Question

Consider an ideal mixture of just 100 molecules, varying in com- position from pure A to pure B. Use a computer to calculate the mixing entropy as a function of N_A, and plot this function (in units of k). Suppose you start with all A and then convert one molecule to type B; by how much does the entropy increase? By how much does the entropy increase when you convert a second molecule, and then a third, from A to B? Discuss.

Short answer

Therefore, the increase in entropy is

$$\begin{gathered} {\left.\frac{\Delta S_{mix}}{k}\right|}_1=5.60 \\ {\left.\frac{\Delta S_{mix}}{k}\right|}_2=4.20 \\ {\left.\frac{\Delta S_{mix}}{k}\right|}_3=3.67 \end{gathered}$$

Step-by-step solution

Given information

Consider an ideal mixture of just 100 molecules, varying in com- position from pure A to pure B. Use a computer to calculate the mixing entropy as a function of N_A, and plot this function (in units of k). Suppose you start with all A and then convert one molecule to type B.

Concept

When two ideal gases are allowed to mix at the same pressure and temperature and the total volume remains unaltered, the change in entropy is:

$\Delta S_{mix}=-Nk\left(x\ln \right(x)+(1-x\left)\ln \right(1-x\left)\right)$

Explanation

Considerasystemof100moleculesinA.WhenweconvertthemoleculesfromAtoB,wemustdeterminetheentropy.

Tobegin,wemustunderstandthatxistheratioofA'smoleculestothetotalnumberofmolecules,andthat1-xisthenumberofmoleculesinB.

For example, if the first molecule needs to be moved, the number of molecules in A is 99, and the value of x is:

$x=\frac{N_A}{N}=\frac{99}{100}=0.99$

The entropy of mixing is

role="math" localid="1646983560291" $$\begin{gathered} {\left.\frac{\Delta S_{mix}}{k}\right|}_{x=0.99}=-\left(100\right)(0.99\ln (0.99)+(1-0.99\left)\ln \right(1-0.99\left)\right) \\ =5.60 \end{gathered}$$

However, because the initial entropy is zero, the rise in entropy is:

${\left.\frac{\Delta S_{mix}}{k}\right|}_1=5.60$

Explanation

If the second molecule must be moved, the number of molecules in A is 98, and the value of x is:

$x=\frac{N_A}{N}=\frac{98}{100}=0.98$

The entropy of mixing is

$$\begin{gathered} {\left.\frac{\Delta S_{mix}}{k}\right|}_{x=0.98}=-\left(100\right)(0.98\ln (0.98)+(1-0.98\left)\ln \right(1-0.98\left)\right) \\ =9.80 \end{gathered}$$

The increase is:

$$\begin{gathered} {\left.\frac{\Delta S_{mix}}{k}\right|}_2={\left.\frac{\Delta S_{mix}}{k}\right|}_{x=0.98}-{\left.\frac{\Delta S_{mix}}{k}\right|}_1=4.20 \\ {\left.\frac{\Delta S_{mix}}{k}\right|}_2=4.20 \end{gathered}$$

Explanation

If the third molecule must be moved, the number of molecules in A is 97, and the value of x is:

$x=\frac{N_A}{N}=\frac{97}{100}=0.97$

The entropy of mixing is:

$$\begin{gathered} {\left.\frac{\Delta S_{\text{mix}}}{k}\right|}_{x=0.97}=-\left(100\right)(0.97\ln (0.97)+(1-0.97\left)\ln \right(1-0.97\left)\right) \\ =13.47 \end{gathered}$$

The increase is

$$\begin{gathered} {\left.\frac{\Delta S_{mix}}{k}\right|}_3={\left.\frac{\Delta S_{mix}}{k}\right|}_{x=0.97}-{\left.\frac{\Delta S_{mix}}{k}\right|}_2=3.67 \\ {\left.\frac{\Delta S_{mix}}{k}\right|}_3=3.67 \end{gathered}$$

We can see that when we add more molecules to B from A, the change in entropy diminishes. This is because we start with a pure mixture in A, so mixing one molecule will create a big increase in entropy, but when we add the second molecule, we are adding it to an already mixed system.

To construct a graph between $N_A\text{and}\Delta S/k$, we must first calculate $\Delta S_{mix}/k$ for each N_A. This data is created using Python, and the code is shown below:

The graph is shown below: