Question
I know that as air warms up it can hold more water before saturating. Why does adding heat allow air to hold more water (and please don't simply tell me it becomes less dense)?
Answer
Hi David, <br /> <br />This is a good question. Let me see if I can give you a good answer. <br /> <br />To understand the dependence of air saturation on temperature, you have to understand the concept of an equilibrium state. As you probably know, equilibrium means a "balance". In this case, it maybe easiest to consider some kind of container with water at the bottom and air at the top as an example. If the air in the box is initially dry, the water will begin to evaporate so that a portion of the liquid is transformed into water vapor. As the evaporation process continues, the water vapor pressure will increase because there are now more molecules of water vapor above the liquid surface. At the same time, some portion of the water vapor is condensing (changing from liquid to gas over the water surface) back at the liquid water-air interface. Eventually, an equilibrium state will be achieved so that the rate of evaporation of molecules from the liquid will be equal to the rate of condensation on the liquid from the increasingly moist air. This is our equilibrium state. When the equilibrium state is realized, the air is said to be saturated with respect to liquid water and the pressure exerted by the water vapor is called the saturation vapor pressure. We usually refer to the saturation vapor pressure as e_s. <br /> <br />Now, we can answer your question because it turns out that e_s depends only on temperature: e_s increases with increasing temperature. We can figure out what e_s is for any given temperature by solving an equation known as the Clausius-Claperyon equation. <br /> <br />I hope this helps, <br /> <br />Rob
Follow up from David: <br /> <br />"e_s increases with increasing temperature" My lack of education in physics gives me the understanding that if the saturation vapor pressure increased then a parcel of air would decrease in the amount of water it could contain. So how does increasing the saturation vapor pressure allow a parcel of air to contain more when it is heated? I can explain to the 8th graders that I'm teaching its like warm water dissolving more sugar than cold water but I can't tell them why it does. David <br /> <br />Reponse from Rob: <br /> <br />Hello David, <br /> <br />This is a difficult concept so don't feel bad if it's hard to grasp. I realize that, as a teacher, it's all the more exasperating to try and relate such a concept to kids. <br /> <br />I think we agree that the saturation vapor pressure represents the maximum pressure that water vapor can exert without condensing. In other words, it's just the maximum amount of vapor that the air can hold at any given temperature. As temperature increases, more molecules are leaving the liquid water interface (higher T means more kinetic energy/molecule) and there are subsequently more vapor molecules in the air before the equilibrium state is achieved (i.e., where the vapor pressure equals the saturation vapor pressure e_s). The same reasoning can be used to understand why, on a given day in the summer, the relative humidity (RH is the ratio of the amount of moisture in the air to the maximum amount of moisture that the air can hold at that temperature) can vary so much between, say, sunrise and noon. At dawn when the temperatures are relatively cool, the vapor molecules have relatively little energy so that the equilibrium state is reached at a relatively low vapor pressure (i.e., the air can't hold as much moisture without condensing) and the RH is high. At noon (assuming there's been no significant change in the air mass), the temperature is warmer, vapor molecules are zooming out of the liquid and into the vapor state very quickly, creating a relatively high vapor pressure before the equilibrium state is reached. The net result is that the air can hold more moisture (before saturation), and RH goes down. <br /> <br />Solving the equation for e_s exactly requires a bit of calculus; however, one can use the following empirical equation to see the change of e_s with temperature. This equation is accurate over the range of -35C < T < 35C. Here, e_s is in millibars and T is in deg. C <br /> <br />e_s = 6.112*exp^{(17.67*T)/(T+243.5)} <br /> <br />I hope this isn't too confusing. <br /> <br />Rob <br /> <br /> <br />
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