Graham's Law

Graham's Law


• Effusion molecules of a gas randomly passthrough a tiny opening in a container. Ex:balloon with hole deflating• Diffusion gradual mixing of two gases due to theirsp0ntaneous, random motion. Ex: smellacross the room

Graham’s Law of Effusion• Defined as: The rates of effusion of gasesat the same temperature andpressure are inverselyproportional to the square rootsof their molar masses.

Graham’s Law of Effusion• Rate of effusion: Depends on: Velocity of gas molecules &Molar Mass of molecules½ M A v A2 = ½ M B v B2M A v A2= M B v B2v A2v B2v Av B==M BM AM BM ASO:Rate of effusion of A (V A )Rate of effusion of B (V B )=M BM A

Application of Graham’s Law1. Lighter gases diffuse faster than heavier gases.2. Lighter gases are those with lower Molar mass ordensities Method for determining molar masses.Rates of effusion of known and unknown gases canbe compared to one another

Hints for working Graham’sLaw Problems• Write the Molar Mass of both gases (if known)• Make the heavier gas “B”• or make the faster gas “A” (since it will be lighter)• Mark in the margins of the given informationwhich gas is A and which gas is B• Rearrange the equation before you put numbers in.• Keep up with the units!

Problem 2A sample of hydrogen effuses through a porous containerabout 9 times faster than an unknown gas. Estimate themolar mass of the unknown gas.Rate of effusion of A (V A )Rate of effusion of B (V B )=M BM ARate of effusion of H 2Rate of effusion of unknown=M unknownM H 292=M unknown2.02 g/mol22.o2 g/mol x81 =M unknown=M unknown2.02 g/molx 2.o2 g/mol

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