1. AOS 101 February 12 or 14 Ideal Gas Law

2. 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!)

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 = o C + 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 1 o C.

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

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 32 o F, 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 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 Definition: The amount of heat to raise the temperature of a substance by one degree.

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 o C x 1 g) ΔT = 48 o C So if the initial temperature of the air is 10 o C, the final temperature is 58 o C.