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Unformatted text preview: ChE 101 2012 Problem Set 6 Solutions 1. (a) In order for the system to be in mechanical equilibrium, the pressure must be constant across the two phases. Next, in order to achieve thermal equilibrium, the temperatures of the two phases must be equal. Finally, the condition of phase equilibrium necessitates that the chemical potential is also equal across phases, since the number of adsorbed molecules will be adjusted until the chemical potentials of both phases precisely match. Thus, we conclude that the three intensive properties conserved in this system are T , p , and μ . (b) For M adsorbed molecules, the total energy is given simply by E ( M ) = M (assuming that is positive). The degeneracy is just the number of ways to choose M bound sites from the N possible adsorption sites, so Γ( M ) = N M = N ! M !( N M )! (c) All microstates with M adsorbed molecules are defined by N i = M and E i = M . For each value of M , there are precisely Γ( M ) different microstates with these values of N i and E i . Thus, based on our results from (b), the Gibbs sum is given by Z = N X M =0 N ! M !( N M )! e βM ( μ + ) (d) We are given the probability distribution of microstates in the system. The average number of particles is obtained by taking an expectation value with respect to this distribution, h M i = X { s i } N i P ( s i ) = 1 Z X { s i } N i e β ( N i μ E i ) Now we notice that the summand can be obtained by differentiation of the terms in Z : N i e β ( N i μ E i ) = 1 β ∂ ∂μ h e β ( N i μ E i ) i Substituting into our expression for h M i , we have h M i = 1 Z X { s i } 1 β ∂ ∂μ h e β ( N i μ E i ) i = 1 β Z ∂ ∂μ X { s i } e β ( N i μ E i ) , 1 with the latter equality obtained by taking the derivative out of the sum. Notice that the bracketed term is precisely the Gibbs sum Z , so we find that h M i = 1 β 1 Z ∂ Z ∂μ Using the chain rule, this can be expressed more elegantly as h M i = 1 β ∂ ∂μ (ln Z ) (e) We can compute Z exactly as a binomial sum, Z = N X M =0 N M h e β ( μ + ) i M · 1 N M Using the binomial theorem, this sum evaluates to Z = h 1 + e β ( μ + ) i N From part (d), we know that h M i = 1 β ∂ ∂μ (ln Z ) In our case, we have ln Z = N ln h 1 + e β ( μ + ) i , so h M i = N β · β e β ( μ + ) 1 + e β ( μ + ) = N 1 + e β ( μ + ) (f) The de Broglie wavelength of a particle with momentum p is given by λ = h/p . We know that the typical thermal energy of a particle is on the order of k B T , and that the energy and momentum of a free particle are related by E = p 2 / 2 m . Thus we have p = √ 2 mE ∼ p mk B T, so λ ∼ h √ mk B T = h 2 mk B T 1 / 2 This precisely matches the form of the given expression for λ ( T ) (neglecting the numer ical prefactor of order unity), so λ ( T ) is indeed a measure of the de Broglie wavelength of a particle in thermal equilibrium....
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This note was uploaded on 03/21/2012 for the course CHE 101 taught by Professor Arnold during the Winter '11 term at Caltech.
 Winter '11
 ARNOLD
 Equilibrium, pH

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