Physics 301 18-Oct-2002 16-1The Sackur-Tetrode Entropy and ExperimentIn 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 calculatedQ/τ. What is 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 law 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 atom could be missing 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 made by assuming that σ(0) = 0. (Compare log N with N log(nQ/n)whenN ∼ 1023!)In fact, a bigger problem is getting to very low temperatures. In practice, one gets aslow as one can and then extrapolates to τ = 0 using a Debye law (assuming an insulatingsolid). So to measure the entropy of a monatomic ideal gas such as neon, one makes heatcapacity measurements and does the integralC(τ) dτ/τ . The heat capacity measure-mentsgotoaslowaτ as needed to get a reliable extrapolation to 0 with the Debye law.According to K&K, the calculation 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) requires 335 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 27.2K. ThisisSboiling− Sfreezing=3.85Jmol K.Finally, 1761 J mol−1is required to boil the neon at the boiling point, and∆Sboil=64.74Jmol K.Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 18-Oct-2002 16-2Now we have a gas to 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 numbers presumably because there is some round-off in the inputnumbers. (For example, using a periodic table 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 than 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 the boiling point,so it’s not clear that the ideal gas law should work all that well.Some other things to note. (1) If we had left out the N! over counting correction, Wewould have to addR(log 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 expression 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 logarithm,so it might not have been so easy to spot!Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 18-Oct-2002 16-3The Ideal Fermi GasConsider a metal like sodium or copper (or the other metals in the same columns in theperiodic table). These metals have one valence 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 valence electrons aren’t bound to the atoms. How do we knowthis? Because the metals 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 × 1019cm−3,= 143˚A3.In other words, the density of electrons is equal to the room temperature quantum con-centration if there is one electron every 43˚A. Now consider copper. It has a density of8.90 g cm−3andanatomicmassof63.54 amu. So the number density of copper atoms isnCu=8.44 × 1022cm−3= 12.3˚A3.The number of electrons in the electron gas (assuming one per copper atom) exceeds thequantum concentration by a factor of 6700. For copper the actual concentration and thequantum concentration are equal at a temperature of about 100,000 K (assuming we couldget solid copper that hot!).The upshot of all this is that we are definitely not in the classical domain when dealingwith an electron gas in metals under normal conditions. We will have to use the Fermi-Dirac distribution function. Low energy states are almost certain to be filled. When thisis true, the system is said to be degenerate. Furthermore, the electron gas is “cold” in thesense that thermal