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Q 6.14

Expert-verifiedFound in: Page 228

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

Let the system be a single air molecule, let S_{1 }be a state with the molecule
at sea level, and let S_{2} be a state with the molecule at height z.

Consider a system with a single air molecule, where S_{1} is the state when the molecule is at sea level and S_{2 }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 S_{1 }and S_{2 }is the potential energy, which is $\Delta E=mgz$ and the ratio of S_{2}state 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:

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

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