110c-lecture7 - Problem Set#2 Due after class on Wed April 14th Answer key posted afterwards on website Office Hours Cancelled for Wed April 14th

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem Set #2 Due after class on Wed April 14 th . Answer key posted afterwards on website. Office Hours Cancelled for Wed April 14 th . Held on Tue April 13 th , 10:30-11:30 am. Practice Midterm 1 (see website) Midterm 1 Wed April 21 st . Material covered through end of Fri April 16 th . Chapters 19, 20, 21, and 22.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 + = Statistical Entropy (pp 829-838)
Background image of page 2
Entropy of a Large Collection of Molecules A collection of N =10 23 molecules distributed among energy levels as ... , , 2 1 0 n n n , (i.e. n 0 = number of molecules with energy ε 0 etc.) !... ! ! ! 2 1 0 n n n N W = !... ln ! ln ! ln ! ln ln 2 1 0 n n n N W = We can simplify ln N! using Stirling’s approximation ( n i assumed to be large numbers): N N N N ln ! ln ... ) ln ( ) ln ( ) ln ( ) ln ( ln 2 2 2 1 1 1 0 0 0 n n n n n n n n n N N N W Since = i n N , all non-ln terms (N and n i ) cancel to give i i n n N N W ln ln ln So that ) ln ln ( ln = = i i B B n n N N k W k S . N = Σ n i ) ln ln ( i i B n n N N k S Σ = See Stirling’s approximation (Math Chapter J pp 809-815) (Stirling’s approximation) ( ) N N N N = ln ! ln
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Deriving T dq dS rev = from Boltzmann Eq (S = k B lnW) For reversible process at constant volume: S = k B ln W k B ( N ln N - n i ln n i ) () ( ) = = i i B i i B B n n d k n n N N d k W d k dS ln 0 ln ln ln (Note, 0 ) (ln ln ln = = = N N N N N dN N d and N N N ) = = i i i B i i B i i B i i B n dn n k n dn k n d n k n dn k dS ln ln ln = = = i i B i i B i B i i B n dn k n dn k dn k n dn k dS ln 0 ln ln = i i
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 16

110c-lecture7 - Problem Set#2 Due after class on Wed April 14th Answer key posted afterwards on website Office Hours Cancelled for Wed April 14th

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online