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# Graham's Law

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Have you ever sprayed an air freshener in a room? At first, you can only smell it directly where you sprayed, but eventually, the particles will spread across the room and can be smelt anywhere.

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!

• 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,

• MA and RateA are the molar mass and rate of effusion/diffusion of gas, A respectively.
• MB and RateB 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, MA is the molar mass of species A, and νrms 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:

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".

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

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:

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!

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 (Cl2) 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 Cl2 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 MA and RateA are the molar mass and rate of diffusion/effusion of gas A, respectively, and MB and RateB are the respective molar mass and rate of diffusion/effusion of gas B

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.

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.

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.

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.

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

Where Mand RateA are the molar mass and rate of gas A, respectively, and MB and RateB are the respective molar mass and rate of gas B

## Graham's Law Quiz - Teste dein Wissen

Question

What is 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.

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Question

What is diffusion?

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

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Question

What is 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.

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Question

What is the equation for Graham's law?

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

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Question

What is the basic principle of Graham's law?

Lighter gases effuse/diffuse faster

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Question

What is Root-mean-square speed? (RMS speed)

The average speed of a gas

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Question

True or False: Gases that effuse/diffuse will eventually reach equilibrium

True

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Question

What is Graham's law derived from?

The equation for kinetic energy

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Question

Why do we use RMS speed instead of velocity for gases?

In a gas, the particles move around randomly with the same speed/energy. Since they move in all directions, the velocity cancels since velocity is a measure of speed in a direction.

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Question

The molar mass of helium gas is 4.00 g/mol, and the molar mass of neon gas is 20.2 g/mol. Which gas will diffuse first?

Helium

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Question

Two containers are connected by a (closed) pipe. One container holds neon, while the other holds krypton. When the pipe is opened, which will diffuse first?

Neon

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Question

Calculate the rate of effusion for nitrogen (N2) and chlorine (Cl2) gas.

The molar mass for N2 is 28 g/mol

The molar mass for Cl2 70.9 g/mol

We are assuming nitrogen is gas A and chlorine is gas B

1.59

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Question

What is the ratio for the rate of diffusion for hydrogen (H2) and oxygen (O2) gas?

The molar mass of H2 is 2.02 g/mol

The molar mass of O2 is 32 g/mol

Assume H2 is gas A and O2 is gas B

3.97

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Question

What is the difference between diffusion and effusion?

Effusion is a similar concept to diffusion, except the gases must go through a small hole

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Question

True or False: Graham's law is different for diffusion and effusion

False

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