This Wikipedia page states that sawdust, cellulose etc. can be used to absorb water from azeotropic ethanol to produce anhydrous ethanol. How does this work? Why would the sawdust not simply absorb any liquid, not just water?
[Physics] How does sawdust selectively absorb water
home-experimentphysical-chemistry
Related Solutions
The melting point of nitrogen is greater than that of oxygen. Is gaseous oxygen really liquid oxygen dissolved in gaseous nitrogen? No. It is gaseous oxygen.
"Water vapour in the air" (your phrase) is just that - vapour. Not liquid droplets "dissolved" in air.
You may be confused by the fact that wet steam does contain water droplets, and the steam from a kettle is wet. However, it is perfectly possible to produce dry steam, and this stuff is enormously dangerous when let loose - it's invisible but will cook flesh almost instantly.
Your biggest mistake is to assume that evaporation is equivalent to a solid being dissolved, specifically salt into water. The difference can be illustrated by the fact that water will evaporate into any gas (with the obvious caveat that it must not react with the gas), and the rate is controlled by temperature and the partial pressure of water vapor in the surrounding atmosphere. Salt, on the other hand, simply will not dissolve in gasoline, and in fact will not dissolve in any non-polar liquid.
You can actually predict the shape of the profile precisely using the arguments you mention above, which are by and large correct. To do so, you can make the following assumptions:
- Neglect viscosity (not a great assumption, but it's a start).
- The pressure is the same everywhere in the fluid—the edges are free surfaces, so this is reasonable.
- The flow is axially symmetric (i.e. the top-down cross section is always circular).
If you do this, and take the location of the faucet as the origin, you can then state the relationship between the gravitational potential energy and the flow speed using Bernoulli's equation as:
$$\rho g h + \rho \frac{1}{2}v^2 = \rho \frac{1}{2}v_0^2$$
where $v$ is the speed of the fluid as a function of height $h$, $\rho$ is density, and $v_0$ is the speed at which the water leaves the faucet.
Solving for $v$, you'll find that:
$$v = \sqrt{v_0^2 - 2gh}$$
As the fluid moves further down (i.e. as $h$ becomes further negative), the speed increases as you'd expect.
Then you can use conservation of mass for the rest. Assuming steady flow, you'll find that
$$A_1 v_1 = A_2 v_2$$
for any two cross-sections of the flow. Using the cross-sections at the faucet and another arbitrary cross-section, and declaring the faucet radius as $r_0$, you'll find:
$$\pi r_0^2 v_0 = \pi r^2 v$$ $$\pi r_0^2 v_0 = \pi r^2 \sqrt{v_0^2 - 2gh}$$
Solving for the radius $r$, you find up getting the following expression:
$$\boxed{r(h) = \frac{r_0 \sqrt{v_0}}{(v_0^2 - 2 g h)^{1/4}}}$$
This drop in the radius as the height decreases is consistent with your illustrations. For example, here is what I analytically determine as the flow profile when I use standard values for a bathroom sink faucet flow ($r_0 = 1.5$ centimeters, $v_0 = 0.134$ meters per second, and $g = 9.81$ meters per second squared):
Notice that the flow profile becomes effectively straight at distances observable in your common bathroom sink (4 inches or so). This is consistent with your observations.
After a certain point, the stream becomes so thin that surface tension effects along with shearing at the air-water interface begin to destabilize the shape and cause it to break up into droplets. In addition, the flow becomes turbulent after a certain distance from the faucet, so this prediction is only accurate for the early stages of such a flow (i.e. for "small" $h$).
Best Answer
This is somewhat conjectural, but I think that sawdust is selective in terms of how polar the liquid is. Water and ethanol are both polar, but water significantly more so (ethanol is significantly less nonpolar at the CH3 end).
To determine if this is indeed the case, I tried pouring some very low-viscosity (so that it tends to ball up less) cooking oil on sawdust, and it rolled right off; high-concentration hydrogen peroxide was absorbed quickly, apparently all of it, so this seems fairly likely.