Question

Using the same method as in the text, calculate the entropy of mixing for a system of two monatomic ideal gases, $A$and $B$, whose relative proportion is arbitrary. Let $N$be the total number of molecules and let$x$ be the fraction of these that are of species$B$ . You should find

$\mathrm{\Delta }{S}_{\text{mixing}}=-Nk[x\ln x+(1-x)\ln (1-x\left)\right]$

Check that this expression reduces to the one given in the text when$x=1/2$ .

Short answer

The entropy mixing system of two monatomic ideal gases is, $\mathrm{\Delta }{S}_{\text{mixing}}=-Nk\left(\right(1-x)\ln (1-x)+x\ln (x\left)\right)$

Step-by-step solution

Step: 1 Equating total volume:

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

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

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.

$N_B=xN{N}_A=(1-x)N$

the total volume fractions is

$V_B=xV{V}_A=(1-x)V$

Step: 2 Volume changes:

From the above equation,

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

The change in entropy process where the volume changes is

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

The volume change for two gases is

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

Step: 3 Finding entropy mixing:

The volume of gas part $A$starts from $V_A=(1-x)V$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 $V.$

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

After mixing the total entropy change is entropy of mixing by

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

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

$\mathrm{\Delta }{S}_{\text{mixing}}=Nk\ln 2$

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