Chapter 11A Gases

The Atmosphere The atmosphere is the thin layer of gases that surround the earth. Air is composed of a mixture of gases: 78 % Nitrogen, N2 21 % Oxygen, O2 .1 % Argon, Ar 365 ppm CO2 0 - 4 % water, H2O

Physical Properties of Gases No definite shape or volume: expand to fill container, take shape of container. Compressible increase pressure, decrease volume. Low Density air at room temperature and pressure: 0.00117 g/cm3. Exert uniform pressure on walls of container. Mix spontaneously and completely. Diffusion (high conc.  low conc.)

Pressure Pressure = force exerted per unit area. P = F/A Atmospheric Pressure = force exerted by earth’s atmosphere. Atmospheric Pressure is measured with a barometer. Aneroid Barometer

Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli. during the 17th century. The device was called a “barometer”. Baro = “weight” Meter = “measure” Mercury Barometer

Pressure Units mm of mercury (mm Hg) 1 mm Hg = 1 torr 760 mm Hg = 760 torr = 1 atm 1 atm is 1 atmosphere of pressure, (aka: standard pressure).

Robert Boyle (1627-1691) Born into an aristocratic Irish family. Became interested in medicine, astronomy, chemistry.  Avid hot air balloonist.

Boyle’s Law As pressure of gas increases, volume decreases: This is an inverse proportion: as one increases the other decreases P1V1 = P2V2

Boyle’s Law Graph Insert figure 12.9

Boyle’s Law Problem #1 A gas has a volume of 5 L at 2 atm pressure and the pressure is increased to 4 atm. Calculate the new volume. P1V1 = P2V2 -> V2 = P1V1 P2 V2 = (2 atm) (5 L) = 2.5 L 4 atm

Boyle’s Law Problem #2 A gas has a pressure of 6.5 atm and a volume of 20 L. Calculate the new pressure if the volume is increased to 80 L. P1V1 = P2V2 -> P2 = P1V1 V2 P2 = (6.5 atm) (20 L) = 1.6 atm 80 L

Jaques Charles (1746-1823) French Physicist. Conducted the first scientific balloon flight in 1783.

CHARLES’ LAW Volume is directly proportional to temperature Volume & temperature both increase or both decrease together V1 = V2 T1 T2

Charles’s Law Graph Insert figure 12.11

Charles’s Problem #1 A certain gas in a closed container has a volume of 0.567 mL at a temperature of 25oC. If there is no change in pressure, calculate the volume at - 25oC. First Convert Temperatures to Kelvin. K = oC + 273 T1 = 25 + 273 = 298 K T2 = -25 + 273 = 248 K

Charles’s Problem #1 (continued) T1 = 298 K T2 = 248 K V1 = 0.567 mL V2 = ? V1 = V2 -> V2 = V1 T2 T1 T2 T1 V2 = (0.567 mL) (248 K) = 0.47 mL 298 K

Charles’s Problem #2 A certain gas in a closed container has a volume of 25 mL at a temperature of 23oC. What volume will it occupy at 250 C? First Convert Temperatures to Kelvin. K = oC + 273 T1 = 23 + 273 = 296 K T2 = 250 + 273 = 523 K

Charles’s Problem #2 (continued) T1 = 296 K T2 = 523 K V1 = 25 mL V2 = ? V1 = V2 -> V2 = V1 T2 T1 T2 T1 V2 = (25 mL) (523 K) = 44.2 mL 296 K

Chapter 11A SUTW Prompt Describe how Boyle’s Law & Charles’ Law relate pressure, volume, and temperature. Complete a 8 -10 sentence paragraph using the SUTW paragraph format. Hilight using green, yellow, and pink. Due Date: Tomorrow (start of class).