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## 12.3 Assembly of distinguishable particles

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**12.3 Assembly of distinguishable particles**• An isolated system consists of N distinguishable particles. • The macrostate of the system is defined by (N, V, U). • Particles interact sufficiently, despite very weakly, so that the system is in thermal equilibrium.**Two restrictive conditions apply here**(conservation of particles) (conservation of energy) where Njis the number of particles on the energy level j with the energy Ej. Example: Three distinguishable particles labeled A, B, and C, are distributed among four energy levels, 0, E, 2E, and 3E. The total energy is 3E. Calculate the possible microstates and macrostates.**Solution: The number of particles and their total energy**must satisfy (here the index j starts from 0)**So far, there are only THREE macrostates satisfying the**conditions provided. Detailed configurations (i.e. microstate) for case 1 Thus, thermodynamic probability for case 1 is 3**Microstates for case 2**Microstates for case 3 Therefore, W1 = 3, W2 = 6, W3 = 1, and Ω = 10**The most “disordered” macrostate is the state with the**highest probability. • The macrostate with the highest thermodynamic probability will be the observed equilibrium state of the system. • The statistical model suggests that systems tend to change spontaneously from states with low thermodynamic probability to states with high thermodynamic probability. • The second law of thermodynamics is a consequence of the theory of probability: the world changes the way it does because it seeks a state of higher probability.**12.4 Thermodynamic Probability and Entropy**Boltzman made the connection between the classical concept of entropy and the thermodynamic probability S = f (w) f (w) is a single-valued, monotonically increasing function (because S increases monotonically) For a system which consists of two subsystems A and B Stotal = SA + SB (S is extensive) Or… f (Wtotal) = f (WA) + f (WB)**Configurations of the total system are calculated as Wtotal**= WA x WB thus: f (WA x WB) = f (WA) + f (WB) The only function for which the above relationship is true is the logarithm. Therefore: S = k · lnW where k is the Boltzman constant with the units of entropy.**12.5 Quantum States and Energy Levels**To each energy level, there is one or more quantum states described by a wave function Ф. When there are several quantum states that have the same energy, the states are said to be degenerate. The quantum state associated with the lowest energy level is called the ground state of the system. Those that correspond to higher energies are called excited states.**The energy levels can be thought of as a set of shelves of**different heights, while the quantum states correspond to a set of boxes on each shelf • E3 g3 = 5 • E2 g2 = 3 • E1 g1 = 1 • For each energy level the number of quantum states is given by the degeneracy gj**a particle in a one-dimensional box with infinitely high**walls • A particle of mass m • The time-independent part of the wave function Ф (x) is a measure of the probability of finding the particle at a position x in the 1-D box. • Ф (x) = A*sin(kx) 0 ≤ x ≤ 1 with k = n (π/L), n = 1, 2, 3 n=1 n=2**The momentum of the particle is**P = · k (de Broglie relationship) where k is the wave number and h is the Planck constant. The kinetic energy is:**Plug-in**Therefore, the energy E is proportional to the square of the quantum number, n.**In a three-dimensional box**E = h2 ( + + ) 8m where any particular quantum state is designated by three quantum numbers x, y, and z. If Lx=Ly=Lz=L then, we say nj2 = nx2 + ny2 + nz2 Where is the total quantum number for states whose energy level is EJ.**An important result is that energy levels depends**only on the values of nj2 and not on the individual values of integers (nx, ny, nz). (nj is the total quantum number for energy level Ej) Since L3 = V L2 = V⅔ therefore, For the ground state, nx = 1, ny = 1, nz = 1 There is only one set of quantum number leads to this energy, therefore the ground state is non-degenerate.**For the excited state, such as nx2 + ny2 + nz2 = 6**it could be nx = 1 ny = 1 nz = 2 or nx = 1 ny = 2 nz = 1 or nx = 2 ny = 1 nz = 1 They all lead to Note that increasing the volume would reduce the energy difference between adjacent energy levels! For a one liter volume of helium gas nJ ≈ 2 x 109**12.6. Density of Quantum States**When the quantum numbers are large and the energy levels are very close together, we can treat n’s and E’s as forming a continuous function. For nx2 + ny2 + nz2 = · V⅔ ≡ R2 The possible values of the nx, ny, nz correspond to points in a cubic lattice in (nx, ny, nz) space**Let g(E) · dE = n (E + dE) – n(E) ≈**Within the octant of the sphere of Radius R; n(E) = ⅛ · πR³ = V( ) · E Therefore,**Note that the above discussion takes into account**translational motion only. Where γs is the spin factor, equals 1 for zero spin bosons and equals 2 for spin one-half fermions**Example (12.8) Two distinguishable particles are to be**distributed among nondegenerate energy levels, 0, ε, 2ε , such that the total energy U = 2 ε. • What is the entropy of the assembly? • If a distinguishable particle with ZERO energy is added to the system, show that the entropy of the assembly is increased by a factor of 1.63.