You will need a partner as well as a calculator The Gas Laws You will need a partner as well as a calculator

Ideal Gas Law PV = nRT Boyle’s Law P1V1 = P2V2 Combined Gas Law P1V1/T1 = P2V2/T2

Pressure Pressure (P): The force of gas molecules as they hit the sides of the container in which they are placed.    Units = Atmospheres (atm) StandardTemperaturePressure = 1atm

Temperature StandardTemperaturePressure = 273 K Temperature (T): Average kinetic energy. The higher the energy, the higher the temperature.   Common units of temperature: Kelvin (K): The ONLY units that can be used when doing numerical problems with gases. *Degrees Celsius (0C): Must be converted to Kelvin before doing problems (by adding 273). StandardTemperaturePressure = 273 K

Volume Volume (V): The amount of space in which a gas is enclosed.   The most commonly used unit of volume is liters (L). Conversion: 1L = 1,000 mL

Which tank contains the greatest number of particles? Number of particles = pressure x volume 10 atm 1 atm 40 atm 25 atm 30 L 60 L 45 L 15 L A B C D

Ideal Gas Law Avogadro’s law: One mole of every gas has the same volume. (22.4 L) STP = standard temperature & pressure Standard temperature = 273 K Standard pressure = 1 atm All gases behave the same at the same temperature, volume and pressure. H2O(g), CO2(g), O2

Ideal gas law: PV = nRT P = pressure (in atm) V = volume (L) n = number of (moles)  R = ideal gas constant 0.0821 L•atm/mol•K T = temperature (Kelvin) = (oC + 273)

“We Do” PV = nRT If I have 10 liters of a gas at a pressure of 1.5 atm and a temperature of 250 C, how many moles of gas do I have? P = ___ atm V = ___ L n = ___ mol R = (always this #) ______ atm L/mol K T = __oC + (273) = ___ K

PV=nRT (P x V) = (n x R x T) 1.5 x 10 = n x 0.0821 x 298 Divide both sides by 24.47 so “n” by itself n = 0.613 moles

“You Do” If I have 23.5 moles of gas at a temperature of 67.89 oC, and a volume of 88.98 liters, what is the pressure of gas? P = V = n = R = 0.0821 (always this #) T = ___oC+ (273) =

Boyle’s Law P1V1 = P2V2 P1 = First Pressure V1 = First Volume P2 = Second Pressure V2 = Second Volume P and V are inversely proportional (when one increases, the other decreases)

Boyles’ Law and Breathing During an inhalation, Lung volume increases. the pressure in the lungs decreases. air flows towards the lower pressure in the lungs. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Boyles’ Law and Breathing During an exhalation, Lung volume decreases. pressure within the lungs increases. air flows from the higher pressure in the lungs to the outside. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

“We Do” P1V1 = P2V2 A high-altitude balloon contains 30.0 L of He at 10.3 atm. What is the new volume when the balloon rises to an altitude where the pressure is only 2.3 atm? P1 = V1 = P2 = V2 =

P1V1 = P2V2 (10.3 atm) (30.0 L) = (2.3 atm) ( V2)

“You Do” P1V1 = P2V2 A gas with a volume of 4.02 L at a pressure of 2.31 atm is allowed to expand to 12.89 L. What is the new pressure of the gas? P1 = V1 = P2 = V2 =

Combined Gas Law P1V1 = P2V2 T1 T2 P1 = First Pressure ****Remember that temperature needs to be in Kelvin (oC = 273K)**** P1 = First Pressure V1 = First Volume T1 = First Temperature P2 = Second Pressure V2 = Second Volume T2 = Second Temperature

“We Do” P1V1/T1 = P2V2/T2 If I initially have a gas at a pressure of 1.3 atm, a volume of 24.6 liters and a temperature of 214K, and then I raise the volume to 46.0 L and increase the temperature to 286 K, what is the new pressure of the gas? P1 = P2 = V1 = V2 = T1 = T2 =

P1V1/T1 = P2V2/T2 (1.5 atm)(24.6 L) = (P2)(46.0 L) 214 K 286 K 0.172 atm L/K = (P2) (0.161 atm/K) P2 = 1.75 atm

“You Do” P1V1/T1 = P2V2/T2 If I have 23.5 L of gas held at STP, what is the volume of gas if I increase the pressure to 45.2 atm and increase the temperature to 26.78 oC? P1 = P2 = V1 = V2 = T1 = T2 =