91影视

The Sackur-Tetrode formula by $3-d$ideal gas,

$S=Nk\left[\ln 鈦�\left(\frac{V}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]$

Where,$V$represents volume, $U$represents energy, $N$represents the number of molecules, $m$represents the mass of a single molecule, and $h$represents Planck's constant.

$N_B=xN{N}_A=(1鈭�x)N$

the total volume fractions is

$V_B=xV{V}_A=(1鈭�x)V$

From the above equation,

$\begin{array}{l}S=Nk\left[\ln 鈦�\left(\frac{V}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\\ S=Nk\left[\ln 鈦�\left(V\right)+\ln 鈦�\left(\frac{1}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\\ \ln 鈦�\left(\frac{1}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)\end{array}$

The change in entropy process where the volume changes is

$\begin{array}{l}\mathrm{螖}S=S_f鈭�S_i=Nk\left(\ln 鈦�\left(V_f\right)鈭�\ln 鈦�\left(V_i\right)\right)\\ \mathrm{螖}S=Nk\ln 鈦�\left[\frac{V_f}{V_i}\right]\end{array}$

The volume change for two gases is

$\begin{array}{r}\mathrm{螖}{S}_A=N_{A}k\ln 鈦�\left[\frac{V_{A,f}}{V_{A,i}}\right]\\ \mathrm{螖}{S}_B=N_{B}k\ln 鈦�\left[\frac{V_{B,f}}{V_{B,i}}\right]\end{array}$

The volume of gas part $A$starts from $V_A=(1鈭�x)V$and ends with total volume container role="math" localid="1650278031887" $V.$

The part $B$starts from $V_B=xV$and ends with total volume container $V.$

$\begin{array}{l}\mathrm{螖}{S}_A=(1鈭�x)Nk\ln 鈦�\left[\frac{V}{(1鈭�x)V}\right]\\ \mathrm{螖}{S}_B=xNk\ln 鈦�\left[\frac{V}{xV}\right]\\ \mathrm{螖}{S}_A=(1鈭�x)Nk\ln 鈦�\left[\frac{1}{(1鈭�x)}\right]=鈭�(1鈭�x)Nk\ln 鈦�(1鈭�x)\\ \mathrm{螖}{S}_B=xNk\ln 鈦�\left[\frac{1}{x}\right]=鈭�xNk\ln 鈦�\left(x\right)\end{array}$

After mixing the total entropy change is entropy of mixing by

$\begin{array}{l}\mathrm{螖}{S}_{\text{mixing}}=\mathrm{螖}{S}_A+\mathrm{螖}{S}_B=鈭�Nk\left(\right(1鈭�x)\ln 鈦�(1鈭�x)+x\ln 鈦�(x\left)\right)\\ \mathrm{螖}{S}_{\text{mixing}}=鈭�Nk\left(\right(1鈭�x)\ln 鈦�(1鈭�x)+x\ln 鈦�(x\left)\right)\end{array}$

Both logarithms are negative so $0<x<1\text{, so}\mathrm{螖}{S}_{\text{mixing}}>0$.If $x=\frac{1}{2}$so,the two equal quantities of gases starting by

$\mathrm{螖}{S}_{\text{mixing}}=Nk\ln 鈦�2$

Since $N$is the number of molecules of each gas,not total number.