Section   19.1Composition and Structure

1. (Paired exercises 3 and 4) If the air temperature on a hot summer day is 27.0° C at sea level, how cold will it be in a hot-air balloon that has ascended to a height of 3000 m?

We know the lapse rate for normal air in the troposphere to be 6.5 C° /km, so the temperature change can be found by multiplying the lapse rate R by the vertical change h, which is:

3000 m - 0 m = 3000 m = 3.0 km

The temperature at this altitude can then be found by subtracting the temperature change from the initial temperature at sea level, T₀.

T1	=T0 - Rh
	=27.0°C - (6.5 C°/km)(3.0 km)
	=27.0°C - 19.5°C
	=7.5°C

Section   19.3Atmospheric Measurements and Observations

2. (Paired exercises 7 and 8) A student finds that on a certain day the dry-bulb reading on a psychrometer is 80°F and the wet-bulb reading is 75°F What is the:

a. relative humidity,
b. maximum moisture capacity of the air,
c. actual moisture content in the air, and
d. dew point?

a. Using Table 1 from Appendix VIII of the textbook, we can see that for an air temperature (dry-bulb reading of 80°F - 75°F = 5°F), the relative humidity is 79%

b.The maximum moisture capacity of the air at 80°F can also be read from Table 1 as 10.9 gr/ft³.

c.We must calculate the actual moisture content of the air by multiplying the maximum moisture capacity by the decimal equivalent of the relative humidity:

(Remember that 79% is equivalent to 0.79).
Actual moisture content = (10.9 gr/ft³)(0.79) = 8.6 gr/ft³

d. The dew point can be found using Table 2 in Appendix VIII of the textbook. For the given air temperature and depression of the wet-bulb reading, at 80°F and 5°F depression, the dew point is 73°F.

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

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

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