2.1: The ideal gas law holds approximately for most gasses in most atmospheric conditions:br /blockquote pV = mRT/blockquotebr /br /where p is the pressure exerted by the gas (or partial pressure for a component of a mixture of gasses), V is the volume occupied, m is the mass, R the gas constant (in this formulation R depends upon the type of gas or gasses present) and T is the temperature in Kelvin.br /span style=”font-style:italic;”br /Personally, I prefer the version I know from high school chemistry: pV = nRT – with n the number of moles of gas, then R is a true constant: R* – but stating things in terms of mass and mass density might make more sense in some situations./spanbr /br /The apparent molecular weight of dry air is 28.97 g/mol, so the gas constant for dry air is:br /br /blockquoteR_{dry air} = R* /M_{dry air} = 287 J/deg/kgbr //blockquotebr /br /Example: calculate the density of water vapour at 20^oC which exerts a partial pressure of 9mbar. br /br /Or, in more useful units 900Pa @ 293Kbr /br /blockquotee alpha = R_v Tbr /900 alpha = R*/(M_v)*293br /alpha = 150m^3/kgbr /rho = alpha^{-1} = 6.67×10^{-3}kg m^{-3}/blockquotebr /br /2.1.1 Virtual Temperaturebr /br /In some cases, when dealing with a mixture of air and water vapour, it is more convenient to retain the gas constant for dry air and use a ficticious temperature that corrects for the effect of the moisture in the gas equations.br /br /Consider a mixture thusly:br /br /blockquoterho = (m_d + m_v)/V = rho_d + rho_m/blockquotebr /br /Applying the ideal gas law to the two components to find the partial pressures:br /blockquotebr /e = R_vrho_vTbr /p_d = R_drho_dTbr /br /p = p_d+e/blockquotebr /br /combining leads to:br /blockquotebr /rho = (p-e)/R_dT + e/R_vTbr /rho = (p/R_dT) * [1-(e/p)*(1-eps)]/blockquotebr /br /(where eps is the mass fraction of moisture to dry air)br /so, we can rewrite this as:br /blockquotebr /p = R_drhoT_v/blockquotebr /br /where br /blockquotebr /T_v = T/[1-(e/p)(1-eps)]/blockquotebr /br /is the virtual temperature – the temperature a sample of dry air would have if it had the same density and pressure as the given sample of moist air. As can be seen, T_v T, so moist air is “warmer” than other wise identical dry air – that is, a sample of moist air is less dense than an otherwise identical sample of dry air.

# Wallace and Hobbs 2.1 – The Gas Laws

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