A Statistical View of Entropy — Question 1

Question 1

Counting arrangements distinguishes a macrostate from a microstate

Six distinguishable, independent particles occupy two equal halves of a box. Each particle is equally likely to be in either half. A microstate specifies every particle location; a macrostate specifies only the number $n$ on the left. Use $S=k_B\ln\Omega$ for a macrostate of multiplicity $\Omega$.

1. Derive the multiplicity and probability of $n=3$, and compare with $n=6$.

2. Find the entropy difference between these macrostates. Explain why a particular balanced microstate is no more probable than a particular all-left microstate.

Question 1 - Solution

Counting arrangements distinguishes a macrostate from a microstate

(a) Method and calculation

$$\Omega(n)=\binom6n=\frac{6!}{n!(6-n)!},\quad\Omega_{\mathrm{total}}=2^6=64.$$$$\Omega(3)=20,\quad P(3)=20/64=.3125,\quad\Omega(6)=1,\quad P(6)=1/64.$$

(b) Interpretation and further calculation

$$S(3)-S(6)=k_B\ln20\approx2.99573k_B.$$$$P(\text{specified microstate})=2^{-6}.$$The larger macrostate probability comes from the number of compatible arrangements, not a larger probability for each arrangement.

Physical check

The pictured balanced arrangement is one of twenty; it does not represent all twenty at once.