Rate of diffusion and molar mass relationship

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

rate of diffusion and molar mass relationship

When we watch diffusion with our eyes, we see it smoothing out concentration differences, like a drop of cream dispersing in a cup of coffee. At the molecular. Graham's law of effusion was formulated by Scottish physical chemist Thomas Graham in Graham found experimentally that the rate of effusion of a gas is inversely Graham's law states that the rate of diffusion or of effusion of a gas is inversely In the same conditions of temperature and pressure, the molar mass is. Sep 3, Graham's law expresses the relationship between the rate of diffusion or effusion of a gas and its molar mass. Here's how the law works in.

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.

Gas Law Problems Combined & Ideal - Density, Molar Mass, Mole Fraction, Partial Pressure, Effusion

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.

2.9: Graham's Laws of Diffusion and Effusion

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. 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. 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.

rate of diffusion and molar mass relationship

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. And this is actually going to be Graham's law. So all I'm doing is rearranging the formula. I've got rate 1. This is the diffusion rate of one molecule divided by the diffusion rate of a second molecule, and then the molecular weight on the other side of the second molecule divided by the molecular weight of the first molecule. And you do a square root of this side. So, that's just a rearrangement of the formula.

But what I've written out for you is now Graham's law. It's basically taking the kinetic energy rule and rearranging it to make sense for molecules. And let me make a little bit of space here. And so that an extension of this would be if you're just thinking about one molecule, then the rate, the diffusion rate-- when I say rate I mean diffusion rate-- is going to be proportional to the square root of the molecular weight. So, let's figure out how to apply this to our little riddle.

We wanted to know whether oxygen or carbon monoxide is going to diffuse faster. And I can now go back to our the periodic table and look up oxygen. And I know that the molecular weight of 16 here and carbon is And that means that O2 is just 16 times 2. So the molecular weight is And carbon dioxide is going to be 2 oxygens plus 12 more. So it's going to be So these are the molecular weights of carbon dioxide and oxygen. So basically what I do is I just plug them in.

And I say, OK, let's plug them into the formula. Let's use this one right here. And I'm going to call rate 1 my oxygen rate. So what's happening with rate 1? We'll say, well, rate 1 is rate of oxygen-- I'm going to write a big o here-- equals the square root of-- let's make sure I stay consistent-- I said 1 was oxygen, so it's going to be 32 down here and 44 up here.

And then that's going to be multiplied by rate of carbon dioxide.

Why Does the Rate of Diffusion Vary With Molecular Weight & Temperature?

And I'll put a c for carbon dioxide. So what does this work out to be? The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusionis the escape of gaseous molecules through a small usually microscopic hole, such as a hole in a balloon, into an evacuated space.

The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties.

The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses: The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy recall that a result of the Kinetic Theory of Gases is that the temperature, in degrees Kelvin, is directly proportional to the average kinetic energy of the molecules.

We can write the expression for the average kinetic energy of two gases with different molar masses: Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume.

rate of diffusion and molar mass relationship

In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber.

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