Mercury Seal Gap Size: Is it Possible?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 3K views
artis
Messages
1,479
Reaction score
977
I am wondering about the physical properties of mercury. I know that different materials have different density and ability to penetrate small openings. I wonder if I have a small container of mercury, say about 100 grams or less, and there is a small flat opening at the bottom of such container which can be adjusted. Given the gap is very small (say 0.5mm) and the mercury weight not large would the mercury be able to escape through the gap?
This is more of a theoretical idea than a practical one although if such gap could be made and sealed with some lubricant it could come in handy as a rotary electrical contact for some of my experiments later on which is partly why I'm asking.

I have a feeling that unlike water or diesel or other liquids that can easily seep through very small opening mercury's high density and very low surface adhesion and wetting would prevent it from seeping through a small opening in liquid form.
 
Engineering news on Phys.org
It wouldn't drop down if you don't shake the container but it will still evaporate at the surface, making that a health risk.
 
Well in theory the atmospheric end of the small opening could be sealed off by ferrofluid or a similar material.
I wonder is there any way of determining what is the largest gap size for any given weight and shape of container of mercury?
 
I meant calculating the gap size, but I assume that would be complicated.
 
The Washburn equation is used in mercury porosimetry:
ΔPr = -2γcosθ
where ΔP is the pressure difference (usually just the pressure, as the sample is evacuated); r is the pore radius, γ is the surface tension of mercury and θ is the contact angle of the mercury with the solid. Using typical values of 140° for θ and 0.480 N/m for γ, we obtain
r = 0.736/ΔP
where r is in µm and ΔP in MPa
If r = 250 µm, ΔP = 2.9 kPa, corresponding to a mercury column of ca. 22 mm. A mercury layer deeper than this would go through the pores under its own weight.
 
  • Like
Likes   Reactions: Nik_2213 and mfb