 ###### Kendal Orenstein

Rutger's University
M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

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# Graham's Law - Concept

Kendal Orenstein ###### Kendal Orenstein

Rutger's University
M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

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Graham s Law states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass.

Alright, so let's talk about Graham's law. Graham's law states that the rate of diffusion for a gas is inversely proportional to the square root of its molar mass. Now, let's break that down into what exactly that means. Okay.

An example of effusion, where gas through a small opening might be a tea kettle. A tea kettle, the gas evolved in the boiling of water in a tea kettle escapes the small hole in the opening making that whistling sound. That is example of effusion. Often may be a gas particle escaping from a nylon balloon, the helium gas in the nylon balloon escaping and the gas shrink and the balloon shrinking. That is an example of effusion also.

So let's talk about what that actually means and how fast these particles go. Alright, so we know the rate of effusion is equal to the square root, the inverse of the square root of the molar mass. So let's actually put that into action. So I'm going to go straight to the, typically when you're talking about the rates, you're going to compare one gas to another. So you're going to compare gas a to gas b and this is actually Graham's law and I'll get to this in just a second. But I want to describe it in a pictorial way down here.

Okay. So let's say you're comparing the rates of hydrogen chloride gas which has a molar mass of 36 grams per mole to the rate of ammonia gas which is 17 grams per mole. On one end at the same exact time, you're going to have this gas enter the tube and this gas enter the tube. And see that when they meet they are actually going to have a reaction, a reaction's going to occur. So let's actually do that' So at the same time we're going to enter, you're going to put in, I'm going to get another marker. You're going to put in hydrogen chloride into one end of the tube and at the same exact time you're putting in ammonia to the other end of the tube.

Now, what's going to happen. Well, the gas particles are going to flow and we decided they're going to flow from one, from high concentration to low concentration. So we're going to start flowing towards each other. This guy is really big and heavy, 36 grams per mole and this guy is really light and skinny 17 grams per mole. So this guy should travel faster than this guy, right? So this guy is going to travel fast fast fast fast, this guy's going to travel slow comparatively because it's really heavy and heavy in weight. So they're actually probably going to meet closer to the hydrogen chloride end, probably around here where we're going to get ammonium chloride, neat. Probably typically around here because this is going to travel a lot faster than this guy. So if you were to compare the rates and actually find the rate, when you compare them to rate of a compared to the rate of b it's equal to the square root of the molar mass of a on the bottom and over the molar mass of b on top. Let's put this into practice.

What is the molar ass of a gas that diffuses three times faster than oxygen under similar conditions? Okay. So oxygen has a molar mass of 32 grams per mole. Okay, and this unknown gas we don't know. We have unknown gas. Okay. This guy travels three times faster. So I'm going to say it travels three moles per second, okay? Sorry. We're going to get three meters per second. It makes more sense. Then oxygen gas travels, if this travels three times faster, this is going to travel on metre per second. Okay.

So if we were to put this the rate of a, I'm going to say, unknown gas is a, oxygen is going to be b. Okay. The rate of a is three meters per second. The rate of b is one metre per second. And this is going to equal the square root of and forget to flip them, molar mass of a, in this case we don't know x of the molar mass is b, in this case it's 32 grams per mole. So we want to solve what x is. Okay. Well, easy way to do it is square both sides and that's going to give me 9=32 over x. So when we multiply x times 9, we get 9x=32. We divide both by 9. x=3.55 grams per mole. This is my molar mass of my unknown gas. Okay, so this is how we can do it mathematically and these are the kind of questions you're probably going to see. Let's actually go and watch a video of Graham's law and action.

I promise to show you something really cool as long as you promise not to try it at home, okay? Okay. Now everyone wants to know why my voice sounds higher when I inhale helium. The simple fact is that helium is six times less denser than air which means sound waves travel through it much faster which makes my voice sound much higher. Now the same effect can be achieved in reverse if I inhale something like sulphur hexaflouride which is six times denser than air. I inhale some of that, and my voice gets really low although somehow I'm still funny. It's scientific.

Alright. so how did Graham's law play a part of that? Well, helium that he inhaled was actually very very light. So it traveled through his throat and traveled through his vocal chords very quickly and so made his made his voice very very high. Sulphur hexaflouride is extremely heavy. He said dense and that's the same idea, heavy or dense and it traveled, it traveled through the vocal chords very very slowly making his voice very very deep.

So that is an actual real life example or a fun example of Graham's law.