# Rate of diffusion and molecular mass relationship

### How does molecular weight affect the rate of diffusion? | Socratic

The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as. Molecular weight will affect the rate of diffusion. All other things being constant, lighter weight molecules will move faster or diffuse faster that heavier molecules. Molecular weight and diffusion are inversely related. The heavier the molecule, the slower the rate of diffusion.

Hence, mylar balloons can retain their helium for days. Naturally occurring uranium is only 0. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically.

How many effusion steps are needed to obtain Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity.

Use a logarithmic expression to compute the number of separation steps required. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: Their atomic masses are 3.

Helium-3 has unique physical properties and is used in the study of ultralow temperatures. How many effusion steps are necessary to yield If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound.

## 2.9: Graham's Laws of Diffusion and Effusion

Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line. Summary Gaseous particles are in constant random motion.

Gaseous particles tend to undergo diffusion because they have kinetic energy. And Planet Graham has these little green molecules. But he has such a special nose that he can actually detect whether he's smelling carbon dioxide or oxygen. So, I'm going to actually take this lid off.

And I'm going to say, hey, if you detect with your special nose either one of these-- let's say that these molecules, one of them goes over and kind of goes into his nose-- if you can detect it, please let me know which one you're smelling. And that's my test. And I want to know which of these molecules, oxygen or carbon dioxide, is going to reach his nose, which is 10 feet away, first. So it's basically a race. And you can make a prediction right now as to which molecule you think is going to get to his nose first, the oxygen or the carbon dioxide.

Now you might think, oh, it's very easy.

There's a direct path. But actually remember, these molecules, these green molecules in the planet atmosphere, are whizzing around. They're going in all sorts of different directions. And as a result, they're going to smack into our carbon dioxide or oxygen molecules as they try to make their way over there. And kind of a random fact, but an interesting one to think about, is that in our atmosphere we have a lot of nitrogen, a lot of nitrogen gas. Now, if you took one nitrogen gas molecule, which is N2, and let it go, and measured its speed and kind of clocked it, it would be going at about 1, miles an hour.

But the only reason it doesn't actually go that speed in reality is because the molecules of nitrogen, they actually will smack into each other and bounce off of each other millions and millions of times every second. And so because they're smacking and colliding constantly, they never really reach those real potential speeds. They go much slower. So really what we're talking about is when molecules are bouncing and clanging into each other and slowly making progress towards our little alien's nose, that is the idea of diffusion.

They're going to kind of rattle around and slowly make their way over to his nose. And maybe if I came back, let's say, 10 minutes later, maybe this little oxygen would be right here. Maybe you might have a little carbon dioxide right here. Slowly making progress towards the nose.

### Graham's law of diffusion (video) | Khan Academy

That's what we're trying to figure out-- which one will get over there first. So, you've had time to think about it. And I'm actually going to tell you how I think we should approach the problem, which is thinking back to kinetic energy.

We're heating this thing up, so we're putting thermal, or heat energy, into the molecules. Both types of molecules are getting the same amount. I've got the oxygen getting some kinetic energy. I'm going to put a little o for oxygen.

And it's going to equal, or should equal, the amount of energy that my carbon dioxide is getting. And I'm going to do that as a little c for carbon dioxide. So these two molecule types should be getting the same amount of energy. Now remember, it's not like it's one molecule we're thinking of.

We're thinking of many, many molecules. So first, I'm going to have to change these units a little bit. Because again, I'm thinking about the individual molecule. So I've got to figure out what these molecules weigh.

Root Mean Square Velocity - Equation / Formula

And v is going to change over to rate, or diffusion rate. And the reason I'm doing that is because, again, I'm thinking about the overall diffusion of the gas. It's not like I'm betting on any one molecule. I'm betting on the entire population of carbon dioxide molecules beating out the population of oxygen molecules, or vice versa, the oxygen molecules beating out the carbon dioxide molecules. But not an individual molecule. So I have to think of the average rate that those molecules are moving.

So, let me rewrite this equation.

### Graham's Laws of Diffusion and Effusion - Chemistry LibreTexts

I'm just going to call it rate. And we'll call it rate 1. And 1 will be for the oxygen. In fact, molecular weight 1 can be for the oxygen as well. And 2 refers to carbon oxide, and rate 2 refers to carbon dioxide as well.

And I really don't need to keep carrying on with these halves. I can just multiply both sides of the equation by 2 and get rid of them. So that makes it a little bit easier. And I almost forgot, I have to square both sides. That would have been a mistake. I forgot to square them earlier.

So now I've squared them.

And let me actually rearrange it to make it a little bit neater in a new color. So let's do this. So let me write it out nice and neat.