Difference between minimum work of separation for a binary ideal mixture and a non-ideal mixture

Question

What is the difference between the minimum required work for separation of components of a binary mixture in constant temperature and pressure to its pure components of an ideal mixture with a non-ideal mixture?

Well answer to this question is $g^E$, which is excess free Gibbs energy and is defined as $g^E=g-g^{id}$.

According to @Chet Miller hints: Let's suppose that a flow with enthalpy $H_1$ enters a C.V. (which is in contact with a constant temperature (T) reservoir) and two separated flows with enthalpies $H_1$ and $H_2$ go out. Minimum work is obtained when process is reversible, so first law gives $$Q-W_{rev}=H_1+H_2-H=- \dot{n} \Delta h_{mix}$$ On the other hand second law gives $$\frac{Q}{T}+S_1+S_2-S=0 \ \ \Longrightarrow \ \ Q=T \dot{n} \Delta s_{mix}$$Putting back in the first law equation gives $$\frac{W_{rev}}{\dot{n}}=\Delta h_{mix}+ T \Delta s_{mix}$$ This is supposed to be $\Delta h_{mix}- T \Delta s_{mix}=\Delta g_{mix}$ !! Where did I go wrong?

Answer 1

I am confused by what exactly you are asking, but here are some basic points:

• The work to separate a mixture to its pure components at constant $T$ and $P$ is $|g^\text{mix}|$ (using absolute values to avoid confusion with sign conventions).
• The excess Gibbs energy $g^E$ is the difference between the work to separate the mixture into its pure components, and the work to separate an ideal solution of the same composition: $$ g^E = g^\text{mix} - g^\text{ideal} = g^\text{mix} - RT(x_1\ln x_1+x_2\ln x_2) $$
• For an ideal solution $g^E=0$ but $g^\text{mix}=RT(x_1\ln x_1+x_2\ln x_2)<0$.

Your question seems to be about the difference between the work to separate the mixture and the work to separate an ideal mixture with the same composition -- then answer is $g^E$. In your derivation, however, you do not subtract the work for the ideal solution and that is why you get $g^\text{mix}$ instead.