Graham’s Law of Effusion of Gases

Graham’s Law of Effusion

Before we talk about Graham’s Law of Effusion, let’s explore what effusion is and distinguish it from diffusion.

Effusion is the movement of one type of gas particles through a hole.

effusion

 

Diffusion, on the other hand, is the movement of particles from areas of high concentration to areas of low concentration.

diffusion

 

Graham’s Law of Effusion talks about the relative rates of effusion as a function of the molar mass of gas particles. It can be applied equally well to the rates of diffusion of a gas, so sometimes the law is called Graham’s Law of Diffusion too. Graham’s Law of Effusion states that the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure:

grahams-law

 

In which:

M1 = molar mass of gas 1

M2 = molar mass of gas 2

Let’s look at a sample problem to learn how to use this equation.

Sample Problem

In an experiment, it takes an unknown gas 1.5 times longer to diffuse than the same amount of oxygen gas, O2. Find the molar mass of the unknown gas.

Let’s let M2 be the molar mass of the unknown gas since it’s in the numerator and therefore easier to solve for. That means that M1 is the molar mass of oxygen gas.

We know that the molecular formula of oxygen gas is O2, so its molar mass is 32.00 g/mol.

M2 = molar mass of unknown

M1 = 32.00 g/mol

The problem also says that the unknown gas takes 1.5 times longer to diffuse than oxygen. This means that oxygen diffuses 1.5 times faster, which means that the rate of diffusion of oxygen is 1.5 times as great. So, if the rate of diffusion of O2 is R (we are not given a magnitude, so we will use a variable in its stead), then the rate of diffusion of the unknown gas is 1.5 times as long or 1.5R.

Therefore, if the rate of diffusion of gas 2 is R, the rate of diffusion of gas 1 oxygen is 1.5R.

Rate of effusion of gas 1 = 1.5R

Rate of effusion of gas 2 = R

Substituting these values into our equation:

grahams-law

 

 

15R

The variable R cancels out on the left hand side of the equation:

15

 

To get M2 out of the square root, we square both sides:

152

 

1.5 raised to the second power is equal to 2.25:

225

 

Solving for M2:

72

 

The unknown gas has a molar mass of 72.00 g/mol.

 

Sample Problem

If it takes 20.0 minutes for 0.350 moles of H2S to effuse from a chamber, how long will it take for 0.175 moles of Kr to effuse from the same chamber?

The rate of effusion is expressed as moles per second, or moles per minute. Therefore, the rate of effusion of H2S is:

h2s-effusion

 

We want to know the length of time it will take for 0.175 mol Kr to effuse from the same chamber. Therefore, the rate of effusion of Krypton can be expressed as:

kr-effusion

 

The molar mass of H2S is 34.076 g/mol, and the molar mass of Kr is 83.80 g/mol.

molar-mass-h2s

 

molar-mass-kr

 

Our equation is:

grahams-law

 

We’ll let Kr be gas 2, so that time t will end up in the numerator:

rate-effusion-h2

 

0.0175 divided by 0.175 is equal to 0.100:

100t

 

And, solving the right-hand side of the equation:

100t157

 

Solving for t:

157min

 

It will take Kr 15.7 minutes to effuse from the same chamber.