Explanation for Raoults Law

Question

The vapour pressure of pure liquids is independent of quantity of substance.

But when in a solution, the vapour pressure of the components are given by $$P_A=P^°_AX_A$$$$P_B=P^°_BX_B$$

Since $A-A$, $A-B$ and $B-B$ interactions are assumed to be equal for an ideal solution:

1. In case of liquid-liquid solutions how does it affect the vapour pressure(like whats is mechanism at molecular level). I presume it has something to do with volumes of both liquids being added as a result of which it is kind of like each liquid occupying a larger volume than permitted so making is somewhat similar to effect on gaseous equilibrium upon adding inert gas at constant pressure But I hope someone might provide a better, more intuitive explanation.
2. In case of non-volatile solid solutes, I cannot think of any explanation. Since solids occupy intermolecular spaces of liquids, there is no change in volume of solution, so it should essentially have no effect on its vapour pressure similar to adding inert gas at constant volume to gaseous equilibrium.

Please explain with intuition why vapour pressure is not independent of concentration.

Answer 1

Starting with the easy answer, a solid can attract and absorb a liquid. For example, $\ce{CaCl2}$ is hygroscopic, and is sold packaged to reduce humidity in enclosed locations. Simply think, water molecules are more attracted to $\ce{CaCl2}$ than to the space above.

For liquids in liquids, consider those same intermolecular forces. For example, 100 mL of ethanol ($\ce{C2H5OH}$) + 100 mL of water ($\ce{H2O}$) combine to take up ~4% less volume. Again, thin of it as ethanol being more "attracted" to water than to air.

BTW, surface tension is similar, in that one measures cohesion of a liquid compared to adhesion to an interface.

Answer 2

The lower molar fraction of the volatile liquid A, regardless of if the other component B is volatile or not, means the lower rate of the liquid A evaporation.

It can be easily imagined that there is a lower number of molecules on the surface unit and therefore lower molecular evaporation rate

\begin{align} \frac{\text{d}N_\text{A}}{\text{d}t \cdot \text{d}A} &= k_\text{A,f} \cdot N_\text{A} &- k_\text{A,b} \cdot p_\text{A}\\ &= k_\text{A,f} \cdot N^\circ_\text{A} \cdot x_\text{A} &-k_\text{A,b} \cdot p_\text{A} \end{align}

At equilibrium, the rate of evaporation and condensation is equal.

\begin{align} k_\text{A,f} \cdot N^\circ_\text{A} \cdot x_\text{A} = k_\text{A,b} \cdot p_\text{A}\\ k_\text{A,f} \cdot N^\circ_\text{A} = k_\text{A,b} \cdot p^\circ_\text{A} \end{align}

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Legend:

• $N_\text{A}$ is molecular surface density of the component A.
• $N^\circ_\text{A}$ is molecular surface density of the purecomponent A.
• $t$ is time.
• $A$ is surface area.
• $k_\text{A,f}$ is the evaporation kinetic constant for given temperature.
• $k_\text{A,b}$ is the condensation kinetic constant for given temperature.
• $p_\text{A}$ is the partial pressure of the component A.
• $x_\text{A}$ is molar fraction of the component A.