During vaporization, molecules in a liquid gain enough kinetic energy to break the bonds holding them to the rest of the molecules in the liquid and enter the gas. This can also occur in reverse: gas molecules can collide with the surface of the liquid and re-enter the liquid.

If you place a liquid in a closed container, at first there are very few molecules of the liquid in vapor form, and so the reverse process doesn't occur very often. As more of the liquid converts to vapor, the rate of the reverse process increases until finally there are the same number of molecules entering and leaving the liquid. Once this equilibrium is reached, the number of molecules in vapor form doesn't change, and we can compute the pressure of the vapor. The equilibrium is temperature dependent: as the temperature rises so does the vapor pressure. For example, at 25^oC, water has a vapor pressure of 24 mmHg, at 100^oC it has a vapor pressure of 760 mmHg.

In an open container, the vapor molecules can spread out to a great distance, and the liquid never reaches the equilibrium that it does in a closed container. This is why puddles left by a rainstorm evaporate over time.

Once you know the vapor pressure of a liquid at a certain temperature, you can use the ideal gas law to relate V and n to the pressure.

Example: Ether (CH₃CH₂)₂O has a vapor pressure of 537mmHg at 25^oC. If 20 grams of liquid ether are placed in a closed, 1 L flask at 25^oC, what is the mass of ether in vapor form when equilibrium is reached?

To begin, we can use the relationship PV=nRT since we know P_ether,V and T. The pressure of 537 mmHg can be converted to atmospheres by the relationship 1 atm = 760 mmHg. Thus P_ether = 537 mmHg * 1atm/760 mmHg = .707 atm.

n_ether = P_etherV/RT
n_ether = 0.707 atm* 1L /0.0821 L atm/mol K * 298 K
n_ether = 0.0289 moles

Ether weighs 74.12 grams/mole, so there are 2.14 grams of ether in the vapor phase.

Information provided by: http://learn.chem.vt.edu