3.3 Phase Diagram for Water Vapor: Clausius Clapeyron Equation

3.3 Phase Diagram for Water Vapor: Clausius–Clapeyron Equation

The Clausius–Clapeyron Equation

We can derive the equation for e_s using two concepts you may have heard of and will learn about later: entropy and Gibbs free energy, which we will not go into here. Instead, we will quote the result, which is called the Clausius–Clapeyron Equation,

$\frac{1}{e_{s}} \frac{d e_{s}}{d T} = \frac{l_{v}}{R_{v} T^{2}}$

[3.9]

where l_v is the enthalpy of vaporization (often called the latent heat of vaporization, about 2.5 x 10⁶ J kg^–1), R_v is the gas constant for water vapor (461.5 J kg^–1 K^–1), and T is the absolute temperature. The enthalpy of vaporization (i.e., latent heat of vaporization) is just the amount of energy required to evaporate a certain mass of liquid water.

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/dT > 0). This positive slope makes sense because we know that as water temperature goes up, evaporation is faster (because water molecules have more energy and thus a greater chance to break the bonds that hold them to other water molecules in a liquid or in ice). At saturation, condensation equals evaporation, and since evaporation is greater, condensation must be greater as well. Much of the higher condensation comes from having more water vapor molecules hitting the liquid surface, which according to the Ideal Gas Law, means that the water vapor pressure is higher.

The temperature sensitivity of e_s is quite high. Plugging in the appropriate values to the right side of [3.9] yields 0.07 K^–1, which means that the saturation vapor pressure increases by 7% for every 1 K increase in temperature. This high sensitivity has profound implications for weather and climate.

Separating variables (e_s and T) in [3.9] and integrating, assuming that l_v is a constant with temperature (it is not quite constant!), yields:

$e_{s} = e_{s o} \exp (\frac{l_{v}}{R_{v} T_{o}}) \exp (\frac{- l_{v}}{R_{v} T})$

[3.10]

Generally T_o is taken to be 273 K and e_so is then 6.11 hPa.

Notes:

e_s depends only on T, the absolute temperature. It is essentially independent of the atmospheric pressure, or any other factors.
l_v is not constant with temperature but instead changes slightly (from 2.501 x 10⁶ J kg^–1 at 0 ^oC to 2.257 x 10⁶ J kg^–1 at 100 ^oC).
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?

graph of Clausius Clapeyron Equation. Curve going up as temperature increase

The equilibrium vapor pressure above a liquid water surface as calculated by the Clausius–Clapeyron Equation. ***The y-axis label is incorrect; it should say "equilibrium vapor pressure (hPa)".*** The line for liquid water can be extended below 273 K, the freezing point, because water can remain liquid at those low temperatures and become a “supercooled” liquid.

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

$e_{s i} = e_{s o} \exp (\frac{l_{s}}{R_{v} T_{o}}) \exp (\frac{- l_{s}}{R_{v} T})$

[3.11]

where e_si is the saturation vapor pressure for the ice vapor equilibrium and l_s is the enthalpy of sublimation (direct exchange between solid water and vapor = 2.834 x 10⁶ J kg^–1).

Equations for e_s and e_si that account for variations with temperature of l_v and l_s, respectively, can be found in Bohren and Albrecht (Atmospheric Thermodynamics, Oxford University Press, New York, 1998, ISBN 0-19-509904-4):

$\begin{array}{l} e_{s} = e_{s o} \exp [(6808 K) (\frac{1}{T_{o}} - \frac{1}{T}) - 5.09 \ln \frac{T}{T_{o}}] \\ e_{s i} = e_{s o} \exp [(6293 K) (\frac{1}{T_{o}} - \frac{1}{T}) - 0.555 \ln \frac{T}{T_{o}}] \\ where T_{o} = 273 K, e_{s o} = 6.11 hPa \end{array}$

[3.12]

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.

Dewpoint Temperature as a Measure of Water Vapor

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,

$w (T) = w_{s} (T_{d})$

[3.13]

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 w_s depends only on T_d at a given pressure, T_d is a good method for designating the absolute amount of water vapor.

The Phase Diagram for Water

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; T = 273.14 K). Along the line for e_s, vapor and liquid are in equilibrium, and evaporation balances condensation. Along the line for e_si, vapor and ice are in equilibrium and sublimation equals deposition. Along the line for e_sm, liquid and ice are in equilibrium and melting balances fusion.

Phase diagram for water for most water pressures and temperatures that are relevant to the atmosphere as described in the text above

Phase diagram for water for most water pressures and temperatures that are relevant to the atmosphere. e_s is the saturation pressure (i.e., equilibrium pressure) between liquid and vapor and e_si (***mislabeled as e_i in the diagram***) is the saturation pressure between vapor and ice. The dashed line is a continuation of e_s into the ice phase and represents supercooled water. All three phases meet at the triple point. This diagram is not to scale.

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 vertical cylinder closed on one end and a sealed piston at the other end. The whole cylinder is immersed in a constant-temperature water 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 e_s and there will be net condensation until equilibrium is obtained again. If we continue to slowly push in the piston, eventually all the cylinder’s volume will be filled with liquid water and the cylinder will contain only one phase: liquid. If we continue to push the piston and the bath keeps the temperature constant, then the water pressure will increase.

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 ^oC because l_s > l_v in the Clausius–Clapeyron Equation. This small difference between e_si and e_s can be very important in clouds, as we will also see in the lesson on cloud physics.

Quiz 3-2: Humidity and relative humidity.

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