Answer:
Attractive forces between particles are inversely proportional to vapour pressure.
Explanation:
Inside a liquid, molecules undergo random motion (thermal motion), but also interact with one another via electromagnetic forces of different kinds, like Van der Waals forces, ion-dipole interactions, hydrogen bonds, etc. These forces keep the liquid together, giving it a definite volume, in distinction to gases, which take the volume of the vessel that contains them.
Now, some molecules in a liquid can attain a high velocity as a random outcome of thermal motion, if this molecule is at the liquid's surface, it might actually escape! actually, many molecules might do that, and form a vapour over the liquid's surface.
Now, we know that liquids exist, therefore this process has to reach an equilibrium, that means, once the vapour becomes <em>dense </em>(or <em>concentrated</em>)<em> </em>enough, it would be as likely for a vapour molecule to re-enter the liquid as it is likely for a liquid molecule to leave the liquid and enter into the vapour.
This is called vapour-liquid equilibrium.
How can we measure how "concentrated" the vapour is? by measuring the pressure above the liquid. We know by the ideal gas law that the number of molecules in a gas is proportional to pressure at constant volume and temperature.
But how does vapour pressure relate to intermolecular forces?
Simply, the stronger the intermolecular forces, the less likely a molecule at the liquid's boundary will be to shoot of into the vapour phase! and viceversa, if intermolecular forces are very weak, the molecules won't hold together much and many molecules will leave the liquid.
As an extreme case imagine a solid, for which intermolecular forces are the strongest, what's the vapour pressure of a solid? Do solids evaporate into the air? The answer is no, solids (with few exceptions) don't evaporate, and their vapour pressure is extremely small.
Cheers!
