pesent in the vaious fluids. The acentic facto (ω) is an empiical measue of the extent to which the P (, T ) e.o.s. of a fluid depats fom that of the pototypical simple fluid A. The confomity (ω~) o lack of confomity in volumetic popeties eflects an undelying similaity o dissimilaity of the effective intemolecula potential govening the inteactions between molecules of a given fluid and the simple spheical ( centic ) inteactions opeative in A. The inteactions between two CH 4 molecule ae nealy spheically symmetic and esemble those between two Xe atoms. This similaity which is eflected in the value of ω of CH 4 being close to zeo. Although the natue of the inteactions between C H 6 molecules is essentially the same as between CH 4 molecules, i.e. in both cases only dispesion inteactions ae impotant, the shape of C H 6 is much less spheical and its effective intemolecula potential theefoe deviates moe fom that of a ae gas atom, esulting in a lage value of ω. CO is also fa fom spheical in shape. In addition, although not dipola, it possesses impotant electical (uadupole-uadupole) inteactions that make its inteactions even less simila to those of a ae-gas atom, which is the eason fo its lage value of ω. Finally, HF possesses a lage dipole moment and is capable of foming stong hydogen bonds between molecules. Its lage value of ω eflects the substantial diffeences in its inteactions compaed to simple spheical systems. These diffeences mean that HF displays P (, T ) behavio which depats significantly fom that of ae-gas like fluids. 4. (3 points) The states and enegies of a model 4-bead chain ae enumeated below: confomation enegy staight single bend double bend ε ε (a) What is the patition function conf (T) associated with the confomations of such a chain, and what ae the pobabilities of finding the chain in the vaious confomations (i.e. having,, o bends)? (b) At what tempeatue does the pobability finding the chain with a single bend exceed the pobability of it being staight? (c) Wite an expession fo the aveage enegy of the chain. How does this aveage enegy behave in the limits of T and T? (d) Use the esults of (d) to sketch the aveage enegy and heat capacity of the chain vesus tempeatue. (No need to calculate anything just depict the ualitative behavio.) (a) The patition function is: conf ( β ) + 3e + e 3
Chem 453/544 Fall 3 /8/3 and the pobabilities of finding a -, -, o -bend confomation ae: p e p 3 p e (b) The pobability of a single bend confome elative to a staight chain is: p p 3e / We want to find T (o β) such that p p which means 3 e >. This condition occus fo / > ( ε / ) βε < ln 3 o T > k.9( ε / k ) ln 3 (c) The aveage enegy is given by: βε d ln conf < Econf > (3εe + εe dβ conf i) as T, β, e, conf so that < E > ) ii) as T, β, e, 5 conf so that < E > (3ε + ε ) ε 5 (d) The sketches I made ae shown on the following page. The main thing I was looking fo in these plots was a coect depiction of the low and high tempeatue limiting behavio. The enegy limits ae descibed above. As discussed in class, the heat capacity will go to zeo both at high and low tempeatues fo a system such as this whee thee ae a finite numbe of enegy states available. 5/5 3/4 Heat Capacity C Tempeatue k T/ε Tempeatue k T/ε In my sketches I show steps in <E> to indicate fist andomization of population between and -bend chains and then incopoation of -bond chains. I wasn t sue how well defined these two steps would be, so I pogammed the euations into Mathcad to find out how close my guess was. A illustated below, the enegies of - and -bend chains ae appaently not diffeent enough to poduce the -level stuctue I depicted in <E> o the esulting double-peaked stuctue in C I d sketched. (The enegies need to diffe by 5--fold fo this type of stuctue to be obvious. 4
Chem 453/544 Fall 3 /8/3. E/eps, Cv/k ET ( ) C ( T).5 6.83 9 3 4.5 T 4 kt/eps 5. (5 points) What is the coect value of the otational patition function of HCl at K? Explain you answe. The classical otational patition function of a diatomic molecule like HCl is ot T / Θot. If one used this expession one would pedict that ot as T, but this would be incoect. This expession fo ot is a high-tempeatue appoximation, which is not valid as T. Fo the coect answe one must look at the expession fo ot befoe making the high-tempeatue appoximation: ot (J + ) exp( J ( J + ) Θot / T ) J + 3 exp( Θ /T ) + 5 exp( 6Θ ot ot /T )... As T only the st tem of this seies suvives and this tem yields the coect limiting behavio ot as T (i.e. thee is effectively a single otational state available at K, the gound otational state, J.) 6. ( points) The patition function of a had-sphee gas of mass m and size paamete b can be witten: N ( Nb) h Q( N,, T ) with N! Λ Λ πmk T (a) Deive expessions fo the aveage enegy and heat capacity of such a gas. (b) Show that this Q leads to the expected volumetic euation of state P( N A b) RT. (c) How would you expect the patition function and the answes to pat (a) to change if the gas wee -dimensional athe than 3-dimensional? / 5
Chem 453/544 Fall 3 /8/3 (a) The aveage enegy is detemined fom the patition function by: ln Q < E > kt T N, To pefom the necessay diffeentiations we can focus only on the tempeatue dependence of Q and wite: Q cλ c T / whee c and c ae independent of tempeatue. Thus, ln Q T The enegy is theefoe ln( T ) + ln( c ) T T N, N, 3 < E > NkT The heat capacity is elated to the enegy by C ( < E > / T ) N, so that 3 C Nk It should not be a supise at this point that these enegetic uantities ae the same as those of an ideal gas. (b) The pessue is given by ln Q P kt N, T Focusing on the volume dependence, Q can be witten: ln Q N ln( Nb) + ln c whee c is independent of. Diffeentiation yields: P kt N ln( Nb) NkT ( Nb) N, T Reaanging and witing this expession and substituting povides the desied esult P( N A b) RT. N nn A and n, and using N A k R (c) Except fo the eplacement of by (-Nb), the patition function of this had-sphee gas is identical to the patition function of an ideal monatomic gas. ased on the deivation of the ideal gas patition function, it is easonable to assume that the Λ tem hee aises fom teating tanslational motions using the paticle-in-a-box model. In the ideal gas the tanslational patition function is a poduct of 6
Chem 453/544 Fall 3 /8/3 thee patition functions, one each fo motion in the x, y, and z diections. If one wee to use the same development fo a -dimensional system, one of these patition functions would be absent and instead of N Λ one would have Λ. The enegy and heat capacity would then be: < E > Nk T, and C Nk 7. ( points) The heat capacity of CO (g) is smalle than that of H O (g) at low tempeatues but the opposite is tue at tempeatues highe than about 3 K. Explain why such behavio is expected. CO is a linea molecule wheeas H O is bent. Although both molecules have 9 nuclea degees of feedom, the geometic diffeence means that CO has one fewe otational and one moe vibational mode than does H O. At tempeatues whee tanslations and otations have eached thei classical limits (which is well below 3 K) but whee vibational degees of feedom ae inactive (i.e. C vib ~) the heat capacity of CO should be less than that of H O by the contibution of otational degee of feedom to C, ½R (i.e. unde these conditions C (H O)~3 R wheeas C (CO )~5/ R). Once vibational modes become active this deficit will be made up and ultimately CO will have the highe heat capacity (3/ R vesus 6 R) by vitue of the fact that vibational degees of feedom contibute R each to C wheeas otational modes contibute only ½ R. 7