📂Thermal Physics

Formula for the Number of Gas Molecules at a Given Height in an Isothermal Atmosphere

Formula¹

Assuming the temperature $T$ is constant, let $N(h)$ denote the number of gas molecules per unit volume $V=1$ at height $h$. If the mass of a gas molecule is $m$ and the gravitational acceleration is $g$, then the following equation holds.

$$ N(h) = N(0) e^{- {{mgh} \over {k_{B} T}} } $$

Explanation

This formula is not particularly remarkable in thermodynamics itself, but what makes it interesting is that the two methods of deriving it are strikingly different.

Derivation

Using a Differential Equation

Consider a layer of air from height $h$ to $h + dh$. Within a unit area there will be $N dh$ gas molecules, and the applied pressure is given by $dp = - N dh \cdot mg$.

Ideal gas equation

$$ pV = N k_{B} T $$

Since the volume is fixed at $V=1$ in the ideal gas equation, we obtain the following equation.

$$ p = N k_{B} T \implies dp = k_{B} T d N $$

Substituting $dp = - N dh \cdot mg$ and rearranging gives the following.

$$ {{1} \over {N}} dN = - {{mg} \over {k_{B} T}} dh $$

Solving the above separable first-order differential equation yields the following.

$$ \begin{align*} && \ln N(h) - \ln N(0) =& - {{mg} \over {k_{B} T}} h \\ \implies && \ln N(h) =& \ln N(0) + \ln e^{-mgh / k_{B} T} \\ \implies && \ln N(h) =& \ln \left( N(0) e^{-mgh / k_{B} T} \right) \end{align*} $$

Removing the logarithms, we obtain the following equation.

$$ N(h) = N(0) e^{ -mgh / k_{B} T } $$

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Using the Boltzmann Distribution

Boltzmann distribution

$$ P(\epsilon) \propto e^{ - \epsilon /k_{B} T } $$

The gravitational potential energy of a gas molecule with mass $m$ at height $h$ is $mgh$. Therefore, according to the Boltzmann distribution, the probability that a gas molecule has energy $mgh$ is as follows.

$$ P(mgh) \propto e^{ -mgh / {k_{B} T} } $$

Here, $P(mgh)$ becomes the probability of finding $N(h)$ gas molecules at height $h$, so

$$ N(h) = N(0) e^{ -mgh / k_{B}T } $$

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