SCIENCE A30
REVIEW PROBLEMS with solutions
Spring 2007
PROBLEM 1.
a) The pressure at the earth's surface is 1.013 x 105 Pa. Use this information and formulas
relating mass, pressure and force to calculate the total mass of the atmosphere. Make use
of any useful information given in this problem. Express your answer in kg. Use the
radius of the Earth as 6.37 x 106m.
Atmospheric pressure at the earth's surface = 1013 mb = 1013x105 Nm-2
The pressure is the force exerted by the atmosphere on the surface per unit surface area.
The total force of the atmosphere on the whole surface of the earth is therefore
Fatmos = (Psurface) x (Surface Area of the earth) = 1.013x105 Nm-2 x 4π x (6.37x106m)2
= 5.165x1019N
This force is the weight of the atmosphere, also equal to the mass times the acceleration due to gravity:
Fatmos = Weightatmos = matmosg. The mass is therefore:
matmos = Fatmos/g = 5.165x1019 kg m s-2 / 9.8 m s-2
matmos = 5.27x1018 kg
b) Assuming a mean surface temperature of about 20° C, calculate the number density of
the atmosphere, at the surface, in molecules/cm3 (number density). The atmospheric
abundance of CO2 is about 360 ppm ("parts per million"). What is the number density of
CO2 near the surface?
Number densities at the surface of the earth:
Tsurf = 20oC = 20 + 273.15 K = 293.15 K
Use the form of the perfect gas law involving the number density:
P=nkT
So that n = P/(kT) = 1.013x105 Nm-2 / [(1.38x10-23 NmK-1)(293.15 K) = 2.5x1025 molecules m-3
n = 2.5x1019 molecules cm-3
The mixing ratio of CO2 is 360 ppm, or parts per million. This means that for every million molecules of
air there are 360 CO2 molecules. To go from the number density of air to the number density of CO2 we
therefore multiply n by 360x10-6.
n(CO2) = nair x 360x10-6
n(CO2) = 9.0x1015 CO2 molecules cm-3
PROBLEM 2.
a) A balloon is filled with air so that the air pressure inside just equals the air pressure
outside. The volume of the filled balloon is 3 m3, the mass of air inside is 3.5 kg, and the
temperature inside is 10° C. Calculate the air pressure.
b) If the atmospheric scale height is about 7 km, at what altitude would you expect to
observe this pressure? The pressure at the surface is 1.013 x 105 N/m2.
a) Air pressure inside a balloon:
V = 3 m3
T = 10oC = 283.15 K
Mass inside the balloon = Nm = 3.5 kg
We know about the mass and the volume in this case, which put together yields the mass density
ρ = Nm/V = 3.5kg/3m3 = 1.167kg m-3
Given what we know, the appropriate form of the Perfect Gas Law is
P = ρR'T = (1.167kg m-3) x (287.5 N m kg-1 K-1) x (283.15 K)
P = 9.5 x 104 Nm-2
b) Altitude of above pressure:
H = 7 km
Barometric Law: P(z) = P0exp(-z/H)
Rearrange the barometric law to find z
exp(-z/H) = P/P0
exp(z/H) = P0/P
ln(exp(z/H)) = ln(P0/P)
z = H ln(P0/P) = 7km ln(1.013x105 Nm-2 / 9.5x104 Nm-2)
z = 0.45 km
PROBLEM 3.
a) During a sunny day on the coast, a significant temperature difference often exists
between the land and the ocean. Assuming the temperature at the beach is 32oC and that
1.5km offshore the temperature is 20 oC, calculate the local scale heights (H) over the
beach and over the ocean. You may assume that the given temperatures are representative
of the 3km air column overlying these locations. Assume Po=1.01x105 N/m2 .
b) Calculate the pressure at 3km altitude over both locations in millibars.
c) Calculate the pressure gradient that will be exerted on a cubic meter of air at 3km
above the coast line ( in N/m 3 ) ( Keep in mind that the pressure gradient equals the
pressure gradient divided by the distance over which that gradient occurs). Assuming that
the Coriolis force is negligible on such local scales, what is the wind direction at 3km
altitude? How about at the surface? Explain why.
d) At night, the land cools faster than the ocean. What would you expect to happen to the
surface wind?
a) H = kT/mg = R'T/g where R' = k/m
Tland = 32° C = 305 K
Tocean = 20°C = 293 K
b) P0 = 1.01x105 N m-2 and z = 3 km
c) The pressure gradient is : ∆P/∆x = 10 mb/1.5 km = 10x102 N m-2/ 1500 m = 0.67 N m-3. As a result of
the pressure gradient force the wind will blow from the beach to the ocean at 3 km and from the ocean to
the beach at the surface.
d) At night the land cools faster than the ocean therefore you would expect the circulation to reverse.
Therefore the surface wind should blow from the beach to the ocean.
PROBLEM 4.
One of the most popular places to ice fish in the United States is beautiful Mille
Lacs, Minnesota. On a winter’s day in Minnesota the temperature outside might be
-10°C. Use the above graph to answer the following questions.
a. What is the saturation vapor pressure of a parcel of air at this temperature?
b. If the air parcel has a vapor pressure of 2 mb, what is the relative humidity?
c. What is the dew point of the parcel?
d. Minnesota ice fishermen don’t typically fish outside. They build wooden “fish
houses” to protect them from the elements. These are typically heated to about
10 °C, so as to be warm enough to keep the television functioning, but cold
enough to keep the beverages frosty (We’re not making this up). There is no
water vapor added as air is heated inside the fish house. Will the relative humidity
be higher, lower, or the same inside the fish house? What is the relative humidity
inside the fish house?
a) Using the graph above, we can estimate the saturation vapor pressure at –10 oC to be 3 mb.
b) We know that Pw = 2 mb, and Ps = 3 mb. We calculate the relative humidity as follows:
RH = (Pw/Ps) * 100 = (2/3)*100 = 66.67%
c) Dew point is the temperature a parcel must be cooled to in order for condensation to begin. Each value of
water vapor pressure has its own unique dew point. To determine dew point, hold the vapor pressure
constant and decrease the temperature until Pw = Ps. We can estimate the dew point using the graph above.
Since our parcel contains 2 mb of water vapor, the dew point corresponds to –15 oC.
d) As temperature goes up, the saturation vapor pressure goes up as well. Since we have not added any
water vapor the vapor pressure within the air stays the same. Remember that relative humidity is the ratio
of actual vapor pressure to saturation vapor pressure, so an increase in saturation vapor pressure while
maintaining constant actual vapor pressure will cause relative humidity to be lower inside the fish house.
Using the graph above, we can approximate the saturation vapor pressure at 10 oC to be 12 mb.
Our parcel still contains 2 mb of vapor pressure, so the relative humidity will be:
RH = (Pw/Ps) * 100 = (2/12)*100 = 16.67%
For comparison, 10% relative humidity is a value typical in the desert.
PROBLEM 5.
A cubic air parcel with sides of 10 cm has a temperature of 25 oC and a density of 1.18
kg/m3. It is sitting at sea level surrounded by air at 0 oC and density of 1.29 kg/m3. What
is the net force acting on the air parcel? Will the parcel rise, sink, or stay where it is?
HINT: The force pushing up on the bottom surface of the parcel should equal the
difference between the weight of the parcel and the weight of air displaced by the parcel.
Net force (FB) = forces up - forces down
Forces up:
F1 = Force due to pressure at the bottom of the parcel (Z1) = P1*area
Forces down:
F2 = force due to pressure at the top of the parcel (Z2) = P2*area
W = Weight of the parcel = ρparcel*V*g
FB = P1*area - P2*area - ρparcel*V*g
FB = (P1 - P2)*area - ρparcel*V*g
FB = ρair*(Z1-Z2)*g*area - ρparcel*V*g
FB = weight of displaced air - weight of parcel
FB = (ρair - ρparcel)*V*g
FB = (1.29 kg/m3 - 1.18 kg/m3)(10-3 m3)(9.81 m/s2) = +1.08x10-3 N → parcel will rise!
PROBLEM 6.
A radiosonde is released at the base of a mountain and sends back the temperature
information shown on the left in the figure below.
a) Calculate the atmospheric lapse rate from the surface up to 3000 m.
b) What type of atmospheric stability does this lapse rate suggest?
Suppose the wind is blowing from the west and a parcel of surface air with a temperature
of 12° C and a dew point of 4° C begins to rise upward along the western side of the
mountain. Assume that the dew point decreases by 2° C per km.
c) As the parcel of air rises, at approximately what elevation would condensation begin
and a cloud start to form?
d) If the moist adiabatic lapse rate for this parcel of air is 6 oC/km, at an altitude of 3000
m, how does the air temperature inside the cloud compare with the air temperature
outside the cloud, as measured by the radiosonde? What type of atmospheric stability
(stable or unstable) does this suggest? Explain.
e) At an elevation of 3000 m, would you expect the cloud to continue to rise? Explain.
a) lapse rate = (change in temperature) / (change in altitude) = -24° C / 3 km = -8° C / km
b) conditional stability -- the atmosphere is cooling less quickly than the dry adiabatic lapse rate (-9.8
K/km) but more quickly than most moist parcels cool as they rise. Thus a moist parcel could rise but a dry
parcel could not in this atmosphere. From the perspective of the moist cloud parcel, the atmosphere is
unstable.
c) Parcel temperature = 12° C - (9.8° C/km) * altitude
dew point = 4° C - (2° C/km) * altitude
condensation occurs when dew point = parcel temperature so set the equations equal...
12° C - (9.8° C/km) * altitude = 4_C - (2_C/km) * altitude
7.8° C/km * altitude = 8° C
altitude = 1 km
d) parcel temperature = 2° C (at 1 km) - (6° C/km) * (3 km - 1 km) = -10° C
air temperature = -14° C
Parcel is warmer than the surrounding atmosphere. This suggests conditional stability because the
condensation in the moist parcel enabled it to stay warmer than the atmosphere, but it would have been
cooler if it had risen at the dry lapse rate.
e) The cloud is warmer than the surrounding atmosphere, so the cloud is less dense and buoyancy should
allow it to continue to rise.
MULTIPLE CHOICE QUESTIONS
1. Deserts are located at around 30° N/S latitude primarily because:
a) These areas receive the greatest amount of incoming solar radiation, which
evaporates away all the moisture.
b) The upward branch of the Hadley circulation is located in these regions, causing
low surface pressures and inhibiting rainfall.
c) The downward branch of the Hadley circulation causes air to warm
adiabatically, bringing down warm, dry air and creating high surface
pressures.
d) The albedo of these regions is large, leading to increased absorption of solar
radiation and higher temperatures.
2. Pedro Martinez’s fastest pitches can reach speeds of about 156 km/hr. If he is playing
the Mariners in Seattle (latitude 48°N), the distance that the baseballs get deflected due to
the Coriolis effect is 0.5 mm (the distance between the mound and the plate is 20 m). If
Pedro goes to play against the Florida Marlins (Miami, latitude 25°N), how will the
deflection of the ball due to the Coriolis force be affected?
a) The deflection will be greater than 0.5 mm because Miami is at a lower latitude
than Seattle.
b) The deflection will be smaller than 0.5 mm because Miami is at a lower
latitude than Seattle.
c) The deflection will be smaller than 0.5 mm because there will be too much
friction to achieve geostrophic balance.
d) There is not enough information to tell.