I am trying to build a system of two connected chambers, one of which is also connected to a cuvette, so I can perform a spectrometry experiment with different concentrations of $\ce{KNO3}$ solution. In its simplest form, the cuvette and the first chamber would hold solution at the same concentration, while the second chamber would hold solution of higher concentration. I would perform one experiment, open a valve between the chambers, and wait for the concentration to equilibrate before performing the next experiment.

My advisor suggested this system and said that the equilibration time would likely be on the order of minutes.

I used this equation to calculate equilibration time between two chambers:

$$ t = \frac{\Delta m \cdot l_{v}}{A_{v} \cdot D \cdot 10^6 \cdot \Delta c} $$

where $t$ = equilibration time in s
$\Delta m$ = the difference in mass of $\ce{KNO3}$ between the chambers in g
$l_{v}$ = the length of the valve in m
$A_{v}$ = the orifice area of the valve in m^2
$D$ = the mutual diffusion constant of $\ce{KNO3}$ and $\ce{H2O}$, which is on the order of 10^-9 m^2/s
$\Delta c$ = the difference in concentration between the chambers in g/cm^3
the factor of 10^6 is to convert between g/cm^3 and g/m^3

For a cylindrical valve 1 cm long and 2 cm in radius (improbably short and wide) and chamber volumes 95 cm^3 and 400 cm^3, this gives an equilibration time of 4.7 days!

This is much more than I was expecting. I can reduce the sizes of the chambers to decrease the time demand, but I would prefer to keep them as large as I can, to increase the final concentration that can be achieved in the cuvette.

Is there something obvious that I'm missing in this calculation? Is there a better way to do it?


The equation above was derived by equating the mass flux $j_{m}$ with its definition from Fick's first law (using $c$ as mass/volume rather than amount/volume):

$$ j_{m}=\frac{\Delta m}{A_{v}\cdot \Delta t}=D\cdot\frac{\Delta c}{\Delta x} $$

I know that strictly the RHS should contain $\frac{\delta c}{\delta x}$ rather than $\frac{\Delta c}{\Delta x}$. Does this approximation make much difference? For now, I am also making the fairly large assumptions that the volumes of solution in each chamber do not change, and that the density of the solution is independent of concentration. I think I can include these considerations later on.

This is an actual system that I need to build (well, ask our machine shop to build for me) and I need to have a good idea of the best valve sizes to use as they have a significant effect on the overall structure.

  • 1
    $\begingroup$ I think your estimate time based on diffusion alone is probably quite accurate: diffusion is a fairly slow phenomenon in mixing phenomenons, and mass transfer through diffusion alone is quite slow. This is a main reason for which mechanical stirring is extremely common is chemistry labs. $\endgroup$
    – PLD
    Jun 2, 2012 at 21:09
  • $\begingroup$ You could, perhaps, determine the diffusion time experimentally by using a small, highly colored compound (permanganate? or copper (II) sulfate absorbs well), and watching the absorbance change to find the maximum. $\endgroup$ Jun 6, 2012 at 15:31
  • $\begingroup$ What is happening is not just diffusion, there is dispersion and free convection (due to concentration gradient). Your advisor is right in that it should be quick, but the reason is not diffusion. refer to "Diffusion" edition 3 by Ed Cussler, Pg 47. $\endgroup$
    – picakhu
    Jun 7, 2012 at 16:04
  • $\begingroup$ @picakhu, I have this book and I opened pg 47 which only speaks about free convection in a vertical tube. These are stirred solutions with a narrow tube connecting them. This is quite different from the situation presented here. Her advisor probably thinks she will be stirring the solution. And I don't think I agree with your definition of dispersion. $\endgroup$
    – Chris
    Jun 8, 2012 at 8:30
  • $\begingroup$ @Chris, I did not define dispersion in my answer. Just said that it is another factor to think about. This problem is far from trivial. I think convection will be the major component. $\endgroup$
    – picakhu
    Jun 8, 2012 at 19:03

1 Answer 1


In the future please specify where your equation came from, justify its use and specify all of your parameters.

When only a concentration gradient is present, the diffusion coefficient, $D$, is typically described by Fick's first law of diffusion. To determine $D$ in these terms we must know the flux, $j$, and the gradient, $d c / dx $. $c=$ concentration of our species divided by the volume and $x$ is the distance this occurs over.

Flux, $j$, is the (flow quantity/time)/area or (flow rate)/area. Fluxes depending on potential gradients are molecular, but bulk flow flux is called convective. Here we have molecular flux present and the units of this are kmol/(m$^2 \cdot$ s). (It's true that diffusion creates its own convection but there is no bulk flow so we ignore this diffusion-induced convection.) Properly our flux is the product of a transport property by the gradient of the driving force, $ j = (D) ($Gradient of driving force$)$. Note that $D$ has units of m$^2$/s. For example, $D$ is $10^{-5}$, $10^{-9}$, $10^{-10}$ m$^2$/s for some gases, liquids and solids respectively at STP.

Now let's think back to basic physics. Say we have a feather at $h$ height here on Earth and we drop it. It will experience an acceleration due to gravity, $g$. Remember that gravity has units m/s$^2$. We can estimate the time it takes to travel distance $h$ crudely by dimensional analysis: knowing that $g$ has units of m/s$^2$ and $h$ has units of m then time is simply $\sqrt{\frac{h}{g}}$. In fact the equation that describes this is $h = \frac{1}{2}gt^2$ and so we actually needed to multiply our original answer by $\sqrt{2}$. In the same way we can crudely estimate the time by dimensional analysis on the diffusion coefficient, $D$ with units m$^2$/s and going a distance with units of m. So we need to take the distance$^2$ divided by $D$ to get seconds. Say ( $0.01$m )$^2$ divided by $10^{-9}$ m$^2$/s gives $10^5$ seconds which is about 1.1 days. This could easily be 4 days if we needed to multiply by 3.7 which is certainly possible if your orifice is small. It is calculating these other things which turns out to be tricky. Although it's obvious the single easiest thing you can do is switch to a gas when considering just diffusion because a gas would take on the order of 10 seconds because $0.01^2/(10^{-5}) = 10$. You can also change the time by introducing another flux, such as by stirring the solution.

A better answer requires differential equations. Are you sure you want to venture down that road?

  • $\begingroup$ Hi Chris, thank you for your answer. I have added some extra information in the question above. Switching to a gas is not possible, since this system is intended to study the KNO3 solution specifically. Stirring might be possible, though it raises its own problems as the total space for the solution chambers is quite constrained. I am willing to consider differential equations if it's necessary! :) $\endgroup$
    – bellisk
    Jun 11, 2012 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.