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Unformatted text preview: Summary of last lecture Chemical Equilibrium, examples of chemical reactions. Chemistry 161: Statistical Thermodynamics
Lecture 18 Quantum Statistics Weakly degenerate gases and Bose Einstein condensation
Key Concepts and Lessons: a) Quantum statistics  a general consideration, density of states counting b) Weakly degenerate BE and FD gases c) BE condensation: BE gases at very low T
Reading: Reif, Ch 9 McQuarrie Ch10 Quantum Statistics of Ideal Gases (Cont) The way out is to relax the particle conservation condition, i.e. consider the Grand Canonical Ensemble! e pV = =
N v E e ( v Nv ) Now totalN constraint is removed and we can handily sum over all occupancy numbers, I.e. n's: =
n1 ,n2 ...n j exp
j ( j nj ) Now we should distinguish between BoseEinstein statistics (for particles with integer spin) which allows ANY occupancy and FermiDirac Statistics (for particles with semiinteger spin, I.e. electrons) Which does not allow more than one particle in a state I.e. for FD n can be either 0 or 1. With that we have: BE =e =e pV =
j nj =0 1 e ( ( j nj ) =
j 1 1 e ( j ) ) pV FD =
j e
nj =0 j nj ) =
j { 1+ e ( j } Quantum Statistics of Ideal Gases (Cont)
pV = ln pV = ln
BE =
j ln 1 ln 1 + e
j e (
j j ) FD = ( ) Where summation is taken over all quantum states. Average occupation numbers in each quantum state are determined in a straightforward way: nj nj BE = = ln ( ln (
j j ) ) = e = e ( ( 1
j ) 1 ) +1 1
j FD These relations contain an unknown quantity  chemical potential. It can be determined from the condition that total number of particles is conserved as N: nj
j = N Quantum Statistics of Ideal Gases: high T limit
At high T all occupancy numbers are very small:
n j << 1 which means that
e ( j ) 1 In order to satisfy this condition for each quantum level j we need to have:
nj = e 1 ( j ) = e e j i.e. we omitted 1 in the expression for occupancies. In this case the expression for chemical potential and occupancy numers simplifies: N e = j e
j nj = N e e
j j j Next, we use a standard Thermodynamic relation: lnQ(N,V,T) = N + ln
Which in highT case becomes ln Q( N ,V ,T ) =
(we expanded logs in full formulas for N +
j e ( ) =
j N + N
j because at high T all e ( ) << 1 and therefore ln 1 e ( ( j ) ) e ( j ) Quantum Statistics of Ideal Gases: high T limit
We note that: = ln N
And therefore ln
j e j ln Q = N ln N + N + N ln
j e j So that finally we get the classical expression of CPF which is valid only at high T when all occupancy numbers are small:
N 1 Q = N! e
j j qN = N! Back to low T Introducing a convenient notation = e We get final relations for both BE ( sign) and FD (+ sign) cases: ( V ,T , )
1 e =
k (
k 1 e k ) 1 N =
k e e
k k nk = E = 1 e e k e
k k k k 1 k pV = kT ln 1 ( e k ) How to sum over quantum levels? `'Particle in a Box'' is a solution: k x ,k y ,k z h2 2 = k x2 + k y + k z2 8mV 2/ 3 ( )
3/ 2 1/ 2 Where k are integers. We derived the conversion in Lecture 4, slide #8 (included below) kx k y kz = () 8m d = V 2 4 h d And we get general closed form expressions for BE and FD ideal gases: N =2 2m h2 3/ 2 V
0 e d 1 e
3/ 2 pV = 2 kT 2m h2 V
0 ln 1 ( e )d We will next study a very rich behavior that these relations predict! Now let us try to evaluate density of states just for N=1 case In this case it is more convenient first to evaluate the function which gives the total number of states below certain energy. In other words we need to evaluate how many integer numbers satisfy the relation: ( ) ny h2 2 2 nx + ny + nz2 8ma 2 { } ; nx 0;ny 0;nz 0 It is clear from the picture that integer points are located equidistantly on the axes (with increment 1). Therefore their number is prportional to the volume of the octant (because they are all nonnegative). The radius of the sphere is:
nx
R = 8ma 2 h2 4 3 and the volume of the octant is R
3 ( 1 )= 8 = 6 8ma 2 h2 3/ 2 This expression gives the number of states with energy at or below . The number of states in the range ( , 4 +d ) is then:
3/ 2 1/ 2 ( +d ) ( )=d d d ( )=d 8ma 2 h2 +O d ( 2 ) Weakly Degenerate FD and BE gases
Now we look at conditions when T is high enough that our ideal gases are almost classical And consider quantum effects as small corrections. Our goal will be to investigate expansion of TD functions over these small corrections. Let us focus first on FD case  the BE case is (almost) identical. 1 and therefore e 1 for any >1 As we discussed above at high T
=2 2m h2
3/ 2 0 e 1
3/ 2 d e ln 1 p = 2 kT where = 2m h2 ( e )d 0 N is density of the gas V The FD case corresponds to the + sign in these equations. Now our program is simple: we expand both equation in posers of small parameter e take integrals over one at a time and get the following series (for the FD case) l +1
= 1
3 l =1 ( 1 ) l l 3/ 2 Where = h 2 mkT 2 1/ 2 p = kT 1
3 l =1 ( 1 ) l +1 l l 5/ 2 is deBroigle wavelength. The powers of l in denominator appear from Integrations over Weakly Degenerate FD and BE gases (Cont) Our goal: to get p as a series expansion in gas density However we are not yet there because we need to determine chemical potential (or better to say, exclude from both equations. In order to do that we from using first equation and then plug it into second one. In other determine words, we seek to invert first equation into the form: = a0 + a1 + a2
a0 = 0 a1 = a2 a3 .......
3 2 +++ We can do that by direct substitution making sure that all terms in proper order are taken care of: 2 a1 23/ 2 a1a2 21/ 2 = 0 +
3 a1 33/ 2 = 0 Weakly Degenerate FD and BE gases (Cont) Now substituting the expansion for get the series for the FD Ideal gas: in the expression for p (second equation) we p = kT + 3 2 2 5/ 2 + 1 8 2 35/ 2 6 3 This expression as general form of a virial expansion: p = kT
In this case + B2 (T ) 2 + B3 (T ) > 0 3 + .... B2 = 3 25 /2 The fact that B2>0 indicates that pressure is higher than that of a classical IG I.e. that there exists effective repulsion between molecules. It is intuitive for the FD gas because particles `'fight'' for available Quantum levels. In the BE case the same expansion would give: pBE = kT 3 2 2 5/ 2 + (derive second order term yourself) Q: derive the degenerate gas expansion for BE gas (esp second order term) and discuss what Does the  sign of B2 means (20pt) 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 e d + 1 e V 1 0 ( ) 2m p = 2 kT h2 3/ 2 V ln 1
0 ( e )d
l 1 ln 1 V ( ) For convenience of notations we define the following function (which is essentially Euler's zetafunction) 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. Next lecture:
Quantum Statistics of Ideal gases: Highly degenerate FD gas  electrons in metals, Fermi level and all that.. (Reif Ch 9.7, 9.16, McQuarrie 102) ...
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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.
 Spring '10
 Shaklovich

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