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According to this post,

$$\mu_i = g_i + RT \ln \frac{N_i}{N_1+N_2}$$

It shows that chemical potential indeed do vary with composition. I have trouble understanding how this equation is derived.

I started from Sakur-Tetrode equation and the definition for Gibbs energy $G=U+PV-TS$ and I got the following for the ideal gas Gibbs energy

$$G=Nk_BT\ln\left[\frac{N}{V}\left(\frac{h^2}{2\pi m k_BT}\right)^\frac32\right]$$

Consider a sealed container of volume $V$ with constant and well-defined temperature and pressure $P,T$ filled with a mixture of two kinds of ideal gases 1 and 2 with particle counts $N_1$ and $N_2$. The Gibbs energy becomes

$$G=(N_1+N_2)k_BT\ln\left[\frac{N_1+N_2}{V}\left(\frac{h^2}{2\pi m k_BT}\right)^\frac32\right]$$

By the ideal gas law,

$$G=(N_1+N_2)k_BT\ln\left[\frac{P}{k_BT}\left(\frac{h^2}{2\pi m k_BT}\right)^\frac32\right]$$

The chemical potentials for the mixture are

$$\mu_1 = \left(\frac{\partial G}{\partial N_1}\right)_{T,P,N_2}=k_BT\ln\left[\frac{P}{k_BT}\left(\frac{h^2}{2\pi m k_BT}\right)^\frac32\right]$$

$$\mu_2 = \left(\frac{\partial G}{\partial N_2}\right)_{T,P,N_1}=k_BT\ln\left[\frac{P}{k_BT}\left(\frac{h^2}{2\pi m k_BT}\right)^\frac32\right]$$

So

$$\mu_1=\mu_2$$

You can see that under the conditions of constant temperature and pressure, chemical potential does not depend on the number of particles $N$, so it is constant. Thus, chemical potentials does not vary on mixture composition since they do not depend on $N_1$ and $N_2$.

What am I missing?

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    $\begingroup$ Entropy of mixing… $\endgroup$
    – Jon Custer
    Apr 15, 2023 at 2:26
  • $\begingroup$ @JonCuster What do you mean? $\endgroup$
    – Ray Siplao
    Apr 15, 2023 at 3:11
  • $\begingroup$ Gibbs's paradox... $\endgroup$
    – Themis
    Apr 15, 2023 at 12:29

1 Answer 1

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You fell in the trap known as Gibbs's paradox: you took the Gibbs energy of pure component and generalized it to mixtures by replacing $N$ with $N_1+N_2$.

We need to start from the partition function of the ideal gas. For a mixture of two components the partition function is $$ \tag{1} Z_{1,2} = \frac{1}{N_1!N_2!}\frac{V^{N_1+N_2}} {\Lambda_1^{3N_1}\Lambda_2^{3N_2}} $$ with $\Lambda_i=h/\sqrt{2\pi m_i k_B T}$. For a single component this reduces to $$\tag{2} Z = \frac{1}{N}\frac{V^{N}} {\Lambda^{3N}} $$ As you see, the two-component partition function is not the partition function of one-component with the substitution $N\to N_1+N_2$. Apart from the $\Lambda$'s, the real issue here is the factorial term $N_1! N_2!$, which is the term that resolves the paradox. More precisely, if this term is omitted we obtain the wrong result (the "paradox").

The Sakur-Tetrode for pure component is obtained from Eq (1). Work out the Gibbs energy for Eq (2) and everything will turn out as it's supposed to.

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    $\begingroup$ I found that $$\mu_1=\left(\frac{\partial F}{\partial N_1}\right)_{T,V,N_2}=k_BT\ln\left[\frac{N_1}{V}\left(\frac{h^2}{2\pi m k_BT}\right)^\frac32\right]$$ where $F=-k_BT \ln Z_{1,2}$. I am still not sure how it is not $$\mu_1 = g_1 + RT \ln \frac{N_1}{N_1+N_2}$$ $\endgroup$
    – Ray Siplao
    Apr 15, 2023 at 22:41
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    $\begingroup$ Your $\mu_1$ is correct. Then show $$ \frac{V}{N_1} = \frac{V_1}{x_1 N_1} $$ where $V_1$ is the volume occupied by pure 1 at same $T$ and $P$ (remember, this is ideal gas) and finally write your result as $$ \mu_1=\mu_1^0 + k_B T\ln x_1 $$ where $\mu_1^0$ is the chemical potential of pure 1 at same $T$ and $P$ as the mixture. $\endgroup$
    – Themis
    Apr 16, 2023 at 0:00
  • $\begingroup$ I see, thank you! So it seems like $\frac{N_1}{V}=\frac{x_1 N_1}{V_1}=\frac{p_1}{k_BT}$ which gives partial pressure $p_1=x_1\frac{N_1 k_BT}{V_1}=x_1 P$ where $P$ is total pressure? $\endgroup$
    – Ray Siplao
    Apr 16, 2023 at 0:22
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    $\begingroup$ Yes, these are all correct results in the ideal gas state. $\endgroup$
    – Themis
    Apr 16, 2023 at 0:55

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