Ideal Gases — Question 7

Question 7

An isothermal atmosphere has an exponential pressure profile

Treat the gas as ideal, $pV=nRT$, and use $R=8.314\,\mathrm{J/(mol\,K)}$. For an ideal gas atmosphere with constant temperature 300 K, molar mass .0290 kg/mol and $g=9.80\,\mathrm{m/s^2}$, use hydrostatic balance $dp/dz=-\rho g$. At $z=0$, pressure is 100 kPa. Assume the idealized atmosphere remains isothermal over the height of interest.

1. Derive the pressure scale height and pressure profile.

2. Find the altitude where pressure falls to one-half its base value. Compare with a constant-density linear extrapolation using the base density.

Question 7 - Solution

An isothermal atmosphere has an exponential pressure profile

(a) Method and calculation

$$\rho=\frac{Mp}{RT},\quad\frac{dp}{p}=-\frac{Mg}{RT}dz\ \Rightarrow\ p(z)=p_0e^{-z/H}.$$$$H=\frac{RT}{Mg}=\frac{8.314(300)}{.0290(9.80)}\approx8776.21\,\mathrm m.$$

(b) Interpretation and further calculation

$$z_{1/2}=H\ln2\approx6083.21\,\mathrm m.$$$$p_{\mathrm{linear}}=p_0(1-z/H)\ \Rightarrow\ z_{1/2,\mathrm{linear}}=H/2\approx4388.11\,\mathrm m.$$Constant-density extrapolation falls too quickly because it does not account for the diminishing weight per unit height.

Physical check

The linear form is only a small-height approximation to the exponential.