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Expert-verified Found in: Page 228 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # 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.)

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

See the step by step solution

## Step 1: Given information

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.

## Step 2: Explanation

Consider a system with a single air molecule, where S1 is the state when the molecule is at sea level and S2 is 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 S1 and S2 is the potential energy, which is $\Delta E=mgz$ and the ratio of S2state probability to state s1 probability is:

role="math" localid="1647369610229" $\frac{P\left({s}_{2}\right)}{P\left({s}_{1}\right)}=\frac{{e}^{-{E}_{2}/kT}}{{e}^{-{E}_{1}/kT}}={e}^{-\Delta E/kT}\phantom{\rule{0ex}{0ex}}\frac{P\left({s}_{2}\right)}{P\left({s}_{1}\right)}={e}^{-mgz/kT}$

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:

$\rho \left(z\right)=\rho \left(0\right){e}^{-mgz/kT}$ ### Want to see more solutions like these? 