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Suppose that a someone decided to calculate, for a given amount of heat added into a planet's atmosphere(an ideal gas), how much is the corresponding temperature rising of the atmosphere, $\Delta T$. Assuming that the whole heat transforms into the internal energy of the gas and distributes uniformly (s)he got $\Delta T=1K$.

A question rises: actually, some of the heat is transformed into gravitational potential energy of the atmosphere(we ignore kinetic energy of the atmosphere). If we take also this into account then how much is the temperature rising of the atmosphere, $\Delta T$ now? Maybe the effect due to the gravity is negligible?

For simplicity let's assume that the pressure change is approximately exponential with height, $h$

$$P=P_0e^{-\frac{h}{H}}$$ where $P_0$ is pressure at $h=0$ and $H$ is called the scale height for pressure.

For definiteness, let's take the Earth's atmosphere. In this case $P_0=101$ $kPa$ is average sea-level pressure. $H$ is about $8$ $km$.

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up vote 5 down vote accepted

The average height of a molecule is $H$ with your exponential distribution, because $\int_0^\infty dt\,t\,\exp(-t)=1$, so its average potential energy is $mgH$ where $m$ is the mass of the molecule. How much does it compare with $3kT/2$ (or $5kT/2$ etc.) in the kinetic energy? And what will happen to $\Delta T$?

The answer is given by the virial theorem (so in advance, I am telling you that the reduction of $\Delta T$ will be of the same order as $\Delta T$ itself) but let me use no pre-derived results.

Instead, just realize that the exponential decrease in $\exp(-h/H)$ is nothing else than the Maxwell-Boltzmann factor $\exp(-mgh/kT)$ which implies that $H=kT/mg$ and the average potential energy is $mgH=kT$. (At this stage, I am neglecting the dependence of the temperature itself on the height etc.)

So if you have a monoatomic gas which has $3kT/2$ in the kinetic energy, $kT=2kT/2$ is ultimately given to the potential energy which is $2/5$ of the total energy. As the temperature rises, the scale height will also have to rise, and you can see that $2/5$ of the energy that is initially pumped to the kinetic energy will be converted to the increased potential energy and only $3/5$ of the original increase of the kinetic energy will stay in the kinetic energy.

So I think that for the monoatomic gas, the answer is $(3/5)\Delta T$. Similarly, it will be $(5/7)\Delta T$ if the kinetic energy of a single molecule is $5kT/2$, and so on.

Just to be sure, this is not equivalent to the "global warming" issue because the greenhouse warming is a permanent change of the energy flows (extra Watts per square meter - Joule in every single second) while the energy needed to change the gravitational potential of the gas when its distribution at different altitudes changes is just a one-time event.

However, the general observation that the ultimate increase of the temperature (or anything else) will be smaller than the most naively calculated one is valid: the extra heat is ultimately redistributed to and consumed by "many consumers" (in this case, potential energy), which means that one particular consumer ultimately gets less. This is the reason why the feedbacks of stable systems tend to be negative - an insight known as the Le Chatelier's principle (in chemistry) or homeostasis (generally) or Lenz's law (in electromagnetism coupled to mechanics) etc.

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Its interesting how the gravitaionally confined heat capacity is the same as the heat capacity at constant pressure. But intuitively this makes sense as the pressure as a function of mass fraction (how much mass of atmosphere is below a given pressure value remains the same). So essentially we heat/cool the atmosphere at constant pressure (on average). In the real world, the oceanic heat content is more than two orders of magnitude greater than the atmosphere, so atmospheric heat storage can be ignored in many contexts. –  Omega Centauri Mar 5 '11 at 15:48
    
š This is a very nice explanation! Thanks! –  Martin Gales Mar 7 '11 at 7:05
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