Overview

Transcript

0:00Introduction

Hello everyone! Okay, two small questions before we start. First, put a closed container of gas on a train moving at a constant velocity, so every molecule is carried along at the train's speed too. Is the gas now hotter? Second, why has the Earth lost most of its hydrogen, but kept its nitrogen? Questions like these come up often in JEE, in statement form. First, how to use this unit's problem bank; then we answer both. So, let's begin.

0:34What is in this chapter

So, first, what is this chapter? It is the problem bank for the whole unit on kinetic theory: mixed solved problems, in three parts. Okay? Part A has seven solved problems, at JEE Main level. Right? Part B has six longer problems, at JEE Advanced level. With me? And Part C has six concept checks, on the ideas often tested in statement form. Both our questions from the start are Part C checks. Clear? And each part ends with one practice problem on the page, for you to solve.

1:09Part A: a 30-minute test

Got it? Now, when should you attempt Part A? On the first day of the unit, just to have a look? No, not on day one: the course says to attempt it only after Chapters one to six. Okay? That is because these are mixed problems: the speeds, the kinetic energy, and the degrees of freedom, all together. So finish every chapter first. Right? Then attempt it as a real thirty minute test: solve each problem yourself first, and only then check the solution. With me?

1:40Part C: our two questions

Now, our first question, as the course asks it. Take a closed container of gas moved at a constant velocity: does T, its temperature, change? The answer is no, it does not change. Temperature measures the random motion of the molecules, relative to the centre of mass of the gas. Okay? Moving the container gives every molecule the same extra velocity, a uniform bulk velocity, which leaves the random motion untouched. The one exception is in the bracket: if the container is stopped suddenly, its ordered kinetic energy becomes internal energy, and the gas gets hotter. Clear? Now, our second question. The Earth has lost most of its hydrogen, but kept its nitrogen. So why did it lose one gas and keep the other? The answer is in the speeds. At the same temperature, speeds go as one over the root of capital M, the molar mass. So v for hydrogen, with the small subscript H two, over v for nitrogen, subscript N two, is the root of capital M of nitrogen over capital M of hydrogen. That is the root of twenty eight over two, root fourteen, about three point seven; the wavy equals sign means approximately. Right? So hydrogen molecules move about root fourteen times faster, and a much larger fraction of them exceeds the escape speed, the speed needed to leave the Earth for good. Hydrogen escapes, and nitrogen stays. And that's it, both of our puzzles are solved!

3:15Example 1: KE and rms speed ratios of gases, M 4 and 64

Now, Part A, an easy one. Gases A and B share a vessel. For gas A, capital M, the molar mass, is four, and for gas B it is sixty four. Find the ratio of their average translational kinetic energies per molecule, and the ratio of their r m s speeds, the root mean square speeds. Pause the video, and try it first. Done? Okay, let's check. The two gases share one vessel, so they are at the same temperature T, and that is the key. With me? Now, the average translational kinetic energy of one molecule, the K E on screen, is three halves k T, with small k Boltzmann's constant. There is no mass in it at all. Right? But look, gas B is sixteen times heavier than gas A, so surely each of its molecules carries more energy, doesn't it? Oh no, careful, that is the trap! Three halves k T depends only on the temperature, so the energies are in the ratio one to one; the colon means ratio. The heavier molecules simply move slower. Clear? So, now the speeds. v r m s, written v with a small subscript r m s, equals the root of three R T over capital M, with R the gas constant. At the same T, it is proportional to one over the root of capital M; the sign like an open alpha means proportional to. With me? So the speeds are in the ratio sixty four to four, that is, sixteen to one, correct? No, no, wait, you forgot the square root! v A over v B, the v's with subscripts A and B, equals the root of capital M of B over capital M of A: the lighter gas is faster. The root of sixty four over four is root sixteen, which is four, so the r m s speeds are in the ratio four to one. In a ratio the units of capital M cancel, so four and sixty four work directly. Got it?

5:18Example 2: T when H2 rms speed equals O2 mean speed at 300 K

Now Part B, a hard one. At what temperature is the r m s speed of hydrogen equal to the mean speed of oxygen at three hundred kelvin? That capital K on screen means kelvin. Pause the video, and try it. Done? Okay, let's check. First, a quick guess: will the answer be above three hundred kelvin, or below it? It will be far below, yes, very good! Look at the molar masses in kilograms per mole, the k g mol to the power minus one on screen: zero point zero zero two for hydrogen, H two, and zero point zero three two for oxygen, O two. Hydrogen is sixteen times lighter, so it needs a much lower temperature. Okay? In symbols, v r m s of hydrogen at the unknown T equals v bar of oxygen at three hundred kelvin. v bar, the v with a bar on top, is the mean speed, also called the average speed. Clear? Now, many students use the r m s formula, the root of three R T over capital M, on both sides. They get three hundred, times two over thirty two, which gives eighteen point seven five kelvin. Is that right? No, careful, that is the trap here! For oxygen, the question asks for the mean speed, v bar, which is the root of eight R T over pi times capital M, not the r m s speed. So square both sides: three R T over zero point zero zero two equals eight R times three hundred, over pi times zero point zero three two. With me? The gas constant R cancels. So T equals eight times three hundred times zero point zero zero two, over three pi times zero point zero three two. On top that is four point eight, and below, zero point zero nine six times pi. Four point eight over zero point zero nine six gives fifty, so T equals fifty over pi. Got it? And fifty over pi is about fifteen point nine kelvin. So the answer is far below three hundred kelvin, just as we guessed, very good!

7:35Quick revision: 3/2 kT for every gas, speeds go as root T/M

So, quick revision. Part A is a thirty minute test after Chapters one to six; Part B is at JEE Advanced level; and Part C has the concept checks. Three halves k T per molecule is the same for every gas at a given temperature. And all speeds go as the root of T over capital M: v r m s is the root of three R T over capital M, and v bar the root of eight R T over pi times capital M, okay? Now set a thirty minute clock, open Part A, and begin. All the best!