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?




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).


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