Problem — Isothermal barometric distribution

Problem

In an isothermal atmosphere at $T = 290$ K, pressure decreases with altitude as $p(z) = p_0\, e^{-z/H}$ with $H = k_B T/(m g)$. Calculate $H$ for air ($M\approx 29$ g/mol) and the altitude $z_{1/2}$ at which $p$ is halved.

Solution

Scale height $H$. It is convenient to use the form with molar mass: multiplying numerator and denominator by $N_A$, since $k_B N_A=R$ and $m N_A=M$, $H=\frac{k_B T}{m g}=\frac{R\,T}{M\,g}$ Substituting $R=8{,}31$ J/(mol·K), $T=290$ K, $M=29\ \text{g/mol}=29\cdot10^{-3}$ kg/mol, $g=9{,}8$ m/s²: $H=\frac{8{,}31\cdot 290}{29\cdot10^{-3}\cdot 9{,}8}$ $H=\frac{2{,}41\cdot10^3}{0{,}284}=8{,}5\cdot10^3\ \text{m}\approx 8{,}5\ \text{km}$ Halving altitude $z_{1/2}$. Setting $p(z_{1/2})=\tfrac12 p_0$: $\frac12=e^{-z_{1/2}/H}$ Taking the logarithm: $-\frac{z_{1/2}}{H}=\ln\frac12=-\ln 2\quad\Longrightarrow\quad z_{1/2}=H\ln 2$ $z_{1/2}=8{,}5\cdot 0{,}693=5{,}9\ \text{km}$ At about 5.9 km altitude the pressure is half of that at sea level, consistent with the fact that at high altitude in the mountains (above 5000 m) one breathes roughly half the air. $\ev{H\approx 8{,}5\ \text{km};\quad z_{1/2}=H\ln 2\approx 5{,}9\ \text{km}}$

Connections

Topics: Teoria cinetica dei gas Concepts: Energia potenziale gravitazionale · Pressione · Distribuzione di Maxwell-Boltzmann