Graham's Law

Graham's Law Definition

Let's start by looking at the definition of Graham's law.

Now, this is a bit of a tricky definition. To help us out, we are first going to look at the equation.

Graham's Law Equation

The equation for graham's law is:$$\frac{Rate_A}{Rate_B}=\sqrt{\frac{M_B}{M_A}}$$

Where,

• M_A and Rate_A are the molar mass and rate of effusion/diffusion of gas, A respectively.
• M_B and Rate_B are the respective molar mass and rate of effusion/diffusion of gas, B.

What this basically tells us is the ratio of rates based on their masses. In simpler terms, the heavier the gas, the slower it's going to go. Here's an example:

To better understand this concept, let's split this law up into two parts: diffusion and effusion.

Graham's Law of Diffusion

Let's first talk a bit more about what diffusion is. Like we talked about in the introduction when you spray an air freshener the spray will only be smelt in the direct area, but the particles will eventually spread across the room.

Here's a diagram of what's happening:

The purpose of diffusion is to create an equilibrium. Equilibrium is essentially a state of balance. In this case, equilibrium means a balanced concentration across the whole container.

So, what happens when there are two gases? That's where Graham's law steps come in.

Our main assumption is that both gases are at the same temperature, and therefore have the same kinetic energy, which is the energy of motion.

In fact, that's where graham's law comes from. Here's the formula for kinetic energy:

$$KE=\frac{1}{2}*M_A*\nu_{rms}^2$$

Where KE is kinetic energy, M_A is the molar mass of species A, and ν_rms is the root-mean-square speed.

You don't really need to worry about the exact definition of RMS speed, so you can just think of it as "speed" for now.

Continuing on with our derivation, we are going to set the kinetic energies of our two gases (A and B) equal to each other:

$$KE=\frac{1}{2}*M_A*\nu_{rms,A}^2=\frac{1}{2}*M_B*\nu_{rms,B}^2$$

Next, we can cancel out the 1/2, since it is on both sides:

$$M_A*\nu_{rms,A}^2=M_B*\nu{rms,B}^2$$

Then, we move both of our molar mass terms onto one side and our RMS speed terms onto the other:

$$\frac{M_A}{M_B}=\frac{\nu_rms,B^2}{\nu_rms,A^2}$$

Lastly, we take the square root of both sides:

$$\sqrt\frac{M_A}{M_B}=\frac{\nu_rms,B}{\nu_rms,A}$$

Now that we know how we got our equation, let's see it in action.Earlier, we calculated that helium would diffuse 2.25x faster than neon, below is a diagram showing this process:

Fig.2-Graham's law of diffusion experiment.

Essentially, helium is going to diffuse faster since it is lighter. This means that more helium particles will reach the neon "side".

However, after a certain amount of time, there will be an equilibrium of gases, i.e. the same concentration on both sides.

Graham's law of effusion

Now let's talk about effusion.

As a reminder, here's the definition of effusion from earlier:

Essentially, effusion is the same as diffusion, except the particles are passing through a hole instead of moving through an open space.Here's what this process looks like:

Fig.3 Gas effusion

Like with diffusion, during effusion, gases are moving to a less concentrated area. Now let's see what effusion looks like with two gases according to Graham's law:

Fig.4-Graham's law of effusion for helium and neon

The two balloons are connected by a small hole, which we assumed to be sealed (i.e. no gas can escape besides the exchange between the two balloons). Since the helium moves faster, it will effuse at a faster rate than the neon. This means the neon balloon will inflate as the helium effuses in, while the helium balloon deflates since it is losing that helium.

Graham's Law Examples

Now that we better understand the concept of effusion and diffusion, let's work on some examples!

Now let's do an example where we actually calculate something:

Since chlorine gas and krypton gas are similar in mass, the difference in the rate of diffusion isn't that large. The greater the difference in mass, the greater the ratio of rates will be.