Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a gas, keep its temperature fixed, and squeeze it until its pressure is four times what it was. Now, are its molecules moving faster than before? Many of you will say yes. Hold that answer; our very first tool today will settle it. This is part A of our unit problem bank, at JEE Main level: three quick tools, each with a short problem from the bank, and then two full problems. So, let's begin.

0:33Tool 1: the rms speed

So, tool one, and here is our puzzle, straight from the bank. The pressure of a gas is made four times at constant temperature. What happens to v r m s? That is the v with a small r m s below, the root mean square speed of the molecules. Okay? First, let us write the formula. v r m s equals the root of three times R T, over capital M, where R is the gas constant, T is the temperature in kelvin, and capital M is the molar mass. It also equals the root of three times p, over rho, with small p the pressure, and rho, the Greek letter, the density. With me? Now look at the second form. Small p sits on top. Four times the pressure, and the root of four is two, so v r m s doubles, right? No, no, careful, that is exactly the trap! At constant temperature, squeezing the gas to four times the pressure makes the density rho four times too. So p over rho does not change at all. Clear? So v r m s is unchanged. It depends on the temperature and the molar mass only. And that settles our puzzle: four times the pressure, but the molecules move exactly as fast as before. Got it?

1:49Tool 2: the same energy per molecule

Now tool two, with another one from the bank. Gases A and B share a vessel, and capital M in the brackets, the molar mass, is four for A and sixty four for B, okay? Find the ratios of their average translational kinetic energies per molecule, and of their r m s speeds. The average translational kinetic energy of one molecule is written epsilon, the Greek letter, with a bar on top for average and a small trans below. It equals half m times v squared bar, the average of v squared, which is three halves k T, right? Small m is the mass of one molecule, and small k is Boltzmann's constant. Now tell me this: each molecule of gas B is sixteen times heavier than one of gas A, so whose molecules carry more kinetic energy on average? Neither of them, they are equal! Three halves k T has only the temperature in it, no mass at all. Same vessel means same temperature, so the energies, with small A and B below, are in the ratio one is to one; the colon means ratio. With me? But the speeds are not equal. From tool one, v r m s is proportional to one over the root of capital M; the sign like an open alpha means proportional to. So v A over v B, the ratio of their r m s speeds, is the root of sixty four over four, the root of sixteen, which is four. So the ratio is four is to one: the lighter gas moves four times as fast. Clear?

3:20Tool 3: equipartition and γ

Now tool three, and the bank asks: a gas has gamma equal to one point four; gamma, the Greek letter, is the ratio of the two specific heats of the gas. How many degrees of freedom do its molecules have? Let us build the tool first, okay? So, here is the law of equipartition. Each independent squared term in the energy of a molecule, like half m v x squared, is counted as one degree of freedom, and it gets half k T per molecule on average, or half R T per mole. Right? Now call the number of degrees of freedom f. Then the internal energy, capital U, of n moles is f over two, times n R T, with small n the number of moles. And gamma equals one plus two over f. Clear? Now you do it. Gamma is one point four, so what is f? Five, yes, exactly! One plus two over f equals one point four, so two over f is zero point four, and f equals five. Got it? And five means three translations plus two rotations. That fits a rigid diatomic gas, one that does not vibrate, like oxygen, O two; the small two below means two atoms. Remember this five, because example two needs it. With me?

4:40Example 1: rms speed of argon at 27 degrees Celsius

Now, example one. Find v r m s of argon, with capital M equal to forty, at twenty seven degrees Celsius; the small circle and the capital C mean degrees Celsius. Pause the video and try it first. Done? Okay, let's check. The temperature is twenty seven, so we put T equals twenty seven into the formula, right? Oh no, careful here, that is the first trap! In kinetic theory the temperature is always in kelvin. Add two hundred and seventy three, and T equals three hundred kelvin; capital K is kelvin. Okay? Now the second trap is in capital M, the molar mass. Forty is in grams per mole, written g mol to the power minus one on screen. The formula needs kilograms per mole, so capital M is zero point zero four zero kilograms per mole, not forty. Clear? Now put everything in, with R equal to eight point three one four. So v r m s is the root of three times eight point three one four times three hundred, over zero point zero four zero. Right? Under the root, that is one hundred and eighty seven thousand and sixty five. Its square root is about four hundred and thirty three metres per second; the wavy equals sign means approximately, and m s to the power minus one means metres per second. Kelvin and kilograms per mole: get those two right, and it is easy. Very good!

6:15Example 2: total kinetic energy of 1 mol of oxygen at 300 K

Now example two, a little harder. Find the total kinetic energy, translational and rotational, of one mole of oxygen at three hundred kelvin; m o l means mole. Pause the video and try it first. Done? Okay, let's check. Kinetic energy of a gas, so it is three halves R T per mole, right? No, no, wait, three halves R T is only the translational part! The question also asks for rotation. So go back to equipartition: per mole, each degree of freedom gets half R T, okay? Oxygen is diatomic, so it has three translations and two rotations, and the question wants exactly these: f equals five, the same five as in tool three. Got it? So the total kinetic energy, capital E, is five halves R T, with me? That is five halves, times eight point three one four, times three hundred. That gives six thousand two hundred and thirty five point five joules, where capital J is joules. So it is about six point two four kilojoules, the k J you see on screen. Count every degree of freedom the question asks for, and you cannot go wrong. Perfect!

7:33Quick revision: rms speed, 3/2 kT, degrees of freedom

So, quick revision. v r m s, the root of three R T over capital M, depends on the temperature and the molar mass only, with the temperature in kelvin and capital M in kilograms per mole. The average translational energy, three halves k T per molecule, is the same for every gas at one temperature. And each degree of freedom gets half k T; gamma is one plus two over f, and a rigid diatomic gas has f equal to five. Now go and try the practice problem on this page. All the best!