1. The average density in a column of air from the surface of the earth to an elevation of 3200 m is 0.95 kg/m³. Use the hydrostatic equation to calculate the pressure at 3200 m knowing the surface pressure is 1020 mb.

Solution

$\frac{\Delta P}{\Delta z} = -\rho g \Rightarrow \Delta P = -\rho g \Delta z = -0.95 \times 9.8 \times 3200 = -29792 \mbox{ Pascals} = -298 \mbox{ mb}$

The decrease in column pressure is 298 mb, so the pressure at the top of the column is 1020 - 298 = 722 mb.

2. Calculate the average density for the layer between 700 mb and 500 mb if the elevation of the 700 mb surface is 2898 m and the elevation of the 500 mb surface is 5370 m.

Solution

$ \frac{\Delta P}{\Delta z} = -\rho g \Rightarrow \rho = -\frac{1}{g}\frac{\Delta P}{\Delta z}$

$ \Delta P = 500 \mbox{ mb} - 700 \mbox{ mb} = -200 \mbox{ mb} = -20000 \mbox{ Pascals} $

$ \Delta z = 5370 \mbox{ m} - 2898 \mbox{ m} = 2472 \mbox{ m}$

$ \bar{\rho} = -\frac{1}{9.8}\frac{-20000}{2472} = 0.83 \mbox{ kg/m}^3$