A Statistical View of Entropy — Question 3

Question 3

A volume doubling gives a huge multiplicity ratio without large particle energy changes

An ideal classical gas with $N=10^6$ particles expands freely from volume $V$ into an evacuated equal volume. For the spatial counting model, independent position-state counts are proportional to volume per particle; momentum-state factors cancel because temperature is unchanged. Use $k_B=1.381\times10^{-23}$ J/K.

1. Find the multiplicity ratio and entropy increase without evaluating the enormous multiplicity itself.

2. At equilibrium, find the probability that every particle lies in the original half at a specified instant. Express its base-ten logarithm and explain the macroscopic implication.

Question 3 - Solution

A volume doubling gives a huge multiplicity ratio without large particle energy changes

(a) Method and calculation

$$\frac{\Omega_f}{\Omega_i}=\left(\frac{2V}{V}\right)^N=2^N,\quad\Delta S=Nk_B\ln2.$$$$\Delta S\approx9.57236\times10^{-18}\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$P_{\mathrm{all\ in\ original\ half}}=2^{-N},\quad\log_{10}P=-N\log_{10}2\approx-301029.996.$$$$P=10^{-301029.996}\quad\text{(approximately)}.$$The probability is nonzero but extremely small. A probability at an instant does not by itself specify a recurrence time; that also requires dynamical information.

Physical check

Temperature and internal energy can remain unchanged while the spatial contribution to entropy increases.