Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You have helium at three hundred kelvin, and you want its atoms to move twice as fast. Twice the speed, so twice the temperature, right? Hmm, are you sure? Hold that answer; by the end you will know the exact heat it takes. First, three quick ideas from Set two of this chapter, then two solved examples. These are JEE Main-pattern questions, so let's begin.

0:29Idea 1: when are two rms speeds equal?

Idea one is a problem from this set. At what temperature is the r m s speed of helium equal to that of oxygen at four hundred kelvin, the capital K on screen? Okay? First, the formula. v r m s, v with the small subscript r m s, is the root mean square speed: the square root of three R T over capital M, with R the gas constant, T in kelvin, and capital M, the molar mass, in kilograms per mole, the k g mol to the power minus one on screen. Right? Now think. Helium atoms are much lighter than oxygen molecules. So, for the same r m s speed, should the helium be hotter than four hundred kelvin, or colder? Yes, the helium must be colder! Equal speeds need equal T over capital M; the double arrow means so. T with the subscript capital H, small e, is helium's temperature, and helium's capital M is four grams per mole, while oxygen's is thirty two. So T for helium, over four, equals four hundred over thirty two; in a ratio the units cancel, so grams per mole are fine here. With me? So T for helium equals four hundred times four over thirty two, which gives fifty kelvin. Very cold, and still the light helium atoms keep pace with oxygen at four hundred kelvin. Clear?

1:51Idea 2: pressure from the molecules

Now for idea two, another problem from this set. Here, a one litre vessel, capital L for litre, holds two times ten to the power twenty two molecules, each of mass five times ten to the power minus twenty six kilograms. Their r m s speed is five hundred metres per second, the m s to the power minus one on screen. Find the pressure. Got it? This uses the kinetic model of an ideal gas: tiny molecules in random motion, colliding elastically, with no other forces between them except during collisions. Okay? On these assumptions, p, the pressure, is one third, times N m over V, times v r m s squared. Capital N is the number of molecules, small m the mass of one molecule, and capital V the volume. So p equals N m v r m s squared, over three V, right? Now, the volume must be in cubic metres. So tell me, one litre is how many cubic metres? It is ten to the power minus three, exactly! And N m, two times ten to the power twenty two, times five times ten to the power minus twenty six, is ten to the power minus three kilograms. Five hundred squared is two point five times ten to the power five. With me? Put it all in: the ten to the power minus three on top cancels the ten to the power minus three below, and two point five times ten to the power five, over three, gives about eight point three times ten to the power four pascals, the P a on screen; the wavy equals sign means approximately. Clear?

3:23Idea 3: the mean free path

Now for idea three, one more problem from this set. The mean free path in a gas at three hundred kelvin and one atmosphere, a t m, is seventy nanometres, n m. Find it at six hundred kelvin and zero point one atmosphere. Okay? In the picture, the dark blue dot is a molecule and each orange dot is a collision. Between collisions it moves along straight free paths, and their average length is the mean free path, lambda, the Greek letter at the top. Right? Lambda equals one, over root two, times pi, times d squared, times n, where d is the molecular diameter and small n the number density. The second form uses the pressure p: k T over root two pi d squared p, with small k Boltzmann's constant. Got it? Same gas, so d does not change, and lambda is proportional to T over p; that sign like an open alpha means proportional to. Clear? Now, the temperature doubles, and the pressure falls ten times, from one atmosphere to zero point one. So what happens to lambda? Both changes make it longer, very good! Doubling T doubles lambda, and p divided by ten makes it ten times longer again, so lambda grows twenty times. With me? So lambda is seventy times twenty, which gives one thousand four hundred nanometres, or one point four micrometres; the Greek letter mu means micro. Okay?

4:50Example 1: rigid diatomic gas heated at constant pressure

Now, Example one, an easy one. Find the heat needed to warm one mole of a rigid diatomic gas by one hundred kelvin, at constant pressure; the m o l on screen means mole. Pause the video and try it first. Done? Okay, let's check. Rigid means no vibration, so a diatomic molecule has five degrees of freedom, f equals five: three for translation and two for rotation. So C v, with the small subscript v, the molar heat capacity at constant volume, is five halves R, right? So the heat is five halves R, times one hundred, isn't it? No, no, careful, this is the trap! The gas is heated at constant pressure, not constant volume, so we need C p, with the small subscript p: C v plus R, which is seven halves R, got it? Now Q, the heat, equals n C p delta T, with n the number of moles and delta T, the Greek capital delta, the rise in temperature. With R as eight point three one four joules per mole per kelvin, Q is one, times seven halves, times eight point three one four, times one hundred. With me? That gives two thousand nine hundred and nine point nine joules, the capital J, which is about two point nine one kilojoules. Clear?

6:08Example 2: heat to double the rms speed of helium

Now the harder one, our puzzle from the start. How much heat doubles the r m s speed of the atoms in two moles of helium at three hundred kelvin, at constant volume? Here is a hint: find the new temperature first. Pause the video and try it. Done? Okay, let's check. Twice the speed, so twice the temperature, which is six hundred kelvin, right? No, no, wait, this is the trap from our puzzle! For the same gas, capital M is fixed, so v r m s is proportional to the square root of T, not to T itself. Okay? So to double v r m s, T must become four times as large; the double arrow means so, and the single arrow means goes to. Three hundred kelvin goes to one thousand two hundred, so delta T is nine hundred kelvin. Right? Helium is monatomic, with f equals three, so C v is three halves R, and the volume is constant, so C v is the right one this time. With me? So Q equals n C v delta T: two, times three halves, times eight point three one four, times nine hundred. Clear? That gives twenty two thousand four hundred and forty seven point eight joules, which is about twenty two point four kilojoules. So the puzzle is solved: twice the speed needs four times the temperature. Got it?

7:30Quick revision: rms speed, pressure, mean free path, Cp

So, quick revision. Speeds go as the root of T over capital M, so twice the speed needs four times the temperature. Pressure is N m v r m s squared over three V, in S I units, and for the same gas, lambda is proportional to T over p. For heat, C v at constant volume, and C p, which is C v plus R, at constant pressure. Now try the practice problem on this page. All the best!