6
$\begingroup$

I feel that there are very few textbooks that treat the chemical potential of mixtures in an understandable clear way, which is why I wanted to ask here about certain things. Although I do not have a single question, but 5 of them, I think that it is merely impossible to separate them thematically, which is why I hope to meet an expert here.

If we have an ideal gas then if we know the chemical potential for a particular $(p,T)$ we have:

$$\mu(p_2,T) \approx \mu(p_1,T) + \left(\frac{\partial \mu}{\partial p} \right)_T(p_1,T) (p_2-p_1) = \mu(p_1,T) + \frac{V}{N}(p_2-p_1).$$

Similarly for changes with respect to temperature we have $$\mu(p,T_2) \approx \mu(p,T_1) + \left(\frac{\partial \mu}{\partial T} \right)_p(p,T_1) (T_2-T_1) = \mu(p,T_1) - \frac{S}{N}(T_2-T_1).$$

Now, if we have either a gas or a liquid (Question 1: is this true, does this law hold for both fluids?) $\ce{A}$ and put another material $\ce{B}$ into it, then we have by Raoult's law:

$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{N_A}{N_A + N_B} \right)$$

Question 2: Now, is the chemical potential of the total fluid given by $A$ and $B$ then the sum $\mu_{\mathrm{A+B}} = \mu_{\mathrm{A,~with~mixed~B}}(p,T) + \mu_{\mathrm{B,~with~mixed~A}}(p,T)$?

Let's imagine that $A$ and $B$ would be gases, would this mean that

Question 3 : $$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_\mathrm{A}(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$

Question 4 : And what if $\ce{A}$ and $\ce{B}$ would be two two miscible liquids? could we still use the vapour pressures to calculate the chemical potential of the liquids mixture?

$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$

Now, imagine that I have a liquid $\ce{A}$ and salt $\ce{B}$ in it. Also, I have a gas layer $\ce{C}$ of $\ce{A}$ above the liquid. What exactly is now in equilibrium: $\mu_{\mathrm{A+B}} = \mu_\mathrm{C}$, or $\mu_{\mathrm{A,~with~mixed~B}} = \mu_\mathrm{C}$?

Question 5: My last question would be: Imagine that you have a liquid (let's say water) and some gaseous water phase above (according to the vapour pressure). Now, you add oxygen to this. What would this mean for the chemical potentials of the liquid/gas water?

$\endgroup$
2
$\begingroup$

It's much simpler than all that.

Pure gas $$\mu_g(T,P)=\mu^0(T)+RT\ln P$$ where $\mu^0(T)$ is the molar free energy of the pure gas at temperature T and pressure 1 Bar, and P is the pressure expressed in bars.

Pure liquid

If $P_{sat}(T)$ is the equilibrium vapor pressure at temperature T, then, at pressure $P_{sat}(T)$, we know that the free energy of the liquid is equal to the free energy of the vapor. So, $$\mu_l(T,P_{sat})=\mu^0(T)+RT\ln P_{sat}(T)$$If we neglect the effect of pressure on the chemical potential of the liquid (because the specific volume is very small), then, for a pure liquid at temperature T and pressure P:$$\mu_l(T,P)=\mu^0(T)+RT\ln P_{sat}(T)$$

Ideal gas mixture

For species A in an ideal gas mixture, the chemical potential of species A is the same as the chemical potential of the pure species, except evaluated at the partial pressure of A in the gas mixture:$$\mu_{Ag}(T,P,y_A)=\mu^0_A(T)+RT\ln p_A=\mu^0_A(T)+RT\ln (Py)_A=\mu_A(T,P)+RT\ln y_A$$ where $p_A$ is the partial pressure of A (bars), $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, P is the total pressure of the mixture (bars), $\mu_A(T,P)$ is the free energy of pure A at the temperature and pressure of the mixture, and $y_A$ is the mole fraction of A in the mixture. So the chemical potential of A in the mixture is equal to the chemical potential of pure A at the same temperature and pressure as the mixture plus the correction term for mixing $RT\ln y_A$.

Ideal liquid mixture

By analogy to an ideal gas mixture, the chemical potential of species A in an ideal liquid solution is given by:

$$\mu_{Al}(T,P,x_A)=\mu_{Al}(T,P)+RT\ln x_A=\mu^0_A(T)+RT\ln P_{A,sat}(T)+RT\ln x_A$$ where $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, $P_{A,sat}(T)$ is the saturation vapor pressure of species A at temperature T (bars), and $x_A$ is the mole fraction of species A in the liquid.

$\endgroup$

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.