Say we are a person sitting in a diving bell and have saturated (acclimated) to an ambient pressure of 5 MPa = 5,000,000 $kg/m/s^2$, breathing a mixture of 99% helium, 0.8% oxygen, and 0.2% impurities with a similar molar mass to surface air. Is our actual mass now higher than it was when we were acclimated to sea level pressure?
Here's an attempt to at least calculate the density of the gas . . . am I on the right track?
Here is my math:
Molar mass = 4 $g/mol_{He}$ * 0.99 $mol_{He}/mol_{air}$ + 32 $g/mol_{O_2}$ * 0.008 $mol_{O_2}/mol_{air}$ * 29 $g/mol_{impurities} * 0.002 $mol_{impurities}/mol_{air} = 4.3 $g/mol_{air}$, approximately. Not accounting for the rounded helium. Ignoring water.
$Density = mass / Volume = mass / (n * R * T / P) = molarmass * Pressure / (R * Temperature) = (4.3g/mol) * 5,000,000kg/m/s^2 / (8,314g*m^2/K/mol/s^2 * 300 K) = (4.3) * 5,000,000kg / (8,314m^3 * 300) = 8.6kg/m^3$
Ideal gas law, still just pretending you can keep things very dry. Which it turns out, you sort of can, because vapor pressure of water at habitable temperatures is much much lower than 50 atm.
If density is higher than surface air (on the order of 1kg/m^3), and solubility of gases always increases with increased pressure, shouldn't the density of our saturated body necessarily also increase? Would we feel the density increase, or would it be balanced out by the additional buoyant force of the increased ambient air density?