0
$\begingroup$

Consider a vessel with a volatile liquid in it. Let's say it has a vapor pressure $P^0$ when it is pure. Now a non-volatile solute is added to it. Its vapor pressure decreases as $P^0\chi_{solvent}$.

Why does this happen at a molecular level?


The reason that I have come across says that less surface area is available for evaporation and hence this effect. But this does not make sense to me as we can have liquid in container of all shapes and sizes and will still get the same Vapor Pressure.


Some users have suggested me a link to a similar question. The answers in the link are remarkable but the explanation I was hoping for was more likely to involve forces (as in Physics) and maybe a sort of "free body diagram" explaining how these forces are acting(words rather than actual diagram will be fine too) rather than thermodynamic picture.

$\endgroup$
  • $\begingroup$ Linked thermodynamic explanation: chemistry.stackexchange.com/questions/22395/… $\endgroup$ – Safdar Sep 22 at 15:02
  • $\begingroup$ @Safdar The answer in the link is talking about the thermodynamic and macroscopic picture of the phenomenon but I am specifically interested in the microscopic interpretation. $\endgroup$ – Tony Stark Sep 22 at 15:10
  • $\begingroup$ For solvent vapour tension not to decrease with the molar fraction, tendency of solvent molecules to leave liquid would have to increase with decreasing it's molar fraction. $\endgroup$ – Poutnik Sep 22 at 17:01
3
$\begingroup$

For solvent at temperature $T$, surface area $A$ and having the molar fraction $x$, the rate of molecules leaving liquid is:

$$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$

Lower molar fraction means lower surface density and lower evaporation rate per surface area.

Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ includes probability of a molecule leaving liquid.

For vapour of the partial pressure $p$, the rate of molecules coming to liquid from vapour is:

$$\frac {\mathrm{d}n}{\mathrm{d}t}=C_2 \cdot A \cdot p \cdot g(T) $$

The $g(T)$ involves temperature dependent relation of pressure and collision rate with liquid surface. Higher temperature means higher kinetic energy and less collisions needed for the same pressure.

For equilibrium, the net rate must be zero, therefore:

$$p= \frac{C_1 \cdot x \cdot f(T)}{C_2 \cdot g(T)}$$

Therefore the equilibrium pressure is proportional to the molar fraction.

It is quite analogical to the steady state of water level(=pressure), when it is simultaneously being filled up(=evaporation) and drained(=condensation). When you dilute a non-volatile solute, you decrease the filling up (= surface concentration of molecules is lower), so draining (=condensation), brings the water level(=pressure) to lower value to match draining rate with the filling.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Sir does that mean the explanation that I have come across is true though not properly depicted? $\endgroup$ – Tony Stark Sep 23 at 8:07
  • $\begingroup$ Secondly the way I am interpreting this is that as soon as non-volatile solute is added, the rate of evaporation decreases. But the rate of vaporisation is still high and hence some vapor comes back in the liquid in the net exchange. This continues until the pressure becomes smaller to an extent both rates again become equal. $\endgroup$ – Tony Stark Sep 23 at 8:13
  • $\begingroup$ The explanation is true, but you may have got it wrong. If some surface area contains 1000 solvent molecules,, the gross rate of evaporation is 2 time bigger than if there is just 500 molecules. $\endgroup$ – Poutnik Sep 23 at 8:27
  • $\begingroup$ Your 2nd comment - you got it. $\endgroup$ – Poutnik Sep 23 at 8:28
2
$\begingroup$

Imagine the converse situation:

  • A sealed chamber has an open beaker of the volatile solute.
  • After some time, equilibrium vapor is established.
  • Now an open beaker of the non-volatile solvent is introduced into the chamber.

Some molecules of the solvent in the chamber's "atmosphere" will dissolve into the open beaker of solvent, and that process continues until all the solute evaporates. Since the solute dissolves into the solvent, it is absorbed from the atmosphere, lowering the pressure.

As an example, consider a desiccator containing $\ce{CaCl2}$. If you start with a container with some water in the desiccator, you end with a puddle of $\ce{CaCl2. n H2O}$ and very little $\ce{H2O}$ in the air.

| improve this answer | |
$\endgroup$
-3
$\begingroup$

If we add non volatile solute to solvent for example water, we decrease the tendency of water to evaporate in gas phase because solute particle obstructs the evaporation of water in gas phase therefore vapour pressure decrease when we add non volatile solute to solvent like water. This image is from ncert:

enter image description here

| improve this answer | |
$\endgroup$
  • $\begingroup$ That is the exact thing that does not make any sense to me. $\endgroup$ – Tony Stark Sep 22 at 16:46

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