Chapter 2.2, Problem 40E

The atmospheric pressure (force per unit area) on a surface at an altitude $z$ is due to the weight of the column of air situated above the surface. Therefore, the drop in air pressure $\rho$ between the top and bottom of a cylindrical volume element of height $\Delta z$ and cross-section area $A$ equals the weight of the air enclosed (density $\rho$ times volume $V=A\Delta z$ times gravity g), per unit area:

$p\left(z+\Delta z\right)-p\left(z\right)=-\frac{\rho \left(z\right)\left(A\Delta z\right)g}{A}=-\rho \left(z\right)g\Delta z$

.

Let $\Delta z\to 0$ to derive the differential equation $dp/dz=-\rho g$. To analyze this further we must postulate a formula that relates pressure and density. The perfect gas law relates pressure, volume, mass $m$, and absolute temperature $T$ according to $pV=mRT/M$, where $R$ is the universal gas constant and $M$ is the molar mass (i.e., the mass of one mole) of the air. Therefore density and pressure are related by $\rho :=m/V=Mp/RT$.

a. Derive the equation $\frac{dp}{dz}=-\frac{Mg}{RT}p$ and solve it for the “isothermal” case where $T$ is constant to obtain the barometric pressure equation

$p\left(z\right)=p\left(z_0\right)\exp \left\{-Mg\left(z-z_0\right)/Rt\right\}$.

b. If the temperature also varies with altitude $T=T\left(z\right)$, derive the solution

$p\left(z\right)=p\left(z_0\right)\exp \left\{-\frac{Mg}{R}{\displaystyle {\int }_{z_0}^z\frac{d\zeta }{T\left(\zeta \right)}}\right\}$.

c. Suppose an engineer measures the barometric pressure at the top of a building to be $99,000\text{ Pa}$ (pascals), and $101,000\text{ Pa}$ at the base $\left(z=z_0\right)$. If the absolute temperature varies as $T\left(z\right)=288-0.0065\left(z-z_0\right)$, determine the height of the building. Take $R=8.31\text{N-m/mol-K}$, $M=0.029\text{kg/mol}$, and $g=9.8{\text{m/sec}}^{\text{2}}$. (An amusing story concerning this problem can be found at http://www.snopes.com/college/exam/barometer.asp)