That's not as big a problem as you'd think. In solution, you don't have molecules of NaCl; you have dissociated ions of Na+ and Cl-, each of which is surrounded by a cluster of rather tightly-bound water molecules. Those clusters are much larger than bare ions or single water molecules, so there's a fair range of pore sizes that will separate the ions from the water.

I think they're abusing the terminology a bit, using "RO" to refer to reverse osmosis conducted with existing membrane technologies. The point at issue is that thermodynamics demands that a certain amount of energy be expended in order to reduce the entropy of a homogeneous salt solution by separating it into pure (or at least low-salinity) water and high-salinity leftovers. This is totally independent of the means by which the molecules are separated. In reverse osmosis, that manifests as a minimum pressure necessary to force salt water through any selectively permeable membrane.

Practical RO systems operate with a pressure drop (and therefore energy consumption per unit volume) that's double or triple the osmotic pressure, in order to achieve useful flow rates across thick membranes with relatively low pore densities. A better filter would allow that excess pressure to be reduced, but can't do anything about the cost of reducing the entropy.

Thank you, anonymous reader, for a confused summary of an idiotic blog post about a moderately dumbed-down article about an interesting article.

What they're talking about is reverse osmosis, and there's no way to make it two or three orders of magnitude more efficient. Commercial systems already hit 30% to 60% of the thermodynamic limit for energy efficiency; all graphene offers in this case is a way to increase the speed, decrease the filter size, or reduce the unnecessarily wasted energy. There's still no getting around that darned osmotic pressure.

John Graham-Cumming is the leading light behind a project...

Leading lights generally work better in front of things. I think your metaphorator might be a bit misaligned...

Yep. Looks like you've got some sinusoidal co-pleneration between the literal input shafts. Gonna have to replace your main spurving bearing, maybe the secondary too. A couple of the marzel vanes on your imagery agitator are looking a pretty worn, might want to get those replaced while you're at it.

There was a forum discussion which someone complained, "so what if I want to talk like a CBer on ham radio? As long as I'm licensed and mention my callsign every 10 min, end of transmission, bla-bla, I can talk in whatever style I want!" However, someone gave example: "That's a big ten-four good buddy and I sure do appreciate that there smokey report on the five oh niner. Well, I'll catch you on the flipper flopper!" Bzzzzztttt. FCC Part 97 prohibits codes and ciphers used to obscure communications.

Which is thoroughly irrelevant to the issue of talking like a CBer. Nothing in your example message is a code or cipher; that's simply slang. All of it is publicly known. The reason hams discourage talking like a CBer is that it makes you sound like one of the drooling shit-flingers who infest CB.

I think this is the comment you're referring to:
12. Bethany Says:
September 21st, 2010 at 8:20 am

Alright, here's what I calculated:
The protons are high energy with lorentz factor of gamma=7500, kinetic energy is about K=7×10^6 eV. The paper cited below says that the stopping power of a proton going 10^6 eV is about 2.5×10^8 eV cm^2 g^-1. Using the density of muscular tissue rho=1g cm^3 and the thickness of my hand of 1 cm, the energy deposited is 2.5×10^8 eV. In other units its 1.07×10^-11 calories, 4.49×10^-11 Joules, and 1×10^-14 grams of TNT. If there are hundred billion protons per bunch in the beam (as the video said) then for every bunch you get 4.49 Joules or 0.001 grams of TNT of energy. (emphasis mine)

There are two beams, each of which contains 2808 bunches. Don't worry about the effect of multiple passes, though, since there won't be any tissue left in the beam's path by the time the first pass is over.

A more informative comment showed up later:
31. Xerxes Says:
September 21st, 2010 at 10:45 am

I think the hand-beam question is best answered by this document: http://lsag.web.cern.ch/lsag/BeamdumpInteraction.pdf

Granted, a carbon block isn't an exact model of the human hand, but it's probably close enough. The key points are:

1) "this energy deposit over 85 s is long enough to change the density of the target material. The density decreases at the inner part of the beam heated region because of the outgoing shock waves in the transverse direction. As an example, after the impact of 200 bunches with a size of = 0.2 mm, a maximum temperature of 7000K and a density decrease by a factor of 4 is expected." The results of heating your hand to 7000K and increasing its volume by a factor of 4 are probably best not imagined. Since a full beam is 2808 bunches instead of 200, you might want to scale that by a factor of 10 too.

2) But on the other hand (hehe): "The beam tunnels through the target and deposits the energy with a penetration depth of 10 m to 15 m" Since your hand is not 10m thick, you won't pick up the full effect. This paper goes into some detail of the spatial distribution of the energy dump: http://cdsweb.cern.ch/record/972357/files/lhc-project-report-930.pdf So at hand-thickness of 2ish cm, you'd only get maybe an eighth of the effects of #1, so your hand will only reach the more modest temperature of 1000K (times 10 for a full 2808 bunches?). The shockwave from the blast will extend several cm in the transverse direction; translation, the rest of your hand will be blown off by the middle of your hand exploding. Probably the part of the accelerator apparatus downstream of your hand picks up the rest of the energy. The rest of you probably wouldn't want to be standing next to it when it blows.

Cool pictures of the effects of a low-energy (450-GeV) beam on copper plates are in http://dx.doi.org/10.1109/PAC.2005.1590851

(I spent so much time looking up references, several other people made the same points. Oh well.)

Note particularly the fact that if one beam hit the solid graphite beam dump without being swept around during the pass, the surface would be at 7000 C, and would be well in the process of exploding, by the time the first 200 bunches had hit. Your hand, having a lower boiling point than graphite, would begin to remove itself from the path of the beam somewhat sooner, and would therefore probably absorb rather less energy. That may be small consolation, though, since it pretty much means that the splattered remnants of your hand wouldn't be as intensely radioactive as the carbon in the beam dump would be.

"Killing all humans" didn't make the list?

Whoops, I just replied to the wrong comment. That should have been in response to the next comment down.