Joule-Thomson Expansion - Hellas

1 - Physical Chemistry IJoule- thomson ExpansionProfessor Paul J. GansSeptember, 1992 Minor Revision: October, 19931. IntroductionWe are at the point where we can make our first analysis of an actual experimentwill be explained first. Then thermodynamics will be used to show that the experiment isisoenthalpic(meaning a constant enthalpy process). With this result, we can then explain the main result of the experi-ment in terms of the properties of the gases The Joule-Thomson ExperimentThe apparatus itself is quite simple. Imagine a tube with a porous plate separating it into two porous plate will allow a gas to go through it, but only slowly. It acts as athrottle. On each side of theplate there is a piston that fits the tube tightly. Each piston can (in principle) be pushed up against theporous plate. The tube itself is insulated so that no heat can enter or leave the experiment is also quite simple. Imagine that some gas is placed between the porous plate andthe piston on the left side of the tube.

2 This is side 1. On the other side the piston is flush against the porousplate. This is side initial volume of gas on side 1 isV1. The pressure isp1and the during the experiment gas is pushed through the porous plate by pushing on the piston on theleft side. At the same time the piston on the right side is pulled in such a way that the pressure on the rightside is the end of the experiment all of the gas has been pushed through the porous plate. The volume onside 1 is zero. The final volume on the right side (side 2) isV2, the pressure isp2and the curious result of this experiment is that careful measurement shows thatT2isnotequal some conditions it is higher, under others it is problem is to decide how to analyse we do need to drag in one result that will not be obtained until later in the is a drawing of the apparatus on page 69 of course,p2must Initial AnalysisThe process starts with volumeV1=V1and volumeV2= 0. It ends with volumeV1= 0 and volumeV2=V2.

3 The work done on the left side is:w1= p1(0 V1). The work done on the right side is:W2= p2(V2 0).The total work done is then:w=p1V1 p2V2(1)Since the process is adiabatic, the total change inUis just the work, or U=U2 U1=p1V1=p2V2(2)U2+p2V2=U1+p1V1(3)so that:H2=H1(4)and the process The Joule-Thomson CoefficientWhat is measured experimentally is JT= p 0limT2 T1p2 p1(5)which becomes, taking into account that the process is isoenthalpic: JT= T p H(6)All that we ve done so far is reduced the experiment to mathematics. The experimental result is that JTis sometimes positive and sometimes negative. In fact it is found that there is a certain temperaturecalled theinversion temperaturesuch that if the initial temperatureT1isabovethe inversion temperature,the final temperature ishigherthan the initial temperature. If the initial temperature isbelowthe inversiontemperature, the final temperature islowerthan the initial temperature. The inversion temperature is found,experimentally, to depend on the understand this we need to relate JTto quantities that are experimentally ability to do this is one of the main points of do this we can start with the total differential ofHassuming thatTandpare the independentvariables:dH= H T pdT+ H p Tdp(7)Now our process is isoenthalpic, sodH=0.

4 This gives:0= H T pdT+ H p Tdp(8)Now I divide through bydpat constantHto get:5 H T p T p H+ H p T(9)Note the subscriptHat the appropriate point in eq. (9). Lastly, let us solve for ( T/ p)Hto get:( T/ p)H= ( H/ p)T( H/ T)p(10)This result simplifies a bit right away. We know that ( H/ T)pis the heat capacityCp, so:( T/ p)H= ( H/ p)TCp(11)but we can t make it any simpler. We need to express ( H/ p)Tin terms that can be measured. Indeed,this can be done, but it requires one slight bit of magic. A result that will be proven later6 H p T=V T V T p(12)which, when used in eq. (11) gives: JT=T( V/ T)p VCp(13)a neat equation in whichallof the quantities are is a touchy point. Pure mathematicians would perhaps object to treating an infinitesimal as if it were afinite quantity. But this procedure of dividing bydx(where herex=p) in thermodyanmic equationscanbejustified mathematically and we do it all the time. Perhaps the best way to think of it is to think ofdp(or what-ev er) as actually representing a very small, but finite change in pressure.

5 Then at the very end let all infinitesi-mal ratios go to their always say this, right?-4-So we come to the end of the theory with a formula but no explanation. It all seems to depend onhow the volume of the gas changes with temperature and on the heat capacity. How can we learn anythingmore? Well, we can look atmodelgases, that is, gases for which we know the equation of state. Lookingat some of these may tell us The Joule-Thomson Coefficient for Model GasesIf we knew the equation of state for a gas, we couldcalculatethe Joule-Thomson coefficient. Forinstance, for an ideal gas ( V/ T)p=V/T, so JT=0ideal gas (14)Ideal gases are they hav e told us something. There isnoJoule- thomson effect for an ideal gas at all. And howdo ideal gases differ from real gases? They hav e no interactions at all. No attractiions and no that s the only way in which they differ. So we conclude thatthe Joule-Thomson effect depends on theinteractions between gas gas obeying van der Waals equation will prove more interesting.

6 Since we can t solve van derWaals equation forVexplicitly7we have to use an approximation. A convenient one is eq in can be writtenZ=1+1RT b aRT p+..(15)which is the first two terms of thevirial equationapproximation for a van der Waals want to take this expression, solve it forV, and then differentiate it to get ( V/ T)p. With thisavailable we can plug into eq. (13) and calculate sinceZ=pV/RT, we can solve eq. (15) for V to get:V=RTp+ b aRT +..(16)What we want is ( V/ T)p. From eq. (16) we get: V T p=Rp+aRT2+..(17)Now this has to be multiplied byT. This gives:87It s a cubic inV, remember?8I m showing all the steps. Nobody expects you to memorize this stuff. Just read it and learnwheretheJoule- thomson effect comes V T p=RTp+aRT+..(18)And now subtractingVas given in eq. (16) we get (after some cancellations:T V T p V= b 2aRT +..(19)and so, finally, the Joule-Thomson coefficient for a gas obeying van der Waals equation is: JT=2aRT bCp+.)

7 (20)OK, you can t tell much from this equation all by itself except that it certainly looks messy. Butthere is a way theoreticians handle things like this: they look at the extremes. Let s see what happens tothe Joule-Thomson coefficient when the temperature gets very low. That is, what is the limit whenTgoesto 0? Well, asTgoes to 0, all terms withTin the denominator get very large. So the term involving2a/RTgets much larger thanb. And thebterm can be neglected at very low temperatures . JT=2aRTCp+..(very low temperature) (21)which is clearlypositivesince every term in it is positive . This agrees with experiment. A positive Joule-Thomson coefficient meanscooling. But also note that the resulting expression only contains the van derWaalsa. All of the terms involvingbhave gone away at low can also look at what happens whenTgets very large. This result is quite simple since all termscontaining aTin the denominator go to zero.

8 The result is: JT= bCp(very high temperature)(22)and that s very interesting. First, it agrees with experiment since it isnegative, meaningheating. Second,notice that all of the terms involving van der Waalsahave now vanished, leaving , it should be pointed out that there is one way in which eq. (20)does notagree with experi-ment. Eq. (20) predicts aninversion temperatureof:Tinversion=2abR(van der Waals gas) (23)which is gotten by setting JTequal to zero in eq. (20). The problem is that this inversion temperature isnot a function of the pressure. Experimentally the inversion temperature is a function of the pressure. Ifwe d carried one more term in eq. (15) we would have had a (much more messy) result that is a function ConclusionsSo what can we conclude? Several things. First, since JTis positive at low temperatures and neg-ative at high temperatures , it must have an inv ersion temperature. Second, the effect seems to depend (aswe expected) on the attractive and repulsive forces acting between molecules.

9 At low temperatures theattractiveforces high temperatures therepulsiveforces the Joule-Thomson effect can be explained this way: At low temperatures the intermolecular attraction is the mostimportant interaction. When the cold gas is expanded, the average distance between molecules isincreased. This means that the molecules are pulled apart. Since they attract each other since the process is adiabatic, the only source of energy is the internal energy of the gas itself. So,with the internal energy reduced, the the other hand, at high enough temperatures the predominant interaction gaswants to separate. It wants to expand. When it does expand energy is obtained as the molecules internal energy of the gas. And the ,aaccounts for the attractiongs between molecules in van der Waals ones associated withb.