📂Thermal Physics

Gas Molecule Count Formula Depending on Height in an Isothermal Atmosphere

Formulas¹

Assuming the temperature $T$ is constant, let’s define the number of gas molecules per unit volume $V=1$ at height $h$ as $N(h)$. If the mass of the gas molecules is $m$ and the gravitational acceleration is $g$, the following formula holds.

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

Explanation

Originally, this formula does not stand out in thermodynamics, but it’s interesting that there are two completely different methods of derivation.

Derivation

Using Differential Equations

Consider a layer of air from height $h$ to $h + dh$. Within a unit area, there would be $N dh$ 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 get the following equation.

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

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

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

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

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

Solving the logarithm we get:

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

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Using the Boltzmann Distribution

Boltzmann Distribution

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

For a gas molecule of mass $m$ at height $h$, its gravitational potential energy is $mgh$. The probability that a gas molecule has energy $mgh$ is given by the Boltzmann distribution as follows.

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

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

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

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