Adiabatic lapse rate is defined as $$ \Gamma_a \equiv -\left(\frac{dT}{dz}\right)_{\rm parcel}=-\frac{R_{\rm a} T}{c_p \,p}\left(\frac{dp}{dz}\right)_{\rm parcel} = \frac{g}{c_p} $$ $R_a$ is gas consntant divided by molar mass. If I have a parcel of dry air, I have dry adiabatic lapse rate.

Now, for standard atmosphere, I do not have $p(z)$ but I have $T(p)$ (in form of point data). Neither have I $c_p(p)$ and neither $g(p)$ and neither. So my idea to get the adiabatic lapse rate in general is: $$ -\left(\frac{dT}{dz}\right)_{\rm parcel}=-\left(\frac{dT}{dp}\right)_{\rm parcel}\left(\frac{dp}{dz}\right)_{\rm parcel}=\left(\frac{\Delta T}{\Delta p}\right)_{\rm parcel} \rho g=\left(\frac{\Delta T}{\Delta p}\right)_{\rm parcel} \frac{p}{R_a T} g $$ Is this a correct approach?

Now to get dry adiabatic lapse rate I have absolutely no idea how to approach that because of I do not know $c_p(p)_{\rm dry-air}$ and neither e.g. $g(p)$. The only thing I know is the molar mass of dry air. Are those data about $T(p)$ for the atmosphere of any use for dry adiabatic lapse rate? What relations am I missing?

  • $\begingroup$ While the same calculations appear in astronomy, the notation and context is different enough I might not know what I'm talking about. Still, feel like your first proposed equality is missing the rest of the chain rule: $dT/dz = (\partial T/\partial p)\vert_\rho (dp/dz) + (\partial T/\partial\rho)\vert_p (d\rho/dz)$. I suppose this all comes down to what $\Delta$ means, and whether you take $p$ to be a reparameterization of altitude (in which case what you wrote makes more sense) or one of necessarily two abstract thermodynamic abscissas to vary. $\endgroup$
    – user10851
    Feb 9, 2016 at 6:01

1 Answer 1


How much does $g$ vary between ground level and $10 \, \mathrm{km}$ altitude. The heat capacity of dry air is also pretty constant over the temperature range of the troposphere. At ground level, the globally averaged temperature is about $25\sideset{^{\circ}}{}{\mathrm{C}} ,$ and at about $10 \, \mathrm{km} ,$ the temperature of the air is on the order of about $-60\sideset{^{\circ}}{}{\mathrm{C}} .$ Look up the heat capacity of air at these temperatures and see how much or little it varies. Then you'll know what to do.

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    $\begingroup$ The above problem relates to Jupiter. I have pressure levels and temperature at those levels. I am confused about the whole concept of dry air. If I have something called "dry air", I would assume, it has some given and, most importantly, constant composition, thus $c_p$ of dry air should be constant (independent of altitude or pressure). Then the lapse rate only changes as gravity changes. So to get adiabatic lapse rate for dry air on jupiter, I should know $c_p$ of dry air on jupiter which I do not know. The only thing I know related to the dry air on jupiter is its molar mass. $\endgroup$
    – atapaka
    Feb 9, 2016 at 12:29
  • $\begingroup$ I don't know much about Jupiter, but I'm pretty sure there is no air there. Whatever gases that are present are at pretty high pressure near the surface, so one would have to take into account non-ideal gas effects. The physical properties of the gases that are there are independent of whether they are on Jupiter or somewhere else. So, if you know the composition of the gases at the surface, you can estimate their heat capacity and other properties. $\endgroup$ Feb 9, 2016 at 13:05

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