One of the most widely used measurements of any atmospheric property is temperature. This quantity not only dominates our daily conversations about the weather but also affects where and when things will grow in the soil. The temperature varies in a uniform way within the troposphere, so let's start with a problem about temperature lapse rates within the atmosphere.

1. An airplane takes off from an airport where the ground temperature is 60°F and climbs to its cruising altitude of 26,000 ft. What is the air temperature outside the plane at this altitude if it is flying in a stable air mass?

Given T₀ = 60° and h = 26,000 ft

T = T₀ - Rh = 60°F - (3.5°F / 1000 ft) (26,000 ft)

= 60°F - 91°F = -31°F

Pressure is also an important property of the atmosphere.


2. A barometer exposed to the air at sea level has a column of mercury (Hg) that is 80 cm high. What is the atmospheric pressure?

Since 76 cm Hg is equal to 1 atm, or 14.7 lb/in², the following conversion factors can be formed:

14.7 lb/in² = 76 cm Hg

1 atm = 76 cm Hg

Therefore, we can express the pressure in this exercise in several different units. For example:

(80 cm Hg)(14.7 lb/in² / 76 cm Hg) = 15 lb/in²

(80 cm Hg)(1 atm / 76 cm Hg) = 1.1 atm

3. If a water barometer were used in the preceding problem, what would be the height of the column of water if the pressure were the same?

We know the density of mercury is 13.6 times greater than the density of water (_Hg = 13.6 g/cm³ and _water = 1.00 g/cm³). Consequently, a water barometer would have a column 13.6 times higher than a mercury barometer for any given pressure. Hence, if a mercury barometer had a column of mercury that was 80 cm high, a water barometer would have a height of:

(80 cm Hg)(13.6 g/cm³ / 1.00 g/cm³) = 1100 cm water

This is equivalent to 11 m, or about the height of a two-story house. It should now be obvious why most barometers are made using mercury instead of water.


4. A 175-lb object rests on the hand of a person whose palm measures 4 in. x 3 in. If the object just covers the palm, what is the approximate pressure, in atmospheres, that the person feels pressing down on this hand?

Pressure is defined as force per unit area; that is,

p = F/A

Since the area of the palm is 4 in. x 3 in. = 12 in², the pressure is:

p = 175 lb / 12 in² = 14.6 lb/in²

Since 1 atm equals 14.7 lb/in², the person feels an additional pressure of about 1 atm pressing down on his hand.


5. A psychrometer shows a dry-bulb reading of 70°F and a wet-bulb reading of 64°F. Find the relative humidity, the dew point, and the actual moisture content of the air.

The depression of the wet bulb is 6°F. From the psychometric Tables 1 and 2 in Appendix VIII of the textbook, the relative humidity is 0.72, or 72%, and the dew point is 61°F. At 70°F, the maximum moisture capacity is 7.8 gr/ft³ (from Table 1).

We also can determine the actual moisture content of the air sample described above. By definition,

Relative humidity = actual moisture content / maximum moisture capacity

orRH = AC / MC

henceAC = (MC)(RH)

= (7.8 gr/ft³)(0.72) = 5.6 gr/ft³

6. In a laboratory experiment to determine the relative humidity, ice is added to water at room temperature (70°F). As the temperature of the water falls, condensation begins to form on the outside of the container when the temperature is 55°F. What is the relative humidity?

The temperature at which condensation forms on the container is the dew point of the air. At the dew point, the actual moisture content (AC) of the air is equal to the maximum moisture capacity. Hence, from Table 1 in Appendix VIII, the AC = 4.8 gr/ft³. Also from Table 1, the air in the room at 70°F has a maximum moisture capacity (MC) of 7.8 gr/ft³. Hence

%RH = (AC / MC) x 100% = (4.8 gr/ft³ / 7.8 gr/ft³) x 100% = 62%

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