Now, what would happen if you sprayed two different air fresheners at the same time in the same general place? While the scents will be mixed, one will be smelt on the opposite end of the room earlier than the other. Why is this? Well, **Graham's law** has the answer, continue reading to find out!

- This article covers
**Graham's law.** - First, we will define Graham's law.
- Next, we will look at Graham's law equation.
- Then we will look at the two parts of Graham's law: Graham's law of
**diffusion**and Graham's law of**effusion.** - Lastly, we will work on some examples using Graham's law.

## Graham's Law Definition

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

**Graham's law **states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of the molar masses of its particles.

**Diffusion **is the process of particles moving from an area of high density to one of low density

**Effusion **is the process of gas particles escaping their container into another container through a small hole. This hole's diameter is much smaller than the particle's **mean free path, **which is the distance a particle travels before it collides with something. Essentially, this means that only one particle can travel through the hole at a time.

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:

**Calculate the rate of effusion/diffusion for a container of both helium (He) and neon (Ne) gas.**

The molar mass of helium is 4.00 g/mol, and the molar mass of neon is 20.2 g/mol. Plugging this in, we get:

$$\frac{Rate_A}{Rate_B}=\sqrt{\frac{M_B}{M_A}}$$

$$\frac{Rate_{He}}{Rate_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{He}}}$$

$$\frac{Rate_{He}}{Rate_{Ne}}=\sqrt{\frac{20.2\frac{g}{mol}}{4.00\frac{g}{mol}}}$$

$$\frac{Rate_{He}}{Rate_{Ne}}=\sqrt{5.05}$$

$$\frac{Rate_{He}}{Rate_{Ne}} =2.25$$

This means that helium will effuse/diffuse 2.25x times faster than neon

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:

Fig.1-Particles diffuse across a space

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**.

**Root-mean-square speed (RMS speed) **is the average speed of a gas. The formula is:

$$\nu_{rms}=\sqrt{3RTM}$$

Where R is the ideal gas constant, T is the temperature, and M is the molar mass of the gas.

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.

For gases, we don't use normal velocity since the net velocity of a gas is zero, since the gases will move in all directions (i.e. they will cancel each other out since velocity is speed + direction)

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:

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

Think of it like rolling a ping-pong ball versus a soccer ball. If I rolled both balls with the exact same amount of energy, the ping-pong ball will travel farther (i.e. have a greater velocity) than the soccer ball since the ping-pong ball is much lighter.

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:

**Effusion** is the process of gas particles escaping their container into another container through a small hole. This hole's diameter is much smaller than the particle's mean free path, which is the distance a particle travels before it collides with something. Essentially, this means that only one particle can travel through the hole at a time.

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:

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:

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!

**A sample of Nitrogen and oxygen gas is in a container with a small hole leading to another, empty container. Which of the gases will effuse first?**

- The molar mass of Nitrogen is 14.00 g/mol and the molar mass of oxygen is 16.00 g/mol.

While we can plug this into our equation, all we really need to do is look at the molar masses. Nitrogen is lighter, so it will effuse faster, and therefore first.

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

**What is the ratio of the rate of diffusion for chlorine (Cl _{2}) gas and krypton (Kr) gas?**

- The molar mass of chlorine is 35.45 g/mol and the molar mass of krypton is 83.80 g/mol.

Since chlorine gas is Cl_{2} and *not *Cl, we need to double the molar mass value when we plug it into our equation:

$$\frac{Rate_A}{Rate_B}=\sqrt{\frac{M_B}{M_A}}$$

$$\frac{Rate_{Cl2}}{Rate_{Kr}}=\sqrt{\frac{M_{Kr}}{M_{Cl2}}}$$

$$\frac{Rate_{Cl2}}{Rate_{Kr}}=\sqrt{\frac{83.80\frac{g}{mol}}{(35.45\frac{g}{mol}*2)}}$$

$$\frac{Rate_{Cl2}}{Rate_{Kr}}=\sqrt{\frac{83.80\frac{g}{mol}}{70.9\frac{g}{mol}}}$$

$$\frac{Rate_{Cl2}}{Rate_{Kr}}=\sqrt{1.18}$$

$$\frac{Rate_{Cl2}}{Rate_{Kr}}=1.09$$

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.

## Graham's Law - Key takeaways

**Graham's law**states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of the molar masses of its particles.- Basically, the heavier gas will move slower than the lighter one

**Diffusion**is the process of particles moving from an area of high density to one of low density**Effusion**is the process of gas particles escaping their container into another container through a small hole. This hole's diameter is much smaller than the particle's**mean free path,**which is the distance a particle travels before it collides with something. Essentially, this means that only one particle can travel through the hole at a time.- 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 diffusion/effusion of gas A, respectively, and M_{B}and Rate_{B}are the respective molar mass and rate of diffusion/effusion of gas B

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##### Frequently Asked Questions about Graham's Law

What does Graham's law state?

**Graham's law **states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of the molar masses of its particles.

Essentially, lighter gases will effuse/diffuse faster than heavier ones.

What is a real life example of Graham's law?

When a balloon has a small hole/leak, the helium will escape faster than the air can get in, since the helium is lighter. This causes the balloon to deflate.

What is the graham's law of diffusion?

**Diffusion **is the process of particles moving from an area of high density to one of low density

Graham's law of diffusion states that the lighter gas will diffuse faster than the heavier gas.

What is Graham's law of effusion?

**Effusion **is the process of gas particles escaping their container into another container through a small hole. This hole's diameter is much smaller than the particle's **mean free path, **which is the distance a particle travels before it collides with something. Essentially, this means that only one particle can travel through the hole at a time.

Graham's law of effusion states that the lighter gas will effuse faster than the heavier one.

What is the equation for Graham's law of effusion?

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 gas A, respectively, and M_{B} and Rate_{B} are the respective molar mass and rate of gas B

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