# Grahams Law

Graham’s Law of Diffusion

To understand the graham's law of diffusion, we must first understand the definition of diffusion and rate of diffusion. Though diffusion and Effusion are interchangeably used many times, they have a completely different meaning. The process through which the particles from one gas move to another gas is diffusion. It causes significant disorder in the entire system. We can also observe diffusion at slower rates in liquids as well as solids. The leading cause of this phenomenon is the concentration levels. When the particles of a material are in an area of low concentration, they tend to move to the content with a high area of concentration. A straightforward example of this process is when we use perfume or a scented spray in one part of the room, and then we can smell it in the entire room. Diffusion is involved in this whole scenario.

(Particles of a soluble material dissolving with time, diffusion)

The change in diffusing molecules over time is called the rate of diffusion. The diffusion rate of a gas is inverse to the square root of the volume (density). Hence, the rate of diffusion formula is:

rate of diffusion ∝ 1/$\sqrt{density}$

We can also rewrite this equation in terms of molar mass since gases with different volumes have the same number of particles.

rate of diffusion ∝ 1/$\sqrt{M}$

Here the molar mass of the gas is denoted by M.

Rate of Effusion

When the gaseous particles move from a tiny opening into the vacuum of space or open container, then the process is called Effusion. This space can be a vacuum, any other gas or atmosphere. In this process, the molecules of material try to escape from a  closed container through the aperture. The best example of you can observe Effusion in balloons. When we make an opening in the balloon, the gas inside starts escaping into the atmosphere, and it deflates. We can call this Effusion of gas into the atmosphere.

The process by which particles of material from closed space escapes with time is known as the rate of Effusion. Now let us discuss the rate of effusion formula. Since the Effusion is inversely proportional to density and the molar mass of the gas (M), we can write the equation as follows:

rate of Effusion ∝ 1/$\sqrt{density}$ ∝ 1/$\sqrt{M}$

Graham’s Law

Graham's law of diffusion was one of the breakthroughs in the field of chemistry. Thomas graham discovered this law in 1848, and it is also known as the graham's law of Effusion. His experimentation with the rate of effusion process unveiled that gas with heavier molecules travels slower than gas with lighter particles. Graham's law of Effusion or diffusion states that when the temperature and pressure are constant than atoms with high molar mass effuse slower than atoms with low molar mass. He also gave the rate at which molecules would escape, i.e. the rate of diffusion.

Moreover, it states that the square root of the molar mass is inversely proportional to the rate of Effusion. This statement gives us the Grahams law of diffusion formula. We can use it to compare two gasses with their rates at constant temperature and pressure. Let us assume r1 and r2 are the rates of Effusion of two gases, and M1 and M2 are the molar masses respectively. Hence, we can write the formula as follows.

r1/r2 = M2/M1

or

r1/r2 = $\sqrt{M2/M1}$

Solved Examples

Example 1: Calculate the molar mass of a given gas whose diffusion rate is 2.92 times the diffusion rate of NH₃.

We know that the diffusion rate is 2.92 times of ammonia; hence we understand that the ratio of diffusion rates of the given gases should be 1/2.92. So,

r1/r2 = 1/2.92

Since we know that the molar mass of ammonia is 17.0307.

We can use Graham’s law. Accordingly,

r1/r2 = $\sqrt{M2/M1}$

Substituting the values

1/2.92 = $\sqrt{M2/17.0307}$

Squaring both sides we have,

0.11728 = M2/17.0307

Hence,

M2 = 2.0 g/mol

M2 is the molar mass of the given gas.

Example 2: find the relative diffusion rate of water (molar mass=18.0152) as compared to hard water (molar mass=20.0276).

We know that,

The molar mass of water(M1)=18.0152

The molar mass of hard water(M2)=20.0276

Let us assume the rate of diffusion formula of heavy water as one since it has a slower diffusion rate.

r2=1

According to graham’s law

r1/r2 = $\sqrt{M2/M1}$

Squaring both sides

r1²/r2² = M2/M1

Substituting the values in the above equation

r1²/1 = 20.0276/18.0152

r1² = 1.111705

r1 = 1.05

Hence the relative diffusion rate is 1.05.