It's possible to use the ideal-gas law to show that the density of the earth's atmosphere decreases exponentially with height. That is, $\rho ={\rho }_0\exp \left(-z/z_0\right)$, where $Z$is the height above sea level, ${\rho }_0$is the density at sea level (you can use the Table $14.1$value), and ${\mathrm{z}}_0$is called the scale height of the atmosphere.

a. Determine the value of ${\mathrm{z}}_0$.

Hint: What is the weight of a column of air?

b. What is the density of the air in Denver, at an elevation of $1600m$? What percent of sea-level density is this?

The density of the earth's atmosphere decreases exponentially with height. The expression for density can be started as

$\rho ={\rho }_0{e}^{\left(-Z/Z_0\right)}$

Where, $Z$is the height above the sea level, ${{\rho }_0}^{}$is the density at the sea level and ${}^{Z_0}$is the scale height of the atmosphere.

Consider a small air column which extends from the sea level to the outer atmosphere. The cross sectional area $\left(A\right)$of the column is $1m^2$.Take an element of thickness $d_z$of the air column.Thus the volume of such an element is

$V=Adz$

$=1\left(m^2\right)dz$

The height of the element

$dw=\left(\rho dV\right)g$

$=\rho g\left(1{\mathrm{m}}^2\right)dz$

$={\rho }_0{e}^{-Z/Z_0}g\left(1{\mathrm{m}}^2\right)dz$

Thus the total weight if the air column with area of $1m^2$is

$w={\int }_0^{\mathrm{\infty }} {\rho }_{0}g\left(1{\mathrm{m}}^2\right)e^{-z_0/z_0}dz$

$=\left[-{\rho }_{0}g\left(1{\mathrm{m}}^2\right)Z_0\right]{\left[e^{-\mathrm{z}/z_0}\right]}_0^{\mathrm{\infty }}$

$={\rho }_{0}g\left(1{\mathrm{m}}^2\right)Z_0$

Substitute $w=101300\mathrm{N},{\rho }_0=1.28\mathrm{kg}/{\mathrm{m}}^3\text{and}g=9.8\mathrm{m}/{\mathrm{s}}^2$

$Z_0=\frac{101300\mathrm{N}}{\left(1.28\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.0{\mathrm{m}}^2\right)}$

$=8.1\times {10}^3\mathrm{m}$

Hence the scale weight is$8.1\times {10}^3\mathrm{m}$

We have an expression for density variation with height as

$\rho ={\rho }_0{e}^{\left(-z/z_0\right)}$

Where $Z$is the height above the sea level ${\rho }_0$is the density at sea level and $Z_O$is the scale height of the atmosphere

The density of air at sea level. Substitute $Z=1600\mathrm{m}\text{and}{Z}_0=8.1\times {10}^3\mathrm{m}$

$\rho =\left(1.28\mathrm{kg}/{\mathrm{m}}^3\right)e^{-1600\mathrm{m}\left(8.1\times {10}^3\mathrm{m}\right)}$

$=1.05\mathrm{kg}/{\mathrm{m}}^3$

The percentage of density as compared to sea level density is

$\text{percentage}=\frac{1.05\mathrm{kg}/{\mathrm{m}}^3}{1.28\mathrm{kg}/{\mathrm{m}}^3}\times 100$

$=82\mathrm{\%}$