8
$\begingroup$

This is quite an interesting situation I had thought of. Consider two beakers, each containing $\pu{180 g}$ of water. Also, suppose one of the two has $\pu{1.8 g}$ of glucose dissolved in it. Let's call the one without glucose as A, and the one with glucose as B. I keep a small bell-jar that is large enough to just cover both of the two vessels. Now I will try to measure the vapor pressure inside the jar. Here's a little diagram to show what I mean:

enter image description here

I do not have access to bell-jars or manometers so I'll try to use Raoult's law to find the vapor pressure inside the vessel. The law in a simplified way would be:

$$P_{\text{solution}}=P_{\text{solvent}}x_{\text{solvent}}$$

Where $x$ denotes mole fraction, and solute is considered non-volatile. Keeping vessel A in mind, I get the vapor pressure as 23.8 mmHg, while applying Raoult's law for vessel B, I get the vapor pressure as:

$$P=23.8\left(\frac{10}{10.01}\right)=23.7 \text{mmHg}$$

Now, both of the values can't be true at the same time, and that's what I believe is the paradox.

How can this paradox be resolved?

My thoughts are that perhaps water from vessel A condenses back onto B, thus slowly increasing water content in B, increasing the vapor pressure. However, since theres solute already in B, the vapor pressure can never reach the value provided by the pure water.

What are your thoughts? Does the paradox ever resolve or does it remain in an unbalanced situation?

$\endgroup$
  • 4
    $\begingroup$ It is not a paradox. It only means those two are not in equilibrium and the solvent slowly moves between the two. $\endgroup$ – Greg Apr 29 '17 at 16:01
  • 1
    $\begingroup$ ... and will reach equilibrium only when all the solvent in A has become vapour and solvent in B. $\endgroup$ – porphyrin Apr 29 '17 at 17:22
6
$\begingroup$

Although "paradox" is not quite the right term, what you have discussed is actually a simple, yet interesting and important phenomenon.

Given the ideal situation as you have presented, your thoughts on what would happen are correct. If the system were to achieve $\pu{100\%}$ humidity with respect to the pure water, that would always be slightly over $\pu{100\%}$ humidity with respect to solution B, so water would condense onto B. This of course would cause the humidity with respect to pure water to drop, so more would evaporate. All things being ideal, eventually all of the water in A would transfer to the solution in B (plus the gas phase). The rate that would happen would depend strongly on the concentration of the solute in B.

An example of this situation playing out in the environment can be seen in the growth of water and ice particles in clouds. The vapor pressure of one particle can differ from that of another due to solute concentration, particle size (small droplets have a greater vapor pressure than larger droplets) and phase (supercooled water droplets have a greater vapor pressure than ice particles at the same temperature). Regardless of the cause of the vapor pressure difference, whether it's solute concentration or one of the other factors, the situation is the same as depicted in your question. And the result is the same; the particles having greater vapor pressure will end up evaporating and then condensing onto the particles having lower vapor pressure (of course we're talking about vapor pressure of water/ice only, not any solute). This is generally referred to as the Wegener–Bergeron–Findeisen process, although this technically only refers to water-to-ice vapor transfer.

$\endgroup$
1
$\begingroup$

Your post suggests (in fact, requires) you believe that Raoult's Law is accurate to 0.5%. It is not. Quantum Mechanics, the foundation of modern chemical theory, was created/discovered in the 1920s, Raoult's Law in the 19th Century. That's pretty good evidence that the law will be, at best, an approximation to the actual state of the system. As greg commented, you assume equilibrium - or at least, you should have understood that Raoult's Law requires equilibrium or near equilibrium. You are correct that since the two liquids have different vapor pressures, then equilibrium can not (yet) exist. The outcome would be (given an infinite amount of time, given a isolated system, given no bacteria nor oxidants present to consume the glucose, that the pure water (pure, right? no dissolved salts, but what about the dissolved air?) that the glucose solution would increase in mass & volume eventually overflowing its container. There are other considerations, which we ignore: surface tension and gravitational potential energy (adding more volume raises the height). One thing I learned fairly recently is that if you picture a cylinder of gas, the thermodynamics I learned in chemistry classes (including graduate level) and the thermodynamics I learned in freshman Physics, says that that cylinder (as a well insulated and closed system) will eventually attain thermal equilibrium. That is wrong. The temperature and density of the gas in the cylinder will depend on height. Thermal equilibrium will "never" be achieved.(although for a sufficiently "short" cylinder, the temperature and density gradient might be so small as to be impossible to measure.) ("never" meaning as long as the planet and cylinder exist. In the long run, the planet will most likely (we don't know for sure) be pulled into the Sun... but I digress...). Chemistry talks about many things, it is much more of a pragmatic science that Physics - which is more theoretical. (Although there may be more pragmatic physics used day-to-day than theoretical chemistry...) One way to divide chemical theory (there are many ways) is between Thermodynamics (equilibrium or near equilibrium) and Kinetics. I advise you always keep in mind whether a particular relationship or process or theory is thermodynamic (usually about energy, about path-independent initial and final states) or kinetics (about speed, shape, and specific process (path)).

$\endgroup$
  • $\begingroup$ You suggest that quantum mechanics trumps Raoult's law, but then you don't explain what you mean (or I missed it). Could you explain how quantum mechanics negates Raoult's law? $\endgroup$ – airhuff Apr 29 '17 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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