Lecture19Ch161April8

Lecture19Ch161April8 - Summary of last lecture Weakly...

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Unformatted text preview: Summary of last lecture Weakly degenerate BE and FD Ideal gases BE Condensation Chemistry 161: Statistical Thermodynamics Lecture 19 Quantum Statistics- Bose Einstein condensation, Strongly degenerate FD gases - electrons in metals Key Concepts and Lessons: a) BE condensation: BE gases at very low T (recap) b) FD gases at `'low'' T - electrons in metals Reading: Reif, Ch 9 McQuarrie Ch10 Strongly degenerate BE gas. At lower T our trick with expansion does not work. Our concern is that at low T the 0 energy level should receive special attention/treatment since at low T there could be disproportionately many particles there. To that end we write the contribution corresponding to zero energy separately from all other contributions and get slightly modified main equations like: =2 2m h2 3/ 2 1/ 2 e e d 0 1 3/ 2 + V 1 e ( ) 2m p = 2 kT h2 V 0 1/ 2 ln 1 ( )d l 1 ln 1 V ( ) For convenience of notations we define the following function (which is essentially Euler's zeta-function) gn ( ) = And write our equations in a compact form: l =1 ln = 1 3 g3/ 2 ( ) + V 1 ( ) ( ) 1 p = 3 g5/ 2 ( ) kT 1 ln 1 V Strongly degenerate BE gas (Cont) The number of particles in 0 energy state is obviously: n0 = 1 Now when is not too close to 1 the contribution from 0 energy states is negligible - it is killed by 1/V term (and V can be as huge as we want!!!). However if we increase density too much or decrease T too low then starts to approach 1 and we cannot ignore the stuff that sits in the ground state. 3 g3/2 ( ) When approaches 1 approaches 2.612 and when exceeds this value the zero energy contribution starts to matter a lot! 3 g3/ 2 (1) = 3 V 1 we get: If we denote = 1 a / V (so that n0 = V / a) a= 3 3 g3/ 2 (1) ( 3/ 2 3 > 2.612 ) Defining a critical temperature as 3 0 = h 2 mkT0 2 = g3/ 2 (1) = 2.612 BE Condensation n0 = N 1 = a =1 T0 T 3/ 2 T < T0 We. encountered a first - and very important - instance of a phase transition For pressure we get another very important relation as well: p = kT p = kT 1 3 g5/ 2 ( ) ( ) < 0 1 3 g5/ 2 1 0 i.e. pressure becomes independent on density after this phase transition. Why? (answer in class - 10 pts) For energy we get: E= 3 kTV g5/ 2 3 2 ( ) 3 kTv E = g5/ 2 3 2 N 3 kTv E = g5/ 2 1 3 2 N ( ) () T > T0 T < T0 BE Condensation (Cont) For heat capacity we have (240 students: derive this relation 20pt) CV 15 T = g Nk 4 3 5/ 2 ( ) () - 9 g 3/2 ( ) T > T0 4 g1/2 ( ) T < T0 CV 15 v = g 1 4 3 5/ 2 N BE condensation is phase transition of 3rd order according to Ehrenfest classification,i.e. third derivative of free energy is discontinuous. FD gases at `'low'' T (Reif 9.16-9.17) Let us start from an old paradox: It has been realized since late 1800's that conductors have their electrons `'freely'' moving inside while insulators had their electrons tightly bound to nuclei. Then one expects freely moving electrons to contribute to heat capacity according to the equipartition rules. In other words the prediction is that, according to Dulong and Petit law, heat capacity pf conductors (per mole) will be anywhere from 50% to factor of 2 greater than that of insulators. However this I not what is observed in reality - significant differences between heat capacities of metals and insulators (above their respective Debye T's) have not been observed. Why? In order to answer this question let us consider ideal gas of electrons in more detail. These are FermiDirac particles and we need to use the FD statistics that we developed earlier. 2m h2 3/ 2 1/ 2 =2 e e 1/ 2 d 0 3/ 2 1+ V 0 p = 2 kT 2m h2 ln 1 + ( e )d FD gases at `'low'' T Now let us consider VERY low T and check quantum state occupancies: nk = e k 1+ e 1+ e At 1 this is almost a perfect step function, i.e.: nk 1 if k k = 1 ( k ) . nk =0 if n k > 1 = F In this case is called `'Fermi energy'' and is denoted often as F . This is a very clear result on a physical ground: at very low T electrons try to fill lowest energy levels as much as possible (i.e. consistent with Fermi statistics). This distribution is essentially a ground energy state for the ideal FD gas. Now we can determine easily the Fermi energy from the condition: FD gases at `'low'' T N =4 2m h2 3/ 2 1/ 2 V 0 d ( 1+ e ) =4 2m h2 3/ 2 F V 0 1/ 2 8 d = 3 2m h2 3/ 2 V ( ) F 3/ 2 Note that we used here 4 instead of 2 We get immediately for Fermi energy: h2 = 2m to account for the fact that electrons have two spin directions. 2/3 2/3 F 3 8 N V 3 Is it low or high for `'normal'' metals? Some numbers: Considering, say, Na with molar volume of 23.7 cm / mole and one valence electron the Fermi energy is about 3.1 eV or roughly 100kT at room temperature. Therefore the condition F kT is perfectly satisfied at room temperature and `'free'' electrons in metal can indeed be viewed as degenerate FD gas. The rest falls in place nicely: E0 = 4 2m h2 3/ 2 F 3/ 2 0 3/ 2 F 3 d = N 5 ln 1+ e F 2m p = 4 kT 2 h 1/ 2 0 ( ( F ) ) d 4 2m h2 3/ 2 F 1/ 2 0 ( F ) d 2 N 5 F /V FD gases: the heat capacity Now let us consider finite T corrections to that picture. In reality the distribution is not exactly a step function: some electrons are thermally excited above Fermi level and some space becomes available below (note: this is true only for metals, in insulators and semiconductors the Fermi levels fall into gap!!!!). Now we note that all essential TD quantities look like: I = f 0 ( )h ( ) d where f ( 1 )= 1+ e ( ) is (almost) step-wise Fermi function Now we take into account that the Fermi function is const almost everywhere except a narrow = That means that its derivative is almost everywhere 0 except for a spike range around = at Therefore all those integrals are convenient to take by parts, I.e.: I = 0 f ' ( )H ( ) d ' Where H ( ) = h ( ')d 0 (Q: prove that integral free term in integration by parts is zero (10pts)) FD gases: the heat capacity Now the contribution to this integral comes from the narrow region around Taylor expansion of H to get the results: H and we may use () =H + ( ) ( ) dH d + = 1 2 ( ) 2 d2H d 2 + .... = Resulting in: dH I = H + L1 d Where: () 1 d2H + L2 2 d 2 = + .... = Lj = 1 j ( x j ex 1 + ex ) 2 E.g. L2 = (1 + e ) x x 2 e x dx 2 = 2 3 (these integrals are obtained by changing variables as ( ) = x and extending the lower limit Of integration to using the condition that T is still low and the value of integrand beyond is negligible x= FD gases: the heat capacity Now our program is as follows: first we determine chemical potential from the equation for N then we plug it in the equation for E and from that determine the heat capacity by taking its derivative with respect to T. We follow general procedure as outlined above and consider equation for N first. For this equation: h( ) = 4 2m h2 3/ 2 V 1/ 2 And we get: 8 N= 3 2m h2 3/ 2 V 3/ 2 1+ ( ) 8 2 2 + ..... 2/ 3 Now recall the expression for the Fermi energy F h2 3 = 2m 8 2/ 3 N V Excluding N from these two equations we get: F = 1+ ( ) 8 2 2/ 3 2 + .... = 1+ ( ) 12 2 2 + .... FD gases: the heat capacity Keep in mind that we remain in the `'low T'' regime I.e. when this case we can solve for the chemical potential F 1 In =1 2 12 ( ) 2 T =1 12 2 2 = It is good enough to set higher order) and we get: F in the right hand side of this equation (correction would be in a = 2 F 1 T F 2 12 + ... Which shows how chemical potential slightly deviates from the Fermi Energy at finite T. Now we turn to the calculation of heat capacity. For energy we use the same trick of integration to get an expression: 8 E= 5 2m h2 3/ 2 V 5/ 2 1 + 5 8 2 ( ) 2 Now we use the expansion for above and get for energy: 5 2 12 T F 2 E = E0 1 FD Gases: the heat capacity Now we take the derivative to get the heat capacity of a FD gas at low T: 2 T NkT CV = = Nk TF 2 2 F / kT 2 ( ) It is clare that contribution of electrons in metals to heat capacity is indeed quite small (about 1% of nuclear heat capacity according to DP law) and could not be observed 100 years ago. Next lecture: Interacting systems - virial expansion and van der Waals equation. (Reif Ch 10.3-10.6, McQuarrie Ch 12 (for 240) ...
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This note was uploaded on 05/04/2010 for the course CHEM 161 taught by Professor Shaklovich during the Spring '10 term at Harvard.

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