By 0x7df, Mon 31 August 2015, in category Atmospheric science

The ideal-gas equation of state is:

$$ p = \rho R T $$

where \( R\) is the individual gas constant for the gas in question, and:

$$ R = \frac{nR^*}{m} = \frac{R^*}{M} $$

where \( R^*\) is the universal gas constant (\( 8.31~ \mathrm{J/mol.K}\)) and \( n, m\) are the number of moles and the mass of the sample of gas, respectively; and \( M = m/n\) is the molecular mass of the gas.

Dry air is typically assumed to be a perfect gas, with an individual gas constant of \( 287~\mathrm{J/kg.K}\) or \( 2.87 \times 10^6~\mathrm{cm^2/s^2.K}\).

Water vapour is also assumed to be a perfect gas, with an individual gas constant of \( 461.5~\mathrm{J/kg.K}\). The ratio of \( R / R_v\):

$$ \sigma = \frac{R}{R_v} = \frac{M_v}{M} = 0.622 $$

where \( R, M\) are the gas constant and molecular mass of dry air, and \( R_v, M_v\) of water vapour.

Moist air is a mixture of dry air and water vapour. To determine the individual gas constant of moist air, consisting of some mixture of dry air and water vapour, we first write the equation of state for water vapour:

$$ e = \rho_v R_v T $$

where \( e\) is the water vapour pressure and \( \rho_v\) is the density of water vapour. The pressure of the moist air is the sum of the partial pressures of the dry air and the water vapour:

$$ p_m = p + e = \rho R T + \rho_v R_v T $$

$$ p_m = \rho R T + \rho_v \frac{R}{\sigma} T $$

$$ p_m = RT\left(\rho + \frac{\rho_v}{\sigma}\right) $$

$$ p_m = RT\frac{1}{V}\left(m + \frac{m_v}{\sigma}\right) $$

$$ p_m = RT \frac{\rho_m}{m + m_v}\left(m +
\frac{m_v}{\sigma}\right) $$

This allows us to introduce the *virtual temperature*, which is the
fictitious temperature that dry air would have to have to achieve
the same (lower) density of the moist air, at the same pressure:

$$ p_m = \rho_m RT_v $$

From comparing the last two equations we can see that:

$$ T_v = \left[m + \frac{m_v}{\sigma}\right]\frac{1}{m+m_v} T $$

$$ T_v = \left[m + m_v + \frac{m_v}{\sigma} - m_v\right]
\frac{1}{m + m_v} T $$

$$ T_v = \left[1 + \frac{m_v}{m +
m_v}\left(\frac{1}{\sigma} - 1\right)\right] T $$

$$ T_v = \left[1 + \frac{m_v}{m + m_v}\frac{1 -
\sigma}{\sigma} \right] \frac{1}{m + m_v} T $$

By definition, the mixing ratio \( \mu = m_v/(m + m_v)\) is the ratio of the mass of water vapour to the mass of moist air:

$$ T_v = \left[1 + \frac{1-\sigma}{\sigma} \mu \right] T $$

The virtual temperature is always slightly higher, by at most a few degrees, than the true temperature of the moist air.

Elsewhere we have seen that the geopotential is:

$$ d\phi = -\frac{1}{\rho} dp $$

Substituting in the equation of state for moist air, involving the virtual temperature:

$$ d\phi = -RT_v \frac{dp}{p} $$

Integrating:

$$ \phi = \phi_0 - R \int_{p_0}^p T_v d\ln p $$