# Entropy of mixing

$$S=k_\mathrm B\ln\mathit\Omega$$ For the mixing of two types of molecules: $$\mathit\Omega=\frac{N!}{N_1!N_2!}$$ Therefore, subbing in and using Stirling’s approximation ($\ln{N!}=N\ln N-N$): $$\Delta S = k_\mathrm B(N\ln N-N_1\ln N_1-N_2\ln N_2)$$ However, I know that the result should be: $$\Delta S =-k_\mathrm B(N_1\ln x_1+N_2\ln x_2)$$ How do I get there? Have I done something wrong?

• You haven't done anything wrong, just bear in mind that $N = N_1 + N_2$ and see whether you can simplify your expression further. – orthocresol Oct 23 '15 at 13:54
• Thank you. However, I only get the number of moles not the mole fraction in the logarithms. i.e i get $\ln \frac {N_i}{N}$ – RobChem Oct 23 '15 at 14:01
• Find an expression for the mole fraction in terms of the symbols you already have (which are basically $N$, $N_1$, and $N_2$), then see whether you can manipulate the equation to get the natural logarithm of that. – orthocresol Oct 23 '15 at 14:04
• I'm sorry, I am really stuck. Would you mind just telling me? Or giving me a large clue? Thank you so much – RobChem Oct 23 '15 at 14:08

The mole fraction $x_1$ is $n_1/n = N_1/N$ (since each mole is exactly $N_\mathrm{A}$ particles). Likewise for $x_2$.

\begin{align} \Delta S &= k_{\mathrm{B}}(N \ln N - N_1 \ln N_1 - N_2 \ln N_2) \\ &= k_{\mathrm{B}}(N_1 \ln N + N_2 \ln N - N_1 \ln N_1 - N_2 \ln N_2) \\ &= k_{\mathrm{B}}[N_1(\ln N - \ln N_1) + N_2(\ln N - \ln N_2)] \\ &= k_{\mathrm{B}}\left[N_1 \ln{\left(\frac{N}{N_1}\right)} + N_2 \ln{\left(\frac{N}{N_2}\right)}\right] \\ &= -k_{\mathrm{B}}\left[N_1 \ln{\left(\frac{N_1}{N}\right)} + N_2 \ln{\left(\frac{N_2}{N}\right)}\right] \\ &= -k_{\mathrm{B}}(N_1 \ln x_1 + N_2 \ln x_2) \end{align}

If you want, you can go further by noting that:

\begin{align} n_1 &= \frac{N_1}{N_\mathrm{A}} & n_2 &= \frac{N_2}{N_\mathrm{A}} & k_{\mathrm{B}} &= \frac{R}{N_\mathrm{A}} \end{align}

which gives

$$\Delta S = -R(n_1 \ln x_1 + n_2 \ln x_2)$$

or

$$\Delta S = -nR(x_1 \ln x_1 + x_2 \ln x_2)$$

whichever you prefer (they're of course all the same).