Use the mixing ratio to calculate the density of the air. In this case, it means that it is 1% water vapor. The mean molecular weight of the air will be about 28.9 g/mole.
Then use the fact that pressure decreases with altitude according to
delta P = -Integral of (density) g dy.
Assume the temperature stays the same with height for 188 feet from the ground, even though this may not be true. It is close enough in this case. You can even ignore the change of density with height and get a good value for the integral.
The density of the 25 C air with mean molecular weight of 28.9 will be about
28.9 g/(24.45 l) = 28.9*10^-3 kg/24.45*10^-3 m^3 = 1.182 kg/m^3
188 ft = 57.3 m
delta P = -(1.182)(9.8)(57.3) = -664 N/m^2 (approximately)
You should convert the pressure decrease to millibars when you are done.
The 1% mixing ratio does not make a lot of difference.
The 15th floor meteorological observatory of the O&M building is 188 ft from the level of the main entrance. Assume that the temperature at the entrance is 25 degrees Celcius, that station pressure is 1000 mb, and the mixing ration is .010 kg/kg. What is the actual (station) pressure at a baramoter in the observatory?
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