Vapor Pressure and Raoult's Law

The molecules at the surface of a liquid are weakly bonded compared to the molecules beneath the surface. For this reason, the molecules at the surface easily vaporize at temperatures lower than the boiling point. This process is called evaporation.

All liquids undergo evaporation, but not forever. They evaporate until the partial pressure of their gas form reaches a certain level, which is known as vapor pressure. Vapor pressure is defined as the partial pressure of a vapor in dynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

To understand vapor pressure, we must get familiar with the concept of dynamic equilibrium. Let's discuss this through an example. Normally if you leave a cup of water outside, the water will completely vaporize if you give it enough time. This is because "outside" is not a closed system, and the pressure of water vapor will never reach the vapor pressure as the evaporated vapor molecules will fly away. However if you put a cup of water in a small closed area, the surface level will decrease at first but will stay still afterwards. At this point, we say that the water and vapor have reached a dynamic equilibrium state, where the rate of condensation and rate of vaporization are equal. In other words, the rate at which vapor condenses into water is equal to the rate at which water vaporizes, and therefore it looks like nothing is happening. The partial pressure of water vapor will stay at some constant value, which is the vapor pressure.

Vapor pressure can be a measure of intermolecular forces. Molecules that bond weakly to each other usually have high vapor pressures, and molecules that interact weakly with each other generally have low vapor pressures.

Vapor pressure tends to increase as temperature. This is because as temperature increases, the molecules have more energy, which means they become more labile. The graph below shows the vapor pressures of some liquids according to temperature. Note that mercury and ethylene glycol have relatively low vapor pressures as they have strong intermolecular attractions.

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Also observe from the graph the transverse line labeled "normal boiling point (1 atm)". Another interesting fact about vapor pressure is that the boiling point is equal to the temperature at which vapor pressure equals atmospheric pressure.

Raoult's law states that the partial vapor pressure of a component of an ideal mixture is the vapor pressure of the pure component multiplied by its mole fraction. An ideal mixture presumes that the intermolecular interactions are equal between all molecules inside the mixture. If there is a mixture of of ethanol and water, the partial vapor pressure that water vapor exerts in this situation is $$p_{\text{water}}=p^*_{\text{water}}x_{\text{water}},$$ where $p^*_{\text{water}}$ is the vapor pressure of pure water and $x_{\text{water}}$ is the mole fraction of water in this mixture.

Suppose you have an ideal mixture of 1 mol each of water and ethanol. If the temperature of the solution is $50^\circ\text{ C}$ and the partial vapor pressure exerted by water vapor is $50\text{ mmHg},$ what is the vapor pressure of pure water at $50^\circ\text{ C}?$

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According to Raoult's law we have $$\begin{aligned} p_{\text{water}}&=p^*_{\text{water}}x_{\text{water}}\\ 50\text{ mmHg}&=p^*_{\text{water}}\times0.5\\ \Rightarrow p^*_{\text{water}}&=100\text{ mmHg}.\ _\square \end{aligned}$$