Three Characteristic Speeds

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Three students are asked how fast oxygen molecules move at three hundred kelvin. One says three hundred and ninety five metres per second, another says four hundred and forty six, and the third says four hundred and eighty four. So who is right? Surprisingly, all three of them are! These speeds come up often in JEE, so let's see how.

0:28Molecules have a spread of speeds

So, the molecules of a gas do not all move at one speed; their speeds spread out in the Maxwell distribution. The horizontal axis is the speed, v, and the vertical axis, one over capital N, times d N by d v, is the fraction of molecules per unit speed, with capital N the number of molecules. Look at the dark blue curve, at the temperature T with a small subscript one. With me? Now tell me, looking at this blue curve, which speed is shared by the largest number of molecules? The speed at the peak, yes, exactly! This is the most probable speed, written v with a small subscript p, which we read as v p, the most common speed in the gas. Clear? But look at the shape. It is not symmetric: it rises steeply, then has a long tail at high speeds. The three dashed lines mark three different speeds; we will name them soon. First, which way do the few very fast molecules in the tail pull the average speed? Correct, to the right, towards speeds higher than the peak! So, from left to right, the dashed lines are v p, then v bar, the v with a bar on top, the average speed, which we read as v average, and then v with the subscript r m s, the root mean square speed, or v r m s. As the less-than signs show, v p is less than v average, which is less than v r m s. Got it?

1:57Heat the gas: the curve changes

Now, what happens when we heat the gas? The orange curve is for T two, greater than T one, as the greater-than sign shows. How has the curve changed? Yes, the peak shifts to the right, and it falls lower. Hotter molecules move faster, so the whole curve moves towards higher speeds. Okay? And the peak falls because the area under the curve stands for all the molecules, a total fraction of one, and heating does not change that. So as the curve spreads out, it must get lower. Right?

2:30Three speeds, three formulas

So, v p equals the square root of two times R T, over capital M, where R is the gas constant and capital M the molar mass. Per molecule, it is the root of two times k T, over small m, with small k Boltzmann's constant and small m the mass of one molecule. With me? Next, v average equals the root of eight times R T, over pi times capital M, and notice the Greek letter pi in the bottom; only this formula has it. Clear? And v r m s equals the root of three times R T, over capital M, and it is also the root of three times p, over rho, with small p the pressure and rho, the Greek letter, the density. Got it? One thing to watch here: capital M must be in kilograms per mole, written k g mol to the power minus one, so for oxygen it is zero point zero three two, not thirty two. And the temperature is in kelvin. Right? Now compare the three. Under the root we have two, eight over pi, and three. Eight over pi is about two point five five, the wavy equals sign meaning approximately, so it sits between two and three. So v p is the smallest and v r m s the largest, just like the graph. Okay? So the ratio is fixed. With the colons meaning ratio, v p, v average and v r m s are in the ratio one, one point one two eight, and one point two two five, in that order. And all three are proportional to the root of T over capital M; the sign like an open alpha means proportional to. Clear?

4:08What the speeds depend on

Now, one common trap. We double the pressure at constant temperature. In the root of three times p, over rho, p is on top, so v r m s must grow, right? No, careful, it does not grow! At constant temperature, doubling the pressure doubles the density rho too. So p over rho stays exactly the same. Okay? So remember this: v r m s depends only on the temperature and the molar mass, not on the pressure. Clear? One last result. The rate of effusion, a gas leaking through a tiny hole, is proportional to v average, so, at the same temperature and pressure, to one over the root of capital M, the molar mass. This is Graham's law: lighter gases effuse faster. Got it?

4:55Example 1: vp, vavg and vrms of oxygen at 300 K

Now, example one, our puzzle from the start. Find v p, v average and v r m s for oxygen at three hundred kelvin; the capital K stands for kelvin. Pause the video and try it first. Done? Okay, let's check. Capital M is zero point zero three two kilograms per mole, not thirty two, and R is eight point three one four joules per mole per kelvin. Right? Now, a topper's trick: all three formulas share the root of R T over capital M, so work it out once. Eight point three one four times three hundred, over zero point zero three two, all under the root, gives about two hundred and seventy nine point two metres per second, the m s to the power minus one. Okay? For v p, multiply by root two. That gives about three hundred and ninety five metres per second. Clear? For v average, multiply by the root of eight over pi. That gives about four hundred and forty six metres per second. With me? And for v r m s, multiply by root three, which gives about four hundred and eighty four metres per second. So those were our three students' answers, and each one was right, for a different speed of the same gas. Very good, that solves our puzzle!

6:18Example 2: T at which mean speed of O2 equals vrms at 300 K

Now example two, one step harder. At what temperature is the mean speed of oxygen equal to its r m s speed at three hundred kelvin? Pause the video and try it first. Done? Okay, let's check. Before any calculation, let me ask you this: will the answer be above three hundred kelvin, or below it? Above, yes, exactly! At the same temperature, v average is less than v r m s, so the mean speed needs a hotter gas to catch up. Okay? Now set them equal. On the left, the root of eight times R T, over pi times capital M, with T unknown. On the right, the root of three times R times three hundred, over capital M, right? It is the same gas, so R and capital M cancel. Square both sides, and eight times T, over pi, equals three times three hundred, which is nine hundred. Got it? So T equals nine hundred times pi, over eight, which gives about three hundred and fifty three kelvin. It is above three hundred, just as we predicted, and that is our answer.

7:29Quick revision: vp, vavg, vrms order and the 2, 8/pi, 3 rule

So, quick revision. v p is less than v average, which is less than v r m s, because of the long tail at high speeds. Under the root we have two, eight over pi, and three, times R T over capital M, with capital M in kilograms per mole. And all three are proportional to the root of T over capital M; pressure alone does not change them. Now try the practice problems on this page. All the best!