Finding Percent of Water & Oil in a Mixture from Surface Tension

[sagigever]

I was looking for on the internet for a while without a success.
If I know that the surface tension of pure oil is $\gamma_o=A$ and I know that the surface tension of pure water is $\gamma_w=B$
so if I have a mixture of water and oil with surface tension $\gamma_m=C$ am I able to know the percent of water and oil in the mixture? if yes how? where the derivation is come from?

 

[dRic2]

Water and oil don't mix at standard temperature and pressure so you don't have a mixture of water and oil, but the oil simply sits on top of the water. Are you asking about the surface tension between oil and water?

 

[Thomas Sturm]

You might find The surface tension of binary liquid mixtures an interesting read. However, what you are dealing with is an emulsion, not a mixture. "Pure Oil" is a mixture (of hydrocarbons) already...and the surface tension (of an emulsion with water) will strongly depend on the chemical composition of the oil and, even in case this should be relatively constant, the physical properties (droplet size, temperature) of the emulsion.
Still, the answer to your question is "Yes". In practice you would measure the surface tension of emulsions with known oil content within the concentration range you are interested in - and use these as "standards" to measure your unknown emulsion against.

 

[etotheipi]

You might find The surface tension of binary liquid mixtures an interesting read.

I found this one as well, but unfortunately it's paywalled.

 

[sagigever]

I think I wasn't clear enough with my question, I will give a numeric example:
Lets assume I am doing an experiment of measuring surface tension, I have some
alcohol liquid and I measure $\gamma_m = 37 \frac{dyne}{cm}$ but I know that the surface tension of pure alchohol is $\gamma_a = 22 \frac{dyne}{cm}$ and on the other hand surface tension of pure water is $\gamma_w = 72 \frac{dyne}{cm}$ so it easy to conclude that I have mixture of water and alchohol, now I am asking how can I know the percent of alchohol and water in the mixture

 

[dRic2]

Well I think it depends whether you are dealing with an ideal solution or not. In an ideal solution the interactions between molecules of specie A and molecules of specie B are the same (i.e. the energy of those interactions is $E_{AA} = E_{BB} = E_{AB}$). In that simple case thermodynamic quantities should be a simple a weighted sum of the ones referred to $A$ and $B$. For example, $G_{mix}^{ideal} = g_ax_a + g_bx_b$ ($G$ is the gibss free energy). Now $\gamma = \frac {\partial G}{\partial A}_{T, P}$. On the other hand, if the solution is not ideal, then the interactions between A and B are generally different from each other, so you have to take that into account. In that case $\gamma = \frac {\partial G}{\partial A}_{T, P}$ still holds, but there is not a simple way to calculate $G$ of the mixture.

Unfortunately I don't think ethanol-water could be approximated as an ideal mixture, but I might be wrong. I think the only way is to measure it or find some sort of empirical correlation between $\gamma$ and the molar fraction of ethanol in water.

PS: To be very clear, when I wrote $G_{mix}^{ideal} = g_ax_a + g_bx_b$ for an ideal mixture, it is important to stress that $g_a$ and $g_b$ are defined as $g_i = g^{pure}_i(T, P) + RT \log(x_i)$. https://en.wikipedia.org/wiki/Ideal_solution

 

[sagigever]

We didn't learned those equation u mentioned so I guess there is another and simplier way for the problem I mentioned