Diffusion and Effusion||Graham’s Law||Separation or Enrichment

Graham's Law of Diffusion

Graham's Law of Diffusion

Graham's Law of Diffusion

Graham’s Law OF Diffusion - Case II :-Diffusion OF Mixture || Case III :- T = Constant || Case IV :- Isotopic Separation Factor or Enrichment Factor

Graham Laws of Diffusion

The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion prop (1)/(sqrt(d)) If r_(1) and r_(2) represent the rates of diffusion of two gases and d_(1) and d_(2) are their respective densities, then r_(1)/(r_(2))=sqrt((d_(2))/(d_(1))) r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2)) (V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1))) V prop n (where n is no of moles) V_(1) prop n_(1) and V_(2) prop n_(2) If some moles of O_(2) diffuse in 18 sec and same moles of other gas diffuse in 45sec then what is the molecular weight of the unknown gas ? .

The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion prop (1)/(sqrt(d)) If r_(1) and r_(2) represent the rates of diffusion of two gases and d_(1) and d_(2) are their respective densities, then r_(1)/(r_(2))=sqrt((d_(2))/(d_(1))) r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2)) (V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1))) V prop n (where n is no of moles) V_(1) prop n_(1) and V_(2) prop n_(2) Helium and argon monoatomic gases and their atomic weights are 4 and 40 respectively Under identical conditions helium will diffuse through a semipermeable membrane .

The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion prop (1)/(sqrt(d)) If r_(1) and r_(2) represent the rates of diffusion of two gases and d_(1) and d_(2) are their respective densities, then r_(1)/(r_(2))=sqrt((d_(2))/(d_(1))) r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2)) (V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1))) V prop n (where n is no of moles) V_(1) prop n_(1) and V_(2) prop n_(2) 2 g of hydrogen diffuse from a container in 10 minutes How many grams of oxygen would diffuse through the same container in the same time under similar conditions ? .

The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion prop (1)/(sqrt(d)) If r_(1) and r_(2) represent the rates of diffusion of two gases and d_(1) and d_(2) are their respective densities, then r_(1)/(r_(2))=sqrt((d_(2))/(d_(1))) r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2)) (V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1))) V prop n (where n is no of moles) V_(1) prop n_(1) and V_(2) prop n_(2) The time taken for a certain volume of gas X to diffuse through a small hole is 2 minutes It takes 5.65 minutes for oxygen to diffuse under the simillar conditions The molecular weight of X is .

Dalton's Law and Graham's Law of Diffusion