I wonder if there is a way to measure the chemical potential of a substance in a two-component mixture and also to account for its dependence on the number of moles. Explaining it further, suppose there is a mixture of a light component (like methane, C1) and a heavier one (like C20). Given temperature and pressure, one could add some amount of the light component to the mixture and that would rise the chemical potential of that first component. Thermodynamics tell us that the chemical potential can be expressed by

$$d \mu_i = - \bar S_i dT + \bar V_i dP + \sum_{m=1}^{n} \left(\frac{\partial \mu_m}{\partial N_i}\right)_{T,P,N_j[i]} d N_m $$

Being pressure and temperature constant, the increase in the chemical potential by adding some amount of light component depends entirely on the summation of $({\partial \mu_m}/{\partial N_i})_{T,P,N_j[i]}$. That is precisely the point: how could I measure or calculate that derivative or, alternatively, variations in the chemical potential by adding mass to the mixture?

Of course, one could use a cubic equation of state or the like and calculate fugacity and chemical potential etc. But I am interested in the more fundamental aspects of thermodynamics. One of them is that, given P-V-T data and specific heats, in addition to the composition, one is supposedly able to calculate all the thermodynamic properties of a mixture or of a component.

Now, the practical side of this question is this: if there is a two-phase mixture of C1 and C20, under ordinary T and P, the vapour phase is almost pure C1, but the liquid phase is a mixture of both. For given P and T, composition can be measured somehow. Given an increase in pressure, say $\Delta P$, with coexisting phases, there must be an increase in the concentration of C1 in the mixture, $\Delta x_1$ (or $\Delta N_1$, if the number of moles is to be used instead of concentration). The change in composition can obviously be calculated by equating the chemical potential of C1 in the vapour and in the liquid phases. While calculating the change in chemical potential in the vapour phase is quite easy because it is pure C1 and $d \mu_1 = \bar V_1 dP$, the chemical potential in the liquid phase depends on the above mentioned derivative:

$$d \mu_1 = \bar V_1 dP + \sum_{m=1}^2 \left(\frac{\partial \mu_m}{\partial N_1}\right)_{T,P,N_2} d N_m $$

Finally, by "measuring" chemical potential I mean any way of obtaining it from other variables that can be measured by ordinary instruments in a lab. And, just to mention, I have used the notation of Modell and Reid, "Thermodynamics and its applications".

  • $\begingroup$ Dear fellows, I've got no answers so far. Is this question too complex or too badly written? Or perhaps it is not suited for this forum? $\endgroup$ – gOliveira May 30 '17 at 22:55

Your Answer

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

Browse other questions tagged or ask your own question.