1. 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.

2. Expected prior knowledge
●
Environmental processes
●
Equilibrium and rate of processes
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Atmospheric pressure
●
Evaporation and condensation
●
Why water is a polar molecule
●
Hydrogen bonds and potential energy
●
Thermodynamics (e.g., TKE, temperature)

3. 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.
>

4. 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.
>

5. 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.
>

6. 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.

7. 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 ….

8. Water is a polar molecule with two Hydrogen atoms and one
Oxygen atom. These atoms share their valence shell electrons in a
covalent bond.
H
(-)
O
H
(+)
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.

9. 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).
H
(+)
(-)
O
H
Liquid

10. 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

11. 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.

12. 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.

13. 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.

14. 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.

15. 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.

16. 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 × %𝐻2 𝑂 𝑣
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

17. 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?

18. 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
34 𝑚𝑏 = 1014 𝑚𝑏 × 0.01 × %𝐻2 𝑂 𝑣
34 𝑚𝑏
= %𝐻2 𝑂 𝑣
1014 𝑚𝑏 × 0.01
3.34 % = %𝐻2 𝑂 𝑣
𝑚𝑏 × 0.01 × %𝐻2 𝑂 𝑣

19. 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.

20. 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.

21. 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.

22. 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.

23. 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.

24. We can calculate the relationship between saturation
vapor pressure and the temperature of a parcel using the
following formula:
17.67 × 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (℃)
𝑒 𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 ℃ + 243.5
Notice, once again, that the only variable in the formula is
the temperature of the air parcel!

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

26. A simple example: The temperature of a parcel of air is
21°C, what is the saturation vapor pressure?
Solution:
17.67 × 21 ℃
𝑒 𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃
= 24.85 𝑚𝑏
21℃ + 243.5

27. 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.

28. 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?

29. 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?
17.67 × 27 ℃
𝑒 𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃
= 35.6 𝑚𝑏
27 ℃ + 243.5
c) How would the rate of evaporation, relative to
condensation change if the temperature of the parcel
were to decrease? It would decrease.

30. 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:
𝑒
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦 = × 100
𝑒𝑠

31. 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?

32. 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?
13 𝑚𝑏
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦 =
× 100 = 76%
17 𝑚𝑏
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.

33. 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:
𝑒
243.5 × ln
6.112
𝑇𝑑 =
𝑒
17.67 − ln
6.112
Where e is the actual vapor pressure.

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

35. A simple example: A parcel of air has an actual vapor
pressure of 14 mb. What is the dew point of the parcel?
Solution:
14.0 𝑚𝑏
243.5 × ln
6.112
𝑇𝑑 =
= 12.0 ℃
14.0 𝑚𝑏
17.67 − ln
6.112
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.

36. 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.

37. 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%).

38. 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?

39. Air temperature = 17 °C
Relative humidity = 47%
Atmospheric pressure = 1019 mb
a) What is the saturation vapor pressure of the parcel?
Solution:
17.67 × 17 ℃
𝑒 𝑠 = 6.112 𝑚𝑏 × 𝐸𝑋𝑃
= 19.3 𝑚𝑏
17 ℃ + 243.5

40. 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)
47 × 19.3𝑚𝑏
× 100 = 47 → 𝑒 =
= 9.07𝑚𝑏
19.3 𝑚𝑏
100
9.07 𝑚𝑏
243.5 × ln
6.112
𝑇𝑑 =
= 5.56 ℃
9.07 𝑚𝑏
17.67 − ln
6.112

41. 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
× 100 = 0.890%
1019 𝑚𝑏