Question

Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere, already derived in Problems 1.16 and 3.37. (Hint: Let the system be a single air molecule, let s1 be a state with the molecule at sea level, and let s2 be a state with the molecule at height z.)

Short answer

Therefore, the exponential formula for the density of an isothermal atmosphere is:$\rho \left(z\right)=\rho \left(0\right)e^{-mgz/kT}$

Step-by-step solution

Given information

Let the system be a single air molecule, let S₁be a state with the molecule at sea level, and let S₂ be a state with the molecule at height z.

Explanation

Consider a system with a single air molecule, where S₁ is the state when the molecule is at sea level and S₂is the state when the molecule is at a height of 2. Assume that the energy is only potential energy, so the difference in energy between the states S₁and S₂is the potential energy, which is $\Delta E=mgz$ and the ratio of S₂state probability to state s1 probability is:

role="math" localid="1647369610229" $$\begin{gathered} \frac{P\left(s_2\right)}{P\left(s_1\right)}=\frac{e^{-E_2/kT}}{e^{-E_1/kT}}=e^{-\Delta E/kT} \\ \frac{P\left(s_2\right)}{P\left(s_1\right)}=e^{-mgz/kT} \end{gathered}$$

This means that the air molecule is less likely to be at height of z than at the see level by a factor of $e^{-mgz/kT}$, and that the number of molecules per unit volume at height of z is also smaller than the see level by the same ratio in the isothermal atmosphere, so:

$\rho \left(z\right)=\rho \left(0\right)e^{-mgz/kT}$