# Statistical mechanics: multiplicity

by SoggyBottoms
Tags: mechanics, multiplicity, statistical
 P: 61 1. The problem statement, all variables and given/known data We have a surface that can adsorb identical atoms. There are N possible adsorption positions on this surface and only 1 atom can adsorb on each of those. An adsorbed atom is bound to the surface with negative energy $-\epsilon$ (so $\epsilon > 0$). The adsorption positions are far enough away to not influence each other. a) Give the multiplicity of this system for $n$ adsorbed atoms, with $0 \leq n \leq N$. b) Calculate the entropy of the macrostate of n adsorbed atoms. Simplify this expression by assuming N >> 1 and n >> 1. c) If the temperature of the system is T, calculate the average number of adsorbed atoms. 3. The attempt at a solution a) $\Omega(n) = \frac{N!}{n! (N - n)!}$ b) $S = k_b \ln \Omega(n) = k_b \ln \left(\frac{N!}{n! (N - n)!}\right)$ Using Stirling's approximation: $S \approx k_B ( N \ln N - N - n \ln n - n - (N - n) \ln (N - n) - (N - n) = k_B ( N \ln N - n \ln n - (N - n) \ln (N - n)$ A Taylor expansion around n = 0 then gives: $S \approx k_B (- \frac{n^2}{2N} + ...)\approx -\frac{k_b n^2}{2N}$ c) I'm not even sure if the previous stuff is correct, but I have no idea how to do this one. Any hints?
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P: 1,391
 Quote by SoggyBottoms 1. The problem statement, all variables and given/known data We have a surface that can adsorb identical atoms. There are N possible adsorption positions on this surface and only 1 atom can adsorb on each of those. An adsorbed atom is bound to the surface with negative energy $-\epsilon$ (so $\epsilon > 0$). The adsorption positions are far enough away to not influence each other. a) Give the multiplicity of this system for $n$ adsorbed atoms, with $0 \leq n \leq N$. b) Calculate the entropy of the macrostate of n adsorbed atoms. Simplify this expression by assuming N >> 1 and n >> 1. c) If the temperature of the system is T, calculate the average number of adsorbed atoms. 3. The attempt at a solution a) $\Omega(n) = \frac{N!}{n! (N - n)!}$ b) $S = k_b \ln \Omega(n) = k_b \ln \left(\frac{N!}{n! (N - n)!}\right)$ Using Stirling's approximation: $S \approx k_B ( N \ln N - N - n \ln n - n - (N - n) \ln (N - n) - (N - n) = k_B ( N \ln N - n \ln n - (N - n) \ln (N - n)$ A Taylor expansion around n = 0 then gives: $S \approx k_B (- \frac{n^2}{2N} + ...)\approx -\frac{k_b n^2}{2N}$ c) I'm not even sure if the previous stuff is correct, but I have no idea how to do this one. Any hints?
Part (a) looks good, but you made a couple of sign mistakes in part (b). In the terms in the denominator you forgot to distribute the negative sign to the second term in n ln n - n and (N-n) ln(N-n). I've corrected the signs here:

$$S \approx k_B ( N \ln N - N - n \ln n + n - (N - n) \ln (N - n) + (N - n) = k_B ( N \ln N - n \ln n - (N - n) \ln (N - n)$$

This change will result in some cancellations that help simplify your expression. Rewriting

$$(N-n)\ln(N-n) = \left(1-\frac{n}{N}\right)N\ln N + (N-n)\ln\left(1-\frac{n}{N}\right)$$