How quickly does sulfur hexafluoride mix with the atmosphere?

Question

A common science demonstration is to float a "boat" on sulfur hexafluoride. How long would the sulfur hexafluoride stay in the container (in the experiment linked above) if the container were to be left uncovered and undisturbed at room temperature?

Also, I'm wondering what would happen if you scaled the experiment up to the size of a lake, ignoring the time it would take to fill said lake and the impossibility of obtaining that much sulfur hexafluoride. Let's also assume that there is no wind to blow it away. Could you have a large permanent pool of sulfur hexafluoride or would it eventually dissipate and mix with the atmosphere?

Would this change if a different dense (non-reactive) gas were used, like xenon?

Answer 1

Up front: this depends on what you mean by "stay in". It would diffuse out of the container and, given the surroundings are closed, obtain a small equilibrium concentration.

Not that I could provide an exact answer to your question (it is rather a physics question), but this paper (paywalled) deals with the diffusion of the $\ce{N2-SF6}$ pair. The binary diffusion coefficient is reported to be 0.0945 cm²/s. In their experiments they used a container of d = 0.5 m, l = 1 m, introduced a small amount (2-5%) of $\ce{SF6}$ at the bottom of the vessel and measured the time it took to obtain the equilibrium concentration of the mixture. For this setup, this was ~15 h, with a reported time constant of 4-5 hours (I believe this is the time it takes to reach a value of [local $\ce{SF6}$ concentration]/[equilibrium concentration] of 0.5).

They used a lengthy formula derived from a one-dimensional model for the comparison of calculated and measured concentrations and in my opinion the measured data agreed quite well with the calculated ones.