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AOS 101

AOS 101. Ideal Gas Law. February 12 or 14. Ideal Gas Law. P = pressure (in Pascals ) ρ = density (in kg/m 3 ) = mass / volume R = gas constant (dry air: R = 287 J/kg K) T = temperaure (in Kelvin !). Reminders.

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AOS 101

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  1. AOS 101 Ideal Gas Law February 12 or 14

  2. Ideal Gas Law • P= pressure (in Pascals) • ρ = density (in kg/m3) = mass / volume • R = gas constant (dry air: R = 287 J/kg K) • T = temperaure (in Kelvin!)

  3. Reminders • In the IGL, pressure must be in units of Pascals (Pa) where 100 Pa = 1 hPa. • To convert, multiply the amount of hPa by 100 to get Pa (for ex: 996 hPa = 996 x 100 = 99600 Pa) • In the IGL, temperature must be in units of Kelvin where K = oC + 273. • Kelvin temperature scale: • All molecular motions stop at 0 K • All temperatures in the universe are > 0 K. • Adding 1 K is the same as adding 1oC.

  4. Latent Heat • Definition: energy released/absorbed during a phase change FOR WATER: 334 J/g released 2260 J/g released SOLID LIQUID GAS 334 J/g absorbed 2260 J/g absorbed SOLID LIQUID GAS LIQUID

  5. Example 1: Getting out of a swimming pool • In the summer, upon exiting a swimming pool you feel cool. Why? • Drops of liquid water are still on your skin after getting out. • These drops evaporate into water vapor, this liquid to gas phase change causes energy to be absorbed from your skin.

  6. Example 2: Citrus farmers • An orange crop is destroyed if temperatures drop below freezing for a few hours. • To prevent this, farmers spray water on the orange trees. Why? • When the temperature drops below 32oF, liquid water freezes into ice. • This liquid to solid phase change causes energy to be released to the fruit. • Thus, the temperature of the orange remains warm enough to prevent ruin.

  7. Example 3: Summer cumulus • Clouds form when water vapor condenses into tiny liquid water drops. • This gas to liquid phase change causes energy to be released to the atmosphere. • The release of latent heat during thunderstorm cloud formation drives many atmospheric processes.

  8. Specific Heat • Definition: The amount of heat to raise the temperature of a substance by one degree. • Q= Amount of heat added to the substance (in calories) • c = specific heat • for water: c = 4.18 J/g K, for air: c = 1.0 J/g K • m = mass of the substance (in grams) • ΔT = change in temperature by the substance

  9. Example: You add 12 calories of heat to 1 gram of air what is the temperature change? • Q = c x m x ΔT ΔT = Q / (c x m) ΔT = 12 cal / (0.24 cal/g oC x 1 g) ΔT = 48oC So if the initial temperature of the air is 10oC, the final temperature is 58oC.

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