When water is spilled on the floor, it is common experience that it will eventually evaporate. But room temperature is far below the boiling point of water; how can this happen? We have already discussed, in the "Chemical Reactions" chapter, that temperature is a representation the average speed of a substance; some molecules will have higher speed, and some will have lower. The molecules with high kinetic energy (speed) will transition to the gas phase. Eventually, every molecule will gain the necessary energy to evaporate. In an open situation, the entire pool will evaporate, given enough time.
However, if the container is closed and the amount of water is large, not all the water will evaporate. Instead, an equilibrium is reached between molecules evaporating and condensing. Because vapor is being added to the gas above the liquid, the vapor will exert a pressure on its container, called the equilibrium vapor pressure, or just vapor pressure for short. The vapor pressure is dependent only on temperature, but the vapor pressure at a specific temperature will vary from liquid to liquid. The vapor pressures of water from 0 to 100 °C are given in our "Reference" section. Since vapor pressure is a representation of how easily liquid molecules can escape into the gas phase, liquids with higher vapor pressures evaporate more easily. The ease with which a liquid can evaporate is called volatility: volatile liquids evaporate easily and have high vapor pressures.
Vapor pressure increases with temperature. When vapor pressure is equal to atmospheric pressure, the liquid molecules have enough energy to overcome the downward pressure of the atmosphere above them and become a gas; so, when vapor pressure equals atmospheric pressure, the liquid will boil. At higher altitudes, where atmospheric pressure is lower, the boiling point of water is also lower.
It seems that there is a direct relationship between the vapor pressure and temperature, but if you look at the entries in the "Reference" section, the relationship is not linear. In the 1800's, a mathematical equation was developed to relate vapor pressure and temperature:
ln P = -ΔHvap/RT + C
This is called the Clausius-Clapeyron equation, after its discoverers. P is the vapor pressure in atmospheres, T is the temperature in kelvins, R is the ideal gas constant (8.314 J/mol-K, a variation of the constant you'll learn about in the next page), ΔHvap is the enthalpy of vaporization of the liquid (the amount of energy necessary to vaporize one mole of liquid) in joules, and C is a constant specific to the liquid under study. A different form of the equation is:
ln (P1/P2) = ΔHvap/R * (1/T2 - 1/T1)
which eliminates the constant. This equation is useful for finding the enthalpy of vaporization, which we'll discuss more in the "Thermodynamics" chapter.
We noted above that small quantities of water will evaporate completely, while larger amounts will not and an equilibrium vapor pressure will be present. So, how do we know whether the liquid will evaporate entirely? We'll have to wait for an answer for this question, because we first need to discuss the Ideal Gas Law, which will allow us to predict the volume and pressure of any gas at any temperature and number of moles. This is the subject of the next page.