Physics 301 15-Oct-2004 16-1The Sackur-Tetrode Entropy and Exp erimentIn this section we’ll be quoting some numbers found in K&K which are quoted fromthe literature.You may recall that I’ve several times asked how one would measure absolute entropy?I suspect that I pretty much gave it away (if you hadn’t figured it out already) in the lastsection. The answer is you have to measure heat transfers from a state of known absoluteentropy to the desired state so that you can calculateRdQ/τ. What i s a state of knownentropy? Answer, at absolute 0, one expects the entropy to be very small and we can takeit to be 0. Actually, there is the third l aw of thermodynamics (not as famous as the firsttwo!) which says that the entropy should go to a constant as τ → 0.At absolute 0, a reasonable system will be in its ground state. In fact the ground statemight not be a single state. For example if we consider a “perfect crystal,” its groundstate is clearly unique. But real crystals have imperfections. Suppose a crystal is missinga single atom from its lattice. If there are N atoms in the crystal there are presumably Ndifferent sites from which the at om could be mi ssing so the entropy is log N. Also, there’spresumably an energy cost for having a missing atom, so the crystal is not really in itsground state. But this might be as close as we can get with a real crystal. The point isthat the energy and the entropy are both very small in this situation and very little erroris ma de by assuming that σ(0) = 0. (Compare log N with N log(nQ/n) when N ∼ 1023!)In fact, a bigger problem is getting to very low temperatures. In practice, one g ets aslow as one can and then extrapolates to τ = 0 using a Debye law (a ssuming an insulatingsolid). So to measure the entropy of a monatomic ideal gas such as neon, one makes heatcapacity measurements and does the integralRC(τ) dτ/τ. The heat capacity measure-ments go to as low a τ as needed to get a reliable extrapolation to 0 with t he Debye law.According to K&K, the calculati on goes like this: solid neon melts at 24.55 K. At themelting point, its entropy (by extrapolation and numerical integration) isSmelting− S0= 14.29Jmol K.To melt the solid (which occurs at a constant temperature) req uires 33 5 J mol−1so theentropy required to melt is∆Smelt= 13.65Jmol K.Again, a numerical integration is required to find the entropy change as the liquid neon istaken from the freezing point to the boiling point at 27.2 K. This isSboiling− Sfreezing= 3.85Jmol K.Finally, 1761 J mol−1is required to boil the neon at the boiling p oint, and∆Sboil= 64.74Jmol K.Copyrightc 2004, Princeton University Physics Department, Edward J. GrothPhysics 301 15-Oct-2004 16-2Now we have a gas t o which we can apply the Sackur-Tetrode expression. AssumingS0= 0, the total isSvapor= ∆Sboil+ (Sboiling− Sfreezing) + ∆Smelt+ (Smelting− S0) = 96.40Jmol K,σ = 6.98 ×1024/mol ,where I have quoted the sum from K&K which differs slightly from the sum you get byadding up the four numb ers presumably because there is some round-off in the inputnumbers. (For example, using a periodic t able on the web, I find the melting and boilingpoints of neon are 24.56 K and 27.07 K.) According to K&K, the Sackur-Tetrode valuefor neon at the boiling point isSSackur−Tetrode= 96.45Jmol K,which is in very good agreement with the observed value.When I plug into the Sackur-Tetrode expression I actually get,SSackur−Tetrode= 96.47Jmol K,still in very good agreement with the observed value. Why did I get a slightly differentvalue t han that quoted in K&K? I usedS = R"log"mkT2π¯h23/2,pkT#+52#,Everything can be looked up, but I’m using p/kT instead of N0/V , which assumes theideal gas law is valid. However, this expression is being applied right at t he boiling point,so i t’s not clear that the ideal gas law should work al l that well.Some other things to note. (1) If we had left out the N! over counting correction, Wewould have to a ddR(lo g N0− 1) = 447Jmol K,to the above. This swamps any slight problems with deviations from the ideal gas law orinaccuracies in the numerical integrations! (2) The Sackur-Tetrode ex pression includes ¯hwhich means it depends on quantum mechanics. So this is an example where measurementsof the entropy pointed towards quantum mechanics. Of course, ¯h occurs inside a logari thm,so i t might not have been so easy to spot!Copyrightc 2004, Princeton University Physics Department, Edward J. GrothPhysics 301 15-Oct-2004 16-3The Ideal Fermi GasConsider a metal like sodium or copper (or t he other metals in the same columns in t heperiodic table). These metals have one val ence electron—an electron which can be easilyremoved from the atom, so these atoms often form chemical bonds as positively chargedions. In the solid metal, the va lence electrons aren’t bound to the atoms. How do we knowthis? Because the meta ls are good conductors of electricity. If the electrons were bound tothe atoms they would be insulators. Of course, there are interactions between the electronsand the ions and between the electrons and other electrons. But, as a first approximationwe can treat all the valence electrons as forming a gas of free (non-interacting) particlesconfined to the metal.Let’s do a little numerology. First, let’s calculate the quantum concentration for anelectron at room temperature,nQ=mekT2π¯h23/2,= (9.108 × 10−28g) (1.380 × 10−16erg K−1) (300 K)2π (1.054 × 10−27erg s)2!3/2,= 1. 26 × 1 019cm−3,=143˚A3.In o ther words, the density of electrons is eq ual to the room temperature quantum con-centra tion if there is one electron every 43˚A. Now consider copper. It has a density of8.90 g cm−3and an atomic mass of 63.54 amu. So the number density of copper atoms isnCu= 8. 44 × 1 022cm−3=12.3˚A3.The number of electrons in the electron gas (assuming one per copper atom) exceeds thequantum concentration by a factor o f 6700. For copper t he actual concentration and thequantum concentration are equal at a temperat ure of about 100,000 K (assuming we couldget solid copper that hot!).The upshot of all this is that we a re definitely not in the classical domain when dealingwith an electron gas i n metals under normal conditions. We will have to use the Fermi-Dirac distribution function. Low energy stat es are almost certain to be filled. When thisis true, the system is said to be degenerate. Furthermore, the