Atmospheric Thermodynamics Atmospheric Composition What is the composition of the Earth’s atmosphere? Gaseous Constituents of the Earth’s atmosphere (dry air) Constituent Molecular Weight Fractional Concentration by Volume of Dry Air Nitrogen (N2) 28.013 78.08% Oxygen (O2) 32.000 20.95% Argon (Ar) 39.95 0.93% Carbon Dioxide (CO2) 44.01 380 ppm Neon (Ne) 20.18 18 ppm Helium (He) 4.00 5 ppm Methane (CH4) 16.04 1.75 ppm Krypton (Kr) 83.80 1 ppm Hydrogen (H2) 2.02 0.5 ppm Nitrous oxide (N2O) 56.03 0.3 ppm Ozone (O3) 48.00 0-0.1 ppm Water vapor is present in the atmosphere in varying concentrations from 0 to 5%. Aerosols – solid and liquid material suspended in the air What are some examples of aerosols? The particles that make up clouds (ice crystals, rain drops, ect) are also considered aerosols, but are more typically referred to as hydrometeors. We will consider the atmosphere to be a mixture of two ideal gases, dry air and water vapor, called moist air.Gas Laws Equation of state – an equation that relates properties of state (pressure, volume, and temperature) to one another Ideal gas equation – the equation of state for gases € pV = mRT p – pressure (Pa) V – volume (m-3) m – mass (kg) R – gas constant (value depends on gas) (J kg-1 K-1) T – absolute temperature (K) This can be rewritten as: € p =mVRTp =ρRT ρ - density (kg m-3) or as: € pVm= RTpα= RT α - specific volume (volume occupied by 1 kg of gas) (m3 kg-1) Boyle’s Law – for a gas at constant temperature € V ∝1 p Charles’ Laws: For a fixed mass of gas at constant pressure € V ∝T For a fixed mass of gas at constant volume € p ∝TMole (mol) – gram-molecular weight of a substance The mass of 1 mol of a substance is equal to the molecular weight of the substance in grams. € n =mM n – number of moles m – mass of substance (g) M – molecular weight (g mol-1) Avogadro’s number (NA) – number of molecules in 1 mol of any substance NA = 6.022x1023 mol-1 Avogadro’s hypothesis – gases containing the same number of molecules (or moles) occupy the same volume at the same temperature and pressure Using the ideal gas law and the definition of a mole gives: € pV = nMRT Using this form of the ideal gas law with Avogadro’s hypothesis indicates that MR is constant for all gases. This constant is known as the universal gas constant (R*). € R*= MR = 8.3145 J K-1 mol-1 With the universal gas constant the ideal gas law becomes: € pV = nR*TBoltzmann’s constant (k) – gas constant for 1 molecule of any gas € k =R*NA Using Boltzmann’s constant the ideal gas law can be written as: € p = n0kT n0 – number of molecules of gas per unit volume € =nNAV      Application of the ideal gas law to dry air € pd=ρdRdT or € pdαd= RdT pd – pressure exerted by dry air ρd – density of dry air Rd – gas constant for dry air αd – specific volume for dry air € Rd=R*Md, where Md – apparent molecular weight of dry air (=28.97 g mol-1) € Md=mdn=mii∑miMii∑=mii∑nii∑ mi – mass of ith constituent of dry air ni – number of moles of ith constituent of dry air Example: Calculate the gas constant for dry airExample: Calculate the density of air on the roof of Duane Physics. Why is the calculated density not exactly correct? Application of the ideal gas law to individual components of air Each gas that makes up the atmosphere obeys the ideal gas law: € pi=ρiRiT For water vapor the ideal gas law is: € e =ρvRvT or € eαv= RvT e – pressure exerted by water vapor (vapor pressure) ρv – density of water vapor αv – specific volume of water vapor Rv – gas constant for water vapor Example: Calculate the gas constant for water vapor Dalton’s law of partial pressure – the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum of the partial pressure of the gases € p = Tρii∑Ri Partial pressure – pressure exerted by a gas at the same temperature as a mixture of gases if it alone occupied all of the volume that the mixture occupies Example: The pressure in a hurricane is observed to be 950 mb. At this time the temperature is 88 deg F and the vapor pressure is 25 mb. Determine the density of dry air alone and the density of water vapor alone.Virtual Temperature How does the gas constant vary as the molecular weight of the gas being considered changes? The molecular weight of dry air is greater than the molecular weight of moist air (i.e. one mole of dry air has a larger mass than one mole of moist air) What does this imply about the gas constant for moist air compared to dry air? For dry and moist air at the same temperature and pressure which will have the smaller density? As the amount of moisture in the air changes the molecular weight of the moist air will also change causing the gas constant for the moist air to vary. The density of moist air is given by: € ρ=md+ mvV=ρd'+ρv', where ρ - density of moist air md - mass of dry air mv - mass of moist air V - volume € ρd' - density that mass md of dry air would have if it occupied volume V € ρv' - density that mass mv of water vapor would have if it occupied volume V € ρd' and € ρv' can be considered partial densities (analogous to partial pressures) Using the ideal gas law the partial pressure of dry air (pd) and water vapor (e) can be calculated as: € pd=ρd'RdT and € e =ρv'RvTFrom Dalton’s law the total pressure exerted by the moist air is: € p = pd+ e=ρd'RdT +ρv'RvT Rewriting in terms of the density of moist air (ρ) gives: € ρ=pdRdT+eRvT=p − eRdT+eRvT=pRdT1 −ep1 −RdRv            =pRdT1 −ep1 −ε( )      where € ε=RdRv=MwMd= 0.622 This equation can be rewritten in terms of the virtual temperature (Tv) as: € ρ=pRdTv or € p =ρRdTv where € Tv≡T1 −ep1 −ε( ) The virtual temperature is the temperature dry air would need to have if it were to have the same density as a sample of moist air at the same pressure How does the magnitude of Tv compare to the magnitude of T?Example: The pressure in a hurricane is observed to be 950 mb. At this time the temperature is 88 deg F and the vapor pressure is 25 mb. Calculate the virtual temperature in the hurricane. How does the observed temperature compare to the virtual