It’s a Gas, Gas, Gas!!!

Defining a Gas For a fixed number of particles Volume is determined by Its temperature Its pressure

What is Pressure? Pressure  Force exerted per unit area Where does the force come from in a gas? The particles colliding with the sides of the container  collisions   pressure and if you  temperature   collisions

kilopascals and atmosphere 1 kilopascal (kPa) = 10 gms / 1 cm2 1 atmosphere = 101.3 kPa

More Pressure Similarly you know that  collisions   pressure                                               Similarly you know that  collisions   pressure So if you  volume   collisions

The Gas Laws

Charles’ Law  temperature   volume for a fixed amount of gas at constant pressure, the temperature and volume are directly proportional:  temperature   volume V T - 273oC P1 P2 V1 V2 T1 T2 =

Guy Lussac’s Law  temperature   pressure for a fixed amount of gas at a constant volume, the temperature and pressure are directly proportional:  temperature   pressure 1 atm 2 atm P T - 273oC V1 V2 P1 P2 T1 T2 =

Boyle’s Law  volume   pressure for a fixed amount of gas at a constant temperature, the volume and pressure are inversely proportional:  volume   pressure P V T1 T2 P1 V1 P2 V2 =

Ideal Gas Law – including moles Suppose we include the number of moles in the Combined Gas Law, then we get: Where n represents the number of moles in the gas. P1 V1 P2 V2 n1 T1 n2 T2 =

Ideal Gas Law – R rated Given this relationship, each side of the equation is a constant value or Where n represents the number of moles in the gas and R represents the constant. P V n T = R

Ideal Gas Law – R value Substituting in the values for a gas at STP, we get Where n represents the number of moles in the gas and R represents the constant. P V n T (101.3kPa) (22.4L) = R = = 8.31 kPa L mol oK (1 mol) (273oK)

Departure from Ideal Gas Law Particles are not spherical Intermolecular Forces are still present Collisions are not completely elastic

Law of Partial Pressures PT = P1 + P2 = …

Graham’s Law RateA molar massB RateB molar massA =

Graham’s Law – Proof! Rate1 molar mass2 Rate2 molar mass1 = KE = ½ mv2 If two gases are at same temperature KE1 = KE2 or ½m1v12 = ½ m2v22 m1v12 = m2v22 v12 / v22 = m2 / m1 Rate1 molar mass2 Rate2 molar mass1 =