The Ideal Gas. One particle in a box:

Transcription

1 IDEAL GAS

2 The Ideal Gas It is an important physical example that can be solved exactly. All real gases behave like ideal if the density is small enough. In order to derive the law, we have to do following: 1. Analyze one free particle in a box.. Analyze N distinguishable particles in a box. 3. Analyze N indistinguishable particles in a box. As it turns out, a complete treatment of the problem, which gives all the pre-factors correctly, has to start from Quantum Mechanics. Luckily, this is not too complicated One particle in a box: The Hamilton operator of single free particle (that has only kinetic energy) is:

3 L L L Hˆ pˆ h # # # " h = = $ & + + ' = $ % m m ( # x # y # z ) m h h = = # " Js m mass of particle Planck action quantum number We have to find the solution ψ(x) to the Schrödinger s equation: Hø ˆ ( x) = Eø ( x) subject to the condition that ψ(x) vanishes on the boundaries of the box. We will now confirm that this solution is:

4 ø = k, l, m { 1,,3,... } kx ly mz klm ( x, y, z ) Asin sin sin L L L Indices k,l,m define the quantum state and A is the normalization constant. Evidently ψ(x) satisfies the boundary conditions. Does it satisfy the Schrödinger s equation?

5 Hˆ ø klm ( x, y, z) = h # # # " = $ % + + & ' ( kx ly mz Asin sin sin m # x # y # z L L L h = ( k + l + m ) ø E ml ø ( x, y, z) = ( x, y, z) klm klm klm Hence, we have found the solution to the Schrödinger s equation. In particular, we have found the energy eigenvalues We can therefore find the canonical partition function: " Eklm # Z( ) = ))) exp % $ k 1 l 1 m 1 & = = = ' ( " h # " h # " h # = ) exp% $ k exp l exp m &) % $ &) % $ & k = 1 ' ml ( l= 1 ' ml ( m= 1 ' ml (

6 These sums are identical, except for the name of the summation index, which is unimportant ( dummy variable ). " " h ## %) exp% k & k = 1 ml & = $ ' ' (( Unfortunately this sum is hard to solve. However, at room temperatures, for usual gas atoms and macroscopic boxes, the summation steps are so finely spaced (check it) that we may replace the sum by an integral: 3 " " h ## &* dk exp& k ' ml ' o 1 $ % = ( ( )) h 3 ml Here we used the known solution of Gaussian integral: So for the partition function we find: +# % $# $ dx e x = " a

7 Z ( ) % db & 3 L " = # $ = ë V ë db where thermal de Broglie wave length is: ë db = h m Where, h is Planck s constant and the denominator has the dimension of momentum, and is called thermal momentum. For example, for a proton at a room temperature we have: ë db " 10 Js o = # 1A -7-3 " 1.67 " 10 kg " 1.38" 10 J K " 300 K (very small) (Remark) One expression that will prove to be useful later on: ë db = " 1 ë db

8 N distinguishable particles in a box If the particles do not interact, they could be imagined as sitting in different boxes. Then we can do the partition functions separately for each box, so to speak. " H s H s H s # Z( ) exp% & ' ( = Z ( ) $ Z ( ) $... $ Z ( ) H( s1 ) H( s ) H( sn ) ( ) + ( ) ( ) )) ) ) ) ) 1 N = $$$ = $ $$$ 1 N 1 1 e e e s s s s s s N N

9 N indistinguishable particles in a box In this case all Z 1,Z,,Z N from above are identical and we would expect Z=(Z 1 ) N. But careful: we would be over-counting states. Any arbitrary permutation of particles would be counted as a new state, while in fact it is the same state The number of such permutations are N Therefore we have to divide by this factor: N V (, V, N ) 3 N ë Z = N db Let s look at the thermodynamics which follows from this (F free energy, p pressure, σ entropy, U internal energy, C V specific heat capacity at const. volume):

10 Free energy: V " F(, V, N ) = # ln Z(, V, N ) = # $ N ln # ln N 3 % & ëdb ',by using Stirling formula: V " õ " = # $ N ln # N ln N + N N ln 1 3 % = # $ + 3 % & ëdb ' & ëdb ' = V N Free energy is proportional to the total number of particles. Pressure: ln N N ln N " N is the specific volume per particle. we get: " F # 1 p = $ & ' = N % pv = NkBT ( V ) V, N So, from deriving the pressure, we get the well known ideal gas equation.

11 # F " õ " # õ " ó = $ % & = $ $ N % ln + 1 N ln 3 & $ 3 ' # ( % V, N ëdb # ë & ' ' ( db ( 3 õ " ë db õ " 1 ëdb " = N % ln + 1 N 3 3 & + % $ $ 4 ëdb õ ë &% db & ' ( ' (' ( õ 5 " = N % ln + 3 & ' ëdb ( This expression for entropy of ideal gas is known as Sackur Tetrode equation. # ln Z # V " % ln ln & # # ' ( U = = N $ N 3 ëdb ë V " 1 ë " 3 = $ $ ) = ' (' ( 3 db db N % 3 U N 4 V ë &% db &

12 This is in accordance with the law of equipartitioning of energy through the degrees of freedom. In case of monoatomic ideal gas, there are 3 motional degrees of freedom (for 3 independent directions in space). Each of them, in thermal equilibrium, has an average energy of (1/)τ and contributes (1/) to system s heat capacity. " ó # ( 3 C ln ) V = % & = $ N ëdb = ' ( V 1 ëdb 3 = $ 3N ) CV = $ N ë db Therefore each particle gives a contribution of 3/ to the heat capacity (in dimensionless case, otherwise it would be (3/)k B )

13 Rotational Levels for Diatomic Molecules By solving the eigen value problem of angular momentum, we get that permitted rotational levels of energies are: å l h l( l + 1) = = = + = I ì m m, I ì d,, l 0,1,,... 1 Where I is moment of inertia, µ is reduced mass, l is orbital quantum number. Each energy level is (l+1) degenerate. The partition function is then given as: Z l= 0 ( l ) = # + " ( + 1) 1 e h l l I

14 h For 1 (low temperature) only the first two elements are important. I Z e h I = F = ln Z = 3 e h I The rotational states with l>0 are frozen out, when temperature reaches zero. The Classical Partition Function Reminder: Quantum partition function is given by: Z H ( s) " e " = = s å g( å) e å

15 Where: s is eigenstate (eigenvector) of Hamiltonian, g(ε) is degeneracy of eigenvalue, ε is eigenvalue of Hamiltonian. This can be formally written (with diagonalized Hamiltonian): Z = ˆ Tre H But, in the classical case, we do not sum over the eigenstates of the Hamiltonian. We integrate over the degrees of freedom in phase space. Hence, if the Hamiltonian is given by: p H = + V r r r r N i r r r 1 N i= 1 mi (,,..., ) Where first part describes system s kinetic energy and second part should specify all interactions within the system. The partition function is then given by:

16 ({ i}{, i} ) r r H p d 1... d N d 1... d Ne N Z p p r r Where we have taken into account the dividing factor, because in classical treatment particles are indistinguishable. But now, Z is not dimensionless and we cannot take the logarithm in order to get the free energy. That s why we introduce a constant A of dimension momentum times length = energy times time = action so that Z becomes dimensionless. Z d p d r """ d p d r = # N A e H N N 3N

17 Now, what would be the choice for A? One answer might be that it doesn t matter so much, because a different will show up as a different prefactor is Z and thus as an additive constant in the free energy. This does not influence thermodynamics. But if we want to get the absolute number right, we can compare the classical partition function of the ideal gas with the quantum partition function. So let s take a look at the ideal gas in a box of volume V=L. r pi H = * m i 3N r r + p1 + p1 L L " " 3 m 3 m 3 3 d 1e d Ne d 1 d " " Z = p ### p r ### r 3N N A r " 3 m p1e $ + p1 % N V = d 3N & N A +++ ' &( " ') = N V N A ( m) 3N N,again using solution of Gaussian integral: N

18 If we define Λ with the dimension of length: A 1 V " N ' # ( # = $ Z = % 3 & m N Which is exactly the same form as we had for the quantum mechanical case. Hence if we equate the two partition functions we get: Zclassical = Zqm " = ëdb A = h The unknown constant of dimension action turns out to be Planck s action quantum number h. So the classical partition function is: Z = # d p d r """ d p d r e H N N 3N N h

19 Some special topics In addition we will cover three important things: 1. Generalized equipartition theorem. Additive Hamiltonian 3. Partition function in generalized coordinates

20 Let X i be either p i or r i : Generalized Equipartition Theorem H H " - # xi = $ ) d p...dr xi % e & x j Z ' x j ( = Z ij ) d p...dr x x = ä ) ij d p...d r Z = ä i j e e H - H - Where Kronecker delta is defined as: ij = { 1 0,if i = j,if i " j

21 Which makes one half k b T for every quadratic degree of freedom. So for the ideal gas: 3 1 U = NkBT = 3N kbt Special case: For every quadratic degree of freedom X i in the partition function, with an energy contribution E=AX i we have: 1 E 1 1 E = Axi = xi = = kbt x i

22 Additive Hamiltonians If the Hamiltonian of the system is a sum of independent terms, the partition function is a product of independent terms, and thus the free energy is again a sum of independent terms. (We used this property in ideal gas case). Additional example: particles with translational, rotational and vibrational degrees of freedom: H = H trans + H rot + H vib Z = = = d... e - ( H + H + H ) trans rot vib e e e d d d -Htrans -Hrot -Hvib trans rot vib Z Z Z trans rot vib F = Ftrans + Frot + Fvib

23 Partition Function in Generalized Coordinates In this case the integrals will contain the Jacobian for the transformation in generalized coordinates. We re not going to look into this very much. Let s just make one example. Dipole in electric field D = r q r +q θ φ The energy is: given by:,where sin rr å å H = D = D cosè -q The rotational partition function is then D cosè Z dö dè sinè e = 0 0 comes from Jacobian for spherical coordinates. å "

24 By substituting: Z = cosè we get: 1-1 Dåz Då " d d # Då Z = ö z = = sinh D D ( ( e e å $ % & ' å Definitions of hyperbolic functions: sinh x e e e + e =, cosh x = x x x x Via parameter differentiation we can now work out the average component of D in the direction of field. D " cosh % ln Z # P D cosè D $ = = = # & å Då $ % # sinh Då$ ' ( å D " D " = D# coth & $ ) P = DL # $ ' Då( ' ( å å

25 Where we have defined Langevin function as: 1 1 x = x x " x x 3 ( ) coth, ( ) LL,for small value of argument. Therefore: P D å 3,where D 3 is susceptibility. Plot of Langevin function: