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Subsections

8.1.2 Thermodynamics of the air

8.1.2.0.1 Ideal gas

For one molecule with n degrees of freedom (e.g. in the rigid-molecule approximation, n=2 (deg. of rotation) +3 (of transl.) = 5 for O2 and N2, and n= 3+3 = 6 for H2O), the internal energy per molecule is

\begin{displaymath}U =
{\rm n} \times \frac{1}{2} k T \end{displaymath}

and for ${\cal N}$ molecules of O2 or N2,

\begin{displaymath}U= \frac{5}{2}{\cal N} k T= \frac{5}{2} n R T\end{displaymath}

where n is the number of moles and R the gas constant.

The Enthalpy is

\begin{displaymath}H = U +pV = U + {\cal N} k T\,\end{displaymath}


\begin{displaymath}c_v=
\frac{dU}{dT}=\frac{5}{2}{\cal N} k ~,~~~ c_p = \frac{dH...
...}{2}{\cal N} k,
~,~~~ \gamma= \frac{c_p}{c_v}= \frac{7}{5}= 1.4\end{displaymath}

For adiabatic expansion, we have the standard relations

\begin{displaymath}c_p
\mathrm{Log}(\frac{T}{T_0})-R \mathrm{Log}(\frac{p}{p_0}) = 0
~~{\rm and}~~~ T= Cst\times p^{m}\end{displaymath}

where $m=\frac{R}{c_p}=\frac{\gamma-1}{\gamma}$is the Poisson constant.

8.1.2.0.2 Gas mixture: Dalton's law

A mixture of ideal gases i= 1, 2, ... in a volume V behaves like an ideal gas:

Partial pressures: $\textstyle p_1 V = {\cal N}_1 k T,$ $\displaystyle p_2 V= {\cal N}_2 k T, ~~...$ (8.1)


Total pressure: $\textstyle p V = (p_1+p_2)V = ({\cal N}_1 + {\cal N}_2) k T+...$ $\displaystyle = {\cal N} kT$ (8.2)

Dry air is a mixture of N2, O2, ... molecules. It behaves indeed very much like an ideal gas: Ra=cpa-cva= 8.3143 J/mol-deg (vs 8.3149 for an ideal gas), $\gamma_a$= 1.404 (vs 1.400 for ideal rigid molecules).

Wet air (without clouds) is a mixture of dry air + H2O molecules. It is customary to denote by e the partial pressure of water vapor, pa that of dry air, and p' the wet air pressure. The specific heats of water vapor are not that different from those of ideal gases: cvw= 25.3 + 2 10-3 (T-273); $\gamma_w= 1.37 $, vs cv=3R= 24.9 and $\gamma=\frac{4}{3}$ for a rigid asymmetric top.

Then, Dalton's law yields:

 \begin{displaymath}c'_v=(1-\frac{e}{p'})c_{v_a}+\frac{e}{p'}c_{v_w} \simeq (1+0.2\frac{e}{p'})c_{v_a}
\end{displaymath} (8.3)

The fractional abundance of water vapor and $\frac{e}{p}$ reaching seldom a few percent, the wet air constants are within a small correction term equal to the dry air constants. In particular, introducing the volume density $\rho$, it is customary to write:

 \begin{displaymath}p'= \frac{R\rho{'}T}{M'}= \frac{R\rho T'}{M_a}
\end{displaymath} (8.4)

where $T'=\frac{M' T}{M_a}= T (1-0.378\frac{e}{p'})^{-1}$, is the virtual temperature.

Then, for the adiabatic expansion of a wet air bubble, one has:

 \begin{displaymath}T'= Cst \times p^{m'}
\end{displaymath} (8.5)

m' is equal to ma within few per mil, so, in practice, the adiabatic curves of dry air can be used for wet air (without clouds), provided one replaces T by T'(the difference could reach a few K and could be important near 0$^\circ $C). In the following, we drop the `prime' signs, except for the virtual temperature T'.


  
Figure 8.1: Column densities of important atmosphere components above a given altitude - after Queney (1974)
\resizebox{9.0cm}{!}{\includegraphics{mg1f1.eps}}


next up previous contents
Next: 8.1.3 Hydrostatic equilibrium Up: 8.1 The Atmosphere Previous: 8.1.1 Constituents of the
S.Guilloteau
2000-01-19