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?

  • $\begingroup$ How is the change in Gibbs free energy in a flow process related to the minimum amount of work required to carry out the process if the control volume is maintained in contact with a constant temperature reservoir? $\endgroup$ Aug 8, 2019 at 16:37
  • $\begingroup$ Can you visualize how such a separation process would be carried out reversibly? $\endgroup$ Aug 9, 2019 at 2:28
  • $\begingroup$ Thanks sir. I did some calculations, but still there is a flaw! Can you kindly take a look at it? $\endgroup$
    – Ghartal
    Aug 9, 2019 at 7:10
  • $\begingroup$ It seems to me there is something wrong with this problem statement. Even for an ideal mixture (featuring zero excess Gibbs free energy), reversible work has to be done to separate the components. $\endgroup$ Aug 9, 2019 at 11:57
  • $\begingroup$ Well If $W_{rev}$ for separation of ideal solution is $\Delta g^{id}_{mix}$ and for non-ideal solution is $\Delta g_{mix}$, then their difference will be $$\Delta g_{mix}-\Delta g^{id}_{mix}=g^E$$ Seems reasonable to me. $\endgroup$
    – Ghartal
    Aug 9, 2019 at 13:53

1 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.


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.