Mean Free Path

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Gas molecules move at hundreds of metres per second. So when someone opens a perfume bottle across the room, you should smell it almost at once, right? But you do not; the smell takes a long time to reach you. So what holds it back? The answer is the mean free path, which comes up often in JEE exams. So, let's begin.

0:26Free paths and the mean free path

So, look at the picture. The dark blue dot is one molecule, and each orange dot is a collision, where it hits another molecule and turns. Between collisions it moves in a straight line, called a free path, like the one marked by the grey arrow. Okay? These free paths are all different. Their average is the mean free path, written with lambda, the Greek letter on screen. Its formula is at the top of the picture, and we will read it soon. With me? Now, think of each molecule as a tiny ball of diameter small d. Let me ask you this: how close must two centres come for a collision? Is it only half of d, one radius? No, it is d itself! One radius plus the other radius makes one full diameter. So our molecule hits any molecule whose centre comes within d of its path. Clear?

1:20The formula for λ

Now let us read the formula. Lambda equals one over, root two times pi times d squared times n. Here n is capital N over capital V: capital N molecules in a volume capital V, so n is the number per cubic metre, called the number density. And the root two comes from allowing for the motion of the other molecules. Right? Now look: the number density n and the diameter d both sit in the bottom of the fraction. So for a more crowded gas, or bigger molecules, is the free path longer or shorter? Shorter, yes, exactly! More molecules in the way, or bigger targets, means an earlier hit. And d is squared, so the size of the molecule matters a lot. Got it? Now, p equals n k T, with p the pressure, small k Boltzmann's constant, and capital T the temperature in kelvin. So n equals p over k T, and lambda becomes k T over, root two pi d squared p. Use whichever form fits the data, okay? Now there are two more results to learn. The v with a bar on top, v bar, is the average speed. Collision frequency, the number of collisions each second, is v bar over lambda. Mean free time, the average time between collisions, is lambda over v bar, just one over the frequency. With me?

2:49How λ changes

Alright, now the pressure form. At fixed temperature, lambda is proportional to one over p; the sign like an open alpha is read as proportional to. So halving the pressure doubles lambda. Okay? At fixed pressure, lambda is proportional to T itself. Why? Because n equals p over k T, so a gas heated at the same pressure spreads out, and n, the number density, falls. Clear? Now here is a tricky one. Heat a gas in a sealed, rigid container. The temperature rises, so lambda rises too, right? No, no, careful, this is the trap! That rule was for fixed pressure. Here the volume is fixed and no molecules escape, so n is fixed, and the first form has no temperature in it. So lambda does not change; the pressure simply rises. Got it?

3:42Example 1: mean free path of nitrogen at STP

Now, a problem from the course. Find the mean free path of nitrogen at STP, standard temperature and pressure. The diameter d is three point seven times ten to the power minus ten metres, and the number density n is two point six nine times ten to the power twenty five per cubic metre, which is what m to the power minus three means. Pause the video and try it first. Done? Okay, let's check. Which form should we use here, the one with n or the one with p? The form with n, because n is given! So lambda is one over, root two pi, times three point seven times ten to the power minus ten, squared, times two point six nine times ten to the power twenty five. Right? Now, d squared is one point three six nine times ten to the power minus nineteen; careful, square the three point seven too. Then root two, times pi, times d squared, times n, comes to about one point six four times ten to the power seven, per metre. The wavy equals sign means approximately. With me? So lambda is one over that, about six point one times ten to the power minus eight metres. That is really tiny, isn't it? Remember it, because it is our first clue. Clear?

5:06Example 2: collision frequency and mean free time of N2

Now a slightly harder one. For the same nitrogen, with lambda equal to six point one times ten to the power minus eight metres, find the collision frequency and the mean free time. Here is a hint: find v bar first. Pause the video and try it. Done? Let's check. The average speed v bar is the square root of eight R T over pi capital M, with R the gas constant and capital M the molar mass. At STP, the temperature is two hundred and seventy three kelvin, with capital K for kelvin. Nitrogen's molar mass is twenty eight, so we put in twenty eight, right? Oh no, careful, not twenty eight! That is in grams per mole, but this formula needs kilograms per mole. So capital M is zero point zero two eight kilograms per mole. This is a very common slip. Okay? So v bar is the square root of eight, times eight point three one four, times two hundred and seventy three, over pi times zero point zero two eight. That is about four hundred and fifty four metres per second, written m s to the power minus one. Got it? Collision frequency is v bar over lambda: four hundred and fifty four, over six point one times ten to the power minus eight. That gives about seven point four times ten to the power nine per second. More than seven billion collisions every second, for each molecule! Right? And mean free time is lambda over v bar: six point one times ten to the power minus eight, over four hundred and fifty four. That is about one point three times ten to the power minus ten seconds, one over the frequency.

6:58Back to the smell puzzle

Clear? Now, back to our puzzle. Look at the zigzag in the picture again. The molecules are so fast, so why is the smell so slow? Because each free path is tiny. The single wavy sign, called a tilde, means of the order of, so each molecule travels only about ten to the power minus seven metres before it is deflected. Okay? And it turns billions of times every second, like the zigzag in the picture. Such a path is called a random walk. With me? The steps point in random directions, so they partly cancel. So the net displacement grows only as the square root of the number of collisions; that is the root sign on screen. Fast molecules, but a slow smell, and that solves our puzzle!

7:45Quick revision: lambda formulas and the p, T, V rules

So, let us do a quick revision. Lambda equals one over, root two pi d squared n, or k T over, root two pi d squared p, with small d the molecular diameter. Collision frequency is v bar over lambda, and mean free time is lambda over v bar. Lambda goes as one over p at fixed temperature, as the temperature at fixed pressure, and stays the same at fixed volume. Now go and try the practice problems on this page. You will find them easy. All the best!