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Diffusion and Effusion: Graham's Law

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  • 0:05 Kinetic Molecular Theory
  • 1:01 Diffusion
  • 1:29 Effusion
  • 2:25 Graham's Law
  • 3:30 Practice Problems
  • 6:13 Lesson Summary
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Lesson Transcript
Instructor: Kristin Born

Kristin has an M.S. in Chemistry and has taught many at many levels, including introductory and AP Chemistry.

Have you ever been in a room where someone has put on perfume or scented lotion and a few minutes later you are able to smell it? What causes you to be able to smell something from so far away? In this lesson, we are going to use the kinetic molecular theory of gases to explain some of their behaviors and determine how we can compare the speeds of different gases.

Kinetic Molecular Theory

One day during his vacation on Ideal Island (the place where all gases behave ideally), Johnny Dalton and his family decided to go for a walk through one of the villages on the Island. While strolling down the sidewalk, Johnny's nose catches something amazing coming from one of the shops: the smell of delicious, fresh-baked chocolate chip cookies! This gets Johnny hungry and thinking! What was it that caused the smell of those chocolate chip cookies to drift out of the bakery and into his nose?

The explanation to this aromatic question lies in the kinetic molecular theory, which describes the movement of ideal gas particles. It states that gas particles are constantly moving in random, rapid motion. So, if you have enough particles (or in this case, cookie smell molecules) in one central location, they will eventually spread out because they are moving randomly and rapidly.

Diffusion

The chemistry word for this phenomenon is diffusion, which is the movement of one substance through another. Usually, it means particles will spread out, or move from a region of high concentration to a region of low concentration. In my chocolate chip cookie example, the gas particles that contain the smells of the cookie are moving through the air from a high concentration (the bakery) to a low concentration (the sidewalk and surrounding area).

Effusion

After Johnny finishes the cookie he purchased, he continues on his walk. A while later, he notices something peculiar about his balloon: it appears to be sinking. What was once a high-flying helium balloon is now just a hovering and slightly smaller balloon. What do you think caused Johnny's balloon to lose some of its helium?

Well, a phenomenon called effusion is to blame. Effusion is when gas particles exit through tiny holes in a container. Effusion occurs for the same reason as diffusion: gas particles are moving in a random, rapid motion. Because helium atoms are so small and light (helium is the lightest non-flammable gas), the particles can easily escape through the microscopic holes in the balloon, causing the balloon to get smaller and more dense, which makes it sink.

Graham's Law

Graham determined that lighter gas molecules travel faster than heavier gas molecules.
Molecules Moving at Different Speeds

In the mid 1800s, Thomas Graham experimented with effusion and discovered a very significant relationship: lighter gas molecules will travel faster than heavier gas molecules.

So, assuming the temperature and pressure remain constant, atoms or molecules with a lower molecular mass will effuse faster than atoms or molecules with a higher molecular mass. Graham even went further by finding out how much faster one molecule would be over another.

This became known as Graham's Law, and it states that the effusion rate of a gas is inversely proportional to the square root of its molecular mass. Usually, this formula is used when comparing the rates of two different gases at equal temperatures and pressures. The ratio of the rate (or speed) of gas A over gas B is equal to the square root of the mass of gas B over the mass of gas A.

Practice Problem 1

So, let's try using this equation with the following scenario: If oxygen molecules effuse at an average rate of 3 mol/s, how fast would hydrogen molecules effuse in the exact same conditions?

Let's start out by filling in the information we know, and let's call hydrogen 'gas A' and oxygen 'gas B'. We know the rate of gas B is 3 mol/s and the molecular mass of oxygen (a diatomic element) is 32 amu. The molecular mass of hydrogen (also diatomic) is 2 amu.

Practice problem
Grahams Law Practice Problem

My first step in solving this is to calculate the right side of the equation. 32 divided by 2 is 16. Notice how the units cancel out. The square root of 16 is 4. To solve for x, we would need to simply multiply both sides by 3 mol/s, which gives us an answer of 12 mol/s. So, if oxygen is effusing at a rate of 3 mol/s, hydrogen would be effusing 4 times faster at a rate of 12 mol/s.

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