Diffusion - the process by which a gas goes through another gas (basically gas particles mixing)

Effusion - the process of a gas passing through a barrier (i.e. balloon's "skin")

- There are tiny holes in the "skin" of the balloon and the gas particles find their way thru to the outside

Both these processes are based on the same principles, that gas particles are always moving and their movement is a function of their average kinetic energy. Before, we have simplified kinetic energy to mean the speed of the gas particle. However, this is not exactly correct because that simplification assumes that all gas particles have the same mass which they don't. So we now need to get more detailed and consider the difference in mass of gas particles with respect to the kinetic average of the particle.

Graham's Law of Diffusion/effusion states how gas particles moves (i.e. their kinetic energy, in this law we call it rate) that takes into account that gas have different masses. The actual law states:

The relative rate of diffusion of a two gases is equal to the inverse of the square root of their molar mass (i.e. molecular weight) or

r

r

r

MW

MW

It is based on the fact that kinetic energy of gas particle is equal to 1/2 mass of particle time the square of the speed of the gas particle or:

KE = 1/2*m*v

A less politically correct way of stating the law is Fat (or high molar mass) things go slow, skinny (low molar mass) things go fast.

- So for non-mathematical problems, it is a simple comparison of molar mass (also called Molecular Weight, MW):
- Greater MW substance diffuses/effuses SLOWER, lower MW substance diffuses/effuses FASTER.

- Greater MW substance diffuses/effuses SLOWER, lower MW substance diffuses/effuses FASTER.

Example of Conceptual Problem:

Which of two substance diffuses Faster (has a higher rate of diffusion), NH

Answer: MW of NH

There are basically two type of problems with respect to Graham's Law. One that asks the relative rate of diffusion between two gasses and one that is specifically asking how fast.slow one gas is compared to another gas.

For relative rates of diffusion problems, there are actually two possible answers since they are relative to each other.

- Example of Relative Rate of Gas Diffusion Problems
- Expected Work for Relative Rate of Gas Diffusion Problems
- You Tube Video of Expected Work for Relative Rate of Gas Diffusion Problems

If the problem specifies how fast/slow on gas is to the other, their is only one possible answer.

- Example of Specific Rate of Gas Diffusion Problems
- Expected Work for Relative Specific Rate of Gas Diffusion Problems
- You Tube Video of Expected Work for Specific Rate of Gas Diffusion Problems

If you only want to remember one way that works all the time for both types of these problems, then always put the first gas in the problem in the numerator and the second gas in the denominator or r