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# Solubility of gases in liquid metal

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## Solubility of gases in liquid metal

(OP)

I am investigating the application of gases dissolved in liquid metals for increasing the electrical resistivity of a liquid metal-gas solution.

Many data on the solubility of gases are often specified in weight percentages, (wt%). For example, the solubility of nitrogen in liquid iron at standard atmospheric pressure is around 0.05 wt% at T 1600 deg.C.

When I need to know the fraction of the gas expressed in a volume-percentage ratio, e.g. by calculating the volume of the gas fraction from the wt% and the density per ccm, I get unrealsitic high volume fractions of the gas. Taken the density of 0.00125g/ccm for N and 7.2g/ccm for liquid Iron, then the mixture with 0.05 wt% fraction of nitrogen would comprise (expressed in volume ratio) 25% Iron and 75% Nitrogen, a resulting volume of 400% compared to a volume of pure Iron.

Where I am mistaking, and how to calculate correctly the solubility in volume ratios from wt%?

Thanks for any help.

### RE: Solubility of gases in liquid metal

Primarily, your problem is that you're comparing apples to go carts.

Gases are NOT dense; they are not solids, so comparing a gas volume to a solids volume is simply makes no sense.

I'm not even sure what the point of such a calculation might be, but a more sounder approach would be to use the atomic ratio and call that a volumetric ratio.

### RE: Solubility of gases in liquid metal

(OP)

IRstuff, the primary goal is to aproximately estimate the increase of the specific resistivity of a metal in the liquid phase when a certain amount of a gas is dissolved in the liquid metal. The electric insulation strength of a gas is given by its low density, independent of its mass. Therefore, knowing just the mass fraction of the gas in the liquid metal will not help.

I agree that the comparison of liquid to gas volumes in general is problematic, but since I could not find any calculation method for the resistivity versus gas fraction in a liquid metal, I assume that rating the volume fractions of the mixture is the only way to at least roughly estimate the change in resistivity.

The volume rating question could also be redescribed as a follows: By how much will the total volume of the mixture increase compared to the original volume of the liquid metal, when 0.05wt% of the gas is dissolved in it?

### RE: Solubility of gases in liquid metal

"The electric insulation strength of a gas is given by its low density, independent of its mass. Therefore, knowing just the mass fraction of the gas in the liquid metal will not help."

That's just makes no sense.  A dissolved gas in a solid is part of the solid, there is no remnant of the gas properties left.

And, apparently, you failed to read my last posting.  I did not suggest using the mass ratio; I suggested using the atomic ratio.  Even at that, whatever calculation there might be will be grossly inaccurate, since the atomic radii are drastically different.

### RE: Solubility of gases in liquid metal

(OP)

Thanks, calculating with the atomic ratio seems indeed promising.

But: "A dissolved gas in a solid is part of the solid, there is no remnant of the gas properties left."

So the solution of liquid iron and dissolved nitrogen would form just liquid Iron nitride (FeN)? Is then the higher electric resistivity in nitrogen or hydrogen enriched compounds never due to the partially gaseous properties? (Nitriding of liquid metals normally increases their resistivity).

### RE: Solubility of gases in liquid metal

The increase in resistivity would normally come from the disruption of the crystal lattice because of the displacement AND the disruption of the Fermi level that allows for the electrons to easily get into the conduction band.

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