Did I Make Conceptual Mistakes in Deriving the Hypsometric Equation?

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SUMMARY

The discussion centers on the derivation of the hypsometric equation, assuming a sea level temperature of 15°C and a temperature lapse rate of 6.5°C per kilometer. The user applied the ideal gas law, hydrostatic equilibrium, and the temperature variation with height to derive the equation. The final expression for pressure is given as P = P₀(1 - 2.26 x 10⁻⁵z)e^(g/(0.0065R)). However, the user identifies a conceptual mistake regarding the dimensional analysis of g/R, questioning the validity of exponentiating this term.

PREREQUISITES
  • Understanding of the ideal gas law (P = ρRT)
  • Familiarity with hydrostatic equilibrium (dP = -ρgdz)
  • Knowledge of temperature lapse rates in atmospheric science
  • Basic skills in dimensional analysis
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  • Review the derivation of the hypsometric equation in atmospheric thermodynamics
  • Study the implications of dimensional analysis in physical equations
  • Learn about the applications of the hypsometric equation in altimetry
  • Explore the relationship between pressure, temperature, and altitude in meteorology
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Students and professionals in atmospheric science, meteorology, and physics, particularly those involved in deriving equations related to pressure and temperature in the atmosphere.

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Homework Statement


Derive the hypsometric equation, assuming a sea level temperature of 15 C, and that the temperature decreases with heigh at a rate of 6.5 C per km.

Homework Equations


Ideal gas law:
P=\rho RT
Hydrostatic equilibrium:
dP = -\rho gdz
Temperature varies with height:
T = 288 - 6.5*10^{-3}z

The Attempt at a Solution


\int_{P_{0}}^{P} \frac{dP}{P} = \int_{0}^{z} \frac {-gdz}{RT(z)}
\int_{P_{0}}^{P} \frac{dP}{P} = \int_{0}^{z} \frac {-gdz}{R*(288 - 6.5*10^{-3}z)}
ln\left(\frac{P}{P_{0}}\right)=\frac{g}{6.5*10^{-3}*R}ln\left(\frac{288-6.5*10^{-3}z}{288}\right)
And exponentiating this:
\frac{P}{P_{0}} = e^{\frac{g}{0.0065R}}\left(1-2.26x10^{-5}z\right)
and finally
P = P_{0}\left(1-2.26x10^{-5}z\right)e^{\frac{g}{0.0065R}}

but from a dimensional analysis, I know g/R is not a dimensionless quantity and therefore I should *not* be exponentiating this. I expect an exponential in the final answer but I'm not sure whether this is correct. (Some of you may recognize the initial conditions set this to be the equation used by a standard altimeter).

Did I make any conceptual mistakes? I'm fairly confident about the math.
 
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The constant 6.5*10^-3 has units as well.
 

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