Originally Posted by Delancey
This is great. RMS took me to Maxwell-Boltzmann Distribution which is what I was really looking for, to understand the mechanism by which gases dissipate. Any idea how long it would take a given quantity of propane to clear from a boat?
Also, for your heavier-than-air viewing pleasure-
Yes. It's part of the Kinetic Theory of Gases. Maxwell-Boltzmann distributions are more of a statistical distribution method which works well with idealized gases.
To answer your question regarding how long would it take for the propane to clear from the boat, then we could consider the RMS Velocity of Propane.
If Propane (C3H8) was in a container by itself, at 1 atmosphere (absolute) pressure at 25C (298K), then the Root-Mean-Square (average, rectified for directionality) velocity of the gas molecules would be:
Firstly, we need the molar mass (kg/mol) for C3H8, which is
(44.0962/mol) * (1kg/1000g) = 0.0440962 Kg/mol
Then the RMS velocity of the propane would be
Vrms = Root(3RT/M) = Root((3*8.31*298)/(0.0440962)) = 410 m/sec
Remember this is a vector corrected mean, such that you'll get a standard distribution (bell curve) of molecular velocities, some slower, some faster.
Note that this appears like a high speed, but bear in mind that we have a more complex mix in the cabin of and actual boat, and only a small percentage of the inner surface area of the cabin is a hole through which the gas can diffuse (and escape). The rate at which the gas diffuses is dependent upon, the volume of the cabin, the geometry of the cabin, the size of the companion-way as a percentage of the overall cabin surface area, the initial volume of the gas (at 1atm), the temperature of the gas, and the rate of throughput (ventilation) of other gases through the volume of the cabin... As you can imagine, there are many, many variables which could give you too many answers. I could have worked through an example calculation using my own boat, but I'm already finding the typing equations into this text-box in 'excel' format a real pain, and the webpage already crashes once while I was halfway through typing.. (Sorry for that)
On that last note regarding molar mass of the gas, you may recall
"The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules."
So, for two gases..
If we say that Gas X is Methane (CH4), and Gas Y is Propane (C3H8), you'll note that Propane's molar mass is much greater than Methane:
Gas X = CH4 = 16.04g/mol
Gas Y = C3H8 = 44.09g/mol
Calculate through and you'll see that Methane diffuses 1.65 times faster than Propane. Indeed, the heavier the molar mass of a gas, the slower the diffusion, due to the lower V(rms).. If you look up at the original RMS calculation, you'll see that the molar mass of the gas is the denominator (at the bottom) of the fraction inside the Root.. the bigger this number, the lower the Velocity..
Back to the diffusion issue - when you consider all of the parameters required to calculate the time to 100% escape for the gas from the cabin, you'll note that statistically, it'd take a very long time for that last (un-trapped) molecule of propane to leave the cabin... the better question to ask is, when will the concentration of gas (compared to air) fall below the Lower Explosive Limit for propane.. and at what point can you reenter the cabin without inhaling pure propane (thereby displacing air from your lungs)... lots of things to think about.
For these reasons, that's why my recommendation was to open what vents you can and then hit the bar for a few hours so you can discuss more interesting stuff!
..hope this went part the way to answer your question.