#### Transcript lecture_2

Lecture 2 Ideal Gas Law Stull: pg 4-9, Wallace & Hobbs: 63-67 Atmospheric Pressure Atmospheric pressure change is approximately exponential with height, z. If temperature were uniform with height (it’s not) then: P Poe ( a / T ) z a= 0.0342 K/m, Po = scaled pressure = 101.325 kPa P Poe z / H p Hp = 7.29 Km = scale height for pressure, it is the e-folding distance for pressure curve. Atmospheric Pressure P Poe z / H p Because pressure decreases with height monotonically (one direction), P can be used as a surrogate for a measure of altitude (Fig. 1.4, Stull). (Example in Stull, pg. 6) Density m /V Density is defined as mass per unit volume. • Density increases as the number and molecular weight of molecules in a volume increase. • Average density at sea level = 1.255 kg m-3 (Table 1.2, Stull) Density Because gases such as air are compressible, air density can vary over a wide range. Density decreases roughly exponentially with height in an atmosphere of uniform temperature. o e z / H p ρo = 1.225 kg/m3, scale height, Hp = 8.55 km. Temperature When a group of molecules (microscopic) move in the same direction, motion is called wind (macroscopic). When they move in random directions, motion is associated with temperature. Higher temperatures, T are associated with greater average molecular speeds, v. T a mwv 2 a = 4 ·10-5 k m-2s2, scale height, mw= molecular weight of most gases. Temperature Absolute units such as K (Kelvin) must be used for temperature in all thermodynamic and radiative laws! Standard average sea-level temperature is: T= 15°C = 288 K = 59 °F Virtual Temperature For moist air, the gas constant changes because water vapor is less dense than dry air. To simplify things, a virtual temperature, Tv can be defined to include the effects of water vapor on density: Tv T (1 0.61 w) where w is the water vapor mixing ratio (g water vapor / kg dry air) and all temperatures are in K. Moist air of temperature T behaves as dry air of temperature Tv. Equation of State Because pressure is caused by the movement of molecules, we might expect the pressure P to be greater where there are more molecules (i.e., greater density ρ), and where they are moving faster (i.e., greater temperature T). The relationship between pressure, density, and temperature is called the Equation of State. For dry air the gases in the atmosphere have a simple equation of state known as the ideal gas law. Ideal Gases Molecules have zero volume No intermolecular forces Elastic collisions with walls of container Works well for the Earth’s atmosphere Molar Form Variables Pressure (p) Volume (V) Temperature (T) Number of moles (n) Ideal Gas Law pV * R (constant) nT Exercise: Determine units of R* Universal Gas Constant * R = 8.314 -1 -1 Jmol K Ideal Gas Law: Molar Form pV nR T * Memorize this! Sealed Container cylinder Gas n fixed Massless, Frictionless Piston piston Gas Weights can be added or removed to change pressure Equilibrium internal pressure = external pressure Gas Keep T fixed, Vary p * nR T constant V p p i.e., pressure and volume are inversely proportional (Boyle’s Law) Increase Pressure Increase Pressure Increase Pressure Ideal gas law V decreases Increase Pressure Ideal gas law V decreases Increase Pressure Ideal gas law V decreases Decrease Pressure Decrease Pressure Decrease Pressure Ideal gas law V increases Decrease Pressure Ideal gas law V increases Decrease Pressure Ideal gas law V increases Keep p fixed, Vary T nR T nR * T V p p * i.e., volume is directly proportional to temperature Increase T Ideal gas law V increases Apply Heat Increase T Ideal gas law V increases Apply Heat Increase T Ideal gas law V increases Apply Heat Decrease T Ideal gas law V decreases Apply Cooling Decrease T Ideal gas law V decreases Apply Cooling Decrease T Ideal gas law V decreases Apply Cooling Fix V nR p V * T i.e., p is directly proportional to T Increase T Ideal gas law p increases Piston doesn’t move Pressure builds up Apply Heat (internal pressure external pressure) Exercise n = 1.00 mol V = 1.00 m3 T = 300 K Calculate p (in Pa) Solution 1.00mol 8.314J mol p 1.00m 2490J m 3 3 2490N m m 2 3 2490N m 2490Pa 1 K 1 300K Partial Pressure Consider a mixture of k gases Volume, V; temperature, T; pressure, p; # moles = n ni = # moles of gas i n1 + n 2 + + n k = n Partial pressure, pi: * ni R T pi V Dalton’s Law k p p i i 1 Proof RT pi ni i 1 i 1 V * nR T V k k * Ratios pi ni p n Ratio of partial pressure to total pressure = mole fraction Partial Volume * ni R T Vi p Volume Ratios Vi ni V n Explains why mole fractions are also called volume fractions. Ideal Gas Law – Mass Version Start with molar version pV nR T * Let m = mass Let M = molecular weight (1) Mass & Moles m n M m n M (2) Substitute (2) into (1) m * pV R T M Rearrange R pV m M * Define T * R R M Finally pV mRT (3) R is called the specific gas constant. It depends on the molecular weight of the gas. Gas Constant for Dry Air Md = 28.97 gmol-1 1 1 1 1 8.314J mol K Rd 1 28.97g mol 0.2870J g K Switch to SI Units Replace g by kg 1000g Rd 0.2870J g K kg 1 1 1 287.0 J kg K 1 Alternate Forms In meteorology, we are not dealing with fixed volumes Divide both sides of (3) by V m p RT V What is m/V? Density. Symbol: Density Form of Ideal Gas Law p RT Exercise p = 1.000 atm = 1013.25 hPa T = 15.00C = 288.15K Calculate the density of dry air Solution p 1013.25hPa Rd T 287.0J kg1 K 1 288.15K 1013.25x102 Pa 1 1 287.0 288.15 J kg K K N m 2 1.225 1 J kg N m 3 1.225kg m 1.225 1 N m kg 2 Yet Another Form of the Ideal Gas Law Divide both sides of (3) by m V p RT m V/m = specific volume Symbol: Specific Volume Form of Ideal Gas Law p RT Don’t Memorize All the Forms! Remember these molar form: pV = nR*T mass form: pV = mRT Remember that = m/V and = V/m. Note that = 1/. Gas Constant for Moist Air Rair = R/Mair Mair Md/(1 + 0.61w) Thus, Rair R* R* M air M d 1 0.61w * R 1 0.61w Md Rd 1 0.61w Gas Law for Moist Air p RairT or p RairT Density of Moist Air p p 1 RairT Rd T 1 0.61w dry 1 0.61w For a given temperature and pressure, the greater the mixing ratio, the smaller the density. Example Given: T = 300K, p = 1000 hPa (Take all zeros to be significant) dry p Rd T 1.000x105 Pa 287.0 J kg1 K 1 300K N m 2 1.16 J kg1 N m 2 1.16 N m kg1 1.16kg m 3 Continued Now, let w = 0.040 (high!) 3 1.16kg m 1 0.61 0.040 3 1.16kg m 1.02 3 1.13kg m Conclusion Density of moist air at most 3% less than density of dry air For most purposes, Rd is OK. Virtual Temperature, Tv Idea: Use Rd for moist air Absorb information about moisture content into T p RT Rd (1 0.61w)T p RT Rd (1 0.61w)T Tv Ideal Gas Law with Tv p Rd Tv Tv > T, but the difference is usually small T vs. Tv Tv T 0.61w T Let w = 0.040 (About as large as w gets) Tv T 0.61 0.040 .024 T (A 2.4% difference)