Pressure Force acting on a unit area of a surface

Standard Atmospheric Pressure Barometric Pressure Air has weight and therefore exerts pressure. A standard Atmosphere (1atm) is the average pressure at sea level. What happens to air pressure as you increase your elevation? How do we measure pressure?

Kinetic Molecular Theory Particles in an ideal gas… –have elastic collisions. –are in constant, random, straight-line motion. –don’t attract or repel each other. –have an avg. KE directly related to Kelvin temperature. Courtesy Christy Johannesson

Characteristics of Ideal Gases Gases expand to fill any container. Gases are fluids (like liquids). Gases have very low densities. Courtesy Christy Johannesson

Characteristics of Gases Gases can be compressed. Gases undergo diffusion. –r–random motion Courtesy Christy Johannesson

Properties of Gases V = volume of the gas (liters, L) T = temperature (Kelvin, K) P = pressure (atmospheres, atm) n = amount (moles, mol) Gas properties can be modeled using math. Model depends on:

Pressure - Temperature - Volume Relationship P T V Gay-Lussac’s P T  CharlesV T  P T V Boyle’s P 1V1V  ___

Boyle’s Law P 1 V 1 = P 2 V 2 (Temperature is held constant)

As the pressure on a gas increases As the pressure on a gas increases - the volume decreases Pressure and volume are inversely related 1 atm 4 Liters 2 atm 2 Liters

Pressure vs. Volume for a Fixed Amount of Gas (Constant Temperature) Pressure Volume PV (Kpa) (mL) , , , , , , , ,500 Volume (mL) Pressure (KPa)

Boyle’s Law Illustrated Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 404

Pressure Units Atmosphere Feet of water mm Hg cm Hg Torr Barr mbarr kPa Pa

PV Calculation (Boyle’s Law) A quantity of gas has a volume of 120 dm 3 when confined under a pressure of 93.3 kPa at a temperature of 20 o C. At what pressure will the volume of the gas be 30 dm 3 at 20 o C? P 1 x V 1 = P 2 x V 2 (93.3 kPa) x (120 dm 3 ) = (P 2 ) x (30 dm 3 ) P 2 = kPa

Volume and Pressure Two-liter flask The molecules are closer together; the density is doubled. The average molecules hits the wall twice as often. The total number of impacts with the wall is doubled and the pressure is doubled. One-liter flask Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 101

Charles' Law This means, for example, that Volume goes up as Temperature goes up. Jacques Charles ( ) Isolated boron and studied gases. Balloonist. A hot air balloon is a good example of Charles's law. VT V and T are directly related. T 1 T 2 V 1 V 2 = (Pressure is held constant)

Raising the temperature of a gas increases the pressure if the volume is held constant. The molecules hit the walls harder. The only way to increase the temperature at constant pressure is to increase the volume. Temperature

VT Calculation (Charles’ Law) At constant pressure, the volume of a gas is increased from 150 dm 3 to 300 dm 3 by heating it. If the original temperature of the gas was 20 o C, what will its final temperature be ( o C)? T 1 = 20 o C = 293 K T 2 = X K V 1 = 150 dm 3 V 2 = 300 dm dm K = 300 dm 3 T 2 T 2 = 586 K o C = 586 K T 2 = 313 o C

Temperature and the Pressure of a Gas High in the mountains, Richard checked the pressure of his car tires and observed that they has kPa of pressure. That morning, the temperature was -19 o C. Richard then drove all day, traveling through the desert in the afternoon. The temperature of the tires increased to 75 o C because of the hot roads. What was the new tire pressure? Assume the volume remained constant. What is the percent increase in pressure? P 1 = kPa P 2 = X kPa T 1 = -19 o C = 254 K T 2 = 75 o C = 348 K kPa 254 K = P K P 2 = 277 kPa % increase = 277 kPa kPa x 100 % kPa or 37% increase

The pressure and absolute temperature (K) of a gas are directly related –at constant mass & volume P T Gay-Lussac’s Law Courtesy Christy Johannesson Temperature (K) Pressure (torr) P/T (torr/K) ,

The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 dm 3. What volume will it occupy at 0 o C and 93.3 kPa? P 1 = kPa T 1 = 273 K V 1 = 500 dm 3 P 2 = 93.3 kPa T 2 = 0 o C = 273 K V 2 = X dm 3 (101.3 kPa) x (500 dm 3 ) = (93.3 kPa) x (V 2 ) 273 K V 2 = dm 3 (101.3) x (500) = (93.3) x (V 2 )

Standard Temperature and Pressure 0 o C 1 atm = 760 mmHg = 760 torr = kPa

Collecting Gasses over a liquid (such as water) Some water vapor will be mixed with the collected gas and will exert pressure. The partial pressure of the water vapor for each temperature is constant. So, take the pressure of the gas and subtract the partial pressure of water at the given temperature.