 Solution

Assume a constant downward gravitational acceleration $g$, constant composition, and an ideal gas law $p=\rho \mathcal{R}T$, where $\mathcal{R}=k_B/(\mu m_u)$ is the gas constant per unit mass. Here $m_u$ is the unified atomic mass unit; this convention makes the specified mean molecular weight dimensionless. Hydrostatic equilibrium givesIntegrating from sea level yieldsDividing by the temperature ratio gives the hydrostatic atmosphere with a constant temperature lapse rate:The model requires $T_0-\alpha z>0$ and is used here only within the troposphere.

Convert the atmospheric lapse rate to $\alpha =6.5\times 10^{-3}\mathrm{K}{\mathrm{m}}^{-1}$. At $11\mathrm{km}$, $T=228.5\mathrm{K}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66054\times 10^{-27}\mathrm{kg}$ and $\mu =28$, the density exponent is $4.0825$. HenceThe mass density is therefore approximately one third of its sea-level value in this model.