Published on *METEO 300: Fundamentals of Atmospheric Science* (https://www.e-education.psu.edu/meteo300)

We can derive the equation for *e _{s}* using two concepts you may have heard of and will learn about later:

$$\frac{1}{{e}_{s}}\frac{d{e}_{s}}{dT}=\frac{{l}_{v}}{{R}_{v}{T}^{2}}$$[1]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaGymaaqaaiaadwgadaWgaaWcbaGaam4CaaqabaaaaOWaaSaaaeaacaWGKbGaamyzamaaBaaaleaacaWGZbaabeaaaOqaaiaadsgacaWGubaaaiabg2da9maalaaabaGaamiBamaaBaaaleaacaWG2baabeaaaOqaaiaadkfadaWgaaWcbaGaamODaaqabaGccaWGubWaaWbaaSqabeaacaaIYaaaaaaaaaa@4182@

where ** l_{v} is the enthalpy of vaporization** (often called the latent heat of vaporization, about 2.5 x 10

What is the physical meaning? The right-hand side of [3.9] is always positive, which means that the saturation vapor pressure always increases with temperature (i.e., *de _{s}*/

This expression can be integrated, assuming that *l _{v}* is a constant with temperature (it is not quite constant!) to give the equation:

$${e}_{s}={e}_{so}\text{\hspace{0.17em}}\mathrm{exp}\left(\frac{{l}_{v}}{{R}_{v}{T}_{o}}\right)\mathrm{exp}\left(\frac{-{l}_{v}}{{R}_{v}T}\right)$$[1]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwgadaWgaaWcbaGaam4CaaqabaGccqGH9aqpcaWGLbWaaSbaaSqaaiaadohacaWGVbaabeaakiaaykW7ciGGLbGaaiiEaiaacchadaqadaqaamaalaaabaGaamiBamaaBaaaleaacaWG2baabeaaaOqaaiaadkfadaWgaaWcbaGaamODaaqabaGccaWGubWaaSbaaSqaaiaad+gaaeqaaaaaaOGaayjkaiaawMcaaiGacwgacaGG4bGaaiiCamaabmaabaWaaSaaaeaacqGHsislcaWGSbWaaSbaaSqaaiaadAhaaeqaaaGcbaGaamOuamaaBaaaleaacaWG2baabeaakiaadsfaaaaacaGLOaGaayzkaaaaaa@4F84@

Generally *T _{o}* is taken to be 273 K and

*e*depends only on_{s}*T*, the absolute temperature. It is essentially independent of the atmospheric pressure, or any other factors.*l*is not constant with temperature but instead changes slightly (from 2.501 x 10_{v}^{6}J kg^{–1}at 0^{o}C to 2.257 x 10^{6}J kg^{–1}at 100^{o}C).- Thus, the most accurate forms of the integrated Clausius–Clapeyron Equation are more complicated but easy to deal with when using a computer.

What does the plot of this equation look like?

What happens between vapor and ice? The same methods can be applied and the same basic equations are obtained, except with a different constant:

$${e}_{si}={e}_{so}\text{\hspace{0.17em}}\mathrm{exp}\left(\frac{{l}_{s}}{{R}_{v}{T}_{o}}\right)\mathrm{exp}\left(\frac{-{l}_{s}}{{R}_{v}T}\right)$$[1]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwgadaWgaaWcbaGaam4CaiaadMgaaeqaaOGaeyypa0JaamyzamaaBaaaleaacaWGZbGaam4BaaqabaGccaaMc8UaciyzaiaacIhacaGGWbWaaeWaaeaadaWcaaqaaiaadYgadaWgaaWcbaGaam4CaaqabaaakeaacaWGsbWaaSbaaSqaaiaadAhaaeqaaOGaamivamaaBaaaleaacaWGVbaabeaaaaaakiaawIcacaGLPaaaciGGLbGaaiiEaiaacchadaqadaqaamaalaaabaGaeyOeI0IaamiBamaaBaaaleaacaWGZbaabeaaaOqaaiaadkfadaWgaaWcbaGaamODaaqabaGccaWGubaaaaGaayjkaiaawMcaaaaa@506C@

where *e _{si}* is the saturation vapor pressure for the ice vapor equilibrium and

Equations for *e _{s}* and

$$\begin{array}{l}{e}_{s}={e}_{so}\mathrm{exp}\left[\left(6808\text{K}\right)\left(\frac{1}{{T}_{o}}-\frac{1}{T}\right)-5.09\mathrm{ln}\frac{T}{{T}_{o}}\right]\hfill \\ \hfill \\ {e}_{si}={e}_{so}\mathrm{exp}\left[\left(\text{6293K}\right)\left(\frac{1}{{T}_{o}}-\frac{1}{T}\right)-0.555\mathrm{ln}\frac{T}{{T}_{o}}\right]\hfill \\ \hfill \\ \text{where}\text{\hspace{0.17em}}{T}_{o}=273\text{K},\text{\hspace{1em}}{e}_{so}=6.11\text{hPa}\hfill \end{array}$$[1]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@9770@

Note that *e*_{so} is the saturation vapor pressure at *T*_{o} and that *e*_{so} = 6.11 hPa and *T*_{o} = 273 K. Note how the constants are slightly different because the latent heat of vaporization for liquid water is different from the latent heat of vaporization for ice. Note that *T* in these equations must be in Kelvin.

Simply put, the dewpoint temperature is the temperature at which the atmosphere’s water vapor would be saturated. It is always less than or equal to the actual temperature. Mathematically,

$$w(T)={w}_{s}({T}_{d})$$[1]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEhacaGGOaGaamivaiaacMcacqGH9aqpcaWG3bWaaSbaaSqaaiaadohaaeqaaOGaaiikaiaadsfadaWgaaWcbaGaamizaaqabaGccaGGPaaaaa@3C79@

which means that the water vapor pressure at some temperature *T* (not multiplied by* T*) equals the water vapor saturation pressure at the dewpoint temperature, *T _{d}*. So we see that because

We can draw the phase diagram for water. There are three equilibrium lines that meet at the triple point, where all three phases exist (*e _{s}* = 6.1 hPa;

Credit: W. Brune

Is it possible to have water in just one phase? Yes!

The simplest case is when all the water is vapor, which occurs when the water vapor pressure is low enough and the temperature (and thus saturation vapor pressure) is high enough that all the water in the system is evaporated and in the vapor phase.

Let’s think about what it would take to have all the water in the liquid phase. Suppose we have a cylinder closed on one end and a sealed piston in the other that is in a temperature bath so that we can hold the cylinder and its contents at a fixed temperature (i.e., isothermal). Initially we fill the cylinder with liquid water and have a small volume of pure water vapor at the top. If we set the bath temperature to, say, 280 K and let the system sit for a while, the vapor will become saturated, which is on the *e _{s}* line. For isothermal compression, in which energy is removed from the system by the bath in order to keep the temperature constant, a push on the piston will slightly raise the vapor pressure above

In the atmosphere, ice or liquid almost always has a surface that is exposed to the atmosphere and thus there is the possibility that water can sublimate or evaporate into this large volume. Note that the presence or absence of dry air has little effect on the condensation and evaporation of water, so it is not the presence of air that is important, but instead, it is the large volume for water vapor that is important.

Conditions can exist in the atmosphere for which the water pressure and temperature are in the liquid or sometimes solid part of the phase diagram. But these conditions are unstable and there will be condensation or deposition until the condensation and evaporation or sublimation and deposition come into equilibrium, just as in the case of the piston above. Thus, more water will go into the liquid or ice phase so that the water vapor pressure drops down to the saturation value. When the water pressure increases at a given temperature to put the system into the liquid region of the water phase diagram, the water vapor is said to be supersaturated. This condition will not last long, but it is essential in cloud formation, as we will see in the lesson on cloud physics.

Note also that the equilibrium line for ice and vapor lies below the equilibrium line for supercooled liquid and vapor for every temperature. Thus *e _{si} < e_{s}* for every temperature below 0

- Find
**Practice Quiz 3-2**in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 3-2**. You will be allowed to take this quiz only**once**. Good luck!

**Links**

[1] mailto:MathType@MTEF