Effusion - Online Tutor, Practice Problems & Exam Prep

Understanding the behavior of gases is essential for comprehending their physical properties. Gases consist of molecules or atoms in constant motion, and their interactions can be described by concepts such as mean free path—the average distance between collisions of gas particles—and the processes of effusion and diffusion. Effusion refers to the escape of gas through a tiny pinhole, typically one molecule at a time, while diffusion is the movement of gas from a region of high concentration to low concentration, regardless of the direction. Graham's law of effusion quantifies this behavior, stating that the rate of effusion is inversely proportional to the square root of the gas's molar mass, meaning lighter gases effuse more rapidly than heavier ones. This law is crucial when comparing the effusion rates of different non-reacting gases, as it helps predict the movement and speed of a gas based on its molecular weight.

Effusion

So when it comes to the state of matter in terms of gases, realize that a gas that exists by itself, in fact it's a collection of molecules or atoms that are in constant motion. And we're going to say with this collection of gaseous molecules or atoms, there's some terms that you need to keep in mind. We have mean free path of fusion versus diffusion.

Now mean free path. This is just the average distance traveled by gas molecules between their collisions. Remember, gas molecules are very far apart and spread apart within a container. But they do come into contact with one another because they're bouncing around, they're going to interact. Now a mean free path is just the distance between 2 gas particles before they collide.

When it comes to effusion and diffusion, there are some differences. Now here we have images for effusion versus diffusion. For effusion, this is just the escape of gaseous molecule gas molecules or atoms through a pinhole. So the opening is very small. Because this opening is very small, that means these gases typically exit the container in a 1 by 1 fashion. It's not all of them coming out at once, it's one by one.

For diffusion though, diffusion is just the motion of a gas mixture from a high to low concentration. So this means a couple of different things. The gas molecules may not necessarily try to leave the container. Remember they're going from an area of high concentration to an area of low concentration. If inside is high concentration, then the gas molecules will escape to go outside. But if outside is higher concentration, then some gas could actually go into the container.

So that's a big difference from a fusion. In a fusion, you're only going out for diffusion. It's all based on concentration. You're always going to the area that has lower concentration, whether that's inside or outside the container. Secondly, for diffusion, the opening is bigger, and because this opening is bigger, more than one gas molecule can exit at a time or enter at a time.

So those are the subtle differences between effusion and diffusion. Remember, a gas particle doesn't exist by itself, it's a collection of gases. There are other gases around the vicinity. OK, so keep this in mind when talking about the behavior of any particular gas that you may read about.

Effusion

The rate of effusion states that the rate of a gas is inversely proportional to the square of their mass. So that means rate of gas equals 1m of molar mass.

But what exactly does that mean? Well in simple terms it means increasing the molar mass of a gas means that we are going to lower its speed and the lower its rate.

So basically the more a gas particle or molecule weighs, then the slower it's going to move. So that's basically the idea behind the rate of a fusion.

Effusion Example 1

Here it says to rank the following in order of increasing rate of effusion. Remember we said that the higher the molar mass than the lower the speed or rate of a gas. So here if we take a look, O₂ has two oxygens. The combined mass would be 32g per mole.

PF₅ has one phosphorus and five fluorines. When you add the mall up together, it gives you approximately 125.97g per mole. Carbon dioxide weighs approximately 44.01 grams per mole. When you add up the one carbon and the two oxygen and then xenon. If you look on the periodic table, it has a weight of approximately 131, oh .29 grams per mole.

We want to do it in terms of increasing rate of effusion, which means we're going to start out with the heaviest, which has the slowest rate, all the way up to the lightest one, which has the highest rate. All right, so who has the highest molar mass or highest weight would be Xenon. Xenon will be the slowest and therefore lowest rate of effusion.

Then we have PF₅, then we have CO₂ and then O₂ weighs the least out of everyone, so it would move the fastest. This would be our order going from the slowest rate of a fusion to the highest rate of fusion based on their individual molar masses.

Effusion

Now, Grand's lava fusion is used when comparing the rate of two different non reacting gases, and we're going to say here that the effusion rate of a gas at its time to travel are directly proportional. So let's say that we're dealing with gases A&B, so gas A and gas B. Since rate and time are directly proportional, that means they'll be on the same level. So the rate of gas A is equal to the time of gas A. The rate of gas B is equal to the time of gas B.

But then remember when we talked about the rate of effusion? The effusion rate of a gas and its molar mass are inversely proportional. So that would mean that the rate of gas A is on top. Therefore the molar mass of gas A would have to be on the bottom. Then the rate of gas B is here on the bottom, and therefore it's molar mass has to be here on top.

So remember, when we're directly proportional, we're on the same level with one another, and when we're inversely proportional, one will be numerator and the other will be the denominator. Understanding these relationships are key to answering questions dealing with Graham's Law of effusion.

Effusion Example 2

Here the example question says calculate the ratio of the effusion rates of helium to methane. All right, so the first gas that's name represents our gas A and the second one will represent our gas B. So here we're going to say rate of helium divided by the rate of methane. We're not given times for them, so we can't use time in terms of this question. But we know the identity of the gases, therefore we know they're molar masses.

So this equals. Remember rate and molar mass are inversely proportional. So if the rate of helium is on the top, then the molar mass of helium has to be on the bottom. Because the rate of methane is on the bottom, the molar mass of methane has to be on top. Here we have one carbon and four hydrogens. When you add all of their masses together from the periodic table, we'll get its mass as 16.042g per mole for methane.

When you look up the atomic mass of helium on the periodic table, it's 4.003g per mole for helium. So divide 16.042 by 4.003 and take the square root and that gives us approximately 2.00187. So this would be the ratio in terms of the rate of effusions for helium to methane.

Now what is this number telling us? Well, what this number is telling us? It's telling us that helium moves twice as fast as methane, and this makes sense because helium weighs less. Remember, we said that the less you weigh, the faster you're going to be able to move as a gas. Because methane weighs more, it makes sense. It's not going to be able to move quite as fast as helium.

So again, when we're figuring out the rate of gas A to gas B, whatever is discussed first is gas A. Second one is gas B. The answer is telling us how much faster the top gas, in this case helium, is to methane.

If H₂ has an effusion rate that is 3.72 times faster than a gas, what is the identity of the unknown gas?

• A. Cl₂
• B. CO₂
• C. N₂O₄
• D. N₂
• E. O₂

How many times faster will H₂ gas pass through a pinhole into an area of vacuum than O₂ gas?

• A. 32
• B. 2
• C. 2.5
• D. 4
• E. 8

It takes 6.3 minutes for 2.3 L argon to effuse through a semipermeable membrane. How long would it take for 2.3 L of chlorine gas to effuse under similar conditions?

• A. 8.4 mins
• B. 12.3 mins
• C. 15.3 mins
• D. 17.1 mins
• E. 21.5 mins

Here’s what students ask on this topic:

How can the rate of effusion be found?

The rate of effusion of a gas can be found using Graham's law of effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:

$\text{Rate of Effusion}\propto \frac{1}{\sqrt{\text{Molar Mass}}}$

When comparing two different gases, the law can be used to find the relative rates of effusion:

$\frac{\text{Rate of Effusion of Gas 1}}{\text{Rate of Effusion of Gas 2}}=\sqrt{\frac{\text{Molar Mass of Gas 2}}{\text{Molar Mass of Gas 1}}}$

To find the actual rate of effusion, you would need to measure how much gas passes through a small hole or a porous barrier in a given amount of time. The rate can be expressed in units of volume or moles per unit time. If you're conducting an experiment, you'll measure the volume of gas that has effused over a certain period and then apply the formula to calculate the rate. Remember that temperature and pressure can also affect the rate of effusion, so these should be kept constant during the comparison of two gases.

How do you calculate the rate of effusion?

The rate of effusion of a gas is calculated using Graham's Law of Effusion, which states that the rate at which a gas will effuse is inversely proportional to the square root of its molar mass. This can be expressed with the formula:

$\text{Rate of Effusion}\propto \frac{1}{\sqrt{\text{Molar Mass}}}$

If you're comparing the rates of effusion between two gases, you can use the formula:

$\frac{\text{Rate of Effusion (Gas 1)}}{\text{Rate of Effusion (Gas 2)}}=\sqrt{\frac{\text{Molar Mass of Gas 2}}{\text{Molar Mass of Gas 1}}}$

For practical calculations, if you know the molar mass of both gases and the rate of effusion of one gas, you can rearrange the formula to solve for the unknown rate:

$\text{Rate of Effusion (Gas 1)}=\text{Rate of Effusion (Gas 2)}\bullet \sqrt{\frac{\text{Molar Mass of Gas 2}}{\text{Molar Mass of Gas 1}}}$

Remember to keep your units consistent, typically using grams per mole (g/mol) for molar mass. The actual rate can be measured in various units, such as moles per second, depending on the context of the problem.

Which of the following would increase the rate of effusion of a gas?

The rate of effusion of a gas - which is how quickly it escapes through a small hole into a vacuum - is governed by Graham's Law of Effusion. According to this law, the rate of effusion is inversely proportional to the square root of the molar mass of the gas particles. Therefore, to increase the rate of effusion, you would need to decrease the molar mass of the gas.

Additionally, increasing the temperature of the gas would also increase the rate of effusion. This is because higher temperatures result in higher average kinetic energy of the gas particles, causing them to move more rapidly and effuse faster.

So, to sum up, the rate of effusion of a gas can be increased by either using a gas with a lower molar mass or by raising the temperature of the gas.