Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Quick puzzle first. We heat a gas from twenty seven degrees Celsius to three hundred and twenty seven degrees Celsius. The temperature number went up more than twelve times! The root mean square speed, or r m s speed, goes as the root of the temperature, so it goes up about three and a half times, right? Hold that answer. This is set one of mixed JEE Main pattern problems on kinetic theory, each with a small trap. So, let's begin.

0:31The toolkit for this set

First, the toolkit. Small p, the pressure, times capital V, the volume, equals n R T, and also capital N k T, where small n is the number of moles, R the gas constant, T the temperature, capital N the number of molecules, and small k is Boltzmann's constant. With me? Two forms of the same law, so let me ask you this: when do we use R, and when do we use k? Yes, exactly! R goes with moles, and k goes with single molecules. In fact, k equals R over N A, where N with the small subscript A is Avogadro's number. Okay? R is eight point three one four joules per mole per kelvin, and k is one point three eight times ten to the power minus twenty three joules per kelvin; J is the joule, mol the mole, capital K the kelvin, and the power minus one means per. Right? And three habits for every question: the temperature in kelvin, capital M, the molar mass, in kilograms per mole, and the pressure in pascals. Got it?

1:30Idea 1: per molecule, use k

Now, idea number one: per molecule. Part a: a gas has three times ten to the power twenty five molecules per cubic metre at three hundred kelvin. Find its pressure. Clear? Look, we are given molecules, not moles. So which constant do we use, R or k? We use k, that is correct, since we count molecules. So p equals capital N over capital V, times k T, which is small n k T, where small n, capital N over capital V, is the number density: molecules per cubic metre, not moles. Okay? Three times one point three eight times three hundred gives one thousand two hundred and forty two, and the powers of ten leave ten squared. So p is about one point two four times ten to the power five pascals, written P a. Got it? Now part b: the average translational kinetic energy of a molecule at five hundred kelvin. But wait, the question does not name the gas. Do we need it? No, we do not need it at all! Epsilon, the Greek letter, with a bar on top for average and trans below for translational, equals three halves k T per molecule, for every gas. Put in k and five hundred kelvin. With me? Three halves of one point three eight is two point zero seven, and two point zero seven times five hundred gives one thousand and thirty five. So that is one point zero three five times ten to the power minus twenty joules, about one point zero four, as the wavy equals sign says, the same for every gas. Right?

3:06Idea 2: per mole, use R and count f

Alright, idea number two: per mole. Capital U, the internal energy, is f over two, n R T, where small n is again the number of moles and f is the number of degrees of freedom. C v, with the small v below, is the molar heat capacity at constant volume, f over two times R, and gamma, the Greek letter, is one plus two over f. Clear? Part a: the internal energy of two moles of helium at four hundred kelvin. First, how many degrees of freedom does helium have? Yes, exactly, three! Helium is monatomic, made of single atoms, so it has only the three of translation. Okay? So capital U equals two, times three halves R, times four hundred, that is one thousand two hundred times R, which gives nine thousand nine hundred and seventy six point eight joules, or about nine point nine eight kilojoules. Got it? Now part b, the other way round: C v is three times R, and we want f and gamma. So first, what is f? Yes, f over two times R equals three times R, and the double arrow means so: f equals six, three of translation and three of rotation, as for a non-linear molecule with no vibration. Right? And gamma equals one plus two over six, which gives four thirds. Clear?

4:25Example 1: ratio of rms speeds at 327 C and 27 C

Now, example one, our puzzle from the start. Find the ratio of the r m s speeds of a gas at three hundred and twenty seven degrees Celsius and at twenty seven degrees Celsius; the small circle with C means degrees Celsius. Pause the video and try it first. Done? Okay, let's check. v r m s, the v with the subscript r m s, is the root of three R T over capital M, and for one gas, R and capital M are fixed. So v r m s is proportional to root T; the sign like an open alpha means proportional to. With me? So the ratio is the root of three hundred and twenty seven over twenty seven, about three point five, as in our puzzle, right? No, no, wait, that is the trap! The formula needs kelvin. T with the subscript one is three hundred and twenty seven plus two hundred and seventy three, which gives six hundred kelvin, and T two, twenty seven plus two hundred and seventy three, is three hundred kelvin. Clear? So v one over v two, the ratio of the two speeds, is the root of six hundred over three hundred, which is root two, about one point four one. So the speed goes up only about one point four one times, not three and a half. Puzzle solved: always use kelvin. Got it?

5:40Example 2: density of nitrogen at 2 atm and 300 K

Now example two, one step harder. Find the density of nitrogen at two atmospheres and three hundred kelvin; the screen writes atmospheres in short. Here is a hint: watch the units. Pause the video and try it first. Done? Okay, let's check. Density is mass per volume. So which form of the gas law has density in it? We want the one with rho in it, so let us build it. Start from p V equals n R T, as usual. The moles, n, are the mass, small m, over capital M, and rho, the Greek letter, is the density, small m over capital V, so p equals rho R T, over capital M, the molar mass. Okay? Turn it round, and rho equals p times capital M, over R T, our formula. The pressure is two, and nitrogen's molar mass is twenty eight, so rho is two times twenty eight, over R T, right? Oh no, careful, that comes out about a hundred times too small! The pressure must be in pascals, and one atmosphere is one point zero one three times ten to the power five pascals. And capital M for nitrogen, N two, is zero point zero two eight kilograms per mole, not twenty eight. With me? So the top gives five thousand six hundred and seventy two point eight, and the bottom, eight point three one four times three hundred, gives two thousand four hundred and ninety four point two. Divide, and rho is about two point two seven kilograms per cubic metre. That's it!

7:14Quick revision: k vs R, U, Cv and gamma, units

So, quick revision. Per molecule, use k: p equals capital N over capital V, times k T, and the energy is three halves k T; per mole, use R, the gas constant. U equals f over two, n R T, with small n in moles, C v equals f over two, R, and gamma equals one plus two over f. And always: kelvin, kilograms per mole, and pascals. Now go and try the practice problems on this page. All the best!