Graham’s Law of Diffusion Formula

Graham’s Law of Diffusion Formula can be expressed as

Rate 1/Rate 2 = 

Rate 1/Rate 2 = 

Where,

  • Rate 1 = Rate of effusion of the first gas,
  • Rate 2 = Rate of effusion of the Second gas,
  • d1 = First gas density,
  • d2 = second gas density,
  • M1 = Molar mass of first gas,
  • M2 = Molar mass of second gas.

Graham’s Law of Diffusion

Graham’s law of diffusion is the relationship between a gas’s rate of diffusion or effusion and its molecular weight. The law of diffusion’s basic tenet is that any gas’s rate of diffusion, at any given temperature and pressure, is inversely proportional to the square root of its density. The mechanism by which a gas can escape from the container is known as effusion, and the ability of a gas to spread and occupy all of the volumes that are available to it is known as diffusion.

Similar Reads

Graham’s Law of Diffusion

In the year 1848, a scientist by the name of Thomas Graham made the initial discovery of Graham’s Law. During his experiments with the effusion of gas, he developed an essential theory that the lighter gas molecules will move through the air more quickly than the heavier gas molecules. Graham’s law of effusion is the common name for this law....

Graham’s Law of Diffusion Formula

Graham’s Law of Diffusion Formula can be expressed as...

Rate of Diffusion

Diffusion of a gas is the term used to describe the net movement of the substance from the area of higher concentration to the area of lower concentration. Each gas particle starts crashing into one another. The maximum particle density zone of a gas causes the particles to begin bouncing off of each other and the border container at a faster pace than the particles in the lower particle density regions....

Rate of Effusion

When air particles escape or leak through a hole that has a width much smaller than the mean free path of the molecules, this action is known as an effusion of a gas. Due to the small amount of molecule-to-molecule collisions that occur in these regions, all particles and molecules that reach the hole will pass through throughout this procedure....

Derivation of Graham’s Law of Diffusion

According to Kinetic Molecular theory, PV = mNc2 / 3 If we have 1 mole of gas then N = NA PV = mNAc2 / 3 Where, P = Pressure, V = Volume, m = Mass, NA = Avogadro’s number of molecule, c2 = Velocity ∴ PV = Mc2 / 3  …(mNA = M (Molar mass)) ∴ c2 = 3PV/M ∴ c2 = 3P/d  …(density (d) = M/V) Take Square root on both side, ∴ √c2 = √(3P/d) ∴ √c2 = Rate of Diffusion ∴ Rate = 1/√d...

Problems based on Graham’s Law of Diffusion

Problem 1: Identify the gas’s molar mass by calculating its diffusion coefficient, which is 4.11 times that of ammonia (NH3)....

Contact Us