Atmospheric moisture: Evaporation and condensation

● The goal of this section is to use previously learned concepts to develop an understanding of relative humidity and thus the processes that determine relative rates of evaporation and condensation in the atmosphere.

Expected prior knowledge

● Environmental processes

● Equilibrium and rate of processes

● Atmospheric pressure

● Evaporation and condensation

● Why water is a polar molecule

● Hydrogen bonds and potential energy

● Thermodynamics (e.g., TKE, temperature)

The rate of many environmental processes is dependent on how far a system is from equilibrium. Consider the diagram below. Ball A has more gravitational potential energy than Ball B because the difference in height from the top to bottom of the ramp is greater. Ball A will therefore be moving at a greater velocity than Ball B along the ramp.

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Previously, we learned that the second law of thermodynamics dictates that thermal energy will only spontaneously travel from warmer to colder objects (increasing entropy). The rate of energy transfer is dependent on the difference in temperature between the two objects. This is why an ice cube feels colder in our hand than a glass of liquid water.

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The second law has also been applied to the spontaneous expansion of gases. Gas molecules will move from areas of high concentration (high density) to areas of low concentration (low density). The rate of transfer is dependent on the difference in density. In the example below, the rate of transfer is greater between the left hand boxes because the difference in density is greater.

Each of the described systems is moving in the direction of equilibrium. Once equilibrium is achieved, there will be no net change to the system unless acted upon by an outside force. For example, thermal energy will be exchanged between objects in the above illustration but the temperature of the objects will not change. Likewise, molecules of air may move between boxes in the lower diagram, but there will be no net change in density.

When considering atmospheric moisture, our primary concern is the relative rates of evaporation and condensation of water.

● Evaporation is the process by which water undergoes a state change from a liquid to a gaseous state (water vapor)

● Condensation is the process by which water undergoes a state change from a gaseous to a liquid state.

Prior to exploring the drivers of each of these rates, we need to review what we know about water ….

O

H

H

( - )

( + )

Water is a polar molecule with two Hydrogen atoms and one Oxygen atom. These atoms share their valence shell electrons in a covalent bond.

Because the Oxygen atom is more electronegative than the hydrogen atom, the valence shell electrons are held closer to the Oxygen atom. This give the Oxygen atom a net negative charge and the Hydrogen atoms a net positive charge.

O

H

H

( - )

( + )

Due to the unequal distribution of charge, intermolecular forces, known as hydrogen bonds, hold together water molecules in liquid or solid states. These bonds are a form of intermolecular potential energy (recall that potential energy is energy related to an object’s position).

Liquid

If the thermal kinetic energy (vibrational, rotational, and random translational motion) of a molecule exceeds the potential energy of the hydrogen bond that holds it in liquid state, the molecule will move from a liquid state to gaseous state (water vapor).

Gas

Liquid

Recall that temperature is a measure of the average thermal kinetic energy of a substance or system. Some of the molecules in the system have relatively low TKE (slow-moving) and some relatively high (fast-moving). The plot below shows the proportion of molecules present within a system at a given TKE.

The diagram below displays a system containing liquid water (solid area under the curve) and water vapor (textured area under the curve). The temperature of the system is 20 degrees Celsius.

The dashed line represents a boundary. Below the boundary, the potential energy of the hydrogen bonds exceeds the TKE, therefore the water is in a liquid state. Above the boundary the TKE of the molecules exceeds the intermolecular potential energy, therefore the water is in a gaseous state.

As displayed, this system is at equilibrium – in meteorology, this equilibrium is known as the saturation point. This is the point at which the evaporation and condensation rates are equal.

At equilibrium, the shaded area represents the partial vapor pressure (e, synonyms: actual vapor pressure, vapor pressure). This is essentially a proxy measurement of the amount of water held in a vapor state.

We can calculate partial pressure by multiplying the proportion of water vapor in an air parcel by the total air pressure. This is expressed by the following formula:

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑒 = 𝐴𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑚𝑏 × 0.01 × %𝐻2𝑂𝑣

For example, if the atmospheric pressure at a given location is 1000 mb and a parcel of air contains 4.0% water vapor:

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑒 = 1000 𝑚𝑏 × 4 × 0.01 = 40 𝑚𝑏

A simple example: A parcel of air has a partial vapor pressure of 34.0 mb at an atmospheric pressure of 1014 mb. Water vapor makes up what percentage of molecules within the parcel?

A simple example: A parcel of air has a partial vapor pressure of 34.0 mb at an atmospheric pressure of 1014 mb. Water vapor makes up what percentage of molecules within the parcel? Solution:

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑒 = 𝐴𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑚𝑏 × 0.01 × %𝐻2𝑂𝑣

34 𝑚𝑏 = 1014 𝑚𝑏 × 0.01 × %𝐻2𝑂𝑣

34 𝑚𝑏

1014 𝑚𝑏 × 0.01= %𝐻2𝑂𝑣

3.34 % = %𝐻2𝑂𝑣

Saturation vapor pressure (SVP, es; synonym: equilibrium vapor pressure) is a proxy measure of the amount of water that can be held in vapor state as a function of a system's temperature. Because the described system is at equilibrium, the shaded area under the curve represents both the actual VP and SVP.

If the temperature of the system were to increase from 20 to 22 °C,

the capacity to hold water molecules in a vapor state (SVP) is increased. If no additional water is added to the system, the system will no longer be at equilibrium and the evaporation rate will exceed the rate of condensation.

Likewise, if the system was saturated at a temperature of 22 °C and you cooled the system to 20 °C, the rate of condensation would

(instantaneously) exceed evaporation and the excess water vapor would condense into liquid water.

Notice how the proportion of molecules that can be held in a vapor state (saturation vapor pressure) increases considerably with temperature. This results in a positive exponential relationship between temperature and saturation vapor pressure.

We can diagram the positive exponential relationship between saturation vapor pressure and temperature using the following plot. Notice that SVP is not dependent on the amount of water vapor present in the air but, rather, the temperature alone.

We can calculate the relationship between saturation vapor pressure and the temperature of a parcel using the following formula:

𝑒𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃17.67 × 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (℃)

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 ℃ + 243.5

Notice, once again, that the only variable in the formula is the temperature of the air parcel!

A simple example: The temperature of a parcel of air is 21°C, what is the saturation vapor pressure?

A simple example: The temperature of a parcel of air is 21°C, what is the saturation vapor pressure?

Solution:

𝑒𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃17.67 × 21 ℃

21℃ + 243.5= 24.85 𝑚𝑏

The greater the difference between the actual vapor pressure and the saturation vapor pressure, the further the system is from equilibrium. Similar to the ball on the ramp, the greater the distance from equilibrium, the faster the process will occur. This leads to high rates of evaporation relative to condensation.

A more complicated example: A parcel of air contains 3.20 % water vapor at an atmospheric pressure of 976 mb.

a) What is the partial pressure of the air parcel?

b) If the temperature of the parcel is 27.0 °C, what is the

saturation vapor pressure?

c) How would the rate of evaporation, relative to condensation change if the temperature of the parcel were to decrease?

A more complicated example: A parcel of air contains 3.20 % water vapor at an atmospheric pressure of 976 mb. Solution:

a) What is the partial pressure of the air parcel?

𝑒 = 976 𝑚𝑏 × 6 × 0.01 = 31.2 𝑚𝑏

b) If the temperature of the parcel is 27.0 °C, what is the

saturation vapor pressure?

𝑒𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃17.67 × 27 ℃

27 ℃ + 243.5= 35.6 𝑚𝑏

c) How would the rate of evaporation, relative to condensation change if the temperature of the parcel were to decrease? It would decrease.

We can summarize the relationship between the actual and saturation vapor pressure of a parcel with relative humidity. Relative humidity is the proportion of actual water vapor (e) relative to the saturation vapor pressure (es), expressed as a percentage. This can be thought of more simply as the amount of water vapor in a parcel of air relative to the amount of water that can be held in a vapor state. Relative humidity is expressed with the formula:

𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦 =𝑒

𝑒𝑠× 100

A simple example: A parcel of air has a partial vapor pressure of 13 mb and a saturation vapor pressure of 17 mb.

a) What is the relative humidity of the parcel?

b) If the temperature of the parcel were increased, how would the relative humidity be altered? Why?

c) If the temperature of the parcel were decreased, how would the rate of evaporation be altered?

A simple example: A parcel of air has a partial vapor pressure of 13 mb and a saturation vapor pressure of 17 mb.

Solution:

a) What is the relative humidity of the parcel?

b) If the temperature of the parcel were increased, how would the relative humidity be altered? Why? Decrease, because the SVP increases exponentially with increasing temperature.

c) If the temperature of the parcel were decreased, how would the rate of evaporation be altered? Why? Decrease, because the relative humidity is increased.

𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦 =13 𝑚𝑏

17 𝑚𝑏× 100 = 76%

The temperature at which the rate of evaporation is equal to condensation (i.e., equilibrium or the saturation point) is the dew point temperature. At constant atmospheric pressure, the dew point of a given air parcel is ONLY determined by the actual vapor pressure.

The dew point temperature can be calculated as follows:

Where e is the actual vapor pressure.

𝑇𝑑 =243.5 × ln

𝑒6.112

17.67 − ln 𝑒

6.112

A simple example: A parcel of air has an actual vapor pressure of 14 mb. What is the dew point of the parcel?

A simple example: A parcel of air has an actual vapor pressure of 14 mb. What is the dew point of the parcel?

Solution:

Recall that this means that, in order to reach equilibrium at this vapor pressure, the parcel would have to cool to a temperature of 12.0 oC.

𝑇𝑑 =243.5 × ln

14.0 𝑚𝑏6.112

17.67 − ln 14.0 𝑚𝑏

6.112

= 12.0 ℃

The difference between the dew point temperature and the air temperature can provide a qualitative measure of relative evaporation and condensation rates. If the air temperature is considerably higher than the dew point, the relative humidity is low and the evaporation rate, relative to the condensation rate, is high.

If the air temperature falls below the dew point, the rate of condensation (instantaneously) exceeds evaporation and the excess water vapor condenses into liquid water. Because the air now contains a lower concentration of water vapor (e decreases), the dew point decreases simultaneously (thus relative humidity will not exceed 100%).

A “less-than-simple” example: You measure the relative humidity and air temperature of a parcel of air using a sling psychrometer (see lab manual, lab six) and the atmospheric pressure with a barometer. The air temperature is 17.0 °C, the relative humidity is 47.0% and

the atmospheric pressure is 1019 mb.

a) What is the saturation vapor pressure of the parcel?

b) What is the dew point temperature of the parcel?

c) What is the percent composition of water vapor in the parcel?

Recall that this means that, in order to reach equilibrium at this vapor pressure, the parcel would have to cool to a temperature of 12.0 oC.

Air temperature = 17 °C

Relative humidity = 47%

Atmospheric pressure = 1019 mb

a) What is the saturation vapor pressure of the parcel?

Solution:

𝑒𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃17.67 × 17 ℃

17 ℃ + 243.5= 19.3 𝑚𝑏

Recall that this means that, in order to reach equilibrium at this vapor pressure, the parcel would have to cool to a temperature of 12.0 oC.

Air temperature = 17 °C

Relative humidity = 47%

Atmospheric pressure = 1019 mb

Saturation vapor pressure = 19.3 mb

b) What is the dew point temperature of the parcel?

Solution:

𝑒 (𝑚𝑏)

19.3 𝑚𝑏× 100 = 47 → 𝑒 =

47 × 19.3𝑚𝑏

100= 9.07𝑚𝑏

𝑇𝑑 =243.5 × ln

9.07 𝑚𝑏6.112

17.67 − ln 9.07 𝑚𝑏

6.112

= 5.56 ℃

Air temperature = 17 °C

Relative humidity = 47%

Atmospheric pressure = 1019 mb

Saturation vapor pressure = 19.3 mb

Partial vapor pressure = 9.07mb

Dew point = 5.56 °C

c) What is the percent composition of water vapor in the parcel?

9.07𝑚𝑏 1019 𝑚𝑏

× 100 = 0.890%

Recall that this means that, in order to reach equilibrium at this vapor pressure, the parcel would have to cool to a temperature of 12.0 oC.