Lets say we have a membrane which separates a tank of pure water from a tank of water saturated with a solute(there are also undissolved chunks of solute which can immediately dissolve if water is added). Is there a theoretical maximum of osmotic pressure that can be achieved? What's the highest we can achieve using known solutes? I know there's an equation for estimating osmotic pressure for low molarity but I'm only interested in fully-saturated solutions. As such, the errors stemming from using that formula in extreme-molarity situations makes its usage here unreasonable. Is there an equation for the fully-saturated osmotic pressure? Does having multiple solutes effect anything? Any help is appreciated.

  • $\begingroup$ The value for any non-ideal concentration (including saturation) will have to be determined experimentally. There is no simple equation describing nonideality. $\endgroup$
    – Karl
    Commented Mar 15, 2022 at 21:02
  • 1
    $\begingroup$ I updated my answer; you might want to take another look. $\endgroup$
    – theorist
    Commented Mar 18, 2022 at 4:59

1 Answer 1


We can do an end-run around the van't Hoff equation, and its needed ideality approximations, by instead looking directly at experimentally-determined vapor pressure lowering in saturated aqueous solutions. From each of these we can determine the activity of water, and from that we can calculate the osmotic pressure.

Conveniently, the online CRC Handbook provides a table of vapor pressures of saturated aqueous salt solutions $[1]$. To find the one that would give the highest osmotic pressure, we want the want the one that yields the greatest vapor pressure lowering, i.e., the one with the lowest vapor pressure at a given temperature. Of the 19 listed salts, the one whose saturated solution has the lowest vapor pressure at $\pu{25 ^{\circ}C}$ is, by a large margin, $\ce{LiCl(aq)}$.

Note that there may be solid substances whose aqueous saturated solutions yield an even lower water vapor pressure than does $\ce{LiCl(aq)}$*; but given how much lower $\ce{LiCl(aq)}$'s vapor pressure is than that of the 2nd place finisher on this list ($\pu{0.384 kPa}$ for $\ce{LiCl(aq)}$ vs. $\pu{1.536 kPa}$ for $\ce{K2CO3(aq)}$), I think we can safely say that $\ce{LiCl(aq)}$ is a strong contender (for even higher osmotic pressures, see my discussion of water-ethanol solutions at the end).

[*One candidate is saturated $\ce{NaOH(aq)}$, but I found it surprisingly hard to find data for its vapor pressure; this MSDS says 50% NaOH has a vapor pressure of $\pu{1.2 hPa}$ at $\pu{20 ^{\circ}C}$, which I assume means $\pu{0.12 kPa}$: http://www.labchem.com/tools/msds/msds/LC24150.pdf . However, I don't know how much I can trust an MSDS for this type of information, particularly since the MSDS doesn't itself cite a source.]

With no approximations,

$$\int_{p'}^{p'+\Pi} \bar{V}(p) dp + R T\ln a =0,$$

where $p'$ is the ambient pressure, $\Pi$ is the osmotic pressure, $\bar{V}(p)$ is the molar volume of water (which is a function of pressure), and $a$ is the activity. This is the same equation that can be seen in the initial steps of the derivation of the van't Hoff equation in any physical chemistry textbook, modified to remove the two approximations typically made up to this point, namely that $\bar{V}$ is independent of pressure, and that $a = x$, where $x$ is the mole fraction. [Many more approximations are made subsequently in that derivation; we will avoid all of them, while introducing one of our own, to simplify the calculation of the pressure-dependence of the molar volume.]

The first step is to determine $a$ from the vapor pressure lowering. To do this we use:

$$p_{vap} = a*p^{\circ}_{vap},$$

where $p^{\circ}_{vap}$ is the vapor pressure of pure water, and $p_{vap}$ is the vapor pressure of the solution. Note that, to account for non-ideality, we are using the activity instead of the mole fraction:

$$a = \gamma x,$$

where $\gamma$ is the activity coefficient ($\gamma =1$ in an ideal solution).

Using $p^{\circ}_{vap} = \pu{3.169 kPa}$ (vapor pressure of pure $\ce{H2O}$ at $\pu{25 ^{\circ}C}$) and $ p_{vap} = \pu{0.384 kPa}$ (vapor pressure of water over saturated $\ce{LiCl(aq)}$ solution at $\pu{25 ^{\circ}C}$) gives:

$$\bf{a = 0.121174}$$

This is an extraordinarily low value. To achieve this in an ideal solution, you'd need to be (by mole fraction) $12\%$ water and $88\%$ $\ce{LiCl}$!

To account for the compression of water at high pressures, we use the compressibility of water: $k =\pu{45.8*10^-6 bar^-1}$ $[2]$. And thus:

$$\bar{V}(p)=\bar{V}^o (1-k*p),$$

where $\bar{V}^o$ is the molar volume of water at $\pu{1 bar}$. Note that this is itself an approximation, since $k$ is a function of pressure. A more sophisticated calculation would make use of an equation of state for water, but I think this is sufficient for our purposes.


$$\int_{p'}^{p'+\Pi} \bar{V}^o (1-k*p) dp + R T\ln a =0$$

Inserting $p' = \pu{1 bar}$, $\bar{V}^o =\pu{0.018068 L/mol}$, $k =\pu{45.8*10^-6 bar^-1}$, $T =\pu{298.15 K}$, and $a=0.121174$, and solving for $\Pi$, we obtain:

$$\bf{\Pi = \pu{3,000\bf{bar}}=\pu{2,900 \bf{atm }}=\pu{43,000 \bf{psi}} }$$

[I've rounded to 2 significant figures because $k$ has two sig figs, and, as explained above, assuming it's pressure-independent is itself an approximation.]

Note that this pressure is high enough to justify the extra work we did to account for the compressibility of water: Its molar volume at this pressure is 13% less than that at $\pu{1 bar}$.

To put this in context, this is approximately three times the pressure at the bottom of Challenger Deep, the deepest point in the world's oceans ($\pu{1127 bar}$ at $\pu{10925 m}$). Or if you're a fan of U-shaped tubes, this would be enough to create a column of water $\pu{30 km}$ high! Indeed, we are beginning to approach the transition to Ice IV, which occurs at $\approx \pu{9500 bar}$ at $\pu{25 ^{\circ}C}$—see Martin Chaplin's phase diagram of water $[3]$.

Furthermore, I could calculate an even higher osmotic pressure by extrapolating the vapor pressure lowering of $\ce{LiCl}$ to a higher temperature.

Alas, such a high osmotic pressure is not achievable, because of the technological limits of currently-available semipermeable membranes, which would begin to leak or collapse at pressures an order of magnitude below this! According to a $2018$ paper by Davenport, Deshmukh, Werber, and Elimelech $[4]$, "At present, only a few studies have assessed RO above $\pu{100 bar}$, with maximum tested pressures of $\pu{200 bar}$." They add that $\pu{300 bar}$ remains "a challenging target from membrane materials and module design perspectives." It appears the highest pressure rating currently available for commercial RO membranes is $\pu{1800 psi}$ ($\pu{124 bar}$) $[5,6]$.

Having said that, if technology were not a limit, we could achieve even higher osmotic pressures than those available from dissolving salts in water, by instead using a mixture of water with some liquid that is miscible with water in all proportions. One example would be ethanol. If you separated pure ethanol and pure water with a membrane permeable to water but not ethanol (and such membranes do exist—they are used by distilleries), the initial calculated osmotic pressure would be infinite, because the mole fraction of water on the ethanol side would be $0$, and thus the chemical potential of the water on the ethanol side would be $-\infty$.

Of course, this calculated osmotic pressure is not a true theoretical osmotic pressure, since the system needs to equilibrate. Thus the water would move from the water side to the ethanol side (leading to a finite mole fraction of water on the ethanol side) until the chemical potential of the water on both sides was the same. If the container is rigid, the theoretical hydrostatic pressure difference (equal to the theoretical osmotic pressure) one could generate would be enormous.

$[1]$ "Vapor Pressure of Saturated Salt Solutions," in CRC Handbook of Chemistry and Physics, 102nd Edition (Internet Version 2021), John R. Rumble, ed., CRC Press/Taylor & Francis, Boca Raton, FL. https://hbcp.chemnetbase.com/faces/documents/06_42/06_42_0001.xhtml

$[2]$ Sears, Zemansky, Young, and Freedman, University Physics, 10th Ed., Section 11-6.

$[3]$ https://water.lsbu.ac.uk/water/water_phase_diagram.html

$[4]$ Douglas M. Davenport, Akshay Deshmukh, Jay R. Werber, and Menachem Elimelech. High-Pressure Reverse Osmosis for Energy-Efficient Hypersaline Brine Desalination: Current Status, Design Considerations, and Research Needs. Environmental Science & Technology Letters 2018 5 (8), 467-475. DOI: 10.1021/acs.estlett.8b00274 https://pubs.acs.org/doi/10.1021/acs.estlett.8b00274

$[5]$ https://www.prnewswire.com/news-releases/first-commercial-order-of-a-next-generation-ultra-high-pressure-reverse-osmosis-uhp-ro-industrial-system-301058563.html

$[6]$ https://www.saltworkstech.com/articles/applying-ultra-high-pressure-reverse-osmosis-in-brine-management/

  • $\begingroup$ Wow! This is an amazing amount of scientific knowledge and effort! Thank you so much! I was originally wondering if one could use osmotic pressure to create a sort of cheap high energy density battery. Reading your initial comment made me realize that that'd basically be like compressing a virtually-incompressible fluid and storing energy in the minute change in volume at astronomical pressure. Something that'd work in theory but is nearly impossible in practice. I don't think any of the issues regarding that lie in the pressure differential across the membrane though. One could have a (1/3) $\endgroup$
    – Laff70
    Commented Mar 20, 2022 at 17:19
  • $\begingroup$ series of membranes separating sections of fluid of increasing concentrations. While the net osmotic pressure differential would be too much for a single membrane to withstand, a series of them could trivially withstand them. The real issue is the pressure on the container(especially if the container's volume would change, like it is in this case). In order to counteract the pressure, an opposing force needs to be generated. This could be supplied by a curved material with a lot of strain. With a spherical shell, the pressure differential that could be generated would be 2σln(b/a) where (2/3) $\endgroup$
    – Laff70
    Commented Mar 20, 2022 at 17:19
  • $\begingroup$ σ is the max stress the material can handle and b and a are the radii of the spherical shell. The other option is by constantly exchanging momentum with particles. An example of this would be a U-shaped pipe with fast fluid flowing through it. The fluid at the U-turn part would be accelerating towards the rest of the pipe and would create an opposing force. Enough of these and several layer of such pipes might allow for the creation of extreme pressure. Still impractical for energy storage. (3/3) $\endgroup$
    – Laff70
    Commented Mar 20, 2022 at 17:20
  • 1
    $\begingroup$ @Laff70: If you wanted to do energy storage on an industrial scale, you wouldn't need to go to the maximum possible osmotic pressure. You could just use the energy to power an industrial reverse osmosis device, and then extract the energy by allowing it to run in the reverse direction. The question is: How robust, cost-efficient, and energy-efficient would this be compared with other industrial-scale energy-storage devices? You can do a search for 'reverse osmois for energy storage'. Here's an example: sciencedirect.com/science/article/abs/pii/S0011916421001594 $\endgroup$
    – theorist
    Commented Mar 21, 2022 at 2:04
  • $\begingroup$ I was hoping for something which could in theory compete with gasoline and lithium ion batteries in terms of energy density. $\endgroup$
    – Laff70
    Commented Mar 21, 2022 at 20:53

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