AT 351, Spring 2010
Lab #1
Atmospheric Structure and Composition & Energy
Question #1: Atmospheric Composition (15 pts)
A. What are the major components of clean dry air? Give the approximate percentage
related to each, along with a source and sink.
Nitrogen (78%): Sources: N2 is returned to the atmosphere mainly through the decay
of plant and animal matter
Sinks: N2 is primarily removed from the atmosphere by biological
processes involving soil bacteria and ocean dwelling plankton
Oxygen (21%): Sources: The addition of oxygen to the atmosphere occurs during
photosynthesis
Sinks: O2 is removed from the atmosphere when organic matter decays
and when oxygen combines with other substances, producing
oxides. It is also removed from the atmosphere by breathing, as
the lungs take in oxygen and release CO2
B. Atmospheric pressure decreases exponentially with height according to:
z
( )
H
PP0 e
Assuming H(scale height) is 8 km, and average sea level pressure (P0) is 1013.25 mb,
calculate the average pressure of Fort Collins (elevation, z = 1522 m).
1
C. Mt. Denali reaches the highest point in North America when measured from sea level
(approximately 6 km). Using what we know about the composition of the
atmosphere, how much oxygen is at the peak of Mt. Denali compared to at sea level?
Assume that concentration of gases is constant to peak of Mt Everest
Use given scale height H = 8km
PEverest

Pslp

e
e
6 km
8 km
0 km

8 km

0.472
 47.2%
1
2
Question #2: Atmospheric Structure (25 pts)
A. The Figure Below shows the standard temperature profile of the earth’s
atmosphere. Please label the 4 layers and the three levels of the atmosphere.
B. What is the standard lapse rate in the troposphere? Please provide a
definition as well as a numerical value (oC/km).
The lapse rate is the rate at which temperature decreases with height in the
troposphere. The standard lapse rate has a numerical value equal to 6.5º C/km
May also accept the dry adiabatic lapse rate, which has a numerical value equal to
10º C/km
3
C. Using the result from part B, calculate the temperature at the summit of Mt.
Denali. The elevation of the summit is approximately 6 km. Assume a surface
temperature of 5 oC.
TDenali Summit= (-6.5 deg C/km)*(6km) +5 deg C
= -34.0 deg C
If using dry adiabatic lapse rate:
TDenali Summit= (-10 deg C/km)*(6km) +5 deg C
= -55.0 deg C
D. Briefly explain why the temperature decreases with height in the troposphere.
Troposphere: Recall that the troposphere is heated from below by infrared
radiation emitted by the earth. The temperature in the troposphere decreases
with height primarily because an increase in elevation in the troposphere is
further away from the heat source (the earth) and the heat is then distributed
over a greater area.
Example: If you place your hand over a hot burner, the further your hand
gets away from the burner the less heat you feel.
E. To what region(s) of the atmosphere is most of the daily weather phenomena
confined?
The daily weather is primarily confined to the troposphere
4
Question # 3 Pressure (15 pts)
A. Pictured below are two imaginary columns of air exerting pressure on the surface
below them. Both columns cover an area of 1 square meter each. Column A has a
mass of 15,000 kg and column B has a mass of 4,500 kg.
Compute the pressure exerted by each column of air on the surface. Please express your
answer in terms of Pascals (Pa) and millibars (mb). Note: 100 Pa = 1 mb, or 1 hPa = 1
mb.
Recall:
Pressure = Force/Area
Force = Mass*Acceleration
Gravitational acceleration on earth is 9.8 ms-2
Column A
ForceA = (15,000 kg)*(9.8 ms-2)
= 147,000 kg ms-2
PressureA = (147,000 kg ms-2)/(1 m2)
= 147,000 kg m-1 s-2
= 147,000 Pa
= 1470 mb
Column B
ForceB = (4,500 kg)*(9.8 ms-2)
= 44,100 kg ms-2
PressureB = (44,100 kg ms-2)/(1 m2)
= 44,100 kg m-1 s-2
= 44,100 Pa
= 441 mb
5
B. On a particular day in January, Fort Collins had a surface pressure of 837 mb. On
the same day, with the exact same weather conditions, the surface pressure in Tulsa,
Oklahoma was 1026 mb. Give a brief explanation of why the pressure is lower in
Fort Collins even though the same weather conditions exist.
Recall that air pressure is the weight of the atmosphere above us. In Fort Collins,
because we are at higher altitude there is less of the atmosphere’s mass above us.
Consequently, the surface pressure is lower over Ft. Collins
Question #4 Surface Plot (15 pts)
Using the information provided in lecture and in Appendix B in the book, look at the
surface plot image below and state what conditions (temp, dewpoint, wind, pressure, weather)
are indicated at stations marked GNV, SGJ and AHN.
GNV
Temp = 64 F
Dewpoint = 64 F
Wind = NW, 15 kt
Pressure =N/A
Weather = Tstorms
SJG
Temp = 64 F
Dewpoint = 63 F
Wind = SW, 5 kt
Pressure =N/A
Weather = Rain, moderate continuous
AHN
Temp = 43 F
Dewpoint = 42 F
Wind = NE, 10 kt
Pressure =1003.8 mb
Weather = fog
6
7