91影视

Let's say we start with a set of $N$identical molecules. This system's entropy is a number $S_0$, which may or may not be straightforward to determine. Assume that we suddenly transform $N_A$of these molecules to a new species at some point in the future (which has similar properties to the original species as mentioned).The amount of different ways we may select to locate these $N_A$molecules among the $N$sites accessible will increase the entropy.

The entropy of mixing is:

$\mathrm{螖}{S}_{\text{mixing}}=k\ln 鈦�\left(\begin{array}{c}N\\ N_A\end{array}\right)$

The Coefficient binomial is

$\begin{array}{c}\mathrm{螖}{S}_{\text{mixing}}=k\ln 鈦�\left(\frac{N!}{N_A!\left(N鈭�N_A\right)!}\right)\\ \end{array}$

where,$N_A=(1-x)N$

$\mathrm{螖}{S}_{\text{mixing}}=k\ln 鈦�\left(\frac{N!}{\left(\right(1鈭�x\left)N\right)!\left(xN\right)!}\right)$

By using Stirling's approximation$n!鈮�\sqrt{2蟺n{n}^n}{e}^{鈭�n}$,the factorial is

$\begin{array}{l}N!鈮�\sqrt{2蟺N}{N}^N{e}^{鈭�N}\\ xN!鈮�\sqrt{2蟺xN}(xN{)}^{xN}{e}^{鈭�xN}\\ (1鈭�x)N!鈮�\sqrt{2蟺(1鈭�x)N}\left(\right(1鈭�x)N{)}^{(1鈭�x)N}{e}^{(x鈭�1)N}\end{array}$

Substitutiong values,we get

$\begin{array}{l}\mathrm{螖}{S}_{\text{mixing}}鈮�k\ln 鈦�\left(\frac{\sqrt{2蟺N}{N}^N{e}^{鈭�N}}{\sqrt{2蟺(1鈭�x)N}\left(\right(1鈭�x\left)N{)}^{(1鈭�x)N}{e}^{(x鈭�1)N}\sqrt{2蟺xN}\right(xN{)}^{xN}{e}^{鈭�xN}}\right)\\ \mathrm{螖}{S}_{\text{mixing}}鈮�k\ln 鈦�\left(\frac{\sqrt{2蟺N}{N}^N}{\sqrt{2蟺(1鈭�x)N}\left(\right(1鈭�x\left)N{)}^{(1鈭�x)N}\sqrt{2蟺xN}\right(xN{)}^{xN}}\right)\\ \mathrm{螖}{S}_{\text{mixing}}鈮�k\ln 鈦�\left(\frac{N^N}{\sqrt{2蟺Nx(1鈭�x)}\left(\right(1鈭�x\left)N{)}^{(1鈭�x)N}\right(xN{)}^{xN}}\right)\end{array}$

Where,$\ln 鈦�\left(ab\right)=\ln 鈦�\left(a\right)+\ln 鈦�\left(b\right);\ln 鈦�\left(\frac{a}{b}\right)=\ln 鈦�\left(a\right)鈭�\ln 鈦�\left(b\right)$

we get,

$\mathrm{螖}{S}_{\text{mixing}}鈮�k\left[N\ln 鈦�N鈭�\frac{1}{2}\ln 鈦�\left(2蟺Nx\right(1鈭�x\left)\right)鈭�\left(\right(1鈭�x\left)N\right)\ln 鈦�\left(\right(1鈭�x\left)N\right)鈭�xN\ln 鈦�\left(xN\right)\right]$

Taking third term part

$\begin{array}{l}鈭�\left(\right(1鈭�x\left)N\right)\ln 鈦�\left(\right(1鈭�x\left)N\right)=鈭�\left(\right(1鈭�x\left)N\right)(\ln 鈦�(1鈭�x)+\ln 鈦�(N\left)\right)\\ 鈭�\left(\right(1鈭�x\left)N\right)\ln 鈦�\left(\right(1鈭�x\left)N\right)=鈭�N\ln 鈦�(1鈭�x)+xN\ln 鈦�(1鈭�x)鈭�N\ln 鈦�\left(N\right)+xN\ln 鈦�\left(N\right)\end{array}$

Taking fourth term part

$鈭�xN\ln 鈦�\left(xN\right)=鈭�xN\ln 鈦�x鈭�xN\ln 鈦�N$

The entropy mixing by

$\begin{array}{l}\mathrm{螖}{S}_{\text{mixing}}鈮�k\left[鈭�\frac{1}{2}\ln 鈦�\left(2蟺Nx\right(1鈭�x\left)\right)鈭�(1鈭�x)N\ln 鈦�(1鈭�x)鈭�xN\ln 鈦�x\right]\\ \mathrm{螖}{S}_{\text{mixing}}鈮�鈭�Nk\left[\right(1鈭�x)\ln 鈦�(1鈭�x)+x\ln 鈦�x]\end{array}$

The first component in the second line has been omitted since it is insignificant in comparison to the following two terms for big$N$.