1
$\begingroup$

Can anyone tell me whether or not air or N$_2$ dissolves easily in the R-113 (liquid solvent) under the saturation pressure of R-113? If yes, what is the solubility roughly? From this link one can only find its solubility in water. It will be better if you could provide a link. Thank you in advance!

PS: In Table V of this data provided by @Jon Custer, I found no solubility data for air (or N_$2$) dissolved in R-113.

Improve this question
$\endgroup$
5
  • $\begingroup$ Well, solubility wouldn't be as good as in perfluorocarbons, but it may be tough to get specific data. $\endgroup$
    – Mithoron
    Oct 7, 2019 at 18:34
  • $\begingroup$ pubs.acs.org/doi/pdfplus/10.1021/cr60242a003?src=recsys has data... $\endgroup$
    – Jon Custer
    Oct 7, 2019 at 20:37
  • $\begingroup$ @Mithoron a little irrelevant... I knew R113 is a chlorofluorocarbon. Did you mean chlorofluorocarbon $\in$ perfluorocarbon or perfluorocarbon $\in$ chlorofluorocarbon or chlorofluorocarbon $=$ perfluorocarbon? $\endgroup$
    – user55777
    Oct 8, 2019 at 2:38
  • $\begingroup$ @JonCuster thank you, sir. I checked Table 5 in that article. But there seems no solubility data of air or N$_2$ in R113 (liquid solvent). $\endgroup$
    – user55777
    Oct 8, 2019 at 2:47
  • $\begingroup$ @JonCuster I mean all fluor beats half-chloro/half-fluoro. $\endgroup$
    – Mithoron
    Oct 8, 2019 at 15:25

1 Answer 1

Reset to default
3
$\begingroup$

While data for air solubility in R-113 is difficult to find, Henry's law constants $k_\text{H}$ are available for $\ce{N2}$ in a series of related refrigerants (Ref. 1) comprised of variously halogenated ethanes. However, these are predicted, not experimental, Henry's law constants: the vapor-liquid equilibrium of binary mixtures containing nitrogen was modeled with an equation-of-state based on a perturbed–Chain (PC) modification of the Statistical Associating Fluid Theory (SAFT). I back-calculated the $k_\text{H}$ from the published temperature coefficients. Assuming a $\ce{N2}$ partial pressure $\pu{p_{\ce{N2}}= 0.78 atm}$ then leads to the following $\ce{N2}$ solubilities (as mole fractions) in the refrigerants in the temperature range $\pu{170-280 K}$:

enter image description here

Evidently the solubilities are generally below $\pu{0.02 mol \%}$, which I believe would qualify as "not high" but not negligible ($\chi=1\times 10^{-4}$ corresponds to $\pu{\approx 1 mM}$ solution concentration).

After more careful examination, I found that Vinš et al. cite a publication by Huang et al. (Ref. 2) which compares simulation and experimental (Ref. 3) Henry's law constants for $\ce{N2}$ in R-113, the value at 298 K being $\pu{k_H= 52.5 MPa}$. This is larger than the values above, which are in the range $\pu{1-15 MPa}$. Since $p_\ce{N2} = k_\text{H} \chi_\ce{N2}$, the larger $k_\text{H}$ for R-113 means nitrogen is even less soluble in this refrigerant.

Note of course the analysis here is restricted to nitrogen, but you can proceed in the same way to analyze other gases for which $k_\text{H}$ are reported.

References

  1. Vinš, V., Hruby, J., Solubility of nitrogen in one-component refrigerants: Prediction by PC-SAFT EoS and a correlation of Henry’s law constants. International Journal of Refrigeration, 34, 2109-2117 (2011).

  2. Huang, Y.L., Miroshnichenko, S., Hasse, H., Vrabec, J., Henry's Law Constant from Molecular Simulation: A Systematic Study of 95 Systems. Int. J. Thermophys. 30, 1791–1810 (2009).

  3. H. Hiraoka, J. H. Hildebrand, J. Phys. Chem. 68, 213 (1964).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for your reply. In the plot, there appears no curve for R113? What is $k_H$ and $T_r$? Most importantly, did you predict that N$_2$ should be not easily dissolved in saturated R113? $\endgroup$
    – user55777
    Oct 8, 2019 at 2:57
  • $\begingroup$ @user55777 I updated my answer to address your question. $\endgroup$
    – Buck Thorn ♦
    Oct 8, 2019 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.