110c-lecture6 - qrev is Path Independent and a State...

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Path A: isothermal ( T = 0) Path B: adiabatic (q B = 0) Path C: const. V ( δ w= ±0) Path D: const. P Path E: const. V δ q rev T is Path Independent and a State Function Path Condition S T q rev = δ A T=const = = + = = 1 2 1 1 ln 0 V V nR S dV V nRT dS T w q dU A B q=0 0 = = T q S B B C V=const * ln ln 1 2 2 1 1 2 1 2 V V nR T T C T dT C dS S TdS q w q dU dT C V T T V T T C V = = = = = = + = = Path A connects Path (B+C) and S A = S B + S C
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Statistical Thermodynamics (pp 829-838) Boltzmann Equation of Entropy (S = k B lnW). W = number of microstates accessible to system (W V N , ideal gas) k B = Boltzmann constant = 1.38 x 10 -23 J K -1 = R/N A S total = S 1 + S 2 = k B (ln W 1 + ln W 2 ) where W total = W 1 *W 2 Boltzmann Entropy for System with 2-fold degeneracy. W = 2 N 2 ln 2 ln B N B system Nk k S = = N particles distributed between two states with equal probability (e.g. N spin-1/2 nuclei or CO in crystal lattice) Entropy of Mixing. {} { } 2 2 1 1 ln ln y R y y R y S mix = 2 1 1 1 n n n y + = 2 1 2 2 n n n y + =
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Boltzmann’s Entropy Applied to Particle-in-a-Box 3D particle in a box, p694: () 2 2 2 2 2 8 , , z y x z y x n n n ma h n n n + + = ε where a=V 1/3 for a cubic box and n x , n y and n z define integral quantum energy levels (1, 2, 3, …). 2 2 2 2 2 8 , , z y x z y x n n n ma h n n n + + = a = side of cube m =mass of particle h =Planck's const E j (N,V) = ε 1 + ε 2 + ε 3 + … + ε N W k S B ln = = = j j a A a a a A a a a W ! ! !... ! ! ! ,...) , , ( 3 2 1 3 2 1 = j j A a 4 ! 3 ! 1 ! 4 ) 3 , 1 ( = = W W V N See problems 18-42 and 20-23 a:bcd, b:acd, c:abd, d:abc
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Number of Microstates (W) for a Simplified System A microstate of N particles is a point in 6N dimensional “phase space” (3 position and 3 momentum coordinates per particle). W ('number of states') is proportional to the region of phase space occupied by the N gas molecules. N particles occupy volume “V” in phase space. Divide V into small bins with volume “v” so there are V/v bins where an individual molecule can be located. If particles are constrained to occupy only half the box (volume = V/2), then each particle can be placed in any one of V/2v bins on one side of the box. The total number of ways of placing N particles in V/2v bins is W = (V/2v) N . S(V/2) = k B ln(V/2v) N = N k B ln(V/2v) S = N k B ln(V/v) – N k B ln(V/2v) What is the change in entropy ( S) upon opening the door to the other side? The entropy will then increase from S(V/2) to a final value S(V) = N k B ln (V/v). S = S(V) – S(V/2) = N k B ln(V/v) – (N k B ln(V/v) - N k B ln 2) S = N k B ln2 W=(V/v) N Nk B (ln(V/v) – ln2)
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Entropy Calculation for Systems with 2-fold Degeneracy For N spins (I=1/2), each spin is oriented in one of two directions (m = ± 1/2) with equal probability.
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This note was uploaded on 06/22/2010 for the course CHEM 21360 taught by Professor Ame during the Spring '09 term at East Los Angeles College.

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110c-lecture6 - qrev is Path Independent and a State...

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