I am planning an experiment in an isolated system that contains two droplets. One droplet is water and the other one is a 40% ethanol-water mixture. The isolated compartment is filled with air at 20°C.

I understand that for ensuring that the water droplet volume remains constant during the course of the experiment, the air in the compartment needs to be fully saturated according to the vapor pressure of water at 20°C (100% relative humidity).

Under these conditions, my understanding is that the ethanol from the 40% ethanol-water droplet is not yet in equilibrium with respect to the evaporation/condensation rate in the fully water-saturated air, thus decreasing the droplet volume. Is that correct? If it is correct, that would mean that I would also have to saturate the air with ethanol before the experiment. I can find the vapor pressure of ethanol for 20°C. But how would I calculate the resulting concentration of ethanol in the air in kg/m^3 or ppm? Also, how would the total air pressure within the compartment be influenced by the saturation?

  • $\begingroup$ Why would you need that? It is not like you can buy yourself some air with such-and-such concentration of ethanol, really. $\endgroup$ Commented Jan 21, 2022 at 14:08
  • 1
    $\begingroup$ There would evaporation of ethanol + condensation of water at 40% ethanol drop AND evaporation of water and condensation of ethanol at the pure water drop. In the and, there would be the same concentration of ethanol in both drops of unsure volumes, in equilibrium with water/ethanol vapor partial pressure ratio. $\endgroup$
    – Poutnik
    Commented Jan 21, 2022 at 14:11
  • $\begingroup$ You can use formal equality of ethanol vapor concentration and pure vapor density: $$[\ce{C2H5OH}]=\rho_{\ce{C2H5OH}}=\frac {p_{\ce{C2H5OH}}M_{\ce{C2H5OH}}} {RT},$$ assuming ideal vapor behavior. $\endgroup$
    – Poutnik
    Commented Jan 21, 2022 at 14:22
  • $\begingroup$ @IvanNeretin: To ensure that the droplet volume remains constant over the course of the experiment. The idea of course is not to buy the air, but to ensure that the equilibirum according to the vapor pressure is reached before the start of the experiment $\endgroup$ Commented Jan 21, 2022 at 16:09
  • $\begingroup$ @Poutnik: That sounds logical. I did not think of this. Thank you! $\endgroup$ Commented Jan 21, 2022 at 16:12


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