$$S=k_\mathrm B\ln\mathit\Omega$$ 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?

  • 2
    $\begingroup$ 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. $\endgroup$ – orthocresol Oct 23 '15 at 13:54
  • $\begingroup$ Thank you. However, I only get the number of moles not the mole fraction in the logarithms. i.e i get $\ln \frac {N_i}{N}$ $\endgroup$ – RobChem Oct 23 '15 at 14:01
  • $\begingroup$ 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. $\endgroup$ – orthocresol Oct 23 '15 at 14:04
  • $\begingroup$ I'm sorry, I am really stuck. Would you mind just telling me? Or giving me a large clue? Thank you so much $\endgroup$ – 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

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


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

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

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.