Einstein relation and stability of isothermal atmosphere — Phénomènes de transport 3 : Diffusion de particules

The atmosphere is described as an isothermal ideal gas. In the terrestrial gravitation field, pressure varies as

where the axis

𝑂_{𝑧}

is vertical and ascending,

𝑧 = 0

at sea level and

𝑘_{𝐵} = 𝑅 / {𝒩︀}_{𝑎}

is the Boltzmann constant.

Coup de pouce 1

Be careful not to confuse the amount of substance and the particle density, both of which are commonly written

𝑛

Corrigé

The ideal gas law gives $𝑃 𝑉 = 𝑛_{mol} 𝑅 𝑇$ where $𝑛_{mol}$ is the amount of substance. By dividing by the volume, we get

𝑃 = \frac{𝑛_{mol}}{𝑉} 𝑅 𝑇 = \frac{𝑛}{{𝒩︀}_{𝑎}} 𝑅 𝑇 = 𝑛 𝑘_{𝐵} 𝑇

Thus

𝑛 (𝑧) = \frac{𝑃 (𝑧)}{𝑘_{𝐵} 𝑇} = \frac{𝑃_{0}}{𝑘_{𝐵} 𝑇} exp (- \frac{𝑚 𝑔 𝑧}{𝑘_{𝐵} 𝑇})

2/ Using Fick law, express the current density vector

{\vec{𝑗}}_{𝑑𝑖𝑓𝑓}

due to the diffusion of particles.

Corrigé

{\vec{𝑗}}_{𝑑𝑖𝑓𝑓} = - 𝐷 \vec{grad} 𝑛 = 𝐷 \frac{𝑃_{0} 𝑚 𝑔}{{(𝑘_{𝐵} 𝑇)}^{2}} exp (- \frac{𝑚 𝑔 𝑧}{𝑘_{𝐵} 𝑇}) {\vec{𝑒}}_{𝑧}

3/ The particles that make up air are in motion at microscopic scale. The collisions between particles are modeled with a drag force

\vec{𝑓} = - \frac{𝑚}{𝜏} \vec{𝑣}

that applies to an average particle. Make an inventory of the forces and deduce the limit speed

\vec{𝑣}

of an average particle. Deduce the current density vector

{\vec{𝑗}}_{𝑚𝑖𝑔}

due to the gravitation.

Coup de pouce 1

The model is close to the Drude model. Apply Newton’s second law in the stationary regime.

Coup de pouce 2

Relate the speed to the particle current density vector.

Corrigé

The forces that apply on an average particle are

the weight $\vec{𝑃} = 𝑚 \vec{𝑔}$
the drag force $\vec{𝑓} = - \frac{𝑚}{𝜏} \vec{𝑣}$

In stationary regime, these forces balance each other:

𝑚 \vec{𝑔} - \frac{𝑚}{𝜏} \vec{𝑣} = 0

Thus, the limit speed is

\vec{𝑣} = 𝜏 \vec{𝑔}

The current density vector due to the gravitation is

{\vec{𝑗}}_{𝑚𝑖𝑔} = 𝑛 \vec{𝑣} = 𝑛 𝜏 \vec{𝑔} = - \frac{𝑃_{0} 𝜏 𝑔}{𝑘_{𝐵} 𝑇} exp (- \frac{𝑚 𝑔 𝑧}{𝑘_{𝐵} 𝑇}) {\vec{𝑒}}_{𝑧}

4/ By making an inventory of the particles on a slice of atmosphere in a stationary state, express a relation between

𝐷

𝜏

𝑘_{𝐵}

𝑇

and

𝑚

. This relation is known as Einstein relation.

Coup de pouce 1

Take a particle balance on an infinitesimal slice of atmosphere. Four particle fluxes cross it: the diffusion one and the gravitation one, each at

𝑧

and at

𝑧 + d 𝑧

Corrigé

We study a slice of atmosphere between $𝑧$ and $𝑧 + d 𝑧$ and of area $𝑆$ . In stationary state, the number of particles that enter this slice is equal to the number of particles that exit it :

\begin{matrix} 0 = 𝛿^{2} 𝑄 = \iint_{𝑆_{𝑧}} {\vec{𝑗}}_{𝑑𝑖𝑓𝑓} (𝑧) \cdot \vec{d 𝑆_{𝑧}} d 𝑡 + \iint_{𝑆_{𝑧 + d 𝑧}} {\vec{𝑗}}_{𝑑𝑖𝑓𝑓} (𝑧 + d 𝑧) \cdot \vec{d 𝑆_{𝑧 + d 𝑧}} d 𝑡 \\ + \iint_{𝑆_{𝑧}} {\vec{𝑗}}_{𝑚𝑖𝑔} (𝑧) \cdot \vec{d 𝑆_{𝑧}} d 𝑡 + \iint_{𝑆_{𝑧 + d 𝑧}} {\vec{𝑗}}_{𝑚𝑖𝑔} (𝑧 + d 𝑧) \cdot \vec{d 𝑆_{𝑧 + d 𝑧}} d 𝑡 \\ = 𝑆 𝑗_{diff} (𝑧) d 𝑡 - 𝑆 𝑗_{diff} (𝑧 + d 𝑧) d 𝑡 + 𝑆 𝑗_{mig} (𝑧) d 𝑡 - 𝑆 𝑗_{mig} (𝑧 + d 𝑧) d 𝑡 \\ = - 𝑆 d 𝑡 (\frac{𝜕 𝑗_{diff}}{𝜕 𝑧} + \frac{𝜕 𝑗_{mig}}{𝜕 𝑧}) d 𝑧 \end{matrix}

\frac{𝜕 𝑗_{diff}}{𝜕 𝑧} + \frac{𝜕 𝑗_{mig}}{𝜕 𝑧} = 0

𝐷 \frac{𝑃_{0} 𝑚 𝑔}{{(𝑘_{𝐵} 𝑇)}^{2}} \frac{- 𝑚 𝑔}{𝑘_{𝐵} 𝑇} exp (- \frac{𝑚 𝑔 𝑧}{𝑘_{𝐵} 𝑇}) - \frac{𝑃_{0} 𝜏 𝑔}{𝑘_{𝐵} 𝑇} \frac{- 𝑚 𝑔}{𝑘_{𝐵} 𝑇} exp (- \frac{𝑚 𝑔 𝑧}{𝑘_{𝐵} 𝑇}) = 0

\frac{𝐷 𝑚}{𝑘_{𝐵} 𝑇} - 𝜏 = 0

𝜏 = \frac{𝐷 𝑚}{𝑘_{𝐵} 𝑇}