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I want to derive an expression for the internal energy of mixing, $\Delta_\mathrm{mix}U$, and entropy of mixing, $\Delta_\mathrm{mix}S$. The framework for this is the Lattice Theory of Ideal Solutions. Specifically, I am mixing $N_A$ molecules of type A with $N_B$ molecules of type B together. I am supposed to find the following two relations:

$$ \Delta_\mathrm{mix} U = 0 \\ \Delta_\mathrm{mix} S = -kM (x_A \ln x_A + x_B\ln x_B) $$


My attempted solution

I have the canonical partition function for an ideal solution of two different components:

$$ Q = \left( q_A e^{-cw_{AA}/2kT} \right)^{N_A} \left( q_B e^{-cw_{BB}/2kT} \right)^{N_B} \frac{(M)!}{N_A!N_B!} $$

Taking the logarithm

$$ \ln Q = N_A\ln q_A - N_A\frac{cw_{AA}}{2kT} + N_B\ln q_B - N_B\frac{cw_{BB}}{2kT} + M \ln M - N_A\ln N_A - N_B \ln N_B $$

where $w_{xx}$ is the pairwise interaction energy, $M=N_A+N_B$ is the total number of molecules = number of lattice sites.

Further, some thermodynamical relations for this system

$$ U = kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N,M} $$

$$ S = \frac{U}{T} + k\ln Q $$

The internal energy of mixing and entropy of mixing should be

$$ \Delta_\mathrm{mix}U = \Delta U_A + \Delta U_B \\ \Delta_\mathrm{mix}S = \Delta S_A + \Delta S_B $$

Generally we have that

$$ \Delta U = U_\mathrm{final} - U_\mathrm{initial} \\ \Delta S = S_\mathrm{final} - S_\mathrm{initial} $$

I am a bit confused about how to compare the final and initial conditions of the system. I assume that we have initially two separated and equal volumes, one containing $N_A$ of A and one containing $N_B$ of B. Then we allow flow between the two volumes. Hence, the final volume is twice the initial volume. $N_A$ and $N_B$ are constant during the mixing. So I should have something like this

$$ \Delta U_A = U_{final} - U_{initial} = kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N_A,M} - kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N_A,M/2} $$

and similarly for the other component. I just replace $M$ with $M/2$ for the expression of the initial state for both components? Most of the terms will be zero when doing the derivations, and the non-zero terms will actually just cancel. This gives that $\Delta_\mathrm{mix} U = 0$, which is the expected result for an ideal solution.

Looking at $\Delta_\mathrm{mix} S$ we have

$$ \Delta_\mathrm{mix} S = \Delta S_A + \Delta S_B = [S_{f,A} - S_{i,A}] + [S_{f,B} - S_{i,B}] \\ = \left[ \left(\frac{U_{f,A}}{T} + k\ln Q_f \right)- \left( \frac{U_{i,A}}{T} + k\ln Q_i \right) \right] + \left[ \left( \frac{U_{f,B}}{T} + k\ln Q_f \right) - \left( \frac{U_{i,B}}{T} + k\ln Q_i \right) \right] \\ =\frac{1}{T} (U_{f,A} - U_{i,A} + U_{f,B} - U_{i,B}) + 2k(\ln Q_f - \ln Q_i) \\ = \frac{\Delta_\mathrm{mix}U}{T} + 2k(\ln Q_f - \ln Q_i) \\ = 2k(\ln Q_f - \ln Q_i) $$

where I used in the last step that $\Delta_\mathrm{mix} U=0$ for an ideal solution, as explained above. However, using this last expression to calculate the entropy of mixing, I find that (most of the terms cancel directly)

$$ \Delta_\mathrm{mix}S = kM \ln (2N_A + 2N_B) $$

which is quite different from the correct expression. Is there something wrong with my procedure? (there likely is!)


Following the commented suggestion:

$$ S = k\ln W \Rightarrow \Delta_\mathrm{mix}S = k\ln \frac{W_\mathrm{mix}}{W_AW_B} = k\ln W_\mathrm{mix} = k\ln \frac{M!}{N_A!N_B!} \\ = kM [\ln M - x_A\ln N_A - x_B \ln N_B] $$

which is almost correct. If I could divide all logarithmic arguments by $M$, I would get the correct answer.

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  • $\begingroup$ If the energy of mixing is zero which you imply in your first equation then you can reach your answer by using the number of distinguishable arrangements of $N_1+N_2$ molecules which is $W_{mix}=(N_1+N_2)!/(N_1!N_2!)$ and $\Delta S=k\ln(W_{mix}/(W_1W_2))$ where $W_1,W_2$ are the numbers for pure components and =1. If its not zero ($w \ne 0$ ) its a really tough problem in regular but non-perfect solution stat mech: :( $\endgroup$ – porphyrin Oct 20 '16 at 13:51
  • $\begingroup$ Okay, so I use that $S=k\ln W$ where $W$ is the number of distinguishable ways we can arrange $N_A$ and $N_B$ particles over $M=N_A+N_B$ sites. See edit. Why is $W_A = $W_B =1$? Ah, it is because all possible arrangements of the pure components are indistinguishable, and so only one possible distinguishable arrangement exists? Is there a formula for these combinatorics, or did you use common sense and logic? $\endgroup$ – Yoda Oct 21 '16 at 12:40
  • $\begingroup$ From $W_{mix}$ let $N_1$ or $N_2 = 0$ then it follows that $W_1=1$ etc. and as you say only one arrangement. You need to let $M=N_1 + N_2$ and use the Stirling approx $\ln(N!)=N\ln(N)-N$ etc. as the numbers are large. The $W_{mix} $ is a permutational expression for the number of arrangements of $N_1$ and $N_2$ objects $\endgroup$ – porphyrin Oct 21 '16 at 13:43
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    $\begingroup$ I do not get the correct expression. I did use Stirling's approximation. $\endgroup$ – Yoda Oct 24 '16 at 8:10
  • $\begingroup$ IMO, porphyrin's answer is pretty much all there is to be said - I think you are complicating it by bringing in partition functions. Within the lattice model and using $S = k\ln W$ you don't need to bring in anything else. Furthermore for an ideal solution $\Delta U$ (and $\Delta H$ for that matter) is zero by definition (all pairwise interactions are the same, so it doesn't matter how you arrange the particles, the total interaction is constant) although porphyrin went the extra mile to give you the equation for a regular solution. $\endgroup$ – orthocresol Oct 24 '16 at 11:13
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In the solution there are two types of molecules $N_1$ and $N_2$. Assume that they do not interact with one another but simply occupy particular 'lattice' sites by blocking them. The total number occupied sites is $N=N_1 + N_2$. The first molecule can be placed at any of the $N$ sites, the second at $N-1$ empty sites and so on. The total number of possibilities is $N(N-1)(N-2)\cdots = (N_1 + N_2)!$ . Now, molecules of the same type are not distinguishable from one another thus this number must be divided by $N_1$ for molecules of the first type and similarly for the other type. Thus the number of distinguishable arrangements is

$$W_\mathrm{mix}= \frac{(N_1 + N_2)!}{N_1!N_2!} $$

The pure compounds have only one type of molecule thus for them $W_1=N_1!/N_1! = 1$ etc.

The entropy is defined as $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements $W_\mathrm{mix}/(W_1W_2)$ thus $$ \Delta S=k\ln\left(\frac{(N_1 + N_2)!}{N_1!N_2!}\right)$$

To simplify this it is convenient to use Stirling's approximation as $N$ is large:

$$\ln(N!) \approx N\ln(N)-N+1$$ and also as $N$ is large the $1$ can be ignored, giving $$\ln(N!) \approx N\ln(N)-N$$

Expanding gives $$ \Delta S = k[(N_1 + N_2)\ln(N_1 + N_2)-N_1\ln(N_1)-N_2\ln(N_2)]$$ expanding, multiplying by $-1$ and simplify gives

$$ \Delta S = -k\left[N_1\ln (\frac{N_1}{N_1 + N_2} )+ N_2\ln(\frac{N_1}{N_1 + N_2})\right] $$

which can be expressed as mole fractions $$ \Delta S= -k[N_1\ln(x_1) + N_2\ln(x_2)]$$

in molar terms, $ \Delta S= -R[n_1\ln(x_1) + n_2\ln(x_2)]$. This is the ideal solution entropy of mixing.

If the molecules do interact then this is a complicated problem in regular solutions, the free energy is $$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$

where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.

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